Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions

Solve simple concepts of Rational Exponents and Radical Functions by practicing problems from Big Ideas Math Algebra 2 Answers? Get free access to download Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions from this page.  You may have doubts about the solutions don’t worry our subject experts created this BIM algebra 2 ch 5 solution key as per the Common core student edition BIM Algebra 2 Textbooks. Improve your preparation level with the help of Big Ideas Math Book Algebra 2 Chapter 5 Answer Key.

Big Ideas Math Book Algebra 2 Answer Key Chapter 5 Rational Exponents and Radical Functions

Get the direct links of Big Ideas Math Book Algebra 2 Answer Key Chapter 5 Rational Exponents and Radical Functions prevailing in order to help the students and teachers. Quick and easy learning is possible with our BIM Algebra 2 Book Solutions of ch 5 Rational Exponents and Radical Functions. With the help of Big Ideas Math Book Algebra 2 Answer Key Chapter 5 Rational Exponents and Radical Functions, you can score good marks in the exams. Test your math skills by solving the questions provided at the end of this chapter. Click on the links such as chapter tests, reviews, practices, quizzes, cumulative assessments, etc to check your answers.

Rational Exponents and Radical Functions Maintaining Mathematical Proficiency

Simplify the expression

Question 1.
y6 • y
Answer: y7
Explanation:
Given,
y6 • y
When bases are equal powers should be added.
am . an = am+n
y6 • y = y6+1 = y7

Question 2.
\(\frac{n^{4}}{n^{3}}\)​
Answer: n
Explanation:
Given,
\(\frac{n^{4}}{n^{3}}\)​
am / an = am-n
= n4-3
= n¹
So, \(\frac{n^{4}}{n^{3}}\)​ = n

Question 3.
\(\frac{x^{5}}{x^{6} \cdot x^{2}}\)
Answer: n-3
Explanation:
Given,
https://ccssanswers.com/wp-admin/post.php?post=40609&action=edit
Question 4.
\(\frac{x^{-6}}{x^{5}}\) • 3x²
Answer: 3x-9
Explanation:
Given,
\(\frac{x^{-6}}{x^{5}}\) • 3x²
am / an = am-n
= x-6-5 .  3x²
= x-11 .  3x²
= 3x-9
So, \(\frac{x^{-6}}{x^{5}}\) • 3x² = 3x-9

Question 5.
\(\left(\frac{4 w^{3}}{2 z^{2}}\right)^{3}\)
Answer: \(\left(\frac{2 w^{3}}{z^{2}}\right)^{3}\)
Explanation:
Given,
\(\left(\frac{4 w^{3}}{2 z^{2}}\right)^{3}\)
4/2 = 2
\(\left(\frac{2 w^{3}}{z^{2}}\right)^{3}\)

Question 6.
\(\left(\frac{m^{7} \cdot m}{z^{2} \cdot m^{3}}\right)^{2}\)
Answer: (m5/z²)²
Explanation:
Given,
\(\left(\frac{m^{7} \cdot m}{z^{2} \cdot m^{3}}\right)^{2}\)
= (m8/z²m³)²
= (m5/z²)²
So, \(\left(\frac{m^{7} \cdot m}{z^{2} \cdot m^{3}}\right)^{2}\) = (m5/z²)²

Solve the literal equation for y.

Question 7.
4x + y = 2
Answer: 2 – 4x
Explanation:
Given,
4x + y = 2
Rewrite the equation so a different variable stands alone on one side of the equal sign.
We have to make y the subject.
y = 2 – 4x
So, the equation for y is 2 – 4x.

Question 8.
x − \(\frac{1}{3}\)y = −1
Answer: 3(x + 1)
Explanation:
Given,
x − \(\frac{1}{3}\)y = −1
Rewrite the equation so a different variable stands alone on one side of the equal sign.
Make y as the subject.
− \(\frac{1}{3}\)y = -1 – x
-(\(\frac{1}{3}\)y = -(1 + x)
\(\frac{1}{3}\)y = x + 1
y = 3(x + 1)
So, the equation for y is 3(x + 1).

Question 9.
2y − 9 = 13x
Answer:
https://ccssanswers.com/wp-admin/post.php?post=40609&action=edit

Question 10.
2xy + 6y = 10
Answer: 5/3 + x
Explanation:
Given,
2xy+6y=10
Take 2 as the common factor.
2(xy+3y)=2(5)
Divide by 2 on both sides.
2(xy+3y)​/2=2(5)/2
xy+3y=5xy+3y
=5xy+3y=5
Rewrite the equation so a different variable stands alone on one side of the equal sign.
Make y as the subject.
y(x+3)=5
y(x+3)​/x+3 = 5/3+x
y = 5/3 + x
So, the equation for y is 5/3 + x

Question 11.
8x − 4xy = 3
Answer: -3/4x+2
Explanation:
Given,
8x − 4xy = 3
Apply inverse operations to isolate 
8x4xy=3
4x(2y)=3
Rewrite the equation so a different variable stands alone on one side of the equal sign.
Make y as the subject.

2y=3/4x
y=3/4x2
y=1(3/4x2)
y=-3/4x+2
So, the equation for y is -3/4x+2

Question 12.
6x + 7xy = 15
Answer: (15 – 6x)/7x
Explanation:
Given,
6x + 7xy = 15
Rewrite the equation so a different variable stands alone on one side of the equal sign.
Make y as the subject.

7xy = 15 – 6x
y = (15 – 6x)/7x
So, the equation for y is (15 – 6x)/7x

Question 13.
ABSTRACT REASONING Is the order in which you apply properties of exponents important? Explain your reasoning.
Answer: Exponents are values that are powers that show how many times we have to multiply the base number by itself. The exponents are important to write the values of numbers in a simplified form.
The properties of the exponent are
1. Product of powers
2. Quotient of powers
3. Power to a power
4. Power of a product
5. Power of a quotient

Rational Exponents and Radical Functions Mathematical Practices

Monitoring Progress

Question 1.
Use the Pythagorean Theorem to find the exact lengths of a, b, c, and d in the figure.

Question 2.
Use a calculator to approximate each length to the nearest tenth of an inch.
Answer:
We can find the length and angle measured to the nearest tenth using the calculator.
a = √2
a = 1.414
a ≈ 1.41
b = √3
b = 1.732
b ≈ 1.73
c = 2
c is the whole number
d = √5
d = 2.2360
d ≈ 2.24

Question 3.
Use a ruler to check the reasonableness of your answers.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 1
Answer:
a = √2
b = √3
c = 2
d = √5
Explanation:
Given,
s = 1 in.
1² + 1² = a²
a = √1 + 1
a = √2
The measure of hypotenuse a is √2.
√2² + 1² = b²
2 + 1 = b²
b² = 3
b = √3
The measure of hypotenuse b is √3.
√3² + 1² = c²
c² = 3 + 1
c² = 4
c = 2
The measure of hypotenuse c is 2.
2² + 1² = d²
d² = 4 + 1
d² = 5
d = √5
The measure of hypotenuse d is √5.

Lesson 5.1 nth Roots and Rational Exponents

Essential Question

How can you use a rational exponent to represent a power involving a radical?
Previously, you learned that the nth root of a can be represented as
\(\sqrt[n]{a}\) = \(a^{1 / n}\) Definition of rational exponent
for any real number a and integer n greater than 1.

EXPLORATION 1
Exploring the Definition of a Rational Exponent
Work with a partner.
Use a calculator to show that each statement is true.
a. \(\sqrt{9}\) = \(9^{1 / 2}\)
Answer:
Given,
\(\sqrt{9}\) = \(9^{1 / 2}\)
In order to rewrite a radical using a fractional exponent, the power to which the radicand becomes the numerator and the root becomes the denominator.
Square is nothing but 2.
So, \(\sqrt{9}\) can be written in the fractional exponent as \(9^{1 / 2}\).
The statement \(\sqrt{9}\) = \(9^{1 / 2}\) is true.

b. \(\sqrt{2}\) = \(2^{1 / 2}\)
Answer:
Given,
\(\sqrt{2}\) = \(2^{1 / 2}\)
In order to rewrite a radical using a fractional exponent, the power to which the radicand becomes the numerator and the root becomes the denominator.
Square is nothing but 2.
So, \(\sqrt{2}\) can be written in the fractional exponent as \(2^{1 / 2}\).
The statement is true.

c. \(\sqrt[3]{8}\) = \(8^{1 / 3}\)
Answer:
Given,
\(\sqrt[3]{8}\) = \(8^{1 / 3}\)
In order to rewrite a radical using a fractional exponent, the power to which the radicand becomes the numerator and the root becomes the denominator.
Cube root is nothing but 1/3.
\(\sqrt[3]{8}\) can be written in the fractional exponent as \(8^{1 / 3}\).
The statement is true.

d. \(\sqrt[3]{3}\) = \(3^{1 / 3}\)
Answer:
Given,
\(\sqrt[3]{3}\) = \(3^{1 / 3}\)
In order to rewrite a radical using a fractional exponent, the power to which the radicand becomes the numerator and the root becomes the denominator.
Cube root is nothing but 1/3.
\(\sqrt[3]{3}\) can be written in the fractional exponent as \(3^{1 / 3}\).
The statement is true.

e. \(\sqrt[4]{16}\) = \(16^{1 / 4}\)
Answer:
Given,
\(\sqrt[4]{16}\) = \(16^{1 / 4}\)
In order to rewrite a radical using a fractional exponent, the power to which the radicand becomes the numerator and the root becomes the denominator.
The fourth root is nothing but 1/4.
\(\sqrt[4]{16}\) can be written in the fractional exponent as \(16^{1 / 4}\).
The statement is true.

f. \(\sqrt[4]{12}\) = \(12^{1 / 4}\)
Answer:
Given,
\(\sqrt[4]{12}\) = \(12^{1 / 4}\)
In order to rewrite a radical using a fractional exponent, the power to which the radicand becomes the numerator and the root becomes the denominator.
The fourth root is nothing but 1/4.
\(\sqrt[4]{12}\) can be written in the fractional exponent as \(12^{1 / 4}\).

EXPLORATION 2
Writing Expressions in Rational Exponent Form

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 2
Work with a partner. Use the definition of a rational exponent and the properties of exponents to write each expression as a base with a single rational exponent. Then use a calculator to evaluate each expression. Round your answer to two decimal places.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 3
a. \((\sqrt{5})^{3}\)
Answer: 11.08
Explanation:
Given,
\((\sqrt{5})^{3}\)
The square root of 5 is 2.23
(\(\sqrt{5}\))³ = (2.23)³ = 11.08

b. \((\sqrt[4]{4})^{2}\)
Answer: 1.99
Explanation:
Given,
\((\sqrt[4]{4})^{2}\)
The fourth root of 4 is 1.414
\((\sqrt[4]{4})^{2}\) = (1.414)² = 1.99

c. \((\sqrt[3]{9})^{2}\)
Answer: 4.32
Explanation:
Given,
\((\sqrt[3]{9})^{2}\)
The cube root of 9 is 2.08
\((\sqrt[3]{9})^{2}\) = (2.08)² = 4.32

d. \((\sqrt[5]{10})^{4}\)
Answer: 6.23
Explanation:
Given,
\((\sqrt[5]{10})^{4}\)
The fifth root of 10 is 1.58
\((\sqrt[5]{10})^{4}\) = (1.58)4 = 6.23

e. \((\sqrt{15})^{3}\)
Answer: 57.96
Explanation:
Given,
\((\sqrt{15})^{3}\)
the square root of 15 is 3.87
\((\sqrt{15})^{3}\) = (\(\sqrt{15}\))³ = (3.87)³ = 57.96

f. \((\sqrt[3]{27})^{4}\)
Answer: 81
Explanation:
Given,
\((\sqrt[3]{27})^{4}\)
The cube root of 27 is 3.
\((\sqrt[3]{27})^{4}\) = (3)4 = 81

EXPLORATION 3
Writing Expressions in Radical Form
Work with a partner.
Use the properties of exponents and the definition of a rational exponent to write each expression as a radical raised to an exponent. Then use a calculator to evaluate each expression. Round your answer to two decimal places.
Sample \(5^{2 / 3}\) = (\left(5^{1 / 3}\right)\(\))2 = (\((\sqrt[3]{5})\))2 ≈ 2.92
a. \(8^{2 / 3}\)
b. \(6^{5 / 2}\)
c. \(12^{3 / 4}\)
d. \(10^{3 / 2}\)
e. \(16^{3 / 2}\)
f. \(20^{6 / 5}\)
Answer:
a. \(8^{2 / 3}\) = (\((\sqrt[3]{8})\))²
The cube root of 8 is 2.
= 2² = 4
b. \(6^{5 / 2}\) = (\((\sqrt[5]{6})\))²
The fifth root of 6 is 1.43
= (1.43)² = 2.04
c. \(12^{3 / 4}\)
The cube root of 12 is 2.28
= (\((\sqrt[3]{12})\))4 = (2.28)4 = 27.02
d. \(10^{3 / 2}\)
The cube root of 10 is 2.15
= (\((\sqrt[3]{10})\))² =(2.15)² = 4.62
e. \(16^{3 / 2}\)
The cube root of 16 is 2.51
= (\((\sqrt[3]{16})\))² =  (2.51)² = 6.3
f. \(20^{6 / 5}\)
The sixth root of 20 is 1.82
= (\((\sqrt[6]{20})\))5 =  (1.82)5 = 19.96

Communicate Your Answer

Question 4.
How can you use a rational exponent to represent a power involving a radical?
Answer:
In order to rewrite a radical using a fractional exponent, the power to which the radicand becomes the numerator and the root becomes the denominator. Any power can be written in the fractional exponent form.

Question 5.
Evaluate each expression without using a calculator. Explain your reasoning.
a. \(4^{3 / 2}\)
b. \(32^{4 / 5}\)
c. \(625^{3 / 4}\)
d. \(49^{3 / 2}\)
e. \(125^{4 / 3}\)
f. \(100^{6 / 3}\)
Answer:
a. \(4^{3 / 2}\)
= 41/2
= 2
= 2³ = 8
So, \(4^{3 / 2}\) = 8
b. \(32^{4 / 5}\)
= 321/5×4
= 25(1/5×4)
= (2)4 = 16
So, \(32^{4 / 5}\) = 16
c. \(625^{3 / 4}\)
= 6251/4×3
= (625)1/4 = 5
= (5)³ = 125
d. \(49^{3 / 2}\)
= 491/2 = 7
= (7)³= 343
So, \(49^{3 / 2}\) = 343
e. \(125^{4 / 3}\)
= 1251/3
= (5)4 = 625
So, \(125^{4 / 3}\) = 625
f. \(100^{6 / 3}\)
= 1001/3
= 4.64
= (4.64)6 = 9979.
So, \(100^{6 / 3}\) = 9979

5.1 Lesson

Monitoring Progress

Question 1.
n = 4, a = 16
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-1.jpg

Question 2.
n = 2, a = −49
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-2.jpg

Question 3.
n = 3, a = −125
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-2.jpg

 

 

 

 

Question 4.
n = 5, a = 243
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-4.jpg

Evaluate the expression without using a calculator.

Question 5.
\(4^{5 / 2}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-5.jpg

Question 6.
\(9^{-1 / 2}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-6.jpg

Question 7.
\(81^{3 / 4}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-7.jpg

Question 8.
\(1^{7 / 8}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-8.jpg

Evaluate the expression using a calculator. Round your answer to two decimal places when appropriate.

Question 9.
\(6^{2 / 5}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-9.jpg

Question 10.
\(64^{-2 / 3}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-10.jpg

Question 11.
\((\sqrt[4]{16})^{5}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-11.jpg

Question 12.
\((\sqrt[3]{-30})^{2}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-12.jpg

Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 13.
8x3 = 64
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-13.jpg

Question 14.
\([\frac{1}{2}/latex]x5 = 512
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-13.jpg
Question 15.
(x + 5)4 = 16
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-15.jpg

Question 16.
(x − 2)3 = −14
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-16.jpg

Question 17.
WHAT IF? In Example 5, what is the annual depreciation rate when the salvage value is $6000?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-17.jpg

nth Roots and Rational Exponents 5.1 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY Rewrite the expression [latex]a^{-s / t}\) in radical form. Then state the index of the radical.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 1

Question 2.
COMPLETE THE SENTENCE For an integer n greater than 1, if bn = a, then b is a(n) ___________ of a.
Answer:

Question 3.
WRITING Explain how to use the sign of a to determine the number of real fourth roots of a and the number of real fifth roots of a.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 3

Question 4.
WHICH ONE DOESN’T BELONG? Which expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 4
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–10, find the indicated real nth root(s) of a.

Question 5.
n = 3, a = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 5

Question 6.
n = 5, a = −1
Answer:

Question 7.
n = 2, a = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 7

Question 8.
n = 4, a = 256
Answer:

Question 9.
n = 5, a = −32
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 9

Question 10.
n = 6, a = −729
Answer:

In Exercises 11–18, evaluate the expression without using a calculator.

Question 11.
\(64^{1 / 6}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 11

Question 12.
\(8^{1 / 3}\)
Answer:

Question 13.
\(25^{3 / 2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 13

Question 14.
\(81^{3 / 4}\)
Answer:

Question 15.
\((-243)^{1 / 5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 15

Question 16.
\((-64)^{4 / 3}\)
Answer:

Question 17.
\(8^{-2 / 3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 17

Question 18.
\(16^{-7 / 4}\)
Answer:

ERROR ANALYSIS In Exercises 19 and 20, describe and correct the error in evaluating the expression.

Question 19.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 19

Question 20.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 6
Answer:

USING STRUCTURE In Exercises 21–24, match the equivalent expressions. Explain your reasoning.

Question 21.
\((\sqrt[3]{5})^{4}\) A. \(5^{-1 / 4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 21

Question 22.
\((\sqrt[4]{5})^{3}\) B. \(5^{4 / 3}\)
Answer:

Question 23.
\(\frac{1}{\sqrt[4]{5}}\) C. \(-5^{1 / 4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 23

Question 24.
\(-\sqrt[4]{5}\) D. \(5^{3 / 4}\)
Answer:

In Exercises 25–32, evaluate the expression using a calculator. Round your answer to two decimal places when appropriate.

Question 25.
\(\sqrt[5]{32,768}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 25

Question 26.
\(\sqrt[7]{1695}\)
Answer:

Question 27.
\(25^{-1 / 3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 27

Question 28.
\(85^{1 / 6}\)
Answer:

Question 29.
\(20,736^{4 / 5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 29

Question 30.
\(86^{-5 / 6}\)
Answer:

Question 31.
\((\sqrt[4]{187})^{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 31

Question 32.
\((\sqrt[5]{-8})^{8}\)
Answer:

MATHEMATICAL CONNECTIONS In Exercises 33 and 34, find the radius of the figure with the given volume.

Question 33.
V = 216 ft3
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 33

Question 34.
V = 1332 cm3
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 8
Answer:

In Exercises 35–44, find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 35.
x3 = 125
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 35

Question 36.
5x3 = 1080
Answer:

Question 37.
(x + 10)5 = 70
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 37

Question 38.
(x − 5)4 = 256
Answer:

Question 39.
x5 = −48
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 39

Question 40.
7x4 = 56
Answer:

Question 41.
x6 + 36 = 100
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 41

Question 42.
x3 + 40 = 25
Answer:

Question 43.
\(\frac{1}{3}\)x4 = 27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 43

Question 44.
\(\frac{1}{6}\)x3 = −36
Answer:

Question 45.
MODELING WITH MATHEMATICS When the average price of an item increases from p1 to p2 over a period of n years, the annual rate of inflation r (in decimal form) is given by r = \(\left(\frac{p_{2}}{p_{1}}\right)^{1 / n}\) − 1. Find the rate of inflation for each item in the table.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 45.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 45.2

Question 46.
HOW DO YOU SEE IT? The graph of y = xn is shown in red. What can you conclude about the value of n? Determine the number of real nth roots of a. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 10
Answer:

Question 47.
NUMBER SENSE Between which two consecutive integers does \(\sqrt[4]{125}\) lie? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 47

Question 48.
THOUGHT PROVOKING In 1619, Johannes Kepler published his third law, which can be given by d3 = t2, where d is the mean distance (in astronomical units) of a planet from the Sun and t is the time (in years) it takes the planet to orbit the Sun. It takes Mars 1.88 years to orbit the Sun. Graph a possible location of Mars. Justify your answer. (The diagram shows the Sun at the origin of the xy-plane and a possible location of Earth.)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 11
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.1-48.jpg

Question 49.
PROBLEM SOLVING A weir is a dam that is built across a river to regulate the flow of water. The flow rate Q (in cubic feet per second) can be calculated using the formula Q= 3.367ℓ\(h^{3 / 2}\), where ℓ is the length (in feet) of the bottom of the spillway and his the depth (in feet) of the water on the spillway. Determine the flow rate of a weir with a spillway that is 20 feet long and has a water depth of 5 feet.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 49

Question 50.
REPEATED REASONING The mass of the particles that a river can transport is proportional to the sixth power of the speed of the river. A certain river normally flows at a speed of 1 meter per second. What must its speed be in order to transport particles that are twice as massive as usual? 10 times as massive? 100 times as massive?
Answer:

Maintaining Mathematical Proficiency

Simplify the expression. Write your answer using only positive exponents. (Skills Review Handbook)

Question 51.
5 • 54

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 51

Question 52.
\(\frac{4^{2}}{4^{7}}\)
Answer:

Question 53.
\(\left(z^{2}\right)^{-3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 53

Question 54.
\(\left(\frac{3 x}{2}\right)^{4}\)
Answer:

Write the number in standard form. (Skills Review Handbook)

Question 55.
5 × 103

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 55

Question 56.
4 × 10−2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-56.jpg
Question 57.
8.2 × 10−1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.1 Question 57

Question 58.
6.93 × 106
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.1-58.jpg
Lesson 5.2 Properties of Rational Exponents and Radicals

Essential Question
How can you use properties of exponents to simplify products and quotients of radicals?

EXPLORATION 1
Reviewing Properties of Exponents
Work with a partner.
Let a and b be real numbers. Use the properties of exponents to complete each statement. Then match each completed statement with the property it illustrates.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 13
Answer:
Big-Ideas-Math-Algebra-2-Answers-Chapter-5-Rational-Exponents-and-Radical-Functions-13
Explanation:
a. Negative Exponent: The negative exponent tells how many times we have to divide the base number.
b. Power of Product: The power of a product states that if a term is being to an exponent, you can distribute the power to each factor.
c. Power of a power: Multiply powers together when raising a power by another exponent.
d. Product of Powers: Add powers together when multiplying like bases.
e. Power of a Quotient: Distribute power to all values in a quotient.
f. Quotient of powers: Subtract powers when dividing the bases.
g. Zero Exponent: Any base raised to the power of 0, it gives the result as 1.

EXPLORATION 2
Simplifying Expressions with Rational Exponents
Work with a partner.
Show that you can apply the properties of integer exponents to rational exponents by simplifying each expression. Use a calculator to check your answers.
a. \(5^{2 / 3}\) • \(5^{4 / 3}\)
b. \(3^{1 / 5}\) • \(3^{4 / 5}\)
c. \(\left(4^{2 / 3}\right)^{3}\)
d. \(\frac{\sqrt{98}}{\sqrt{2}}\)
e. \(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\)
f. \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)

Answer:
a. \(5^{2 / 3}\) • \(5^{4 / 3}\) = 25
Explanation:
Given,
\(5^{2 / 3}\) • \(5^{4 / 3}\)
When bases are equal powers should be added.
52/3+4/3 = 6/3 = 2
5² = 25
b. \(3^{1 / 5}\) • \(3^{4 / 5}\) = 3
Explanation:
Given,
\(3^{1 / 5}\) • \(3^{4 / 5}\)
=31/5+4/5 = 5/5 = 1
=3
c. \(\left(4^{2 / 3}\right)^{3}\) = 16
Explanation:
Given,
\(\left(4^{2 / 3}\right)^{3}\)
42/3×3 = 2
=4²
= 16
d. \(\frac{\sqrt{98}}{\sqrt{2}}\) = 7
Explanation:
Given,
\(\frac{\sqrt{98}}{\sqrt{2}}\)
√(98/2) = √49 = 7
e. \(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\) = 5
Explanation:
Given,
\(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\)
\(\frac{\sqrt[4]}\)(4/1024) = \(\frac{\sqrt[4]}\)1/231
f. \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)
\(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)
= 5

EXPLORATION 3
Simplifying Products and Quotients of Radicals
Work with a partner.
Use the properties of exponents to write each expression as a single radical. Then evaluate each expression. Use a calculator to check your answers.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 14
a. \(\sqrt{3}\) • \(\sqrt{12}\)
b. \(\sqrt[3]{5}\) • \(\sqrt[3]{25}\)
c. \(\sqrt[4]{27}\) • \(\sqrt[4]{3}\)
d. \(\frac{\sqrt{98}}{\sqrt{2}}\)
e. \(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\)
f. \(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)
Answer:
a.
Given,
\(\sqrt{3}\) • \(\sqrt{12}\)
\(\sqrt{a}\) • \(\sqrt{b}\) = \(\sqrt{ab}\)
= \(\sqrt{36}\)
= 6
b.
Given,
\(\sqrt[3]{5}\) • \(\sqrt[3]{25}\)
\(\sqrt{a}\) • \(\sqrt{b}\) = \(\sqrt{ab}\)
= \(\sqrt[3]{125}\)
= 5
c.
Given,
\(\sqrt[4]{27}\) • \(\sqrt[4]{3}\)
\(\sqrt{a}\) • \(\sqrt{b}\) = \(\sqrt{ab}\)
\(\sqrt[4]{81}\)
= 3
d.
Given,
\(\frac{\sqrt{98}}{\sqrt{2}}\)
√(98/2) = √49 = 7
e. \(\frac{\sqrt[4]{4}}{\sqrt[4]{1024}}\)
\(\frac{\sqrt[4]}\)(4/1024) = \(\frac{\sqrt[4]}\)1/231
f.
Given,
\(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\)
\(\frac{\sqrt[3]{625}}{\sqrt[3]{5}}\) = 5

Communicate Your Answer

Question 4.
How can you use properties of exponents to simplify products and quotients of radicals?
Answer: In order to rewrite radical expression applying first factor of expression which is under root making perfect power.
Property A: xm . xn = xm+n
Example:
52 . 52 = 54
Property B: am/bn = am-n
52/51 = 52-1= 5
Property C:
(am)n = amn
(52)1 = 52

Question 5.
Simplify each expression.
a. \(\sqrt{27}\) • \(\sqrt{6}\)
Answer:
Given,
\(\sqrt{27}\) • \(\sqrt{6}\)
\(\sqrt{162}\)
b. \(\frac{\sqrt[3]{240}}{\sqrt[3]{15}}\)
Answer:
Given,
\(\frac{\sqrt[3]{240}}{\sqrt[3]{15}}\)
\(\frac{\sqrt[3]{240/15}\) = 16
c. (\(5^{1 / 2}\) \(16^{1 / 4}\))²
Answer:
Given,
(\(5^{1 / 2}\) \(16^{1 / 4}\))²
(\(5^{1 / 2}\) \(16^{1 / 4}\))²
5 × \(16^{1 / 2}\)
5 × 4 = 20

5.2 Lesson

Monitoring Progress

Simplify the expression.

Question 1.
\(2^{3 / 4}\) • \(2^{1 / 2}\)
Answer:
Given,
\(2^{3 / 4}\) • \(2^{1 / 2}\)
xm . xn = xm+n
23/4+1/2 = 5/4
25/4
So, \(2^{3 / 4}\) • \(2^{1 / 2}\) = 25/4

Question 2.
\(\frac{3}{3^{1 / 4}}\)
Answer:

Question 3.
\(\left(\frac{20^{1 / 2}}{5^{1 / 2}}\right)^{3}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-3.jpg

Question 4.
(\(5^{1 / 3}\) • \(7^{1 / 4}\))3

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-4.jpg

 

Simplify the expression

Question 5.
\(\sqrt[4]{27}\) • \(\sqrt[4]{3}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-4.jpg

Question 6.
\(\frac{\sqrt[3]{250}}{\sqrt[3]{2}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-6.jpg

Question 7.
\(\sqrt[3]{104}\)
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-7.jpg

Question 8.
\(\sqrt[5]{\frac{3}{4}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-8.jpg

Question 9.
\(\frac{3}{6-\sqrt{2}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-9.jpg

Question 10.
\(7 \sqrt[5]{12}\) – \(\sqrt[5]{12}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-10.jpg

Question 11.
4(\(9^{2 / 3}\)) + (\(9^{2 / 3}\))
Answer:

Question 12.
\(\sqrt[3]{5}\) + \(\sqrt[3]{40}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-12.jpg

Simplify the expression. Assume all variables are positive.

Question 13.
\(\sqrt[3]{27 q^{9}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-13.jpg

Question 14.
\(\sqrt[5]{\frac{x^{10}}{y^{5}}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-14.jpg

Question 15.
\(\frac{6 x y^{3 / 4}}{3 x^{1 / 2} y^{1 / 2}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-15.jpg

Question 16.
\(\sqrt{9} w^{5}\) – \(w \sqrt{w^{3}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.2-16.jpg

Properties of Rational Exponents and Radicals 5.2 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING How do you know when a radical expression is in simplest form?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 1

Question 2.
WHICH ONE DOESN’T BELONG? Which radical expression does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 15
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, use the properties of rational exponents to simplify the expression.

Question 3.
\(\left(9^{2}\right)^{1 / 3}\)
Answer:

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 3

Question 4.
\(\left(12^{2}\right)^{1 / 4}\)
Answer:

Question 5.
\(\frac{6}{6^{1 / 4}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 5

Question 6.
\(\frac{7}{7^{1 / 3}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-6.jpg

Question 7.
\(\left(\frac{8^{4}}{10^{4}}\right)^{-1 / 4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 7

Question 8.
\(\left(\frac{9^{3}}{6^{3}}\right)^{-1 / 3}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-8.jpg

Question 9.
(\(3^{-2 / 3}\) • \(3^{1 / 3}\))-1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-9.jpg
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 9

Question 10.
(\(5^{1 / 2}\) • \(5^{-3 / 2}\))-1/4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-10.jpg

Question 11.
\(\frac{2^{2 / 3} \cdot 16^{2 / 3}}{4^{2 / 3}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 11

Question 12.
\(\frac{49^{3 / 8} \cdot 49^{7 / 8}}{7^{5 / 4}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-12.jpg

In Exercises 13–20, use the properties of radicals to simplify the expression.

Question 13.
\(\sqrt{2}\) • \(\sqrt{72}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 13

Question 14.
\(\sqrt[3]{16}\) • \(\sqrt[3]{32}\)
Answer:

Question 15.
\(\sqrt[4]{6}\) • \(\sqrt[4]{8}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 15

Question 16.
\(\sqrt[4]{8}\) • \(\sqrt[4]{8}\)
Answer:

Question 17.
\(\frac{\sqrt[5]{486}}{\sqrt[5]{2}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 17

Question 18.
\(\frac{\sqrt{2}}{\sqrt{32}}\)
Answer:

Question 19.
\(\frac{\sqrt[3]{6} \cdot \sqrt[3]{72}}{\sqrt[3]{2}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 19

Question 20.
\(\frac{\sqrt[3]{3} \cdot \sqrt[3]{18}}{\sqrt[6]{2} \cdot \sqrt[6]{2}}\)
Answer:

In Exercises 21–28, write the expression in simplest form.

Question 21.
\(\sqrt[4]{567}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 21

Question 22.
\(\sqrt[5]{288}\)
Answer:

Question 23.
\(\frac{\sqrt[3]{5}}{\sqrt[3]{4}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 23

Question 24.
\(\frac{\sqrt[4]{4}}{\sqrt[4]{27}}\)
Answer:

Question 25.
\(\sqrt{\frac{3}{8}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 25

Question 26.
\(\sqrt[3]{\frac{7}{4}}\)
Answer:

Question 27.
\(\sqrt[3]{\frac{64}{49}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 27

Question 28.
\(\sqrt[4]{\frac{1296}{25}}\)
Answer:

In Exercises 29–36, write the expression in simplest form.

Question 29.
\(\frac{1}{1+\sqrt{3}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 29

Question 30.
\(\frac{1}{2+\sqrt{5}}\)
Answer: √5 + 2

Explanation:
Given,
\(\frac{1}{2+\sqrt{5}}\) × \(\frac{2-\sqrt{5}{2-\sqrt{5}}\)
= \(\frac{2-\sqrt{5}{4-5}\)
= 2-√5/-1
= √5 + 2
Thus, \(\frac{1}{2+\sqrt{5}}\) = √5 + 2

Question 31.
\(\frac{5}{3-\sqrt{2}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 31

Question 32.
\(\frac{11}{9-\sqrt{6}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-32.jpg

Question 33.
\(\frac{9}{\sqrt{3}+\sqrt{7}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 33

Question 34.
\(\frac{2}{\sqrt{8}+\sqrt{7}}\)
Answer:

Question 35.
\(\frac{\sqrt{6}}{\sqrt{3}-\sqrt{5}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 35

Question 36.
\(\frac{\sqrt{7}}{\sqrt{10}-\sqrt{2}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-36.jpg

In Exercises 37–46, simplify the expression.

Question 37.
\(9 \sqrt[3]{11}\) + \(3 \sqrt[3]{11}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 37

Question 38.
\(8 \sqrt[6]{5}\) – \(12 \sqrt[6]{5}\)
Answer:

Question 39.
\(3\left(11^{1 / 4}\right)\) + \(9\left(11^{1 / 4}\right)\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 39

Question 40.
\(13\left(8^{3 / 4}\right)\) – \(4\left(8^{3 / 4}\right)\)
Answer:

Question 41.
\(5 \sqrt{12}\) – \(19 \sqrt{3}\)
Answer:

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 41

Question 42.
\(27 \sqrt{6}\) + \(7 \sqrt{150}\)
Answer: 34\(\sqrt{6}\)

Explanation:
Given,
\(27 \sqrt{6}\) + \(7 \sqrt{6.25}\)
= \(27 \sqrt{6}\) + \(35 \sqrt{6}\)
Take \( \sqrt{6}\) as common factor.
= \( \sqrt{6}\)(27+7)
= 34\(\sqrt{6}\)
So, \(27 \sqrt{6}\) + \(7 \sqrt{150}\) = 34\(\sqrt{6}\)

Question 43.
\(\sqrt[5]{224}\) + \(3 \sqrt[5]{7}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 43

Question 44.
\(7 \sqrt[3]{2}\) – \(\sqrt[3]{128}\)
Answer:
Given,
\(7 \sqrt[3]{2}\) – \(\sqrt[3]{128}\)
\(\sqrt[3]{128}\) = 2\(\sqrt[3]{16}\)
\(7 \sqrt[3]{2}\) – 2\(\sqrt[3]{16}\)
So, \(7 \sqrt[3]{2}\) – \(\sqrt[3]{128}\) = \(7 \sqrt[3]{2}\) – 2\(\sqrt[3]{16}\)

Question 45.
\(5\left(24^{1 / 3}\right)\) – 4(\(\left(3^{1 / 3}\right)\))
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 45

Question 46.
\(5^{1 / 4}\) + 6(\([405^{1 / 4}/latex])
Answer:

Question 47.
ERROR ANALYSIS Describe and correct the error in simplifying the expression.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 47

Question 48.
MULTIPLE REPRESENTATIONS Which radical expressions are like radicals?
A. [latex]\left(5^{2 / 9}\right)^{3 / 2}\)
B. \(\frac{5^{3}}{(\sqrt[3]{5})^{8}}\)
C. \(\sqrt[3]{625}\)
D. \(\sqrt[3]{5} 145\) – \(\sqrt[3]{875}\)
E. \(\sqrt[3]{5}\) + \(3 \sqrt[3]{5}\)
F. \(7 \sqrt[4]{80}\) – \(2 \sqrt[4]{405}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-48.jpg

In Exercises 49–54, simplify the expression.

Question 49.
\(\sqrt[4]{81 y^{8}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 49

Question 50.
\(\sqrt[3]{64 r^{3} t^{6}}\)
Answer:
Given,
\(\sqrt[3]{64 r^{3} t^{6}}\)
= 4r\(\sqrt{t^{6}}\)

Question 51.
\(\sqrt[5]{\frac{m^{10}}{n^{5}}}\)
Answer:

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 51

Question 52.
\(\sqrt[4]{\frac{k^{16}}{16 z^{4}}}\)
Answer:

Question 53.
\(\sqrt[6]{\frac{g^{6} h}{h^{7}}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 53

Question 54.
\(\sqr

Question 55.
ERROR ANALYSIS Describe and correct the error in simplifying the expression.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 17
Answer:

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 55

Question 56.
OPEN-ENDED Write two variable expressions involving radicals, one that needs absolute value in simplifying and one that does not need absolute value. Justify your answers.
Answer:

In Exercises 57–64, write the expression in simplest form. Assume all variables are positive.

Question 57.
[latex]\sqrt{81 a^{7} b^{12} c^{9}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 57

Question 58.
\(\sqrt[3]{125 r^{4} s^{9} t^{7}}\)
Answer:

Question 59.
\(\sqrt[5]{\frac{160 m^{6}}{n^{7}}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 59

Question 60.
\(\sqrt[4]{\frac{405 x^{3} y^{3}}{5 x^{-1} y}}\)
Answer:

Question 61.
\(\frac{\sqrt[3]{w} \cdot \sqrt{w^{5}}}{\sqrt{25 w^{16}}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 61

Question 62.
\(\frac{\sqrt[4]{v^{6}}}{\sqrt[7]{v^{5}}}\)
Answer:

Question 63.
\(\frac{18 w^{1 / 3} v^{5 / 4}}{27 w^{4 / 3} v^{1 / 2}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 63

Question 64.
\(\frac{7 x^{-3 / 4} y^{5 / 2} z^{-2 / 3}}{56 x^{-1 / 2} y^{1 / 4}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-64.jpg

In Exercises 65–70, perform the indicated operation. Assume all variables are positive.

Question 65.
\(12 \sqrt[3]{y}\) + \(9 \sqrt[3]{y}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 65

Question 66.
\(11 \sqrt{2 z}\) – \(5 \sqrt{2 z}\)
Answer:
Given,
\(11 \sqrt{2 z}\) – \(5 \sqrt{2 z}\)
Taking \( \sqrt{2 z}\) as common
(11 – 5)\( \sqrt{2 z}\)
6\( \sqrt{2 z}\)
So, \(11 \sqrt{2 z}\) – \(5 \sqrt{2 z}\) = 6\( \sqrt{2 z}\)

Question 67.
\(3 x^{7 / 2}\) – 5\(x^{7 / 2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 67

Question 68.
\(7 \sqrt[3]{m^{7}}\) + \(3 m^{7 / 3}\)
Answer:
Given,
\(7 \sqrt[3]{m^{7}}\) + \(3 m^{7 / 3}\)
\(7 m^{7 / 3}\) + \(3 m^{7 / 3}\)
Taking \(m^{7 / 3}\) as common.
(7 + 3)\(m^{7 / 3}\)
10\(m^{7 / 3}\)
So, \(7 \sqrt[3]{m^{7}}\) + \(3 m^{7 / 3}\) = 10\(m^{7 / 3}\)

Question 69.
\(\sqrt[4]{16 w^{10}}\) + \(2 w \sqrt[4]{w^{6}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 69

Question 70.
(p1/2 • p1/4) – \(\sqrt[4]{16 p^{3}}\)
Answer:

MATHEMATICAL CONNECTIONS In Exercises 71 and 72, find simplified expressions for the perimeter and area of the given figure.

Question 71.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 18
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 71

Question 72
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 19
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-72.jpg

Question 73.
MODELING WITH MATHEMATICS The optimum diameter d (in millimeters) of the pinhole in a pinhole camera can be modeled by d = 1.9[(5.5 × 10−4)ℓ]1/2, where ℓ is the length (in millimeters) of the camera box. Find the optimum pinhole diameter for a camera box with a length of 10 centimeters.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 20
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 73

Question 74.
MODELING WITH MATHEMATICS The surface area S(in square centimeters) of a mammal can be modeled by S = km2/3, where m is the mass (in grams) of the mammal and k is a constant. The table shows the values of k for different mammals.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 21
a. Find the surface area of a bat whose mass is 32 grams.
b. Find the surface area of a rabbit whose mass is 3.4 kilograms (3.4 × 103 grams).
c. Find the surface area of a human whose mass is 59 kilograms.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-74.jpg

Question 75.
MAKING AN ARGUMENT Your friend claims it is not possible to simplify the expression 7\(\sqrt{11}\) − 9 \(\sqrt{44}\) because it does not contain like radicals. Is your friend correct? Explain your reasoning.
Answer:

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 75

Question 76.
PROBLEM SOLVING The apparent magnitude of a star is a number that indicates how faint the star is in relation to other stars. The expression f(x) = \(\frac{2.512^{m_{1}}}{2.512^{m_{2}}}\) tells how many times fainter a star with apparent magnitude m1 is than a star with apparent magnitude m2.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 22
a. How many times fainter is Altair than Vega?
b. How many times fainter is Deneb than Altair?
c. How many times fainter is Deneb than Vega?
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 23
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-76.jpg

Question 77.
CRITICAL THINKING Find a radical expression for the perimeter of the triangle inscribed in the square shown. Simplify the expression.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 24
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 77

Question 78.
HOW DO YOU SEE IT? Without finding points, match the functions f(x) = \(\sqrt{64 x^{2}}\) and g(x) = \(\sqrt[3]{64 x^{6}}\) with their graphs. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 25
Answer:

Question 79.
REWRITING A FORMULA You have filled two round balloons with water. One balloon contains twice as much water as the other balloon.
a. Solve the formula for the volume of a sphere, V = \(\frac{4}{3}\)πr3, for r.
b. Substitute the expression for r from part (a) into the formula for the surface area of a sphere, S = 4πr2. Simplify to show that S = (4π)1/3(3V)2/3.
c. Compare the surface areas of the two water balloons using the formula in part (b).
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 79.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 79.2

Question 80.
THOUGHT PROVOKING Determine whether the expressions (x2)1/6 and (x1/6)2 are equivalent for all values of x.
Answer:

Question 81.
DRAWING CONCLUSIONS Substitute different combinations of odd and even positive integers for m and n in the expression \(\sqrt[n]{x^{m}}\). When you cannot assume x is positive, explain when absolute value is needed in simplifying the expression.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 81
Maintaining Mathematical Proficiency

Identify the focus, directrix, and axis of symmetry of the parabola. Then graph the equation. (Section 2.3)

Question 82.
y = 2x2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-82.jpg

Question 83.
y2 = −x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 83.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 83.2

Question 84.
y2 = 4x
Answer:

Write a rule for g. Describe the graph of g as a transformation of the graph of f. (Section 4.7)

Question 85.
f(x) = x4 − 3x2 − 2x, g(x) = −f(x)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 85

Question 86.
f(x) = x3 − x, g(x) = f(x) − 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-86.jpg

Question 87.
f(x) = x3 − 4, g(x) = f(x − 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.2 Question 87

Question 88.
f(x) = x4 + 2x3 − 4x2, g(x) = f(2x)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.2-88.jpg

Lesson 5.3 Graphing Radical Functions

Essential Question

How can you identify the domain and range of a radical function?

EXPLORATION 1
Identifying Graphs of Radical Functions
Work with a partner.
Match each function with its graph. Explain your reasoning. Then identify the domain and range of each function.
a. f(x) = \(\sqrt{x}\)
b. f(x) = \(\sqrt[3]{x}\)
c. f(x) = \(\sqrt[4]{x}\)
d. f(x) = \(\sqrt[5]{x}\)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 26
Answer:
a.
f(x) = \(\sqrt{x}\) suits with graph D.
So, D is the correct answer.
b. f(x) = \(\sqrt[3]{x}\) suits with graph C.
So, Graph C is the correct answer.
c. f(x) = \(\sqrt[4]{x}\) suits with graph A.
So, Graph A is the correct answer.
d. f(x) = \(\sqrt[5]{x}\) suits with graph B.
So, Graph B is the correct answer.

EXPLORATION 2
Identifying Graphs of Transformations
Work with a partner.
Match each transformation of f(x) = \(\sqrt{x}\) with its graph. Explain your reasoning. Then identify the domain and range of each function.
a. g(x) = \(\sqrt{x+2}\)
b. g(x) = \(\sqrt{x-2}\)
c. g(x) = \(\sqrt{x}+2-2\)
d. g(x) = −\(\sqrt{x+2}\)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 27

Communicate Your Answer

Question 3.
How can you identify the domain and range of a radical function?
Answer:

  • The domain and range of a radical function f(x) having an even order are all positive numbers and zero as the radicand takes only positive values.
  • The domain and range of a radical function f(x) having an odd order are all real numbers as radicand can take the value of any real number. f(x) = √x.

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 28
Question 4.
Use the results of Exploration 1 to describe how the domain and range of a radical function are related to the index of the radical.
Answer:

  • The functions \(\sqrt{x}\) and \(\sqrt[4]{x}\) have the domain and the range as [0, ∞)
  • The functions \(\sqrt[3]{x}\) and \(\sqrt[5]{x}\) have the domain and the range as (-∞, ∞)
  • If the index of the given radical is odd, its domain and range will not have any restriction. Whereas if it is even, they will have restrictions.

5.3 Lesson

Monitoring Progress

Question 1.
Graph g(x) = \(\sqrt{x+1}\). Identify the domain and range of the function.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.3-1.jpg

Question 2.
Describe the transformation of f(x) = \(\sqrt[3]{x}\) represented by g(x) = −\(\sqrt[3]{x}\) − 2. Then graph each function.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.3-2.jpg

Question 3.
WHAT IF? In Example 3, the function N(d ) = 2.4 • E(d) approximates the number of seconds it takes a dropped object to fall d feet on the Moon. Write a rule for N. How long does it take a dropped object to fall 25 feet on the Moon?
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.3-3.jpg

Question 4.
In Example 4, is the transformed function the same when you perform the translation followed by the horizontal shrink? Explain your reasoning.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.3-3.jpg

Question 5.
Use a graphing calculator to graph −4y2 = x + 1. Identify the vertex and the direction that the parabola opens.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.3-5.jpg

Question 6.
Use a graphing calculator to graph x2 + y2 = 25. Identify the radius and the intercepts.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-monitoring-5.3-6.jpg

Graphing Radical Functions 5.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE Square root functions and cube root functions are examples of __________ functions.
Answer:
Square root functions and cube root functions are examples of radical functions.

Question 2.
COMPLETE THE SENTENCE When graphing y = a\(\sqrt[3]{x-h}\) + k, translate the graph of y = a\(\sqrt[3]{x}\)h units __________ and k units __________.
Answer:
Square root functions and cube root functions are examples of radical functions

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, match the function with its graph.

Question 3.
f(x) = \(\sqrt{x}+3\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 3

Question 4.
h(x) = \(\sqrt{x}\) + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-4.jpg

Question 5.
f(x) = \(\sqrt{x-3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 5

Question 6.
g(x) = \(\sqrt{x}\) − 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-6.jpg

Question 7.
h(x) = \(\sqrt{x+3}\) − 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 7

Question 8.
f(x) = \(\sqrt{x-3}\) + 3
Answer:

Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 28.1

In Exercises 9–18, graph the function. Identify the domain and range of the function.

Question 9.
h(x) = \(\sqrt{x}\) + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 9

Question 10.
g(x) = \(\sqrt{x}\) − 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-10.jpg

Question 11.
g(x) = − \(\sqrt[3]{2 x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 11

Question 12.
f(x) = \(\sqrt[3]{-5 x}\)
Answer:

Question 13.
g(x) = \(\frac{1}{5} \sqrt{x}-3\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 13

Question 14.
f(x) = \(\frac{1}{2} \sqrt[3]{x}+6\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-14.jpg

Question 15.
f(x) = \((6 x)^{1 / 2}\) + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 15

Question 16.
g(x) = −3(x + 1)1/3

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-16.jpg

Question 17.
h(x) = −\(\sqrt[4]{x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 17

Question 18.
h(x) = \(\sqrt[5]{2 x}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-18.jpg

In Exercises 19–26, describe the transformation of f represented by g. Then graph each function.

Question 19.
f(x) = \(\sqrt{x}\), g(x) = \(\sqrt{x+1}\) + 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 19

Question 20.
f(x) = \(\sqrt{x}\), g(x) = 2\(\sqrt{x-1}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-20.jpg

Question 21.
f(x) = \(\sqrt[3]{x}\), g(x) = −\(\sqrt[3]{x}\) − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 21

Question 22.
f(x) = \(\sqrt[3]{x}\), g(x) = \(\sqrt[3]{x+4}\) − 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-22.jpg

Question 23.
f(x) = \(x^{1 / 2}\), g(x) = \(\frac{1}{4}(-x)^{1 / 2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 23

Question 24.
f(x) = \(x^{1 / 3}\), g(x) = \(\frac{1}{3} x^{1 / 3}\) + 6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-24.jpg

Question 25.
f(x) = \(\sqrt[4]{x}\), g(x) = \(2 \sqrt[4]{x+5}\) − 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 25

Question 26.
f(x) = \(\sqrt[5]{x}\), g(x) = \(\sqrt[5]{-32 x}\) + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-26.jpg

Question 27.
ERROR ANALYSIS Describe and correct the error in graphing f(x) = \(\sqrt{x-2}\) − 2.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 29
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 27

Question 28.
ERROR ANALYSIS Describe and correct the error in describing the transformation of the parent square root function represented by g(x) = \(\sqrt{\frac{1}{2} x}\) + 3.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 30
Answer:
The graph of g is a horizontal stretch by a factor of 2 and translation 3 units up of the parent square root function.

USING TOOLS In Exercises 29–34, use a graphing calculator to graph the function. Then identify the domain and range of the function.

Question 29.
g(x) = \(\sqrt{x^{2}+x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 29

Question 30.
h(x) = \(\sqrt{x^{2}-2 x}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-30.jpg

Question 31.
f(x) = \(\sqrt[3]{x^{2}+x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 31

Question 32.
f(x) = \(\sqrt[3]{3 x^{2}-x}\)
Answer:

Question 33.
f(x) = \(\sqrt{2 x^{2}+x+1}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 33

Question 34.
h(x) = \(\sqrt[3]{\frac{1}{2} x^{2}-3 x+4}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-34.jpg

ABSTRACT REASONING In Exercises 35–38, complete the statement with sometimes, always, or never.

Question 35.
The domain of the function y = a\(\sqrt{x}\) is ______ x ≥ 0.
Answer: The domain of the function y = a\(\sqrt{x}\) is always x ≥ 0.

Question 36.
The range of the function y = a\(\sqrt{x}\) is ______ y ≥ 0.
Answer: The range of the function y = a\(\sqrt{x}\) is sometimes y ≥ 0.

Question 37.
The domain and range of the function y = \(\sqrt[3]{x-h}\) + k are ________ all real numbers.
Answer: The domain and range of the function y = \(\sqrt[3]{x-h}\) + k are always all real numbers.

Question 38.
The domain of the function y = a\(\sqrt{-x}\) + k is ________ x ≥ 0.
Answer: The domain of the function y = a\(\sqrt{-x}\) + k is never x ≥ 0.

Question 39.
PROBLEM SOLVING The distance (in miles) a pilot can see to the horizon can be approximated by E(n) = 1.22\(\sqrt{n}\), where n is the plane’s altitude (in feet above sea level) on Earth. The function M(n) = 0.75E(n) approximates the distance a pilot can see to the horizon n feet above the surface of Mars. Write a rule for M. What is the distance a pilot can see to the horizon from an altitude of 10,000 feet above Mars?
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 31
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 39

Question 40.
MODELING WITH MATHEMATICS The speed (in knots) of sound waves in air can be modeled by v(K) = 643.855\(\sqrt{\frac{K}{273.15}}\) where K is the air temperature (in kelvin). The speed (in meters per second) of sound waves in air can be modeled by s(K) = \(\frac{v(K)}{1.944}\). Write a rule for s. What is the speed (in meters per second) of sound waves when the air temperature is 305 kelvin?
Answer:

In Exercises 41–44, write a rule for g described by the transformations of the graph of f.

Question 41.
Let g be a vertical stretch by a factor of 2, followed by a translation 2 units up of the graph of f(x) = \(\sqrt{x}\) + 3.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 41

Question 42.
Let g be a reflection in the y-axis, followed by a translation 1 unit right of the graph of f(x) = \(2 \sqrt[3]{x-1}\).
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-42.jpg

Question 43.
Let g be a horizontal shrink by a factor of \(\frac{2}{3}\), followed by a translation 4 units left of the graph of f(x) = \(\sqrt{6 x}\).
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 43

Question 44.
Let g be a translation 1 unit down and 5 units right, followed by a reflection in the x-axis of the graph of f(x) = −\(-\frac{1}{2} \sqrt[4]{x}+\frac{3}{2}\)
Answer:

In Exercises 45 and 46, write a rule for g.

Question 45.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 32
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 45

Question 46.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 33
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-46.jpg

In Exercises 47–50, write a rule for g that represents the indicated transformation of the graph of f.

Question 47.
f(x) = 2\(\sqrt{x}\), g(x) = f(x + 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 47

Question 48.
f(x) = \(\frac{1}{3} \sqrt{x-1}\), g(x) = −f(x) + 9
Answer:

Question 49.
f(x) = −\(\sqrt{x^{2}-2}\), g(x) = −2f(x + 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 49

Question 50.
f(x) = \(\sqrt[3]{x^{2}+10 x}\), g(x) =\(\frac{1}{4}\)f(−x) + 6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-50.jpg

In Exercises 51–56, use a graphing calculator to graph the equation of the parabola. Identify the vertex and the direction that the parabola opens.

Question 51.
\(\frac{1}{4}\)y2 = x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 51

Question 52.
3y2 = x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-52.jpg

Question 53.
−8y2 + 2 = x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 53

Question 54.
2y2 = x − 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-54.jpg

Question 55.
x + 8 = \(\frac{1}{5}\)y2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 55

Question 56.
\(\frac{1}{2}\)x = y2 − 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-56.jpg

In Exercises 57–62, use a graphing calculator to graph the equation of the circle. Identify the radius and the intercepts.

Question 57.
x2 + y2 = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 57

Question 58.
x2 + y2 = 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-58.jpg

Question 59.
1 − y2 = x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 59

Question 60.
64 − x2 = y2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-60.jpg

Question 61.
−y2 = x2 − 36
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 61

Question 62.
x2 = 100 − y2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-62.jpg

Question 63.
MODELING WITH MATHEMATICS The period of a pendulum is the time the pendulum takes to complete one back-and-forth swing. The period T (in seconds) can be modeled by the function T = 1.11\(\sqrt{\ell}\), where ℓ is the length (in feet) of the pendulum. Graph the function. Estimate the length of a pendulum with a period of 2 seconds. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 34
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 63

Question 64.
HOW DO YOU SEE IT? Does the graph represent a square root function or a cube root function? Explain. What are the domain and range of the function?
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 35
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-64.jpg

Question 65.
PROBLEM SOLVING For a drag race car with a total weight of 3500 pounds, the speed s(in miles per hour) at the end of a race can be modeled by s = 14.8\(\sqrt[3]{p}\), where p is the power (in horsepower). Graph the function.
a. Determine the power of a 3500-pound car that reaches a speed of 200 miles per hour.
b. What is the average rate of change in speed as the power changes from 1000 horsepower to 1500 horsepower?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 65

Question 66.
THOUGHT PROVOKING The graph of a radical function f passes through the points (3, 1) and (4, 0). Write two different functions that could represent f(x + 2) + 1. Explain.
Answer:

Question 67.
MULTIPLE REPRESENTATIONS The terminal velocity vt(in feet per second) of a skydiver who weighs 140 pounds is given by vt = 33.7\(\sqrt{\frac{140}{\Lambda}}\)
where A is the cross-sectional surface area (in square feet) of the skydiver. The table shows the terminal velocities (in feet per second) for various surface areas (in square feet) of a skydiver who weighs 165 pounds.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 36
a. Which skydiver has a greater terminal velocity for each value of A given in the table?
b. Describe how the different values of A given in the table relate to the possible positions of the falling skydiver.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 67

Question 68.
MATHEMATICAL CONNECTIONS The surface area S of a right circular cone with a slant height of 1 unit is given by S = πr + πr2, where r is the radius of the cone.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 37
a. Use completing the square to show that r = \(\frac{1}{\sqrt{\pi}} \sqrt{S+\frac{\pi}{4}}-\frac{1}{2}\).
b. Graph the equation in part (a) using a graphing calculator. Then find the radius of a right circular cone with a slant height of 1 unit and a surface area of \(\frac{3 \pi}{4}\) square units.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-68.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.3-68.jpg

Maintaining Mathematical Proficiency

Solve the equation. Check your solutions. (Skills Review Handbook)

Question 69.
| 3x + 2 | = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 69

Question 70.
∣4x + 9 ∣ = −7
Answer:

Question 71.
| 2x − 6 ∣ = ∣ x ∣
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 71

Question 72.
| x + 8 | = | 2x + 2 |
Answer:
Given,
| x + 8 | = | 2x + 2 |
x + 8 = 2x + 2
x – 2x = 2 – 8
-x = -6
x = 6
x + 8 = -2x – 2
x + 2x = -2 – 8
3x = -10
x = -10/3

Solve the inequality. (Section 3.6)

Question 73.
x2 + 7x + 12 < 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 73

Question 74.
x2 − 10x + 25 ≥ 4
Answer:

Question 75.
2x2 + 6 > 13x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.3 Question 75

Question 76.
\(\frac{1}{8}\)x2 + x ≤ −2
Answer:

Rational Exponents and Radical Functions Study Skills : Analyzing Your Errors

5.1–5.3 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 38

Core Concepts
Section 5.1
Real nth Roots of a, p. 238
Rational Exponents, p. 239

Section 5.2
Properties of Rational Exponents, p. 244
Properties of Radicals, p. 245

Section 5.3
Parent Functions for Square Root and Cube Root Functions, p. 252
Transformations of Radical Functions, p. 253

Mathematical Practices

Question 1.
How can you use definitions to explain your reasoning in Exercises 21–24 on page 241?

Question 2.
How did you use structure to solve Exercise 76 on page 250?

Question 3.
How can you check that your answer is reasonable in Exercise 39 on page 257?

Question 4.
How can you make sense of the terms of the surface area formula given in Exercise 68 on page 258?

Study Skills
Analyzing Your Errors
Application Errors
What Happens: You can do numerical problems, but you struggle with problems that have context.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 39
How to Avoid This Error: Do not just mimic the steps of solving an application problem. Explain out loud what the question is asking and why you are doing each step. After solving the problem, ask yourself, “Does my solution make sense?”

Rational Exponents and Radical Functions 5.1 – 5.3 Quiz

5.1–5.3 Quiz

Find the indicated real nth root(s) of a. (Section 5.1)

Question 1.
n = 4, a = 81
Answer: 3 and -3

Explanation:
Given,
\(\sqrt[n]{a}\)
where, n = 4, a = 81
\(\sqrt[4]{81}\)
The nth root is even so its root will be negative or positive.
The root of this is a number when multiplied by itself 4 times will give 81.
81 is odd and is a multiple of 3 and 9 so it is likely 3.
3 × 3 × 3 × 3 = 81
-3 × -3 × -3 × -3 = 81
The real nth root are 3 and -3

Question 2.
n = 5, a = −1024
Answer: -5

Explanation:
Given,
\(\sqrt[5]{-1024}\)
The nth root when n is, even so, the root will be negative or positive.
The root of this is a number when multiplied by itself 5 times will give 1024.
1024 is even and is multiple.
-5 × -5 × -5 × -5 × -5 = -1024
The real nth root of a is -5.

Question 3.
Evaluate (a) 163/4 and (b) 1252/3 without using a calculator. Explain your reasoning. (Section 5.1)
Answer:
(a) 163/4
\(\sqrt[4]{16^3}\)
= \(\sqrt[4]{4096}\)
= 8
(b) 1252/3
\(\sqrt[3]{125^2}\)
\(\sqrt[3]{15625}\)
= 25

Find the real solution(s) of the equation. Round your answer to two decimal places.(Section 5.1)

Question 4.
2x6 = 1458
Answer: x = 3

Explanation:
Given,
2x6 = 1458
x6 = 1458/2
x6 = 729
x6 = 36
x = 3
The real solution of the equation is x = 3.

Question 5.
(x + 6)3 = 28
Answer: x(x² + 18x + 108) = -188

Explanation:
Given,
(x + 6)3 = 28
This is in the form of (a + b)³ = a³ + b³ + 3a²b + 3ab²
(x)³ + (6)³ + 3(x)²(6) + 3(x)(6)² = 28
x³ + 216 + 18x² + 108x = 28
x³ + 18x² + 108x = 28 – 216
x³ + 18x² + 108x = – 188
x(x² + 18x + 108) = -188
So, the equation is x(x² + 18x + 108) = -188

Simplify the expression.(Section 5.2)

Question 6.
\(\left(\frac{48^{1 / 4}}{6^{1 / 4}}\right)^{6}\)

Question 7.
\(\sqrt[4]{3}\) • \(\sqrt[4]{432}\)

Question 8.
\(\frac{1}{3+\sqrt{2}}\)

Question 9.
\(\sqrt[3]{16}\) – \(5 \sqrt[3]{2}\)

Question 10.
Simplify \(\sqrt[8]{x^{9} y^{8} z^{16}}\). (Section 5.2)

Write the expression in simplest form. Assume all variables are positive.(Section 5.2)

Question 11.
\(\sqrt[3]{216 p^{9}}\)

Question 12.
\(\frac{\sqrt[5]{32}}{\sqrt[5]{m^{3}}}\)

Question 13.
\(\sqrt[4]{n^{4} q}\) + \(7 n \sqrt[4]{q}\)

Question 14.
Graph f(x) = 2\(2 \sqrt[3]{x}\) + 1. Identify the domain and range of the function. (Section 5.3)

Describe the transformation of the parent function represented by the graph of g. Then write a rule for g.(Section 5.3)

Question 15.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 40

Question 16.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 41

Question 17.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 42

Question 18.
Use a graphing calculator to graph x = 3y2 − 6. Identify the vertex and direction the parabola opens. (Section 5.3)

Question 19.
A jeweler is setting a stone cut in the shape of a regular octahedron. A regular octahedron is a solid with eight equilateral triangles as faces, as shown. The formula for the volume of the stone is V= 0.47s3, where s is the side length (in millimeters) of an edge of the stone. The volume of the stone is 161 cubic millimeters. Find the length of an edge of the stone. (Section 5.1)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 43

Question 20.
An investigator can determine how fast a car was traveling just prior to an accident using the model s = 4\(\sqrt{d}\), where s is the speed (in miles per hour) of the car and d is the length (in feet) of the skid marks. Graph the model. The length of the skid marks of a car is 90 feet. Was the car traveling at the posted speed limit prior to the accident? Explain your reasoning. (Section 5.3)
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 44

Lesson 5.4 Solving Radical Equations and Inequalities

Essential Question
How can you solve a radical equation?

EXPLORATION 1
Solving Radical Equations
Work with a partner.
Match each radical equation with the graph of its related radical function. Explain your reasoning. Then use the graph to solve the equation, if possible. Check your solutions.
a. \(\sqrt{x-1}\) – 1 = 0
b. \(\sqrt{2 x+2}\) – \(\sqrt{x+4}\) = 0
c. \(\sqrt{9-x^{2}}\) = 0
d. \(\sqrt{x+2}\) – x = 0
e. \(\sqrt{-x+2}\) – x = 0
f. \(\sqrt{3 x^{2}+1}\) = 0
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 45

EXPLORATION 2
Solving Radical Equations
Work with a partner.
Look back at the radical equations in Exploration 1. Suppose that you did not know how to solve the equations using a graphical approach.
a. Show how you could use a numerical approach to solve one of the equations. For instance, you might use a spreadsheet to create a table of values.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 46
b. Show how you could use an analytical approach to solve one of the equations. For instance, look at the similarities between the equations in Exploration 1. What first step may be necessary so you could square each side to eliminate the radical(s)? How would you proceed to find the solution?

Communicate Your Answer

Question 3.
How can you solve a radical equation?

Question 4.
Would you prefer to use a graphical, numerical, or analytical approach to solve the given equation? Explain your reasoning. Then solve the equation.
\(\sqrt{x+3}\) – \(\sqrt{x-2}\) = 1

Monitoring Progress

Solve the equation. Check your solution.

Question 1.
\(\sqrt[3]{x}\) – 9 = -6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-1.jpg

Question 2.
\(\sqrt{x+25}\) = 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-2.jpg

Question 3.
2\(\sqrt[3]{x-3}\) = 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-3.jpg

Question 4.
WHAT IF? Estimate the air pressure at the center of the hurricane when the mean sustained wind velocity is 48.3 meters per second.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-.5JPG.jpg

Solve the equation. Check your solution(s).

Question 5.
\(\sqrt{10 x+9}\) = x + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-.5JPG-1.jpg

Question 6.
\(\sqrt{2 x+5}\) = \(\sqrt{x+7}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-6.jpg

Question 7.
\(\sqrt{x+6}\) – 2 = \(\sqrt{x-2}\)
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-7.jpg

Solve the equation. Check your solution(s).

Question 8.
(3x)1/3 = −3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-8.jpg

Question 9.
(x + 6)1/2 = x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-9.jpg

Question 10.
(x + 2)3/4 = 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-10.jpg

Question 11.
Solve
(a) 2\(\sqrt{x}\) − 3 ≥ 3 and
(b) 4\(\sqrt[3]{x+1}\) < 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.4-11.jpg

Solving Radical Equations and Inequalities 5.4 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY Is the equation 3x − \(\sqrt{2}\) = \(\sqrt{6}\) a radical equation? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 1

Question 2.
WRITING Explain the steps you should use to solve \(\sqrt{x}\) + 10 < 15
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation. Check your solution.

Question 3.
\(\sqrt{5 x+1}\) = 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 3

Question 4.
\(\sqrt{3 x+10}\) = 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-.4JPG.jpg

Question 5.
\(\sqrt[3]{x-16}\) = 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 5

Question 6.
\(\sqrt[3]{x}\) − 10 = −7
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-6.jpg

Question 7.
−2\(\sqrt{24 x}\) + 13 = −1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 7

Question 8.
8\(\sqrt[3]{10 x}\) − 15 = 17
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-8.jpg

Question 9.
\(\frac{1}{5} \sqrt[3]{3 x}\) = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 9

Question 10.
\(\sqrt{2 x}-\frac{2}{3}\) = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-10.jpg

Question 11.
\(2 \sqrt[5]{x}\) + 7 = 15
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 11

Question 12.
\(\sqrt[4]{4 x}\) − 13 = −15
Answer:

Question 13.
MODELING WITH MATHEMATICS Biologists have discovered that the shoulder height h (in centimeters) of a male Asian elephant can be modeled by h = 62.5\(\sqrt[3]{t}\) + 75.8, where t is the age (in years) of the elephant. Determine the age of an elephant with a shoulder height of 250 centimeters.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 47
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 13

Question 14.
MODELING WITH MATHEMATICS In an amusement park ride, a rider suspended by cables swings back and forth from a tower. The maximum speed v (in meters per second) of the rider can be approximated by v = \(\sqrt{2 g h}\), where h is the height (in meters) at the top of each swing and g is the acceleration due to gravity (g ≈ 9.8 m/sec2). Determine the height at the top of the swing of a rider whose maximum speed is 15 meters per second.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-14.jpg

In Exercises 15–26, solve the equation. Check your solution(s).

Question 15.
x− 6 = \(\sqrt{3 x}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 15

Question 16.
x − 10 = \(\sqrt{9 x}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-16.jpg

Question 17.
\(\sqrt{44-2 x}\) = x − 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 17

Question 18.
\(\sqrt{2 x+30}\) = x + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-18.jpg

Question 19.
\(\sqrt[3]{8 x^{3}-1}\) = 2x − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 19

Question 20.
\(\sqrt[4]{3-8 x^{2}}\) = 2x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-20.jpg

Question 21.
\(\sqrt{4 x+1}\) = \(\sqrt{x+10}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 21

Question 22.
\(\sqrt{3 x-3}\) – \(\sqrt{x+12}\) = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-22.jpg

Question 23.
\(\sqrt[3]{2 x-5}\) – \(\sqrt[3]{8 x+1}\) = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 23

Question 24.
\(\sqrt[3]{x+5}\) = 2\(\sqrt[3]{2 x+6}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-24.jpg

Question 25.
\(\sqrt{3 x-8}\) + 1 = \(\sqrt{x+5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 25.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 25.2

Question 26.
\(\sqrt{x+2}\) = 2 – \(\sqrt{x}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-26.jpg

In Exercises 27–34, solve the equation. Check your solution(s).

Question 27.
2x2/3 = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 27

Question 28.
4x3/2 = 32
Answer:
Given,
4x3/2 = 32
x = 32/4
x3/2 = 8
\(\sqrt{x³}\) = 8
Squaring on both sides
x³ = 64
x³ = 4³
x = 4

Question 29.
x1/4 + 3 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 29

Question 30.
2x3/4 − 14 = 40
Answer:

Question 31.
(x + 6)1/2 = x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 31

Question 32.
(5 − x)1/2 − 2x = 0
Answer:

Question 33.
2(x + 11)1/2 = x + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 33.1

Question 34.
(5x2 − 4)1/4 = x
Answer:

ERROR ANALYSIS In Exercises 35 and 36, describe and correct the error in solving the equation.

Question 35.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 48
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 35.1

Question 36.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 49
Answer:

In Exercises 37–44, solve the inequality.

Question 37.
\(2 \sqrt[3]{x}\) − 5 ≥ 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 37

Question 38.
\(\sqrt[3]{x-4}\) ≤ 5
Answer:
Given,
\(\sqrt[3]{x-4}\) ≤ 5
Apply Cubing on both sides
(x – 4) ≤ 5³
x-4 ≤ 125
x ≤ 125 + 4
x ≤ 129

Question 39.
\(4 \sqrt{x-2}\) > 20
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 39

Question 40.
7\(\sqrt{x}\) + 1 < 9
Answer:

Question 41.
2\(\sqrt{x}\) + 3 ≤ 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 41

Question 42.
\(\sqrt[3]{x+7}\) ≥ 3
Answer:

Question 43.
\(-2 \sqrt[3]{x+4}\) < 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 43

Question 44.
−0.25\(\sqrt{x}\) − 6 ≤ −3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-44.jpg

Question 45.
MODELING WITH MATHEMATICS The length ℓ (in inches) of a standard nail can be modeled by ℓ = 54d3/2, where d is the diameter (in inches) of the nail. What is the diameter of a standard nail that is 3 inches long?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 45

Question 46.
DRAWING CONCLUSIONS “Hang time” is the time you are suspended in the air during a jump. Your hang time t (in seconds) is given by the function t = 0.5\(\sqrt{h}\), where h is the height (in feet) of the jump. Suppose a kangaroo and a snowboarder jump with the hang times shown.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 50
a. Find the heights that the snowboarder and the kangaroo jump.
b. Double the hang times of the snowboarder and the kangaroo and calculate the corresponding heights of each jump.
c. When the hang time doubles, does the height of the jump double? Explain.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-46.jpg

USING TOOLS In Exercises 47–52, solve the nonlinear system. Justify your answer with a graph.

Question 47.
y2 = x − 3
y = x − 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 47

Question 48.
y2 = 4x + 17
y = x + 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-48.jpg

Question 49.
x2 + y2 = 4
y = x – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 49

Question 50.
x2 + y2 = 25
y = \(-\frac{3}{4}\)x + \(\frac{25}{4}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-50.jpg

Question 51.
x2 + y2 = 1
y = \(\frac{1}{2}\)x2 – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 51

Question 52.
x2 + y2 = 4
y2 = x + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-52.jpg
Question 53.
PROBLEM SOLVING The speed s (in miles per hour) of a car can be given by s = \(\sqrt{30 f d}\), where f is the coefficient of friction and d is the stopping distance (in feet). The table shows the coefficient of friction for different surfaces.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 51
a. Compare the stopping distances of a car traveling 45 miles per hour on the surfaces given in the table.
b. You are driving 35 miles per hour on an icy road when a deer jumps in front of your car. How far away must you begin to brake to avoid hitting the deer? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 53.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 53.2

Question 54.
MODELING WITH MATHEMATICS The Beaufort wind scale was devised to measure wind speed. The Beaufort numbers B, which range from 0 to 12, can be modeled by B = 1.69\(\sqrt{s+4.25}\) − 3.55, where s is the wind speed (in miles per hour).
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 52
a. What is the wind speed for B = 0? B = 3?
b. Write an i
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-54.jpg

Question 55.
USING TOOLS Solve the equation x − 4 = \(\sqrt{2 x}\). Then solve the equation x − 4 = − \(\sqrt{2 x}\).
a. How does changing \(\sqrt{2 x}\) to −\(\sqrt{2 x}\) change the solution(s) of the equation?
b. Justify your answer in part (a) using graphs.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 55.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 55.2
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 55.3

Question 56.
MAKING AN ARGUMENT Your friend says it is impossible for a radical equation to have two extraneous solutions. Is your friend correct? Explain your reasoning.
Answer:

Question 57.
USING STRUCTURE Explain how you know the radical equation \(\sqrt{x+4}\) = −5 has no real solution without solving it.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 57

Question 58.
HOW DO YOU SEE IT? Use the graph to find the solution of the equation 2\(\sqrt{x-4}\) = −\(\sqrt{x-1}\) + 4. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 53
Answer:

Question 59.
WRITING A company determines that the price p of a product can be modeled by p = 70 − \(\sqrt{0.02 x+1}\), where x is the number of units of the product demanded per day. Describe the effect that raising the price has on the number of units demanded.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 59

Question 60.
THOUGHT PROVOKING City officials rope off a circular area to prepare for a concert in the park. They estimate that each person occupies 6 square feet. Describe how you can use a radical inequality to determine the possible radius of the region when P people are expected to attend the concert.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 54
Answer:

Question 61.
MATHEMATICAL CONNECTIONS The Moeraki Boulders along the coast of New Zealand are stone spheres with radii of approximately 3 feet. A formula for the radius of a sphere is
r = \(\frac{1}{2} \sqrt{\frac{S}{\pi}}\)
where S is the surface area of the sphere. Find the surface area of a Moeraki Boulder.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 61

Question 62.
PROBLEM SOLVING You are trying to determine the height of a truncated pyramid, which cannot be measured directly. The height h and slant heightℓof the truncated pyramid are related by the formula below.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 55
In the given formula, b1 and b2 are the side lengths of the upper and lower bases of the pyramid, respectively. When ℓ = 5, b1 = 2, and b2 = 4, what is the height of the pyramid?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.4-62.jpg

Question 63.
REWRITING A FORMULA A burning candle has a radius of r inches and was initially h0inches tall. After t minutes, the height of the candle has been reduced to h inches. These quantities are related by the formula
r = \(\sqrt{\frac{k t}{\pi\left(h_{0}-h\right)}}\)
where k is a constant. Suppose the radius of a candle is 0.875 inch, its initial height is 6.5 inches, and k = 0.04.
a. Rewrite the formula, solving for h in terms of t.
b. Use your formula in part (a) to determine the height of the candle after burning 45 minutes.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 63

Maintaining Mathematical Proficiency

Perform the indicated operation. (Section 4.2 and Section 4.3)

Question 64.
(x3 − 2x2 + 3x + 1) + (x4 − 7x)
Answer:
x4 − 7x + x3 − 2x2 + 3x + 1
= x4 + x3 − 2x2 − 7x  + 3x + 1
= x4 + x3 − 2x2 − 4x + 1
So, (x3 − 2x2 + 3x + 1) + (x4 − 7x) = x4 + x3 − 2x2 − 4x + 1

Question 65.
(2x5 + x4 − 4x2) − (x5 − 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 65

Question 66.
(x3 + 2x2 + 1)(x2 + 5)
Answer:

Question 67.
(x4 + 2x3 + 11x2 + 14x − 16) ÷ (x + 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 67

Let f(x) = x3 – 4x2 + 6. Write a rule for g. Describe the graph of g as a transformation of the graph of f.(Section 4.7)

Question 68.
g(x) = f(−x) + 4
Answer:

Question 69.
g(x) = \(\frac{1}{2}\)f(x) − 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.4 Question 69

Question 70.
g(x) = −f(x − 1) + 6
Answer:

Lesson 5.5 Performing Function Operations

Essential Question
How can you use the graphs of two functions to sketch the graph of an arithmetic combination of the two functions?
Just as two real numbers can be combined by the operations of addition, subtraction, multiplication, and division to form other real numbers, two functions can be combined to form other functions. For example, the functions f(x) = 2x − 3 and g(x) = x2 − 1 can be combined to form the sum, difference, product, or quotient of f and g.
f(x) + g(x) = (2x − 3) + (x2 − 1) = x2 + 2x − 4 sum
f(x) − g(x) = (2x − 3) − (x2 − 1) = −x2 + 2x − 2 difference
f(x) • g(x) = (2x − 3)(x2 − 1) = 2x3 − 3x2 − 2x + 3 product
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 56

EXPLORATION 1
Graphing the Sum of Two Functions
Work with a partner.
Use the graphs of f and g to sketch the graph of f + g. Explain your steps.
Sample Choose a point on the graph of g. Use a compass or a ruler to measure its distance above or below the x-axis. If above, add the distance to the y-coordinate of the point with the same x-coordinate on the graph of f. If below, subtract the distance. Plot the new point. Repeat this process for several points. Finally, draw a smooth curve through the new points to obtain the graph of f + g.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 57
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 58

Communicate Your Answer

Question 2.
How can you use the graphs of two functions to sketch the graph of an arithmetic combination of the two functions?

Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 59
Question 3.
Check your answers in Exploration 1 by writing equations for f and g, adding the functions, and graphing the sum.

5.5 Lesson

Monitoring Progress

Question 1.
Let f(x) = −2x2/3 and g(x) = 7x2/3. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(8) and (f − g)(8).
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.5-1.jpg

Question 2.
Let f(x) = 3x and g(x) = x1/5. Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate (fg)(32) and (\(\frac{f}{g}\))(32).
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.5-2.jpg

Question 3.
Let f(x) = 8x and g(x) = 2x5/6. Use a graphing calculator to evaluate (f + g)(x), (f − g)(x), (fg)(x), and (\(\frac{f}{g}\)) (x) when x = 5. Round your answers to two decimal places.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.5-3.jpg

Question 4.
In Example 5, explain why you can evaluate (f + g)(3), (f − g)(3), and (fg)(3) but not (\(\frac{f}{g}\))(3).
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.5-4.jpg

Question 5.
Use the answer in Example 6(a) to find the total number of heartbeats over the lifetime of a white rhino when its body mass is 1.7 × 105 kilograms.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-Monitoring-5.5-5.jpg

Performing Function Operations 5.5 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Let f and g be any two functions. Describe how you can use f, g, and the four basic operations to create new functions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 1

Question 2.
WRITING What x-values are not included in the domain of the quotient of two functions?
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, find (f + g)(x) and (f – g)(x) and state the domain of each. Then evaluate f + g and f – g for the given value of x.

Question 3.
f(x) = \(-5 \sqrt[4]{x}\), g(x) = 19\(\sqrt[4]{x}\); x = 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 3

Question 4.
f(x) = \(\sqrt[3]{2 x}\), g(x) = −11\(\sqrt[3]{2 x}\) ; x = −4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.5-4.jpg

Question 5.
f(x) = 6x − 4x2− 7x3, g(x) = 9x2− 5x; x = −1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 5

Question 6.
f(x) = 11x + 2x2, g(x) = −7x − 3x2 + 4; x = 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.5-6.jpg

In Exercises 7–12, find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate fg and \(\frac{f}{g}\) for the given value of x.

Question 7.
f(x) = 2x3, g(x) = \(\sqrt[3]{x}\) ; x = −27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 7

Question 8.
f(x) = x4, g(x) = \(3 \sqrt{x}\) ; x = 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.5-.8JPG.jpg

Question 9.
f(x) = 4x, g(x) = 9x1/2; x = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 9

Question 10.
f(x) = 11x3, g(x) = 7x7/3; x = −8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.5-10.jpg

Question 11.
f(x) = 7x3/2, g(x) =−14x1/3; x = 64
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 11

Question 12.
f(x) = 4x5/4, g(x) = 2x1/2; x = 16
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.5-12.jpg

USING TOOLS In Exercises 13–16, use a graphing calculator to evaluate (f + g)(x), (f − g)(x), (fg)(x), and (\(\frac{f}{g}\))(x) when x = 5. Round your answers to two decimal places.

Question 13.
f(x) = 4x4; g(x) = 24x1/3

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 13

Question 14.
f(x) = 7x5/3; g(x) = 49x2/3
Answer:

Question 15.
f(x) =−2x1/3; g(x) = 5x1/2

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 15

Question 16.
f(x) = 4x1/2; g(x) = 6x3/4
Answer:

ERROR ANALYSIS In Exercises 17 and 18, describe and correct the error in stating the domain.

Question 17.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 60
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 17

Question 18.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 61
Answer:

Question 19.
MODELING WITH MATHEMATICS From 1990 to 2010, the numbers (in millions) of female F and male M employees from the ages of 16 to 19 in the United States can be modeled by F(t) =−0.007t2 + 0.10t + 3.7 and M(t) = 0.0001t3 − 0.009t2 + 0.11t + 3.7, where t is the number of years since 1990.
a. Find (F + M)(t).
b. Explain what (F + M)(t) represents.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 19

Question 20.
MODELING WITH MATHEMATICS From 2005 to 2009, the numbers of cruise ship departures (in thousands) from around the world W and Florida F can be modeled by the equations
W(t) = −5.8333t3 + 17.43t2 + 509.1t + 11496
F(t) = 12.5t3 − 60.29t2 + 136.6t + 4881
where t is the number of years since 2005.
a. Find (W − F )(t).
b. Explain what (W − F )(t) represents.
Answer:

Question 21.
MAKING AN ARGUMENT Your friend claims that the addition of functions and the multiplication of functions are commutative. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 21

Question 22.
HOW DO YOU SEE IT? The graphs of the functions f(x) = 3x2 − 2x − 1 and g(x) = 3x + 4 are shown. Which graph represents the function f + g? the function f − g? Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 62
Answer:

Question 23.
REASONING The table shows the outputs of the two functions f and g. Use the table to evaluate (f + g)(3), (f − g)(1), (fg)(2), and (\(\frac{f}{g}\))(0).
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 63
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 23

Question 24.
THOUGHT PROVOKING Is it possible to write two functions whose sum contains radicals, but whose product does not? Justify your answers.
Answer:

Question 25.
MATHEMATICAL CONNECTIONS A triangle is inscribed in a square, as shown. Write and simplify a function r in terms of x that represents the area of the shaded region.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 64
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 25

Question 26.
REWRITING A FORMULA For a mammal that weighs w grams, the volume b (in milliliters) of air breathed in and the volume d (in milliliters) of “dead space” (the portion of the lungs not filled with air) can be modeled by
b(w) = 0.007w and d(w) = 0.002w.
The breathing rate r (in breaths per minute) of a mammal that weighs w grams can be modeled by
r(w) = \(\frac{1.1 w^{0.734}}{b(w)-d(w)}\).
Answer:

Question 27.
PROBLEM SOLVING A mathematician at a lake throws a tennis ball from point A along the water’s edge to point B in the water, as shown. His dog, Elvis, first runs along the beach from point A to point D and then swims to fetch the ball at point B.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 65
a. Elvis runs at a speed of about 6.4 meters per second. Write a function r in terms of x that represents the time he spends running from point A to point D. Elvis swims at a speed of about 0.9 meter per second. Write a function s in terms of x that represents the time he spends swimming from point D to point B.
b. Write a function t in terms of x that represents the total time Elvis spends traveling from point A to point D to point B.
c. Use a graphing calculator to graph t. Find the value of x that minimizes t. Explain the meaning of this value.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 27

Maintaining Mathematical Proficiency

Solve the literal equation for n. (Skills Review Handbook)

Question 28.
3xn − 9 = 6y
Answer:
Given,
3xn − 9 = 6y
Take 3 as a common factor
xn – 3 = 2y
xn = 2y + 3
n = (2y + 3)/x

Question 29.
5z = 7n + 8nz
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 29

Question 30.
3nb = 5n − 6z
Answer:
Given,
3nb = 5n − 6z
3nb – 5n = -6z
Take n as the common factor to make n as the subject.
n(3b – 5) = -6z
n = -6z/(3b – 5)

Question 31.
\(\frac{3+4 n}{n}\) = 7b
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 31

Determine whether the relation is a function. Explain. (Skills Review Handbook)

Question 32.
(3, 4), (4, 6), (1, 4), (2, −1)
Answer:

Question 33.
(−1, 2), (3, 7), (0, 2), (−1, −1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 33

Question 34.
(1, 6), (7, −3), (4, 0), (3, 0)
Answer:

Question 35.
(3, 8), (2, 5), (9, 5), (2, −3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.5 Question 35

Lesson 5.6 Inverse of a Function

Essential Question
How can you sketch the graph of the inverse of a function?

EXPLORATION 1
Graphing Functions and Their Inverses
Work with a partner.
Each pair of functions are inverses of each other. Use a graphing calculator to graph f and g in the same viewing window. What do you notice about the graphs?
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 65.1
a. f(x) = 4x + 3
g(x) = \(\frac{x-3}{4}\)
b. f(x) = x3 + 1
g(x) = \(\sqrt[3]{x-1}\)
c. f(x) = \(\sqrt{x-3}[latex]
g(x) = x2 + 3, x ≥ 0
d. f(x) = [latex]\frac{4 x+4}{x+5}\)
g(x) = \(\frac{4-5 x}{x-4}\)

EXPLORATION 2
Sketching Graphs of Inverse Functions
Work with a partner.
Use the graph of f to sketch the graph of g, the inverse function of f, on the same set of coordinate axes. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 66

Communicate Your Answer

Question 3.
How can you sketch the graph of the inverse of a function?

Question 4.
In Exploration 1, what do you notice about the relationship between the equations of f and g? Use your answer to find g, the inverse function of
f(x) = 2x − 3.
Use a graph to check your answer.

5.6 Lesson

Monitoring Progress

Solve y = f(x) for x. Then find the input(s) when the output is 2.

Question 1.
f(x) = x − 2
Answer:
Given,
f(x) = x − 2
The values of x are the input values of the function.
f(x) = 2
If the output value of the function f is 2.
2 = x – 2
2 + 2 = x
x = 4

Question 2.
f(x) = 2x2

Answer:
f(x) = 2x²
The values of x are the input values of the function.
f(x) = 2
If the output value of the function f is 2.
2 = 2x²
x² = 1
x = ±1

Question 3.
f(x) = −x3 + 3
Answer:
Given,
f(x) = −x3 + 3
The values of x are the input values of the function.
f(x) = 2
If the output value of the function f is 2.
2 = -x³ + 3
-x³ = 3 – 2
-x³ = 1
-x = ±1

Find the inverse of the function. Then graph the function and its inverse.

Question 4.
f(x) = 2x

Question 5.
f(x) = −x + 1

Question 6.
f(x) = \(\frac{1}{3}\)x − 2

Find the inverse of the function. Then graph the function and its inverse.

Question 7.
f(x) = −x2, x ≤ 0

Question 8.
f(x) = −x3 + 4

Question 9.
f(x) = \(\sqrt{x+2}\)

Determine whether the functions are inverse functions.

Question 10.
f(x) = x + 5, g(x) = x − 5

Question 11.
f(x) = 8x3, g(x) = \(\sqrt[3]{2 x}\)

Question 12.
The distance d (in meters) that a dropped object falls in t seconds on Earth is represented by d = 4.9t2. Find the inverse of the function. How long does it take an object to fall 50 meters?

Inverse of a Function 5.6 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY In your own words, state the definition of inverse functions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 1

Question 2.
WRITING Explain how to determine whether the inverse of a function is also a function.
Answer:

Question 3.
COMPLETE THE SENTENCE Functions f and g are inverses of each other provided that f(g(x)) = ____ and g(f(x)) = ____.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 3

Question 4.
DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 67
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, solve y = f(x) for x. Then find the input(s) when the output is -3.

Question 5.
f(x) = 3x + 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 5

Question 6.
f(x) = −7x − 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-6.jpg

Question 7.
f(x) = \(\frac{1}{2}\)x − 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 7

Question 8.
f(x) = −\(\frac{2}{3}\)x + 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-8.jpg
Question 9.
f(x) = 3x3

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 9

Question 10.
f(x) = 2x4 − 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-10.jpg

Question 11.
f(x) = (x − 2)2 − 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 11

Question 12.
f(x) = (x − 5)3 − 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-12.jpg

In Exercises 13–20, find the inverse of the function. Then graph the function and its inverse.

Question 13.
f(x) = 6x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 13.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 13.2

Question 14.
f(x) = −3x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-12-1.jpg

Question 15.
f(x) = −2x + 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 15.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 15.2

Question 16.
f(x) = 6x − 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-16.jpg

Question 17.
f(x) = −\(\frac{1}{2}\)x + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 17.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 17.2

Question 18.
f(x) = \(\frac{1}{3}\)x − 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-18.jpg

Question 19.
f(x) = \(\frac{2}{3}\)x − \(\frac{1}{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 19.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 19.2

Question 20.
f(x) = −\(\frac{4}{5}\)x + \(\frac{1}{5}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-20.jpg

Question 21.
COMPARING METHODS Find the inverse of the function f(x) = −3x + 4 by switching the roles of x and y and solving for y. Then find the inverse of the function f by using inverse operations in the reverse order. Which method do you prefer? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 21

Question 22.
REASONING Determine whether each pair of functions f and g are inverses. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 68
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-22.jpg

In Exercises 23–28, find the inverse of the function. Then graph the function and its inverse.

Question 23.
f(x) = 4x2, x ≤ 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 23

Question 24.
f(x) = 9x2, x ≤ 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-24.jpg

Question 25.
f(x) = (x − 3)3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 25

Question 26.
f(x) = (x + 4)3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-26PG.jpg

Question 27.
f(x) = 2x4, x ≥ 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 27

Question 28.
f(x) = −x6, x ≥ 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-28.jpg

ERROR ANALYSIS In Exercises 29 and 30, describe and correct the error in finding the inverse of the function.

Question 29.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 69
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 29

Question 30.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 70
Answer:

USING TOOLS In Exercises 31–34, use the graph to determine whether the inverse of f is a function. Explain your reasoning.

Question 31.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 71
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 31

Question 32.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 72
Answer:
As per the horizontal line test, the horizontal line intersects the graph more than once. f foes do not have the inverse function. therefore the inverse of the function is not a function.

Question 33.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 73
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 33

Question 34.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 74
Answer:
As per the horizontal line test, the horizontal line intersects the graph only once, thus the inverse of the function is a function.

In Exercises 35–46, determine whether the inverse of f is a function. Then find the inverse.

Question 35.
f(x) = x3 − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 35

Question 36.
f(x) = −x3 + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-36.jpg

Question 37.
f(x) = \(\sqrt{x+4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 37

Question 38.
f(x) = \(\sqrt{x-6}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-38.jpg

Question 39.
f(x) = \(2 \sqrt[3]{x-5}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 39

Question 40.
f(x) = 2x2 − 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-40.jpg

Question 41.
f(x) = x4 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 41

Question 42.
f(x) = 2x3 − 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-42.jpg

Question 43.
f(x) = \(3 \sqrt[3]{x+1}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 43

Question 44.
f(x) = \(-\sqrt[3]{\frac{2 x+4}{3}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-44.jpg

Question 45.
f(x) = \(\frac{1}{2}\)x5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 45

Question 46.
f(x) = \(-3 \sqrt{\frac{4 x-7}{3}}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-46.jpg

Question 47.
WRITING EQUATIONS What is the inverse of the function whose graph is shown?
A. g(x) = \(\frac{3}{2}\)x − 6
B. g(x) = \(\frac{3}{2}\)x + 6
C. g(x) = \(\frac{2}{3}\)x − 6
D. g(x) = \(\frac{2}{3}\)x + 12
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 75
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 47

Question 48.
WRITING EQUATIONS What is the inverse of f(x) = −\(-\frac{1}{64}\)x3?
A. g(x) = −4x3
B. g(x) = 4\(\sqrt[3]{x}\)
C. g(x) = −4\(\sqrt[3]{x}\)
D. g(x) = \(\sqrt[3]{-4 x}\)
Answer:

In Exercises 49–52, determine whether the functions are inverses.

Question 49.
f(x) = 2x − 9, g(x) = x — 2 + 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 49

Question 50.
f(x) = \(\frac{x-3}{4}\), g(x) = 4x + 3
Answer:

Question 51.
f(x) = \(\sqrt[5]{\frac{x+9}{5}}\), g(x) = 5x5 − 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 51.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 51.2

Question 52.
f(x) = 7x3/2 − 4, g(x) = (\(\frac{x+4}{7}\))3/2

Answer:

Question 53.
MODELING WITH MATHEMATICS The maximum hull speed v (in knots) of a boat with a displacement hull can be approximated by v = 1.34\(\sqrt{\ell}\), where ℓ is the waterline length (in feet) of the boat. Find the inverse function. What waterline length is needed to achieve a maximum speed of 7.5 knots?
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 76
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 53.1

Question 54.
MODELING WITH MATHEMATICS Elastic bands can be used for exercising to provide a range of resistance. The resistance R (in pounds) of a band can be modeled by R = \(\frac{3}{8}\)L − 5, where L is the total length (in inches) of the stretched band. Find the inverse function. What length of the stretched band provides 19 pounds of resistance?
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 77
Answer:

ANALYZING RELATIONSHIPS In Exercises 55–58, match the graph of the function with the graph of its inverse.

Question 55.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 78
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 55

Question 56.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 79
Answer:

Question 57.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 80
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 57

Question 58.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 81
Answer:

Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 82

Question 59.
REASONING You and a friend are playing a number-guessing game. You ask your friend to think of a positive number, square the number, multiply the result by 2, and then add 3. Your friend’s final answer is 53. What was the original number chosen? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 59

Question 60.
MAKING AN ARGUMENT Your friend claims that every quadratic function whose domain is restricted to nonnegative values has an inverse function. Is your friend correct? Explain your reasoning.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-60.jpg

Question 61.
PROBLEM SOLVING When calibrating a spring scale, you need to know how far the spring stretches for various weights. Hooke’s Law states that the length a spring stretches is proportional to the weight attached to it. A model for one scale is ℓ = 0.5w + 3, whereℓ is the total length (in inches) of the stretched spring and w is the weight (in pounds) of the object.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 83
a. Find the inverse function. Describe what it represents.
b. You place a melon on the scale, and the spring stretches to a total length of 5.5 inches. Determine the weight of the melon.
c. Verify that the function ℓ = 0.5w + 3 and the inverse model in part (a) are inverse functions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 61.1
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 61.2

Question 62.
THOUGHT PROVOKING Do functions of the form y = xm/n, where m and n are positive integers, have inverse functions? Justify your answer with examples.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-5-Rational-Exponents-and-Radical-Functions-5.6-62.jpg

Question 63.
PROBLEM SOLVING At the start of a dog sled race in Anchorage, Alaska, the temperature was 5°C. By the end of the race, the temperature was −10°C. The formula for converting temperatures from degrees Fahrenheit F to degrees Celsius C is C = \(\frac{5}{9}\)(F − 32).
a. Find the inverse function. Describe what it represents.
b. Find the Fahrenheit temperatures at the start and end of the race.
Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 84
c. Use a graphing calculator to graph the original function and its inverse. Find the temperature that is the same on both temperature scales.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 63.1

Question 64.
PROBLEM SOLVING The surface area A (in square meters) of a person with a mass of 60 kilograms can be approximated by A = 0.2195h0.3964, where h is the height (in centimeters) of the person.
a. Find the inverse function. Then estimate the height of a 60-kilogram person who has a body surface area of 1.6 square meters.
b. Verify that function A and the inverse model in part (a) are inverse functions.
Answer:

USING STRUCTURE In Exercises 65–68, match the function with the graph of its inverse.

Question 65.
f(x) = \(\sqrt[3]{x-4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 65.1

Question 66.
f(x) = \(\sqrt[3]{x+4}\)
Answer:

Question 67.
f(x) = \(\sqrt{x+1}\) – 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 67.1

Question 68.
f(x) = \(\sqrt{x-1}\) + 3
Answer:

Big Ideas Math Answer Key Algebra 2 Chapter 5 Rational Exponents and Radical Functions 85

Question 69.
DRAWING CONCLUSIONS Determine whether the statement is true or false. Explain your reasoning.
a. If f(x) = xn and n is a positive even integer, then the inverse of f is a function.
b. If f(x) = xn and n is a positive odd integer, then the inverse of f is a function.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 69.1

Question 70.
HOW DO YOU SEE IT? The graph of the function f is shown. Name three points that lie on the graph of the inverse of f. Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 86
Answer:

Question 71.
ABSTRACT REASONING Show that the inverse of any linear function f(x) = mx + b, where m ≠ 0, is also a linear function. Identify the slope and y-intercept of the graph of the inverse function in terms of m and b.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 71

Question 72.
CRITICAL THINKING Consider the function f(x) = −x.
a. Graph f(x) = −x and explain why it is its own inverse. Also, verify that f(x) = −x is its own inverse algebraically.
b. Graph other linear functions that are their own inverses. Write equations of the lines you graphed.
c. Use your results from part (b) to write a general equation describing the family of linear functions that are their own inverses.
Answer:

Maintaining Mathematical Proficiency

Simplify the expression. Write your answer using only positive exponents.(Skills Review Handbook)

Question 73.
(−3)−3

Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 73

Question 74.
23 • 22
Answer:
Given
23 • 22
When bases are equal powers should be added.
23+2 = 25

Question 75.
\(\frac{4^{5}}{4^{3}}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 75

Question 76.
(\(\frac{2}{3}\))4
Answer:
Given,
(\(\frac{2}{3}\))4
(\(\frac{a}{b}\))n= an/bn
= 24/34
= 16/81

Describe the x-values for which the function is increasing, decreasing, positive, and negative.

Question 77.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 87
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 77

Question 78.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 88
Answer:

Question 79.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 89
Answer:
Big Ideas Math Algebra 2 Answers Chapter 5 Rational Exponents and Radical Functions 5.6 Question 79

Rational Exponents and Radical Functions Performance Task: Turning the Tables

5.4–5.6 What Did You Learn?

Core Vocabulary
radical equation, p. 262
extraneous solutions, p. 263
inverse functions, p. 277

Core Concepts
Section 5.4

Solving Radical Equations, p. 262
Solving Radical Inequalities, p. 265

Section 5.5
Operations on Functions, p. 270

Section 5.6
Exploring Inverses of Functions, p. 276
Inverses of Nonlinear Functions, p. 278
Horizontal Line Test, p. 278

Mathematical Practices

Question 1.
How did you find the endpoints of the range in part (b) of Exercise 54 on page 267?

Question 2.
How did you use structure in Exercise 57 on page 268?

Question 3.
How can you evaluate the reasonableness of the results in Exercise 27 on page 274?

Question 4.
How can you use a graphing calculator to check your answers in Exercises 49–52 on page 282?

Performance Task

Turning the Tables

In this chapter, you have used properties of rational exponents and functions to find an answer to the problem. Using those same properties, can you find a problem to the answer? How many problems can you find?
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 90
To explore the answers to these questions and more, go to BigIdeasMath.com

Rational Exponents and Radical Functions Chapter Review

Evaluate the expression without using a calculator.

Question 1.
87/3

Answer:
Given,
87/3
We can write the given radical as
87 . 81/3
= 2097152 × 2
= 4194304
So, 87/3 = 4194304

Question 2.
95/2

Answer:
Given,
95/2
We can write the given radical as
95 × 91/2
= 59049 × 3
= 177147
So, 95/2 = 177147

Question 3.
(−27)−2/3

Answer:
Given,
(-27)−2 × (-27)1/3
= -3 × √27

Find the real solution(s) of the equation. Round your answer to two decimal places when appropriate.

Question 4.
x5 + 17 = 35
Answer:
Given,
x5 + 17 = 35
x5 = 35 – 17
x5 = 18
Applying the fifth root on both sides
x = 1.79

Question 5.
7x3 = 189
Answer:
Given,
7x3 = 189
x3 = 189/7
x3 = 27
Cubing on both sides
x = 3

Question 6.
(x + 8)4 = 16
Answer:
Given,
(x + 8)4 = 16
16 can be written as 24
(x + 8)4 = 24
x + 8 = 2
x = 2 – 8
x = -6

Simplify the expression.

Question 7.
(\(\frac{6^{1 / 5}}{6^{2 / 5}}\))3

Answer:
(\(\frac{6^{1 / 5}}{6^{2 / 5}}\))³
(\(\frac{6^{3 / 5}}{6^{6 / 5}}\))

Question 8.
\(\sqrt[4]{32} \cdot \sqrt[4]{8}\)
Answer:
Given,
\(\sqrt[4]{32} \cdot \sqrt[4]{8}\)
= \(\sqrt[4]{32×8}\)
= \(\sqrt[4]{256}\)
= \(\sqrt[4]{4^4}\)
= 4

Question 9.
\(\frac{1}{2-\sqrt[4]{9}}\)
Answer:
\(\frac{1}{2-\sqrt[4]{9}}\)
\(\sqrt[2]{9}\) = 1.73
= 1/(2 – 1.73)
= 1/0.27
= 3.7

Question 10.
\(4 \sqrt[5]{8}\) + \(3 \sqrt[5]{8}\)
Answer:
Given,
\(4 \sqrt[5]{8}\) + \(3 \sqrt[5]{8}\)
= \(\sqrt[5]{8}\)(4 + 3)
= 7\(\sqrt[5]{8}\)
So, \(4 \sqrt[5]{8}\) + \(3 \sqrt[5]{8}\) = 7\(\sqrt[5]{8}\)

Question 11.
\(2 \sqrt{48}\) – \(\sqrt{3}\)
Answer:
Given,
\(2 \sqrt{48}\) – \(\sqrt{3}\)
48 = 16 × 3
\(2 \sqrt{16×3}\) – \(\sqrt{3}\)
2 × 4 \(\sqrt{3}\) – \(\sqrt{3}\)
\(\sqrt{3}\)(8 – 1)
= 7\(\sqrt{3}\)
\(2 \sqrt{48}\) – \(\sqrt{3}\) = 7\(\sqrt{3}\)

Question 12.
(\(5^{2 / 3} \cdot 2^{3 / 2}\))1/2

Answer:
Given,
(5^2/3)1/2 × (2^3/2)1/2
=51/3 × 23/4
= \(\sqrt[3]{5}\) . \(\sqrt[4]{8}\)
= 2\(\sqrt[3]{5}\) . \(\sqrt[2]{2}\)
So, (\(5^{2 / 3} \cdot 2^{3 / 2}\))1/2 =2\(\sqrt[3]{5}\) . \(\sqrt[2]{2}\)

Simplify the expression. Assume all variables are positive.

Question 13.
\(\sqrt[3]{125 z^{9}}\)

Question 14.
\(\frac{2^{1 / 4} z^{5 / 4}}{6 z}\)

Question 15.
\(\sqrt{10 z^{5}}-z^{2} \sqrt{40 z}\)

Describe the transformation of f represented by g. Then graph each function.

Question 16.
f(x) = \(\sqrt{x}\), g(x) = −2\(\sqrt{x}\)

Question 17.
f(x) = \(\sqrt[3]{x}\), g(x) = \(\sqrt[3]{-x}\) − 6

Question 18.
Let the graph of g be a reflection in the y-axis, followed by a translation 7 units to the right of the graph of f(x) = \(\sqrt[3]{x}\). Write a rule for g.

Question 19.
Use a graphing calculator to graph 2y2 = x − 8. Identify the vertex and the direction that the parabola opens.

Question 20.
Use a graphing calculator to graph x2 + y2 = 81. Identify the radius and the intercepts.

Solve the equation. Check your solution.

Question 21.
\(4 \sqrt[3]{2 x+1}\) = 20

Question 22.
\(\sqrt{4 x-4}\) = \(\sqrt{5 x-1}\) – 1

Question 23.
(6x)2/3 = 36

Solve the inequality.

Question 24.
\(5 \sqrt{x}\) + 2 > 17

Question 25.
\(2 \sqrt{x-8}\) < 24

Question 26.
7\(\sqrt[3]{x-3}\) ≥ 21

Question 27.
In a tsunami, the wave speeds (in meters per second) can be modeled by s(d ) = \(\sqrt{9.8 d}\), where d is the depth (in meters) of the water. Estimate the depth of the water when the wave speed is 200 meters per second.

Question 28.
Let f(x) = 2\(\sqrt{3-x}\) and g(x) = 4\(\sqrt[3]{3-x}\). Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate(fg)(2) and (\(\frac{f}{g}\))(2).

Question 29.
Let f(x) = 3x2 + 1 and g(x) = x + 4. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(−5) and (f − g)(−5).

Question 30.
f(x) = −\(\frac{1}{2}\)x + 10

Question 31.
f(x) = x2 + 8, x ≥ 0

Question 32.
f(x) = −x3 − 9

Question 33.
f(x) = 3\(\sqrt{x}\) + 5

Determine whether the functions are inverse functions.

Question 34.
f(x) = 4(x − 11)2, g(x) = \(\frac{1}{4}\)(x + 11)2

Question 35.
f(x) = −2x + 6, g(x) = −\(\frac{1}{2}\)x + 3

Question 36.
On a certain day, the function that gives U.S. dollars in terms of British pounds is d = 1.587p, where d represents U.S. dollars and p represents British pounds. Find the inverse function. Then find the number of British pounds equivalent to 100 U.S. dollars.

Rational Exponents and Radical Functions Chapter Test

Question 1.
Solve the inequality \(\) − 2 ≤ 13 and the equation 5\(\) − 2 = 13. Describe the similarities and differences in solving radical equations and radical inequalities.

Describe the transformation of f represented by g. Then write a rule for g.

Question 2.
f(x) = \(\sqrt{x}\)
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 92

Question 3.
f(x) = \(\sqrt[3]{x}\)
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 93

Question 4.
f(x) = \(\sqrt[5]{x}\)
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 94

Simplify the expression. Explain your reasoning.

Question 5.
642/3

Question 6.
(−27)5/3

Question 7.
\(\sqrt[4]{48 x y^{11} z^{3}}\)

Question 8.
\(\frac{\sqrt[3]{256}}{\sqrt[3]{32}}\)

Question 9.
Write two functions whose graphs are translations of the graph of y = \(\sqrt{x}\). The first function should have a domain of x ≥ 4. The second function should have a range of y ≥ −2.

Question 10.
In bowling, a handicap is a change in score to adjust for differences in the abilities of players. You belong to a bowling league in which your handicap h is determined using the formula h = 0.9(200 − a), where a is your average score. Find the inverse of the model. Then find the average for a bowler whose handicap is 36.

Question 11.
The basal metabolic rate of an animal is a measure of the amount of calories burned at rest for basic functioning. Kleiber’s law states that an animal’s basal metabolic rate R (in kilocalories per day) can be modeled by R = 73.3w3/4, where w is the mass (in kilograms) of the animal. Find the basal metabolic rates of each animal in the table.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 95

Question 12.
Let f(x) = 6x3/5 and g(x) = −x3/5. Find (f + g)(x) and (f − g)(x) and state the domain of each. Then evaluate (f + g)(32) and (f − g)(32).

Question 13.
Let f(x) = \(\frac{1}{2}\)x3/4 and g(x) = 8x. Find (fg)(x) and (\(\frac{f}{g}\))(x) and state the domain of each. Then evaluate (fg)(16) and (\(\frac{f}{g}\))(16).

Question 14.
A football player jumps to catch a pass. The maximum height h (in feet) of the player above the ground is given by the function h = \(\frac{1}{64}\)s2, where s is the initial speed (in feet per second) of the player. Find the inverse of the function. Use the inverse to find the initial speed of the player shown. Verify that the functions are inverse functions.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 96

Rational Exponents and Radical Functions Cumulative Assessment

Question 1.
Identify three pairs of equivalent expressions. Assume all variables are positive. Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 5 Rational Exponents and Radical Functions 97

Question 2.
The graph represents the function Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 98. Choose the correct values to complete the function.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 98.1

Question 3.
In rowing, the boat speed s (in meters per second) can be modeled by s= 4.62\(\sqrt[9]{n}\), where n is the number of rowers.
a. Find the boat speeds for crews of 2 people, 4 people, and 8 people.
b. Does the boat speed double when the number of rowers doubles? Explain.
c. Find the time (in minutes) it takes each crew in part (a) to complete a 2000-meter race.

Question 4.
A polynomial function fits the data in the table. Use finite differences to find the degree of the function and complete the table. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 99

Question 5.
The area of the triangle is 42 square inches. Find the value of x.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 100
Answer:
Given,
The area of the triangle is 42 square inches.
b = (x + 8)
h = x
We know that,
Area of the triangle = 1/2 × b × h
A = 42
42 = 1/2 × (x + 8) × x
84 = x² + 8x
x² + 8x – 84 = 0

Question 6.
Which equations are represented by parabolas? Which equations are functions? Place check marks in the appropriate spaces. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 101

Question 7.
What is the solution of the inequality 2\(\sqrt{x+3}\) − 1 < 3?
A. x < 1
B. −3 < x < 1
C. −3 ≤ x < 1
D. x ≥ −3

Question 8.
Which function does the graph represent? Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 102

Question 9.
Your friend releases a weather balloon 50 feet from you. The balloon rises vertically. When the balloon is at height h, the distance d between you and the balloon is given by d = \(\sqrt{2500+h^{2}}\), where h and d are measured in feet. Find the inverse of the function. What is the height of the balloon when the distance between you and the balloon is 100 feet?

Question 10.
The graphs of two functions f and g are shown. Are f and g inverse functions? Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 5 Rational Exponents and Radical Functions 103

Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics

Big Ideas Math Algebra 2 Answers Chapter 11

Are you looking for the best guide for the questions avail in Algebra 2 Data Analysis and Statistics? If yes, then no more searching, here you can find Big Ideas Math Algebra 2 Answer Key Ch 11 Data Analysis and Statistics. All the students can get detailed solutions provided for all the BIM Algebra 2 Chapter 11 Data Analysis and Statistics Questions ensuring they understand the concepts simply. So, Download Big Ideas Math Algebra 2 Answers in order to learn the concepts in Chapter 11 Data Analysis and Statistics in depth.

Big Ideas Math Book Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics

Check out the topics covered in Big Ideas Math Textbook Algebra 2 Chapter 11 Data Analysis and Statistics before you start your preparation. All the questions are prepared as per the Big Ideas Math Algebra 2 Textbooks. Here the students can get the solutions for Exercises, Chapter Tests, Review Tests, Cumulative Assessments, Practice Tests, etc. Click on the links available below to start your preparation. Also, you can test your knowledge by solving the questions in Quizzes, Chapter Reviews and Chapter tests and know how much skills you have improved from here.

Data Analysis and Statistics Maintaining Mathematical Proficiency

Find the mean, median, and mode of the data set. Then determine which measure of center best represents the data. Explain.
Question 1.
36, 82, 94, 83, 86, 82
Answer:
Given data is 36, 82, 94, 83, 86, 82
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
Mean = \(\frac { 36 + 82 + 94 + 83 + 86 + 82 }{ 6 } \)
= 77.16
So, Mean = 77.16
The median best represents the data. It is the middle value of the given list of data when arranged in order.
Arrange the data in ascending order.
36, 82, 82, 83, 86, 94
Median = \(\frac { 82 + 83 }{ 2 } \)
= 82.5
So, Median = 82.5
Mode is defined as the most frequent number.
36, 82, 82, 83, 86, 94
82 is repeated 2 times.
So, the mode is 82.
Mode = 82

Question 2.
74, 89, 71, 70, 68, 70
Answer:
Given data is 74, 89, 71, 70, 68, 70
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
Mean = \(\frac {74 + 89 + 71 + 70 + 68 + 70  }{ 6 } \)
= 73.6
The median best represents the data. It is the middle value of the given list of data when arranged in order.
Arrange the data in ascending order.
68, 70, 70, 71, 74, 89
Median = \(\frac { 70 + 71 }{ 2 } \)
= 70.5
Mode is defined as the most frequent number.
68, 70, 70, 71, 74, 89
70 is repeated 2 times.
Mode = 70

Question 3.
1, 18, 12, 16, 11, 15, 17, 44, 44
Answer:
Given data is 1, 18, 12, 16, 11, 15, 17, 44, 44
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
Mean = \(\frac { 1 + 18 + 12 + 16 + 11 + 15 + 17 + 44 + 44 }{ 9} \)
= 19.7
The median best represents the data. It is the middle value of the given list of data when arranged in order.
Arrange the data in ascending order.
1, 11, 12, 15, 16, 17, 18, 44, 44
Median = 16
Mode is defined as the most frequent number.
44 is repeated 2 times.
Mode = 44

Find and interpret the standard deviation of the data set.
Question 4.
43, 48, 41, 51, 42
Answer:
Given data is 43, 48, 41, 51, 42
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
Mean = \(\frac { 43 + 48 + 41 + 51 + 42 }{ 5} \)
= 45
So, mean = 45

x μ (x – μ) (x – μ)²
43 45 -2 4
48 45 3 9
41 45 -4 16
51 45 6 36
42 45 -3 9

Variance = \(\frac { 4 + 9 + 16 + 36 + 9 }{ 5 } \)
= 14.8
Standard deviation = √(14.8) = 3.84

Question 5.
28, 26, 21, 44, 29, 32
Answer:
Given data is 28, 26, 21, 44, 29, 32
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
Mean = \(\frac { 28 + 26 + 21 + 44 + 29 + 32 }{ 6} \)
= 30

x μ (x – μ) (x – μ)²
28 30 -2 4
26 30 -4 16
21 30 -9 81
44 30 14 196
29 30 -1 1
32 30 2 4

Variance = \(\frac { 4 + 16 + 81 + 196 + 1 + 4 }{ 6 } \)
= 50.3
Standard deviation = √(50.3) = 7.09

Question 6.
65, 56, 49, 66, 62, 52, 53, 49
Answer:
Given data is 65, 56, 49, 66, 62, 52, 53, 49
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
Mean = \(\frac { 65+ 56+ 49+ 66+ 62+ 52+ 53+ 49 }{ 8} \)
= 56.5

x μ (x – μ) (x – μ)²
65 56.5 8.5 72.25
56 56.5 -0.5 0.25
49 56.5 -7.5 56.25
66 56.5 9.5 90.25
62 56.5 5.5 30.25
52 56.5 -4.5 20.25
53 56.5 -3.5 12.25
49 56.5 -7.5 56.25

Variance = \(\frac { 72.25 + 0.25 + 56.25 + 90.25 + 30.25 + 20.25 + 12.25 + 56.25 }{ 8 } \)
= 42.25
Standard deviation = √(42.25) = 6.5

Question 7.
ABSTRACT REASONING
Describe a data set that has a standard deviation of zero. Can a standard deviation be negative? Explain your reasoning.
Answer:
A dataset having a standard deviation of zero would indicate that every data value would be equal to the mean. If all the observations are equal then the standard deviation is 0.
Standard Deviation cannot be negative because when we have at least two numbers in the data set which are not exactly equal to one another, the standard deviation has to be positive.

Data Analysis and Statistics Mathematical Practices

Mathematically proficient students use diagrams and graphs to show relationships between data. They also analyze data to draw conclusions.

Monitoring Progress

Use the Internet or some other reference to determine which age pyramid is that of Canada, Japan, and Mexico. Compare the mean, median, and mode of the three age pyramids.
Question 1.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 1
Answer:
For Male,
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
Mean = 30.65347
The median best represents the data. It is the middle value of the given list of data when arranged in order.
Median class = 24.5 – 29.5 = 27.57
Modal class = 4.5 – 9.5
Mode is defined as the most frequent number.

Mode = 8.3274
For Female,
Mean = Sum of observations/ no. of observations
Mean = 32.2377
The median best represents the data. It is the middle value of the given list of data when arranged in order.
Median class = 29.5 – 34.5
Median = 29.7866
Modal class = 14.5-19.5
Mode is defined as the most frequent number.
Mode = 16.7714

Question 2.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 2
Answer:
Given, Pyramid
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
a. The mean is less than that of Japan’s but higher than Mexico’s
The median best represents the data. It is the middle value of the given list of data when arranged in order.
Arrange the data in ascending order.
b. Median = 14.8
Mode is defined as the most frequent number
c. Mode = 45 years

Question 3.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 3
Answer:

Lesson 11.1 Using Normal Distributions

Essential Question In a normal distribution, about what percent of the data lies within one, two, and three standard deviations of the mean?
Recall that the standard deviation σ of a numerical data set is given by
σ = \(\sqrt{\frac{\left(x_{1}-\mu\right)^{2}+\left(x_{2}-\mu\right)^{2}+\cdots+\left(x_{n}-\mu\right)^{2}}{n}}\)
where n is the number of values in the data set and μ is the mean of the data set.

EXPLORATION 1

Analyzing a Normal Distribution
Work with a partner. In many naturally occurring data sets, the histogram of the data is bell-shaped. In statistics, such data sets are said to have a normal distribution. For the normal distribution shown below, estimate the percent of the data that lies within one, two, and three standard deviations of the mean. Each square on the grid represents 1%.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 1

EXPLORATION 2

Analyzing a Data Set
Work with a partner. A famous data set was collected in Scotland in the mid-1800s. It contains the chest sizes (in inches) of 5738 men in the Scottish Militia. Do the data fit a normal distribution? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 2

Communicate Your Answer

Question 3.
In a normal distribution, about what percent of the data lies within one, two, and three standard deviations of the mean?
Answer:
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
68% of the normal distribution lies within one standard deviation of the mean.
95% of the normal distribution lies within two standard deviations of the mean.
99.7% of the normal distribution lies within three standard deviations of the mean.

Question 4.
Use the Internet or some other reference to find another data set that is normally distributed. Display your data in a histogram.
Answer:

A normal distribution has mean μ and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution.

Question 1.
P(x ≤ μ)
Answer:
Given,
P(x ≤ μ)
Because of the symmetry of the normal curve, half of the values are on each side of the mean μ.
Thus, P(x ≤ μ)= 0.5

Question 2.
P(x ≥ μ)
Answer:
Given,
P(x ≥ μ)
Because of the symmetry of the normal curve, half of the values are on each side of the mean μ.
P(x ≥ μ) = 0.5

Question 3.
P(μ ≤ x ≤ μ + 2σ)
Answer:
Given,
P(μ ≤ x ≤ μ + 2σ)
The probability that a randomly selected x-value lies between μ and μ + 2σ is the marked area under the normal curve.
P(μ ≤ x ≤ μ + 2σ) = 0.34 + 0.135 = 0.475
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 1

Question 4.
P(μ − σ ≤ x ≤ μ)
Answer:
Given,
P(μ − σ ≤ x ≤ μ)
The total area under the related normal curve is 1
About 68% of the area lies within μ ± σ
About 95% of the area lies within μ ± 2σ
About 99.7% of the area lies within μ ± 3σ
compute the area under the curve bounded from the left with μ − σ and from the right with μ. thus, using the properties of the distribution
P(μ − σ ≤ x ≤ μ) = \(\frac { 0.68 }{ 2 } \) = 0.34
Thus P(μ − σ ≤ x ≤ μ) = 0.34

Question 5.
P(x ≤ μ − 3σ)
Answer:
Given,
P(x ≤ μ − 3σ)
The probability that the randomly selected x-value lies between μ and μ + 2σ is the marked area under the normal curve
P(x ≤ μ − 3σ) = 0.0015
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 2

Question 6.
P(x ≥ μ + σ)
Answer:
P(x ≥ μ + σ)
The probability that the randomly selected x-value lies between μ – σ and μ is the marked area under the normal curve
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 3
P(x ≥ μ + σ) = 0.135 + 0.0235 + 0.0015 = 0.16

Question 7.
WHAT IF?
In Example 2, about what percent of the people taking the test have scores between 43 and 79?
Answer:
Given,
The test has scored between 43 and 79
Let’s suppose that a normal distribution has a mean μ = 55 and a standard deviation σ = 12. Compute
P(43 ≤ x ≤ 79)
consider the following
μ – 43 = 55 – 43 = 12 = σ
25 = μ – σ
79 – μ = 79 – 55 = 24 = 2 . 12 = 2σ
7 = μ + 2σ
P(43 ≤ x ≤ 79) = P(μ – σ ≤ x ≤ μ + 2σ)
= P(μ – σ ≤ x ≤ μ) + P(μ ≤ x ≤ μ + 2σ)
= \(\frac { .68 }{ 2 } \) + \(\frac { 0.95 }{ 2 } \)
= 0.34 + 0.475
= 0.815
So, P(43 ≤ x ≤ 79) = 0.815

Question 8.
WHAT IF?
In Example 3, what is the probability that the infant weighs 3990 grams or more?
Answer:
If a normal distribution has a mean μ = 3270 and standard deviation σ = 600.
Compute
P(x ≥ 3990)
z = \(\frac { x – μ }{ σ } \)
P(x ≥ 3990) = P(\(\frac { x – 3270 }{ 600 } \) ≥\(\frac { 3990 – 3270 }{ 600 } \)) = P(z ≥ 1.2) = 1 – P(z<1.2)
P(z<1.2) = 0.8849
P(x ≥ 3990) = 1 – 0.8849 = 0.1151
The probability that the infant weighs 3990 grams or more is 0.1151

Question 9.
Explain why it makes sense that P(z ≤ 0) = 0.5.
Answer:
Given,
P(z ≤ 0) = 0.5
The area under the curve of the normal distribution is 1.
Thus, there are left of the mean will be half of it i.e 0.5. Compute this probability by using the z-values
z = \(\frac { x – μ }{ σ } \)
The z value for a particular x-value is called the z score for the x-value and is the number of standard deviations the x-value lies above or below the mean μ. The z-scores will be read directly from the table.
P(x≤μ) = P(\(\frac { x – μ }{ σ } \)<\(\frac { μ – μ }{ σ } \)) = P(z≤0)
P(x≤μ) = P(z≤0)
Since P(x≤μ) = 0.5
P(z ≤ 0) = 0.5

Question 10.
Determine whether the histogram has a normal distribution.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 3

Answer:
The histogram is skewed left. So, the histogram does not have a normal distribution and you cannot use the normal distribution to interpret the histogram.

Using Normal Distributions 11.1 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe how to use the standard normal table to find P(z ≤ 1.4).
Answer:
Find the value where row 1 and column 4 intersect. The value is the probability is P(z ≤ 1.4).

Question 2.
WHICH ONE DOESN’T BELONG?
Which histogram does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 4

Answer:
The first diagram does not belong with the other three. Because the other diagrams are symmetric in the bell-shaped curve and the first diagram is not.

Monitoring Progress and Modeling with Mathematics

ATTENDING TO PRECISION In Exercises 3–6, give the percent of the area under the normal curve represented by the shaded region(s).
Question 3.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 3

Question 4.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 6

Answer:
The total area under the related normal curve is 1
About 68% of the area lies within μ ± σ
About 95% of the area lies within μ ± 2σ
About 99.7% of the area lies within μ ± 3σ
Compute the area under the curve bounded from the left with μ − σ and from the right with μ. thus, using the properties of the distribution
\(\frac { (μ – 3σ) – (μ – σ) }{ 2 } \) = \(\frac { 99.7 – 68 }{ 2 } \) = 15.85%

Question 5.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 5

Question 6.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 8

Answer:
The total area under the related normal curve is 1
About 68% of the area lies within μ ± σ
About 95% of the area lies within μ ± 2σ
About 99.7% of the area lies within μ ± 3σ
compute the area under the curve bounded from the left with μ − σ and from the right with μ. thus, using the properties of the distribution.
(μ – 2σ) – (μ – σ) = 95 – 68 = 27%

In Exercises 7–12, a normal distribution has mean μ and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution.
Question 7.
P(x ≤ μ − σ)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 7

Question 8.
P(x ≥ μ − σ)
Answer:
The total area under the related normal curve is 1
About 68% of the area lies within μ ± σ
About 95% of the area lies within μ ± 2σ
About 99.7% of the area lies within μ ± 3σ
P(x ≥ μ − σ) = 1 – P(x < μ − σ)
= 1 – P(x ≤ μ − σ)
= 1 – 0.16
= 0.84
So, P(x ≥ μ − σ) = 0.84

Question 9.
P(x ≥ μ + 2σ)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 9

Question 10.
P(x ≤ μ + σ)
Answer:
The total area under the related normal curve is 1
About 68% of the area lies within μ ± σ
About 95% of the area lies within μ ± 2σ
About 99.7% of the area lies within μ ± 3σ
\(\frac { 1 + μ + σ }{ 2 } \) = \(\frac { 1 + 0.68 }{ 2 } \) = 0.84
So, P(x ≤ μ + σ) = 0.84

Question 11.
P(μ − σ ≤ x ≤ μ + σ)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 11

Question 12.
P(μ − 3σ ≤ x ≤ μ)
Answer:
The total area under the related normal curve is 1
About 68% of the area lies within μ ± σ
About 95% of the area lies within μ ± 2σ
About 99.7% of the area lies within μ ± 3σ
P(μ − 3σ ≤ x ≤ μ) = \(\frac { 0.997 }{ 2 } \)
= 0.4985
So, P(μ − 3σ ≤ x ≤ μ) = 0.4985

In Exercises 13–18, a normal distribution has a mean of 33 and a standard deviation of 4. Find the probability that a randomly selected x-value from the distribution is in the given interval.
Question 13.
between 29 and 37
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 13

Question 14.
between 33 and 45
Answer:
Let the normal distribution has a mean μ = 33 and standard deviation σ = 4. Compute
P(33 ≤ x ≤ 45)
μ – 33 = 33 – 33 = 0
So, 33 = μ
45 – μ = 45 – 33 = 12 = 3 . 4 = 3σ
So, 45 = μ + 3σ
P(33 ≤ x ≤ 45) = P(3μ ≤ x ≤ μ + 3σ) = \(\frac { 0.997 }{ 2 } \) = 0.4985
P(33 ≤ x ≤ 45) = 0.4985

Question 15.
at least 25
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 15

Question 16.
at least 29
Answer:
Let the normal distribution has a mean μ = 33 and standard deviation σ = 4. Compute
P(x ≥ 29)
μ – 29 = 33 – 29 = 4 = σ
So, 29 = μ – σ
P(x ≥ 29) = P(x ≥ μ – σ) = 0.5 + \(\frac { 0.68 }{ 2 } \) = 0.5 + 0.34
= 0.84
So, P(x ≥ 29) = 0.84

Question 17.
at most 37
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 17

Question 18.
at most 21
Answer:
Let’s suppose that a normal distribution has a mean μ = 33 and standard deviation σ = 4. Compute
P is less than or equal to 21.
P(x ≤ 21)
μ – 21 = 33 – 21 = 12 = 3 . 4 = 3σ
So, 21 = μ + 3σ
P(x ≤ 21) = P(x ≤ μ – 3σ) = \(\frac { 1 – 0.997 }{ 2 } \) = 0.0015
P(x ≤ 21) = 0.0015

Question 19.
PROBLEM SOLVING
The wing lengths of houseflies are normally distributed with a mean of 4.6 millimeters and a standard deviation of 0.4 millimeter.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 9
a. About what percent of houseflies have wing lengths between 3.8 millimeters and 5.0 millimeters?
b. About what percent of houseflies have wing lengths longer than 5.8 millimeters?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 19

Question 20.
PROBLEM SOLVING
The times a fire department takes to arrive at the scene of an emergency are normally distributed with a mean of 6 minutes and a standard deviation of 1 minute.
a. For about what percent of emergencies does the fire department arrive at the scene in 8 minutes or less?
Answer:
The time a fire department takes to arrive at the scene of an emergency is normally distributed with a mean of 6 minutes, σ = 1 min. Compute what percent of emergencies the fire department arrives at the scene in 8 minutes or less
P(x≤8)
8 – μ = 8 – 6 = 2 = 2 . 1 = 2σ
So, 8 = μ + 2σ
P(x≤8) = P(x≤μ + 2σ) = 0.5 + \(\frac { 0.95 }{ 2 } \)
= 0.5 + 0.475 = 0.975
About 97.5% fire department arrives at the scene in 8 minutes or less

b. The goal of the fire department is to reach the scene of an emergency in 5 minutes or less. About what percent of the time does the fire department achieve its goal?
Answer:
Compute what percent of the time the fire department achieves its goal
P((x≤5)
μ – 5 = 6 – 5 = 1 = σ
So, 5 = μ – σ
P((x≤5) = P((x≤μ – σ) = \(\frac {1 – 0.68 }{ 2 } \)
= 0.16
About 16% fire department reaches the goal.

ERROR ANALYSIS In Exercises 21 and 22, a normal distribution has a mean of 25 and a standard deviation of 2. Describe and correct the error in finding the probability that a randomly selected x-value is in the given interval.
Question 21.
between 23 and 27
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 21

Question 22.
at least 21
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 11
Answer:
P(x ≥ 21) = P(x ≥ μ − 2σ) = 0.475 + 0.5 = 0.975

Question 23.
PROBLEM SOLVING
A busy time to visit a bank is during its Friday evening rush hours. For these hours, the waiting times at the drive-through window are normally distributed with a mean of 8 minutes and a standard deviation of 2 minutes. You have no more than 11 minutes to do your banking and still make it to your meeting on time. What is the probability that you will be late for the meeting?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 23

Question 24.
PROBLEM SOLVING
Scientists conducted aerial surveys of a seal sanctuary and recorded the number x of seals they observed during each survey. The numbers of seals observed were normally distributed with a mean of 73 seals and a standard deviation of 14.1 seals. Find the probability that at most 50 seals were observed during a randomly chosen survey.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 12

Answer:
The number of seals observed were normally distributed with μ = 73 seals, σ = 14.1 seal
z = \(\frac { x – μ }{ σ } \)
P(x≤50) = P(\(\frac { x – 73 }{ 14.1 } \)≤\(\frac { 50 – 73 }{ 14.1 } \)) = P(z≤-1.6)
P(x≤50) = P(z≤-1.6) = 0.0548
The probability that at most 50 seals were observed during a randomly chosen survey is 0.0548

In Exercises 25 and 26, determine whether the histogram has a normal distribution.
Question 25.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 13
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 25

Question 26.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 14

Answer:
The histogram is bell-shaped and fairly symmetric. Thus, the histogram has an approximately normal distribution.

Question 27.
ANALYZING RELATIONSHIPS
The table shows the numbers of tickets that are sold for various baseball games in a league over an entire season. Display the data in a histogram. Do the data fit a normal distribution? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 15
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 27

Question 28.
PROBLEM SOLVING
The guayule plant, which grows in the southwestern United States and in Mexico, is one of several plants that can be used as a source of rubber. In a large group of guayule plants, the heights of the plants are normally distributed with a mean of 12 inches and a standard deviation of 2 inches.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 16
a. What percent of the plants are taller than 16 inches?
Answer:
The heights of the plants are normally distributed with μ = 12 in, σ = 2 in
z = \(\frac { x – μ }{ σ } \)
P(x≥16) = P(\(\frac { x – 12 }{ 2 } \)≥\(\frac { 16 – 12 }{ 2 } \)) = P(z≥2)
= 1 – P(z<2)
P(x≥16)= 1 – P(z<2) = 1 – 0.9772 = 0.0228
The percent of plants that are taller than 16 inches is about 2.28%

b. What percent of the plants are at most 13 inches?
Answer:
What percent of plants are at mos 13 inches = P(x≤13)
z = \(\frac { x – μ }{ σ } \)
P(x≤13) = P(\(\frac { x – 12 }{ 2 } \)≥\(\frac { 13 – 12 }{ 2 } \)) = P(z≤0.5)
P(x≤13) = P(z≤0.5) = 0.6915
The percent of plants that are at most 13 inches is about 69.15%

c. What percent of the plants are between 7 inches and 14 inches?
Answer:
What percent of plants are between 7 and 14 inches = P(7≤x≤14)
P(7≤x≤14) = P(\(\frac { 7 – 12 }{ 2 } \) ≤ \(\frac { x – 12 }{ 2 } \) ≤ \(\frac { 14 – 12 }{ 2 } \))
= P(-2.5≤z≤1)
= P(z≤1) – P(z≤-2.5)
= 0.8413 – 0.0062 = 0.8351
The percent of plants between 7 and 14 inches is about 83.51%

d. What percent of the plants are at least 3 inches taller than or at least 3 inches shorter than the mean height?
Answer:
What percent of plants are at least 3 inches taller than or at least 3 inches shorter than the mean weight
P(μ – 3 ≥ x ≥ μ + 3) = P(12 – 3 ≥ x ≥ 12 + 3)
= P(9 ≥ x ≥ 15)
= 1 – P(9≤x≤15)
P(9≤x≤15) = P(\(\frac { 9 – 12 }{ 2 } \) ≤ \(\frac { x – 12 }{ 2 } \) ≤ \(\frac { 15 – 12 }{ 2 } \))
= P(-1.5≤z≤1.5)
= P(z≤1.5) – P(z≤-1.5)
P(9 ≥ x ≥ 15) = 1 – P(z≤1.5) + P(z≤-1.5)
= 1 – 0.9332 + 0.0668
= 0.1336
The percent of plants that are at least 3 inches taller than or at least 3 inches shorter than the mean height is about 13.36%

Question 29.
REASONING
Boxes of cereal are filled by a machine. Tests show that the amount of cereal in each box varies. The weights are normally distributed with a mean of 20 ounces and a standard deviation of 0.25 ounce. Four boxes of cereal are randomly chosen.
a. What is the probability that all four boxes contain no more than 19.4 ounces of cereal?
b. Do you think the machine is functioning properly? Explain.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 29

Question 30.
THOUGHT PROVOKING
Sketch the graph of the standard normal distribution function, given by
f (x) = \(\frac{1}{\sqrt{2} \pi}\)e−x2/2.
Estimate the area of the region bounded by the x-axis, the graph of f, and the vertical lines x =−3 and x = 3.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 4
P(-3≤z≤3) = P(z≤3) – P(z≤-3)
P(z≤3) = 0.9987
P(z≤-3) = 0.0013
P(-3≤z≤3) = P(z≤3) – P(z≤-3)
= 0.9987 – 0.0013 = 0.997

Question 31.
REASONING
For normally distributed data, describe the value that represents the 84th percentile in terms of the mean and standard deviation.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 31

Question 32.
HOW DO YOU SEE IT?
In the figure, the shaded region represents 47.5% of the area under a normal curve. What are the mean and standard deviation of the normal distribution?
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 17

Answer:
The shaded area represents 47.5% and since about 95% of the area lies within 2 standard deviations of the mean
Mean is the average of the numbers.
Mean = Sum of observations/ no. of observations
47.5% = \(\frac { 95% }{ 2 } \)
μ + 2σ= 16
μ + σ = 13
Subtract both equations
μ + 2σ – μ – σ = 16 – 13
σ = 3
Substitute σ = 3 in one equation
μ + 3 = 13
μ = 10

Question 33.
DRAWING CONCLUSIONS
You take both the SAT (Scholastic Aptitude Test) and the ACT (American College Test). You score 650 on the mathematics section of the SAT and 29 on the mathematics section of the ACT. The SAT test scores and the ACT test scores are each normally distributed. For the SAT, the mean is 514 and the standard deviation is 118. For the ACT, the mean is 21.0 and the standard deviation is 5.3.
a. What percentile is your SAT math score?
b. What percentile is your ACT math score?c. On which test did you perform better? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 33

Question 34.
WRITING
Explain how you can convert ACT scores into corresponding SAT scores when you know the mean and standard deviation of each distribution.
Answer:
If we know the ACt z-score and change it into the z-formula for SAT test, we can obtain an equivalent score by solving the equation we get.
Let’s suppose that SAT and ACT test scores are normally distributed with mean μs, μa and standard deviations σs, σa. Let’s convert the ACT scores into corresponding SAT scores when you know the mean and SD of each distribution.
z-score of ACT is z<act>
P(xa < a) = P(\(\frac { xa – μa }{ σa } \) ≤ \(\frac { a – μa }{ σa } \)) = z<act>
P(xs < x) = z<act>
P(\(\frac { xs – μs }{ σs } \) ≤ \(\frac { X – μs }{ σs } \)) = z<act>
By solving the equation for x, we get the equivalent SAT score

Question 35.
MAKING AN ARGUMENT
A data set has a median of 80 and a mean of 90. Your friend claims that the distribution of the data is skewed left. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 35

Question 36.
CRITICAL THINKING
The average scores on a statistics test are normally distributed with a mean of 75 and a standard deviation of 10. You randomly select a test score x. Find P (∣x−μ∣≥ 15 ).
Answer:
Let’s suppose that statistical tests are normally distributed with μ = 75, σ = 101
P(|x – μ| ≥ 15) = 1 – P(|x – μ| ≤ 15)
= 1 – P(-15 ≤ x – μ ≤ 15)
= 1 – P(μ – 15 ≤ x ≤ μ + 15)
= 1 – P(x≤ μ + 15) – P(x≤ μ – 15)
= 1 – P(x≤ 75 + 15) – P(x≤ 75 – 15)
= 1 – P(x≤90) + P(x≤60)
z = \(\frac { x – μ }{ σ } \)
P(x≤90) = P(\(\frac { x – 75 }{ 10 } \) ≤\(\frac { 90 – 75 }{ 10 } \))
= P(z≤1.5)
= 0.9332
P(x≤60) = P(\(\frac { x – 75 }{ 10 } \) ≤\(\frac { 60 – 75 }{ 10 } \))
= P(z≤-1.5)
= 0.0668
P(|x – μ| ≥ 15) = 1 – P(x≤90) + P(x≤60)
= 1 – 0.9332 + 0.0668
= 0.1336

Maintaining Mathematical Proficiency

Graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.
Question 37.
f(x) = x3 − 4x2+ 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 37

Question 38.
g(x) = \(\frac{1}{4}\)x4 − 2x2−x − 3

Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 5
The x-intercepts are x = -2.8, x = 3.2
The local maximum value is -2.87, -5.07
The local minimum values are -1.86, -9.06
x∈(-∞, -1.86) U (-0.25, 2.11) function decreases
x ∈ (-1.86, -0.25) U(2.11, ∞) function increases.

Question 39.
h(x) = −0.5x2+ 3x + 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 a 39

Question 40.
f(x) = −x4 + 6x2 − 13
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.1 6
There are no x-intercepts
The local maximum values are -4, 4
The local minimum values are -173, 1.73
x ∈ (-∞, -1.73) U (0, 1.73) function increases
x ∈ (-1.73, 0) U(1.73, ∞) function decreases

Lesson 11.2 Populations, Samples, and Hypotheses

Essential Question How can you test theoretical probability using sample data?

EXPLORATION 1

Using Sample Data
Work with a partner.
a. When two six-sided dice are rolled, what is the theoretical probability that you roll the same number on both dice?
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 1
b. Conduct an experiment to check your answer in part (a). What sample size did you use? Explain your reasoning.
c. Use the dice rolling simulator at BigIdeasMath.com to complete the table. Do your experimental data check the theoretical probability you found in part (a)? Explain. What happens as you increase the sample size?
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 2

EXPLORATION 2

Using Sample Data
Work with a partner.
a. When three six-sided dice are rolled, what is the theoretical probability that you roll the same number on all three dice?
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 3
b. Compare the theoretical probability you found in part (a) with the theoretical probability you found in Exploration 1(a).
c. Conduct an experiment to check your answer in part (a). How does adding a die affect the sample size that you use? Explain your reasoning.
d. Use the dice rolling simulator at BigIdeasMath.com to check your answer to part (a). What happens as you increase the sample size?

Communicate Your Answer

Question 3.
How can you test theoretical probability using sample data?
Answer:
Theoretical probability is the ratio of the number of favorable outcomes to the total number of possible outcomes.
P(event) = number of favorable outcomes/total number of possible outcomes.
Sample Data:
A coin is tossed
P(head) = 1/2
P(tail) = 1/2.

Question 4.
Conduct an experiment to determine the probability of rolling a sum of 7 when two six-sided dice are rolled. Then find the theoretical probability and compare your answers.
Answer:
Total number of outcomes = 6 × 6 = 36
Favorable outcomes = (1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)
P(A) = 6/36 = 1/6
So, the probability of getting a sum as 7 when two dice are thrown is 1/6.

Monitoring Progress

In Monitoring Progress Questions 1 and 2, identify the population and the sample.
Question 1.
To estimate the retail prices for three grades of gasoline sold in the United States, the Energy Information Association calls 800 retail gasoline outlets, records the prices, and then determines the average price for each grade.
Answer:
In the population of all retail gasoline outlets, a sample of 800 retail gasoline outlets is chosen to participate in the survey.

Question 2.
A survey of 4464 shoppers in the United States found that they spent an average of $407.02 from Thursday through Sunday during a recent Thanksgiving holiday.
Answer:
In the population of all shoppers in the United States, a sample of 4464 shoppers are chosen to participate in the survey.

Question 3.
A survey found that the median salary of 1068 statisticians is about $72,800. Is the median salary a parameter or a statistic? Explain your reasoning.
Answer:
Since the median salary is calculated for a sample of 1068 statisticians, not the entire population, this median salary cannot represent the whole population. Thus, it is a statistic.

Question 4.
The mean age of U.S. representatives at the start of the 113th Congress was about 57 years. Is the mean age a parameter or a statistic? Explain your reasoning.
Answer:
Since the average age is computed for all representatives, this average represents a parameter.

Question 5.
WHAT IF?
In Example 3, what should you conclude when you roll the actual die 50 times and get (a) 24 odd numbers and (b) 31 odd numbers?
Answer:
a. Getting 24 odd numbers in 50 rolls corresponds to a proportion of \(\frac { 24 }{ 50 } \) = 0.48. In the simulation, this result has a relative frequency of 0.12. Most of the results are close to 0.50. Because this result can easily occur by chance, you can conclude that the maker’s claim is most likely true.

b. Getting 35 odd numbers in 50 rolls corresponds to a proportion of \(\frac { 31 }{ 50 } \) = 0.62. In the simulation, this result occurs with a small frequency which is about 0.02. Because getting 31 odd numbers is highly unlikely to occur by chance, you can conclude that the maker’s claim is most likely false.

Populations, Samples, and Hypotheses 11.2 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A portion of a population that can be studied in order to make predictions about the entire population is a(n) ___________.
Answer:
A portion of a population that can be studied in order to make predictions about the entire population is a sample.

Question 2.
WRITING
Describe the difference between a parameter and a statistic. Give an example of each.
Answer:
Parameters are numbers that describe data for an entire population. Example: population mean.
Statistics are numbers that describe data from a sample i.e some subset of the entire population. Example: Sample mean.

Question 3.
VOCABULARY
What is a hypothesis in statistics?
Answer:
The hypothesis is a type of statistical analysis in which you put your assumptions about a parameter to the test. A hypothesis in statistics is a claim about a characteristic of a population.

Question 4.
WRITING
Describe two ways you can make an incorrect decision when analyzing a hypothesis.
Answer:
A type I error occurs if an investigator rejects a null hypothesis that is actually true in the population.
A type II error occurs if the investigator fails to reject a null hypothesis that is actually false in the population.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, determine whether the data are collected from a population or a sample. Explain your reasoning.
Question 5.
the number of high school students in the United States
Answer:
It is a population mean because data are collected from a population because every high school is counted.

Question 6.
the color of every third car that passes your house
Answer:
It is a sample mean because it was counted as a subset of the entire population.

Question 7.
a survey of 100 spectators at a sporting event with 1800 spectators
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 7

Question 8.
the age of each dentist in the United States
Answer:
It is a population because it is counted as the age of each dentist in the United States.

In Exercises 9–12, identify the population and sample. Describe the sample.
Question 9.
In the United States, a survey of 1152 adults ages 18 and over found that 403 of them pretend to use their smartphones to avoid talking to someone.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 9

Question 10.
In the United States, a survey of 1777 adults ages 18 and over found that 1279 of them do some kind of spring cleaning every year.
Answer:
Given,
In the population of all adults ages 18 and over in the United States, a sample of 1777 adults is surveyed.
In the survey, 1279 of them do some kind of spring cleaning every year and
1777 – 1279 = 489 do not.

Question 11.
In a school district, a survey of 1300 high school students found that 1001 of them like the new, healthy cafeteria food choices.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 11

Question 12.
In the United States, a survey of 2000 households with at least one child found that 1280 of them eat dinner together every night.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 5

Answer:
Given,
In the United States, a survey of 2000 households with at least one child found that 1280 of them eat dinner together every night.
In the population of all households with at least one child in the United States, a sample of 2000 households is surveyed.

In Exercises 13–16, determine whether the numerical value is a parameter or a statistic. Explain your reasoning.
Question 13.
The average annual salary of some physical therapists in a state is $76,210.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 13

Question 14.
In a recent year, 53% of the senators in the United States Senate were Democrats.

Answer:
The numerical value is a parameter. Since we have that 47 percent of all senators in the United States senate were Democrats, it represents a parameter.

Question 15.
Seventy-three percent of all the students in a school would prefer to have school dances on Saturday.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 15

Question 16.
A survey of U.S. adults found that 10% believe a cleaning product they use is not safe for the environment.
Answer:
Given,
A survey of U.S. adults found that 10% believe a cleaning product they use is not safe for the environment.
Since we have a survey where it’s found that 10% of some adults, not the entire population, believe that a cleaning product they use is not safe for the environment. Hence it is a statistic.

Question 17.
ERROR ANALYSIS
A survey of 1270 high school students found that 965 students felt added stress because of their workload. Describe and correct the error in identifying the population and the sample.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 17

Question 18.
ERROR ANALYSIS
Of all the players on a National Football League team, the mean age is 26 years. Describe and correct the error in determining whether the mean age represents a parameter or statistic. Because the mean age of 26 is based only on one football team, it is a statistic.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 8
Answer:
It is not a statistic parameter. Since we have that the mean age of all the players on a National Football League team is 26.

Question 19.
MODELING WITH MATHEMATICS
You flip a coin 4 times and do not get a tails. You suspect this coin favors heads. The coin maker claims that the coin doesnot favor heads or tails. You simulate flipping the coin 50 times by repeatedly drawing 200 random samples of size 50. The histogram shows the results. What should you conclude when you flip the actual coin 50 times and get (a) 27 heads and (b) 33 heads?
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 19

Question 20.
MODELING WITH MATHEMATICS
Use the histogram in Exercise 19 to determine what you should conclude when you flip the actual coin 50 times and get (a) 17 heads and (b) 23 heads.

Answer:
a. Let’s suppose that we flip the actual coin 50 times and get 17 heads.
Getting 17 heads in 50 flipping corresponds to a proportion of \(\frac { 17 }{ 50 } \) = 0.34.
In the simulation, this result occurs with a small frequency.
Because getting 17 heads is highly unlikely to occur by chance, you can conclude that the maker’s claim is most likely false.

b. Let’s suppose that we are flip the actual coin 50 times and get 23 heads.
Getting 23 heads in 50 flipping corresponds to a proportion of \(\frac { 23 }{ 50 } \) = 0.46.
In the simulation, this result had a relative frequency of 0.10. In fact, most of the results are close to 0.50.
Because getting this result can easily occur by chance, you can conclude that the makers claim is most likely true.

Question 21.
MAKING AN ARGUMENT
A random sample of five people at a movie theater from a population of 200 people gave the film 4 out of 4 stars. Your friend concludes that everyone in the movie theater would give the film 4 stars. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 21

Question 22.
HOW DO YOU SEE IT?
Use the Venn diagram to identify the population and sample. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 10
Answer:
A population will be a major of students at a university and the sample is the majors of students at a university that takes chemistry.

Question 23.
OPEN-ENDED
Find a newspaper or magazine article that describes a survey. Identify the population and sample. Describe the sample.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 23

Question 24.
THOUGHT PROVOKING
You choose a random sample of 200 from a population of 2000. Each person in the sample is asked how many hours of sleep he or she gets each night. The mean of your sample is 8 hours. Is it possible that the mean of the entire population is only 7.5 hours of sleep each night? Explain.
Answer:
Given,
You choose a random sample of 200 from a population of 2000.
Mean = 8 hours
Yes, it is possible because the mean of your sample is 8 hours, which is very close to 7.5 hours. Also, the random sample of 200 people is enough big size of the sample to estimate the mean of the population.

Question 25.
DRAWING CONCLUSIONS
You perform two simulations of repeatedly selecting a marble out of a bag with replacement that contains three red marbles and three blue marbles. The first simulation uses 20 random samples of size 10, and the second uses 400 random samples of size 10. The histograms show the results. Which simulation should you use to accurately analyze a hypothesis? Explain.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 25

Question 26.
PROBLEM SOLVING
You roll an eight-sided die five times and get a four every time. You suspect that the die favors the number four. The die maker claims that the die does not favor any number.
a. Perform a simulation involving 50 trials of rolling the actual die and getting a four to test the die maker’s claim. Display the results in a histogram.
b. What should you conclude when you roll the actual die 50 times and get 20 fours? 7 fours?

Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.2 1
Let’s suppose that we are rolling the eight-side die 50 times and get 20 fours. Getting 20 fours in 50 rollings corresponds to a proportion of \(\frac { 20 }{ 50 } \) = 0.4. In the simulation, this result does not occur. Because getting 20 fours is highly unlikely to occur by chance, you can conclude that the maker’s claim is most likely false.
Let’s suppose that we are rolling the eight side die 50 times and get 7 fours. Getting 7 fours in 50 rollings corresponds to a proportion of \(\frac { 7 }{ 50 } \) = 0.14.
In the simulation, this result had a relative frequency of 0.16. Because this result can easily occur by chance, you can conclude that the maker’s claim is most likely true.

Maintaining Mathematical Proficiency

Solve the equation by completing the square.
Question 27.
x2− 10x − 4 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 27

Question 28.
3t2 + 6t = 18
Answer:
Given,
3t² + 6t = 18
3t² + 6t – 180 = 0
3(t² + 2t) – 18 = 0
3(t² + 2t) – 18 + 3 – 3 = 0
3(t² + 2t + 1) – 21 = 0
3(t + 1)² – 21 = 0
3(t + 1)² = 21
(t + 1)² = 7
t + 1 = ±√7
t = ±√7 – 1
The solutions are -1 + √7, √7 + 1

Question 29.
s2 + 10s + 8 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 29

Solve the equation using the Quadratic Formula.
Question 30.
n2+ 2n + 2 = 0
Answer:
Given,
n2+ 2n + 2 = 0
n² + 2n + 1 + 1 = 0
(n + 1)² + 1 = 0
(n + 1)²= -1
n + 1 = ±i
n = ±i – 1
The solutions are -1 + i, -1 – i

Question 31.
4z2 + 28z = 15
Answer:
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics 11.2 a 31

Question 32.
5w − w2 = −11
Answer:
Given,
w = \(\frac { -5 ± √(5² – 4(-1)(11)) }{ 2(-1) } \)
= \(\frac { -5 ± √(25 + 44) }{ -2 } \)
= \(\frac { -5 ± √69 }{ -2 } \)
The solutions of the equations are \(\frac { -5 + √69 }{ -2 } \), \(\frac { -5 – √69 }{ -2 } \)
The solutions of the equations are \(\frac { -5 + √69 }{ -2 } \), \(\frac { -5 – √69 }{ -2 } \)

Lesson 11.3 Collecting Data

Essential Question What are some considerations when undertaking a statistical study?
Answer:
The objective of the statistical study is to collect the data and then utilize the data to make a decision. Any decision you make utilizing the results of a statistical study is essential as the process used to get the data. If the process is flawed, then the resulting decision is problematic.

EXPLORATION 1

Analyzing Sampling Techniques
Work with a partner. Determine whether each sample is representative of the population. Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 1
a. To determine the number of hours people exercise during a week, researchers use random-digit dialing and call 1500 people.
b. To determine how many text messages high school students send in a week, researchers post a survey on a website and receive 750 responses.
c. To determine how much money college students spend on clothes each semester,a researcher surveys 450 college students as they leave the university library.
d. To determine the quality of service customers receive, an airline sends an e-mail survey to each customer after the completion of a flight.

EXPLORATION 2

Analyzing Survey Questions
Work with a partner. Determine whether each survey question is biased. Explain your reasoning. If so, suggest an unbiased rewording of the question.
a. Does eating nutritious, whole-grain foods improve your health?
b. Do you ever attempt the dangerous activity of texting while driving?
c. How many hours do you sleep each night?
d. How can the mayor of your city improve his or her public image?

EXPLORATION 3

Analyzing Survey Randomness and Truthfulness
Work with a partner. Discuss each potential problem in obtaining a random survey of a population. Include suggestions for overcoming the problem.
a. The people selected might not be a random sample of the population.
b. The people selected might not be willing to participate in the survey.
c. The people selected might not be truthful when answering the question.
d. The people selected might not understand the survey question.

Communicate Your Answer

Question 4.
What are some considerations when undertaking a statistical study?
Answer:
The 4 considerations when undertaking a statistical study are as follows
1. what is the research question?
2. What is the study design?
3. What is the scale of measurement?
4. Any data issues?

Question 5.
Find a real-life example of a biased survey question. Then suggest an unbiased rewording of the question.
Answer:

Monitoring Progress

Question 1.
WHAT IF?
In Example 1, you divide the students in your school according to their zip codes, then select all of the students that live in one zip code. What type of sample are you using?
Answer:
Divide the students in our school according to their zip codes, then select all of the students that live in one zip code. Thus, we use a cluster-type sample.

Question 2.
Describe another method you can use to obtain a stratified sample in Example 1.
Answer:
Divide subjects into subgroups called strata based on the characteristics they share. For example gender, and educational accomplishment. Once divided, then each subgroup is randomly sampled using another probability sampling method.

Question 3.
The manager of a concert hall wants to know how often people in the community attend concerts. The manager asks 45 people standing in line for a rock concert how many concerts they attend per year. Identify the type of sample the manager is using and explain why the sample is biased.
Answer:
Given,
The manager of a concert hall wants to know how often people in the community attend concerts.
The manager asks 45 people which stand in line for a rock concert it is a convenience sample because only members of a population who are easy to reach are asked.

Question 4.
In Example 3, what is another method you can use to generate a random sample of 50 students? Explain why your sampling method is random.
Answer:
You have 246 seniors and you have to randomly select 50 students.
We can write the names of all the students on paper and put them in a hat.
Take 50 papers out of the hat and form a pattern.
It’s a random sample because we don’t know which piece of paper we are going to pull out of the hat.

Identify the method of data collection the situation describes.
Question 5.
Members of a student council at your school ask every eighth student who enters the cafeteria whether they like the snacks in the school’s vending machines.
Answer:
Given that,
Members of a student council at your school ask every eighth student who enters the cafeteria whether they like the snacks in the school’s vending machines.
After every eighth student who enters the cafeteria is asked for an opinion, we have a survey.

Question 6.
A park ranger measures and records the heights of trees in a park as they grow.
Answer:
Given that,
A park ranger measures and records the heights of trees in a park as they grow.
We have a park ranger that measures and record the heights of trees in a park as they grow, and we have an observational study.

Question 7.
A researcher uses a computer program to help determine how fast an influenza virus might spread within a city.
Answer:
Given that,
A researcher uses a computer program to help determine how fast an influenza virus might spread within a city.
We have that researcher collecting data by using the computer, we have a simulation.

Question 8.
Explain why the survey question below may be biased or otherwise introduce bias into the survey. Then describe a way to correct the flaw.
“Do you agree that our school cafeteria should switch to a healthier menu?”
Answer:
The question is biased because it encourages the answer yes.

Collecting Data 11.3 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
Describe the difference between a stratified sample and a cluster sample.
Answer:
It is a stratified sample, after the groups are formed, a random sample is selected from each group are formed, and all the members of one or more groups are randomly selected.

Question 2.
COMPLETE THE SENTENCE
A sample for which each member of a population has an equal chance of being selected is a(n) __________ sample.
Answer:
A sample for which each member of a population has an equal chance of being selected is random sample.

Question 3.
WRITING
Describe a situation in which you would use a simulation to collect data.
Answer: You would use a simulation to collect data to determine how quickly an oil spill would spread through a lake.

Question 4.
WRITING
Describe the difference between an unbiased sample and a biased sample. Give one example of each.
Answer: In a biased sample, one or more parts of the population are favored over others, whereas in an unbiased sample, each member of the population has an equal chance of being selected. An unbiased statistic sample would be expected to have a difference of 0 over time.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–8, identify the type of sample described.
Question 5.
The owners of a chain of 260 retail stores want to assess employee job satisfaction. Employees from 12 stores near the headquarters are surveyed.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 5

Question 6.
Each employee in a company writes their name on a card and places it in a hat. The employees whose names are on the first two cards drawn each win a gift card.
Answer:
Given that,
Each employee in a company writes their name on a card and places it in a hat. The employees whose names are on the first two cards drawn each win a gift card.
Since each of the employees in a company has the same chance to be drawn the samples is a random.

Question 7.
A taxicab company wants to know whether its customers are satisfied with the service. Drivers survey every tenth customer during the day.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 7

Question 8.
The owner of a community pool wants to ask patrons whether they think the water should be colder. Patrons are divided into four age groups, and a sample is randomly surveyed from each age group.
Answer:
Given that,
The owner of a community pool wants to ask patrons whether they think the water should be colder. Patrons are divided into four age groups, and a sample is randomly surveyed from each age group.
You are dividing the patrons into four age groups. So the sample is a stratified sample.

In Exercises 9–12, identify the type of sample and explain why the sample is biased.
Question 9.
A town council wants to know whether residents support having an off-leash area for dogs in the town park. Eighty dog owners are surveyed at the park.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 2
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 9

Question 10.
A sportswriter wants to determine whether baseball coaches think wooden bats should be mandatory in collegiate baseball. The sportswriter mails surveys to all collegiate coaches and uses the surveys that are returned.
Answer:
Given data, A sportswriter wants to determine whether baseball coaches think wooden bats should be mandatory in collegiate baseball. The sportswriter mails surveys to all collegiate coaches and uses the surveys that are returned.
The participants in the survey are collegiate coaches and using the surveys that are returned, it is a self-selected sample.

Question 11.
You want to find out whether booth holders at a convention were pleased with their booth locations. You divide the convention center into six sections and survey every booth holder in the fifth section.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 11

Question 12.
Every tenth employee who arrives at a company health fair answers a survey that asks for opinions about new health-related programs.
Answer:
Given that,
Every tenth employee who arrives at a company health fair answers a survey that asks for opinions about new health-related programs.
Since every tenth employee who arrives at a company health fair answers a survey, it is a systematic sample.

Question 13.
ERROR ANALYSIS
Surveys are mailed to every other household in a neighborhood. Each survey that is returned is used. Describe and correct the error in identifying the type of sample that is used.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 3
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 13

Question 14.
ERROR ANALYSIS
A researcher wants to know whether the U.S. workforce supports raising the minimum wage. Fifty high school students chosen at random are surveyed. Describe and correct the error in determining whether the sample is biased.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 4

Answer:
Given data A researcher wants to know whether the U.S. workforce supports raising the minimum wage. Fifty high school students chosen at random are surveyed.
Since the sample is not chosen from the entire population it cannot be a non-biased sample. Thus, a sample is biased.

In Exercises 15–18, determine whether the sample is biased. Explain your reasoning.
Question 15.
Every third person who enters an athletic event is asked whether he or she supports the use of instant replay in officiating the event.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 15

Question 16.
A governor wants to know whether voters in the state support building a highway that will pass through a state forest. Business owners in a town near the proposed highway are randomly surveyed.
Answer:
Participants in the survey are only random business owners in a town near the proposed highway and they cannot represent the whole population of the town.
It is a biased sample.

Question 17.
To assess customers’ experiences making purchases online, a rating company e-mails purchasers and asks that they click on a link and complete a survey.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 17

Question 18.
Your school principal randomly selects five students from each grade to complete a survey about classroom participation.
Answer:
Given data, Your school principal randomly selects five students from each grade to complete a survey about classroom participation.
Since we chose 5 students from each grade to complete a survey about classroom participation, the sample represents the whole population.
Thus, the sample is not biased.

Question 19.
WRITING
The staff of a student newsletter wants to conduct a survey of the students’ favorite television shows. There are 1225 students in the school. Describe a method for selecting a random sample of 250 students to survey.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 19

Question 20.
WRITING
A national collegiate athletic association wants to survey 15 of the 120 head football coaches in a division about a proposed rules change. Describe a method for selecting a random sample of coaches to survey.
Answer:
Given,
A national collegiate athletic association wants to survey 15 of the 120 head football coaches in a division about a proposed rules change.
We can divide 120 coaches into three groups with 40 coaches, and from each group choose 5 coaches.
For example, we can denote each coach from the group by some integer number from 1 to 40, and by using the random integer generator chooses 5 coaches.
In the described way, from all 3 groups, we can choose 15 coaches.

In Exercises 21–24, identify the method of data collection the situation describes.
Question 21.
A researcher uses technology to estimate the damage that will be done if a volcano erupts.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 6
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 21

Question 22.
The owner of a restaurant asks 20 customers whether they are satisfied with the quality of their meals.
Answer:
The owner of a restaurant asks 20 customers whether they are satisfied with the quality of their meals which is an investigation of one or more characteristics of a population. So, this situation is a survey.

Question 23.
A researcher compares incomes of people who live in rural areas with those who live in large urban areas.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 23

Question 24.
A researcher places bacteria samples in two different climates. The researcher then measures the bacteria growth in each sample after 3 days.
Answer:
Given data, A researcher places bacteria samples in two different climates. The researcher then measures the bacteria growth in each sample after 3 days.
We have a treatment on individuals in order to collect data on their responses to the treatment. The situation describes an experiment.

In Exercises 25–28, explain why the survey question may be biased or otherwise introduce bias into the survey. Then describe a way to correct the flaw.
Question 25.
“Do you agree that the budget of our city should be cut?”
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 25

Question 26.
“Would you rather watch the latest award-winning movie or just read some book?”
Answer:
“Would you rather watch the latest award-winning movie or just read some book?”
The question favors the answer that you rather watch the movie.

Question 27.
“The tap water coming from our western water supply contains twice the level of arsenic of water from our eastern supply. Do you think the government should address this health problem?”
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 27

Question 28.
A child asks, “Do you support the construction of a new children’s hospital?”
Answer:
A child asks, “Do you support the construction of a new children’s hospital?” The question favors the answer yes.

In Exercises 29–32, determine whether the survey question may be biased or otherwise introduce bias into the survey. Explain your reasoning.
Question 29.
“Do you favor government funding to help prevent acid rain?”
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 29

Question 30.
“Do you think that renovating the old town hall would be a mistake?”
Answer:
“Do you think that renovating the old town hall would be a mistake?”
The question is biased because it encourages the answer yes.

Question 31.
A police officer asks mall visitors, “Do you wear your seat belt regularly?”
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 31

Question 32.
“Do you agree with the amendments to the Clean Air Act?”
Answer:
The question “Do you agree with the amendments to the Clean Air Act?” is not biased because it is asked without much information in it so that the right opinion of the survey participants can be obtained.

Question 33.
REASONING
A researcher studies the effect of fiber supplements on heart disease. The researcher identified 175 people who take fiber supplements and 175 people who do not take fiber supplements. The study found that those who took the supplements had 19.6% fewer heart attacks. The researcher concludes that taking fiber supplements reduces the chance of heart attacks.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 7
a. Explain why the researcher’s conclusion may not be valid.
b. Describe how the researcher could have conducted the study differently to produce valid results.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 33

Question 34.
HOW DO YOU SEE IT?
A poll is conducted to predict the results of a statewide election in New Mexico before all the votes are counted. Fifty voters in each of the state’s 33 counties are asked how they voted as they leave the polls.
a. Identify the type of sample described.
b. Explain how the diagram shows that the polling method could result in a biased sample.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 8

Answer:
a. The situation describes the Stratified sample.
b. With the date of the damp, we can notice that different countries have different populations. The sample consists of an equal number of respondents from each country and the number of respondents from the country is not proportional to the size of the country. Therefore, a small sample from a country with a large population cannot be considered credible.

Question 35.
WRITING
Consider each type of sample listed on page 610. Which of the samples are most likely to lead to biased results? Explain.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 35

Question 36.
THOUGHT PROVOKING
What is the difference between a “blind experiment” and a “double-blind experiment?” Describe a possible advantage of the second type of experiment over the first.
Answer:
In a blind or blind experiment, information that may influence the participants of the experiment is withheld until after the experiment is complete.
A double-blind experiment is one in which neither the participant nor the experimenters know who is receiving a particular treatment. this procedure is utilized to prevent bias in research results.
Double-blind experiments are particularly useful for preventing bias due to demand characteristics or the placebo effect which represent a possible advantage of double-blind experiments.

Question 37.
WRITING
A college wants to survey its graduating seniors to find out how many have already found jobs in their field of study after graduation.
a. What is the objective of the survey?
b. Describe the population for the survey.
c. Write two unbiased questions for the survey.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 37

Question 38.
REASONING
About 3.2% of U.S. adults follow a vegetarian-based diet. Two randomly selected groups of people were asked whether they follow such a diet. The first sample consists of 20 people and the second sample consists of 200 people. Which sample proportion is more likely to be representative of the national percentage? Explain.
Answer:
Given data is
About 3.2% of U.S. adults follow a vegetarian-based diet. Two randomly selected groups of people were asked whether they follow such a diet. The first sample consists of 20 people and the second sample consists of 200 people.
The larger sample (second sample) will better represent the whole population because some unusual responses will not affect the results much as with the smaller sample (first sample).

Question 39.
MAKING AN ARGUMENT
The U.S. Census is taken every10 years to gather data from the population. Your friend claims that the sample cannot be biased. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 39

Question 40.
OPEN-ENDED
An airline wants to know whether travelers have enough leg room on its planes.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 9
a. What method of data collection is appropriate for this situation?
b. Describe a sampling method that is likely to give biased results. Explain.
c. Describe a sampling method that is not likely to give biased results. Explain.
d. Write one biased question and one unbiased question for this situation.
Answer:
a. An airline wants to know whether travelers have enough legroom on its planes. The situation describes the survey method.
b. We can conduct the survey by mail. Only people with strong opinions about the legroom on its planers will take part in that survey and return the survey by mail.
c. For example, in order to obtain a non-biased survey, an airline may grant a certain discount on tickets if it participates in the survey in all class seats. That way more people will take part in the survey and we will get an unbiased sample.
d. Biased question: “Do you agree that the legroom is too small?”
Non-biased question: “Do you have enough legroom?”

Question 41.
REASONING
A website contains a link to a survey that asks how much time each person spends on the Internet each week.
a. What type of sampling method is used in this situation?
b. Which population is likely to respond to the survey? What can you conclude?
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 41

Maintaining Mathematical Proficiency

Evaluate the expression without using a calculator.
Question 42.
45/2

Answer:
Given,
45/2 = (22)\(\frac { 5 }{ 2 } \)
= 2 2 . \(\frac { 5 }{ 2 } \)
= 25
= 32
So, 45/2 = 32

Question 43.
272/3
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 43

Question 44.
−641/3

Answer:
Given,
−641/3 = -((4)3)1/3
= -43 . 1/3
= -4
So, −641/3 is -4.

Question 45.
8-2/3
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 45

Simplify the expression.
Question 46.
(43/2 • 41/4)4

Answer:
(43/2 • 41/4)4 = (43/2 + 1/4)4
=(47/4>/sup>)4
=47/4 . 4
= 47
= 16384

Question 47.
(61/3 • 31/3)-2
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 47

Question 48.
\(\sqrt [ 3 ]{ 4 }\) • \(\sqrt [ 3 ]{ 16 }\)
Answer:
\(\sqrt [ 3 ]{ 4 }\) • \(\sqrt [ 3 ]{ 16 }\) = \(\sqrt [ 3 ]{ 4 }\) • \(\sqrt [ 3 ]{ 4² }\)
= 41/3 + 42/3
= 41/3 + 2/3
= 4
So, \(\sqrt [ 3 ]{ 4 }\) • \(\sqrt [ 3 ]{ 16 }\) = 4

Question 49.
\(\frac{\sqrt[4]{405}}{\sqrt[4]{5}}\)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 a 49

Data Analysis and Statistics Study Skills: Reworking Your Notes

11.1–11.3 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 10

Core Concepts
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 11

Mathematical Practices

Question 1.
What previously established results, if any, did you use to solve Exercise 31 on page 602?
Answer:

Question 2.
What external resources, if any, did you use to answer Exercise 36 on page 616?
Answer:

Study Skills: Reworking Your Notes

It’s almost impossible to write down in your notes all the detailed information you are taught in class. A good way to reinforce the concepts and put them into your long-term memory is to rework your notes. When you take notes, leave extra space on the pages. You can go back after class and fill in:

  • important definitions and rules
  • additional examples
  • questions you have about the material
    Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics 11.3 12

Data Analysis and Statistics 11.1–11.3 Quiz

A normal distribution has a mean of 32 and a standard deviation of 4. Find the probability that a randomly selected x-value from the distribution is in the given interval.

Question 1.
at least 28
Answer:
Given,
A normal distribution has a mean of 32 and a standard deviation of 4
Mean μ = 32
σ = 4
We have to determine P(x > 28)
28 is one standard deviation to the left of the mean
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 1
P(x > 28) = P(x > μ – σ)
= 34% + 34% + 13.5% + 2.35% + 0.015%
= 84%
So, P(x > 28) = 84%

Question 2.
between 20 and 32
Answer:
A normal distribution has a mean of 32 and a standard deviation of 4
Mean μ = 32
σ = 4
μ – 20 = 32 – 20 = 12 = 3 . 4 = 3σ
20 = μ – 3σ
32 – μ = 32 – 32 = 0
32 = μ
P(20≤x≤32) = P(μ – 3σ≤x≤μ)
= P(x≤μ) – P(x≤μ – 3σ)
= 0.5 – 0.0015
= 0.4985
The probability is 0.4985

Question 3.
at most 26
Answer:
A normal distribution has a mean of 32 and a standard deviation of 4
Mean μ = 32
σ = 4
Now compute P(x≤26)
z = \(\frac { x – μ }{ σ } \)
P(x≤26) = P(\(\frac { x – 32 }{ 4 } \)≤ \(\frac { 26 – 32 }{ 4 } \))
= P(z≤-1.5)
= 0.0668
The probability is 0.0668. (6.68%)

Question 4.
at most 35
Answer:
A normal distribution has a mean of 32 and a standard deviation of 4
Mean μ = 32
σ = 4
Now compute P(x≤35)
z = \(\frac { x – μ }{ σ } \)
P(x≤35) = P(\(\frac { x – 32 }{ 4 } \)≤ \(\frac { 35 – 32 }{ 4 } \))
= P(z≤0.8)
= 0.7881
The probability is 0.7881 (78.8%)

Determine whether the histogram has a normal distribution.
Question 5.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics q 1

Answer:
The variable that is normally distributed has a histogram that is bell-shaped with only one peak and symmetric around the mean.
The above graph has one peak so we can say that the histogram has a normal distribution.

Question 6.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics q 2

Answer:
The variable that is normally distributed has a histogram that is bell-shaped with only one peak and symmetric around the mean.
The above graph does not have a normal distribution.

Question 7.
A survey of 1654 high school seniors determined that 1125 plan to attend college. Identify the population and the sample. Describe the sample.
Answer:
Given,
A survey of 1654 high school seniors determined that 1125 plan to attend college. Identify the population and the sample.
Sample: The 1654 high school seniors surveyed. The sample consists of 1125 high school seniors who plan to attend college and 529 high school seniors who do not.
Population: Al high school students

Question 8.
A survey of all employees at a company found that the mean one-way daily commute to work of the employees is 25.5 minutes. Is the mean time a parameter or a statistic? Explain your reasoning.
Answer:
Given,
A survey of all employees at a company found that the mean one-way daily commute to work of the employees is 25.5 minutes.
The mean time is a parameter.
The entire population was surveyed.

Question 9.
A researcher records the number of bacteria present in several samples in a laboratory. Identify the method of data collection.
Answer:
Given,
A researcher records the number of bacteria present in several samples in a laboratory. The records of samples in the lab will be observational.
The method of data collection is an Observational study.

Question 10.
You spin a five-color spinner, which is divided into equal parts, five times and every time the spinner lands on red. You suspect the spinner favors red. The maker of the spinner claims that the spinner does not favor any color. You simulate spinning the spinner 50 times by repeatedly drawing 200 random samples of size 50. The histogram shows the results. Use the histogram to determine what you should conclude when you spin the actual spinner 50 times and the spinner lands on red (a) 9 times and (b) 19 times.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics q 3

Answer:
a) The marker’s claim is most likely true
b) The marker’s claim is most likely false

Question 11.
A local television station wants to find the number of hours per week people in the viewing area watch sporting events on television. The station surveys people at a nearby sports stadium.
a. Identify the type of sample described.
Answer: The type of sample is Convenience sample

b. Is the sample biased? Explain your reasoning.
Answer: Yes, people at a sports stadium are more likely to watch more sports on television than people who never go to a sports stadium.

c. Describe a method for selecting a random sample of 200 people to survey.
Answer:
Randomly pick 200 members from the seating area by placing all seat numbers in a hat.

Lesson 11.4 Experimental Design

Essential Question How can you use an experiment to test a conjecture?

EXPLORATION 1

Using an Experiment
Work with a partner. Standard white playing dice are manufactured with black dots that are indentations, as shown. So, the side with six indentations is the lightest side and the side with one indentation is the heaviest side.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 1
You make a conjecture that when you roll a standard playing die, the number 6 will come up more often than the number 1 because 6 is the lightest side. To test your conjecture, roll a standard playing die 25 times. Record the results in the table. Does the experiment confirm your conjecture? Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 2

EXPLORATION 2

Analyzing an Experiment
Work with a partner. To overcome the imbalance of standard playing dice, one of the authors of this book invented and patented 12-sided dice, on which each number from 1 through 6 appears twice (on opposing sides). See BigIdeasMath.com.
As part of the patent process, a standard playing die was rolled 27,090 times. The results are shown below.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 3
What can you conclude from the results of this experiment? Explain your reasoning.

Communicate Your Answer

Question 3.
How can you use an experiment to test a conjecture?
Answer:

Question 4.
Exploration 2 shows the results of rolling a standard playing die 27,090 times to test the conjecture in Exploration 1. Why do you think the number of trials was so large?
Answer:

Question 5.
Make a conjecture about the outcomes of rolling the 12-sided die in Exploration 2. Then use the Internet to find a 12-sided die rolling simulator. Use the simulator to complete a table similar to the one shown below. How many times did you simulate rolling the die? Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 4
Answer:

Monitoring Progress
Question 1.
Determine whether the study is a randomized comparative experiment. If it is, describe the treatment, the treatment group, and the control group. If it is not, explain why not and discuss whether the conclusions drawn from the study are valid.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 5

Answer:
The study is a randomized comparative experiment because we have a treatment that is accident reduction, the treatment group is people who woke up before 6:30, and the control group is people who never woke up before 6:30.

Question 2.
Determine whether the following research topic is best investigated through an experiment or an observational study. Then describe the design of the experiment or observational study.
You want to know whether flowers sprayed twice per day with a mist of water stay fresh longer than flowers that are not sprayed.
Answer:
It is better to use an experimental study. The treatment will be, spraying twice per day with a mist of water, the control group will be flowers that don’t spray, and the treatment group will be flowers that will be sprayed.

Question 3.
In Example 3, the company identifies 250 people who are overweight. The subjects are randomly assigned to a treatment group or a control group. In addition, each subject is given a DVD that documents the dangers of obesity. After 3 months, most of the subjects placed in the treatment group have lost weight. Identify a potential problem with the experimental design. Then describe how you can improve it.
Answer:
The company identifies 250 people who are overweight
A potential problem is that each subject is given a DVD documenting the dangers of obesity because in that case, we have no difference between the control and treatment groups.
To improve the experiment, A DVD documenting the dangers of obesity should only be given to obese people in the treatment group.

Question 4.
You design an experiment to test the effectiveness of a vaccine against a strain of influenza. In the experiment, 100,000 people receive the vaccine and another 100,000 people receive a placebo. Identify a potential problem with the experimental design. Then describe how you can improve it.
Answer:
A potential problem is in choosing people who have been vaccinated. For example, some people may have a stronger immune system and will not need the vaccine or have a weak immune system and get sick more easily.
You can develop the experiment by randomly selecting 200,000 people with similar immune systems and forming two groups of them one of which will be vaccinated and the other will receive a placebo.

Experimental Design 11.4 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
Repetition of an experiment under the same or similar conditions is called _________.
Answer:
Repetition of an experiment under the same or similar conditions is called replication.

Question 2.
WRITING
Describe the difference between the control group and the treatment group in a controlled experiment.
Answer:
The treatment group is also called the experimental group. It receives treatment whose effect the researchers want to study, whereas the control group does not receive any treatment. You should include in the experiment a control group that is identical to the group that treats in every way, except that it does not receive experimental treatment.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 and 4, determine whether the study is a randomized comparative experiment. If it is, describe the treatment, the treatment group, and the control group. If it is not, explain why not and discuss whether the conclusions drawn from the study are valid.
Question 3.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 3

Question 4.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 7

Answer:
The study mentioned above is not a randomized comparative experiment since the students were not randomly assigned to the control group and treatment group.
The conclusion may or may not be valid because there may be other reasons why the can students have fewer cavities, such as better oral health practices.

ERROR ANALYSIS In Exercises 5 and 6, describe and correct the error in describing the study.

A company’s researchers want to study the effects of adding shea butter to their existing hair conditioner. They monitor the hair quality of 30 randomly selected customers using the regular conditioner and 30 randomly selected customers using the new shea butter conditioner.
Question 5.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 8
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 5

Question 6.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 9

Answer:
The study cannot be an observational study because the variable is the control group because it does not receive any treatment. Thus, it is a randomized comparative experiment.

In Exercises 7–10, explain whether the research topic is best investigated through an experiment or an observational study. Then describe the design of the experiment or observational study.

Question 7.
A researcher wants to compare the body mass index of smokers and nonsmokers.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 7

Question 8.
A restaurant chef wants to know which pasta sauce recipe is preferred by more diners.
Answer:
A restaurant chef wants to know which pasta sauce recipe is preferred by more diners.
The treatment of different pasta sauce recipes is possible for the diner. So it is ethical to assign each dinner at the restaurant to a control or treatment group.
I will use an experimental study randomly choose one group of diners given one type of pasta sauce recipe. Randomly choose one group of diners given another type of pasta sauce recipe. Monitor each diner in both groups at dining.

Question 9.
A farmer wants to know whether a new fertilizer affects the weight of the fruit produced by strawberry plants.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 9

Question 10.
You want to know whether homes that are close to parks or schools have higher property values.
Answer:
Researchers continue to see that homes located near a park have an 8% to 20% increase in value compared to other properties. Thus, we cannot randomly form two groups and because of that, we will use an observational study. First, we can randomly choose homes close to the park and school and form one group and the other group will be randomly selected homes that are not close to the park and school. Find the property values for each home in both groups.

Question 11.
DRAWING CONCLUSIONS
A company wants to test whether a nutritional supplement has an adverse effect on an athlete’s heart rate while exercising. Identify a potential problem, if any, with each experimental design. Then describe how you can improve it.
a. The company randomly selects 250 athletes. Half of the athletes receive the supplement and their heart rates are monitored while they run on a treadmill. The other half of the athletes are given a placebo and their heart rates are monitored while they lift weights. The heart rates of the athletes who took the supplement significantly increased while exercising.

b. The company selects 1000 athletes. The athletes are divided into two groups based on age. Within each age group, the athletes are randomly assigned to receive the supplement or the placebo. The athletes’ heart rates are monitored while they run on a treadmill. There was no significant difference in the increases in heart rates between the two groups.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 11

Question 12.
DRAWING CONCLUSIONS
A researcher wants to test the effectiveness of reading novels on raising intelligence quotient (IQ) scores. Identify a potential problem, if any, with each experimental design. Then describe how you can improve it.
a. The researcher selects 500 adults and randomly divides them into two groups. One group reads novels daily and one group does not read novels. At the end of 1 year, each adult is evaluated and it is determined that neither group had an increase in IQ scores.
Answer:
A researcher selects 500 adults and randomly divides them into two groups. One group reads novels and the other doesn’t, we do not have a potential problem.

b. Fifty adults volunteer to spend time reading novels every day for 1 year. Fifty other adults volunteer to refrain from reading novels for 1 year. Each adult is evaluated and it is determined that the adults who read novels raised their IQ scores by 3 points more than the other group.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 10

Answer:
From the given, we can note that the groups are not similar. One group is formed by the people who volunteer to read a novel and the other one doesn’t. Also, there are other habits that increase IQ scores which may have one of the volunteers. Thus the groups must be formed randomly because in that way groups will be similar and then compare the IQ scores between the groups.

Question 13.
DRAWING CONCLUSIONS
A fitness company claims that its workout program will increase vertical jump heights in 6 weeks. To test the workout program, 10 athletes are divided into two groups. The double bar graph shows the results of the experiment. Identify the potential problems with the experimental design. Then describe how you can improve it.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 11
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 13

Question 14.
WRITING
Explain why observational studies, rather than experiments, are usually used in astronomy.
Answer:
Astronomy is referred to as observational science and not primarily experimental science because it focuses on the acquisition of data observations of astronomical objects. The universe is still unexplored and cannot be controlled by objects or generally, in most cases, we cannot create controlled conditions, so they rather use an observational study than experiments.

Question 15.
MAKING AN ARGUMENT
Your friend wants to determine whether the number of siblings has an effect on a student’s grades. Your friend claims to be able to show causality between the number of siblings and grades. Is your friend correct? Explain.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 15

Question 16.
HOW DO YOU SEE IT?
To test the effect political advertisements have on voter preferences, a researcher selects 400 potential voters and randomly divides them into two groups. The circle graphs show the results of the study.
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 12
a. Is the study a randomized comparative experiment? Explain.
b. Describe the treatment.
c. Can you conclude that the political advertisements were effective? Explain.
Answer:
a. Since the voters are randomly divided into two groups, none of the groups is formed based on bias and we have a control and treatment group, it is a randomized comparative treatment.
b. The treatment was the ads for Candidate B.
c. Since there are significantly fewer differences between the graphs, we can conclude that there is no important change for candidate B.

Question 17.
WRITING
Describe the placebo effect and how it affects the results of an experiment. Explain how a researcher can minimize the placebo effect.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 17

Question 18.
THOUGHT PROVOKING
Make a hypothesis about something that interests you. Design an experiment that could show that your hypothesis is probably true.
Answer:
A hypothesis is essentially an idea about the population that you think might be true, but cannot prove to be true. Corn treated with the supplement is higher than corn that is not. By random selection, we form two groups with small corn plants. One group is treated with a supplement and the other is not. Keppal the other variables constant and compare the results after measure.

Question 19.
REASONING
Will replicating an experiment on many individuals produce data that are more likely to accurately represent a population than performing the experiment only once? Explain.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 19

Maintaining Mathematical Proficiency

Draw a dot plot that represents the data. Identify the shape of the distribution.
Question 20.
Ages: 24, 21, 22, 26, 22, 23, 25, 23, 23, 24, 20, 25
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.4 1
The shape of a distribution is symmetric.

Question 21.
Golf strokes: 4, 3, 4, 3, 3, 2, 7, 5, 3, 4
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 21

Tell whether the function represents exponential growth or exponential decay. Then graph the function.
Question 22.
y = 4x

Answer:

x -2 -1 0 1 2 3
y 1/16 1/4 1 4 16 64

Identify the value of the base. The base 4 is greater than 1 so the function represents exponential growth.
Plot the points on the graph from the table. Draw from left to right, a smooth curve that begins just above the x-axis, passes through the plotted points, and moves up to the right.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.4 2

Question 23.
y = (0.95)x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 23

Question 24.
y = (0.2)x

Answer:
Identify the value of the base. The base 0.2 is less than 1 so the function represents exponential decay.

x -2 -1 0 1 2 3
y 25 5 1 0.2 0.04 0.008

Draw the points from right to left, a smooth curve that begins just above the x-axis, passes through the plotted points and moves up to the left.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.4 3

Question 25.
y = (1.25)x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 11 Data Analysis and Statistics 11.4 a 25

Lesson 11.5 Making Inferences from Sample Surveys

Essential Question How can you use a sample survey to infer a conclusion about a population?

EXPLORATION 1

Making an Inference from a Sample
Work with a partner. You conduct a study to determine what percent of the high school students in your city would prefer an upgraded model of their current cell phone. Based on your intuition and talking with a few acquaintances, you think that 50% of high school students would prefer an upgrade. You survey 50 randomly chosen high school students and find that 20 of them prefer an upgraded model.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 1
a. Based on your sample survey, what percent of the high school students in your city would prefer an upgraded model? Explain your reasoning.
b. In spite of your sample survey, is it still possible that 50% of the high school students in your city prefer an upgraded model? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 2
c. To investigate the likelihood that you could have selected a sample of 50 from a population in which 50% of the population does prefer an upgraded model, you create a binomial distribution as shown below. From the distribution, estimate the probability that exactly 20 students surveyed prefer an upgraded model. Is this event likely to occur? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 3
d. When making inferences from sample surveys, the sample must be random. In the situation described above, describe how you could design and conduct a survey using a random sample of 50 high school students who live in a large city.

Communicate Your Answer

Question 2.
How can you use a sample survey to infer a conclusion about a population?
Answer:
A population is a group of people or things. A sample is part of the population.
In order to make conclusions about the population, You can use the sample.
A complete set of possible cases.
A subset of the population from which data is collected.
Inference statistics is a process of using sample statistics to draw conclusions about population parameters.
In other words, the data from the sample is used to make inferences about the population.

Question 3.
In Exploration 1(c), what is the probability that exactly 25 students you survey prefer an upgraded model?
Answer:
There are 50,000 high school students
Sample = 50 high school students
The probability that exactly 25 students you survey prefer an upgraded model = 25/50 = 1/2 = 0.5
Thus the probability that exactly 25 students you survey prefer an upgraded model is 0.5

Monitoring Progress

Question 1.
The data from another random sample of 30 teen users of the social networking website are shown in the table. Estimate the population mean μ.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 4

Answer:
The data from another random sample of 30 teen users of the social networking website are shown in the table.
To estimate the mean, we will sum all values from the table and then divide it with the count of members that participate in the sum.
μ = \(\frac { 1 }{ 30 } \)(305 + 237 + . . . + 287) = \(\frac { 1 }{ 30 } \) . 8820 = 294
So, the mean, μ = 294

Question 2.
Two candidates are running for class president. The table shows the results of four surveys of random students in the class. The students were asked whether they will vote for the incumbent. Do you think the incumbent will be reelected? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 5

Answer:
As a sample size increases, we can notice that percentage of the percent of voters for incumbents decreases.
Thus, since for the sample size of 40, we see that support is less than 50%.
Thus the circumbent will not be rejected.

Question 3.
WHAT IF?
In Example 3, what can you conclude about the accuracy of the claim that the population proportion is 0.34 when 21 adults in your random sample say mathematics is the most valuable subject?
Answer:
21 out of 50 corresponds to a sample proportion of \(\frac { 21 }{ 50 } \) = 0.42.
In the simulation, this result occured in 5 of the 80 random samples.
It is unlikely that 21 adults out of 50 would say math is the most valuable subject when the true population percentage is 34%.
The company’s claim is probably not accurate.

Question 4.
In a survey of 1028 people in the U.S., 87% reported using the Internet. Give an interval that is likely to contain the exact percent of all people in the U.S. who use the Internet.
Answer:
Since in the survey participiate n = 1028 people from the U.S, the marigin error will be
\(\frac { 1 }{ √n } \) = \(\frac { 1 }{ √1028 } \) = 0.031
Thus, the interval that is likely to contain the exact percent of all people in the U.S who use the internet will be
(87% – 31%, 87% – 31%) = (83.9%, 90.1%)
Thus, the interval between 83.9% to 90.1%

Making Inferences from Sample Surveys 11.5 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
The ___________ gives a limit on how much the responses of the sample would differ from the responses of the population.
Answer: The margin of error gives a limit on how much the responses of the sample would differ from the responses of the population.

Question 2.
WRITING
What is the difference between descriptive and inferential statistics?
Answer:
Descriptive statistics uses the data to provide descriptions of the population, either through numerical calculations or graphs, or tables. Descriptive statistics summarize the characteristics of a data set. Inferential statistics allow you to test a hypothesis or assess whether your data is generalizable to the broader population.

Monitoring Progress and Modeling with Mathematics

Question 3.
PROBLEM SOLVING
The numbers of text messages sent each day by a random sample of 30 teen cellphone users are shown in the table. Estimate the population mean μ.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 3

Question 4.
PROBLEM SOLVING
The incomes for a random sample of 35 U.S. households are shown in the table. Estimate the population mean μ.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 7

Answer:
The incomes for a random sample of 35 U.S. households are shown in the table.
The mean will be estimated as a sum of elements from the table divided with the 5. Hence
μ = \(\frac { 1 }{ 35 } \)(14300 + 72800 + . . + 63100) = $51029
So, the mean μ = 51029

Question 5.
PROBLEM SOLVING
Use the data in Exercise 3 to answer each question.
a. Estimate the population proportion ρ of teen cellphone users who send more than 70 text messages each day.
b. Estimate the population proportion ρ of teen cellphone users who send fewer than 50 text messages each day.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 5

Question 6.
WRITING
A survey asks a random sample of U.S. teenagers how many hours of television they watch each night. The survey reveals that the sample mean is 3 hours per night. How condent are you that the average of all U.S. teenagers is exactly 3 hours per night? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 8

Answer:
Not at all confident.
We have that the sample mean is 3 hours per night and we cannot expect that the average of all US teenagers is exactly 3 hours per night, maybe the result can be close to 3 but not exactly 3.

Question 7.
DRAWING CONCLUSIONS
When the President of the United States vetoes a bill, the Congress can override the veto by a two-thirds majority vote in each House. Five news organizations conduct individual random surveys of U.S. Senators. The senators are asked whether they will vote to override the veto. The results are shown in the table.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 9
a. Based on the results of the first two surveys, do you think the Senate will vote to override the veto? Explain.
b. Based on the results in the table, do you think the Senate will vote to override the veto? Explain.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 7

Question 8.
DRAWING CONCLUSIONS
Your teacher lets the students decide whether to have their test on Friday or Monday. The table shows the results from four surveys of randomly selected students in your grade who are taking the same class. The students are asked whether they want to have the test on Friday.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 10
a. Based on the results of the first two surveys, do you think the test will be on Friday? Explain.
Answer:
Based on the results of the first two surveys, the test will be on Friday because there we have more than half the majority.

b. Based on the results in the table, do you think the test will be on Friday? Explain.
Answer:
From the table, as the sample size increases, the percentage of voters for test on Friday decreases, and in the last survey, we have 45% which is not enough.

Question 9.
MODELING WITH MATHEMATICS
A national polling company claims that 54% of U.S. adults are married. You survey a random sample of 50 adults.
a. What can you conclude about the accuracy of the claim that the population proportion is 0.54 when 31 adults in your survey are married?
b. What can you conclude about the accuracy of the claim that the population proportion is 0.54 when 19 adults in your survey are married?
c. Assume that the true population proportion is 0.54. Estimate the variation among sample proportions for samples of size 50.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 9

Question 10.
MODELING WITH MATHEMATICS
Employee engagement is the level of commitment and involvement an employee has toward the company and its values. A national polling company claims that only 29% of U.S. employees feel engaged at work. You survey a random sample of 50 U.S. employees.
a. What can you conclude about the accuracy of the claim that the population proportion is 0.29 when 16 employees feel engaged at work?
Answer: The company’s claim is probably correct

b. What can you conclude about the accuracy of the claim that the population proportion is 0.29 when 23 employees feel engaged at work?
Answer: The company’s claim is probably incorrect
c. Assume that the true population proportion is 0.29. Estimate the variation among sample proportions for samples of size 50.
Answer: 0.15 to 0.4

In Exercises 11–16, find the margin of error for a survey that has the given sample size. Round your answer to the nearest tenth of a percent.
Question 11.
260
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 11

Question 12.
1000
Answer:
Use the margin of error formula
The margin of error = ±\(\frac { 1 }{ n } \) = ±\(\frac { 1 }{ √1000 } \) = ±0.031
The margin of error for the survey is about ±3.1%.

Question 13.
2024
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 13

Question 14.
6400
Answer:
Use the margin of error formula
The margin of error = ±\(\frac { 1 }{ n } \) = ±\(\frac { 1 }{√6400 } \) = ±0.0125
The margin of error for the survey is about ±1.25%.

Question 15.
3275
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 15

Question 16.
750
Answer:
Use the margin of error formula
The margin of error = ±\(\frac { 1 }{ n } \) = ±\(\frac { 1 }{√750} \) = ±0.036
The margin of error for the survey is about ±3.6%

Question 17.
ATTENDING TO PRECISION
In a survey of 1020 U.S. adults, 41% said that their top priority for saving is retirement.
a. What is the margin of error for the survey?
b. Give an interval that is likely to contain the exact percent of all U.S. adults whose top priority for saving is retirement.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 17

Question 18.
ATTENDING TO PRECISION
In a survey of 1022 U.S. adults, 76% said that more emphasis should be placed on producing domestic energy from solar power.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 11
a. What is the margin of error for the survey?
Answer:
The margin error of the sample size n = 1020 will be ±\(\frac { 1 }{ √n } \) = ±\(\frac { 1 }{ √1020 } \) = 0.031
The margin error for the survey is about ±3.1%

b. Give an interval that is likely to contain the exact percent of all U.S. adults who think more emphasis should be placed on producing domestic energy from solar power.
Answer:
To find the interval, subtract and add 3.1% to the percent of people surveyed who said that more emphasis should be placed on producing domestic energy from solar power (76%)
76% – 3.1% = 72.9%
76% + 3.1% = 79.1%
It is likely that the exact percent of all people in the U.S who said that is likely to contain the exact percent of all U.S adults whose top priority for saving is retirement is between 72.9% and 79.1%

Question 19.
ERROR ANALYSIS
In a survey, 8% of adult Internet users said they participate in sports fantasy leagues online. The margin of error is ±4%. Describe and correct the error in calculating the sample size.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 12
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 19

Question 20.
ERROR ANALYSIS
In a random sample of 2500 consumers, 61% prefer Game A over Game B. Describe and correct the error in giving an interval that is likely to contain the exact percent of all consumers who prefer Game A over Game B.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 13

Answer:
In a random sample of 2500 consumers, 61% prefer Game A over Game B.
To find, the interval subtracts and add margin error to the percent of people surveyed who prefer Game A over Game B (61%). Thus since the margin error is 2%, the interval is 61% – 2% = 59%
61% + 2% = 63%
hence, it is likely that the exact percent of all consumers who prefer Game A over Game B is between 59% and 63%.

Question 21.
MAKING AN ARGUMENT
Your friend states that it is possible to have a margin of error between 0 and 100 percent, not including 0 or 100 percent. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 21

Question 22.
HOW DO YOU SEE IT?
The figure shows the distribution of the sample proportions from three simulations using different sample sizes. Which simulation has the least margin of error? the greatest? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 14

Answer:
The margin error is proportional to the standard deviation. Thus, from the graph, we can conclude that simulation a has the least standard deviation which implies a has the least marginal error.
The greatest standard deviation has simulation c which implies that simulation c has the greatest marginal error.
The least margin of error has simulation a. The greatest margin of error has simulation c.

Question 23.
REASONING
A developer claims that the percent of city residents who favor building a new football stadium is likely between 52.3% and 61.7%. How many residents were surveyed?
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 15
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 23

Question 24.
ABSTRACT REASONING
Suppose a random sample of size n is required to produce a margin of error of ±E. Write an expression in terms of n for the sample size needed to reduce the margin of error to ±1—2E. How many times must the sample size be increased to cut the margin of error in half? Explain.
Answer:
E = ±\(\frac { 1 }{ √n } \)
To compute how many times must the sample size be increased to cut the margin of error in half
\(\frac { 1 }{ 2 } \)E = ±\(\frac { 1 }{ √x } \)
\(\frac { 1 }{ 2 } \) . (±\(\frac { 1 }{ √n } \)) = ±\(\frac { 1 }{ √x } \)
(±\(\frac { 1 }{ 2√n } \))² = (±\(\frac { 1 }{ √x } \))²
\(\frac { 1 }{ 4n } \) = \(\frac { 1 }{ x } \)
x = 4n
The sample must be four times greater than the size of the sample n with the margin error E.

Question 25.
PROBLEM SOLVING
A survey reported that 47% of the voters surveyed, or about 235 voters, said they voted for Candidate A and the remainder said they voted for Candidate B.
a. How many voters were surveyed?
b. What is the margin of error for the survey?
c. For each candidate, find an interval that is likely to contain the exact percent of all voters who voted for the candidate.
d. Based on your intervals in part (c), can you be confident that Candidate B won? If not, how many people in the sample would need to vote for Candidate B for you to be condent that Candidate B won? (Hint: Find the least number of voters for Candidate B so that the intervals do not overlap.)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 25

Question 26.
THOUGHT PROVOKING
Consider a large population in which ρ percent (in decimal form) have a certain characteristic. To be reasonably sure that you are choosing a sample that is representative of a population, you should choose a random sample of n people where n > 9\(\left(\frac{1-\rho}{\rho}\right)\) .
a. Suppose ρ= 0.5. How large does n need to be?
Answer:
If the proportion population is p = 0.5, let’s compute the size of the population
n > 9(\(\frac { 1 – p }{ p } \)
For p = 0.5
n > 9(\(\frac { 1 – 0.5 }{ 0.5 } \)
n > 9(0.5/0.5)
n > 9

b. Suppose ρ= 0.01. How large does n need to be?
Answer:
If proportion population p = 0.01
n > 9(\(\frac { 1 – 0.01 }{ 0.01 } \)
n > 9 . (\(\frac { 0.99 }{ 0.01 } \))
n > 9 . 99
n > 891

c. What can you conclude from parts (a) and (b)?
Answer:
From parts a and b, you can conclude that the smaller percentage of people with a certain characteristic, the larger the sample size is necessary for the appropriate representation of the population.

Question 27.
CRITICAL THINKING
In a survey, 52% of the respondents said they prefer sports drink X and 48% said they prefer sports drink Y. How many people would have to be surveyed for you to be confident that sports drink X is truly preferred by more than half the population? Explain.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 27

Maintaining Mathematical Proficiency

Find the inverse of the function.
Question 28.
y= 10x-3

Answer:
Given,
y= 10x-3
To find the inverse function, interchange x and y
x = y
x = 10y-3
logx = log10y-3
logx = y – 3
y = logx + 3

Question 29.
y= 2x – 5
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 29

Question 30.
y= ln (x+ 5)

Answer:
Given,
y= ln (x+ 5)
To find the inverse function, interchange x and y
x = y
x = ln(y + 5)
ex = eln(y+5)
ex = y + 5
y = ex – 5

Question 31.
y= log6x− 1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 31

Determine whether the graph represents an arithmetic sequence or a geometric sequence. Then write a rule for the nth term.
Question 32.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 16
Answer:
From the graph, we have the following points (1, 17), (2, 14), (3, 11), (4, 8)
Now, let’s denote the following participants of the sequence
a1 = 17
a2 = 14
a3 = 11
a4 = 8
Now, to decide whether to do is an arithmetic or geometric sequence, let’s consider the following terms
d = a<n-1> – a<n-2> = a<k-1> – a<k-2>
q = \(\frac { a<n> }{ a<n-1> } \) = \(\frac { a<k> }{ a<k-1> } \)
a2 – a1 = a3 – a2 = a3 – a4 = -3
a<n> = a<1> + (n – 1)d
Hence, by using the formula, the form of the sequence is a<n> = 17 + (n – 1)(-3)
= 17 – 3n + 3
= 20 – 3n

Question 33.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 17
Answer:
Big Ideas Math Answers Algebra 2 Chapter 11 Data Analysis and Statistics 11.5 a 33

Question 34.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.5 18

Answer:
From the graph, we have the following points (1, 32), (2, 16), (3, 8), (4, 4)
Now, let’s denote the following participants of the sequence
a1 = 32
a2 = 16
a3 = 8
a4 = 4
Now, to decide whether to do is an arithmetic or geometric sequence, let’s consider the following terms
d = a<n-1> – a<n-2> = a<k-1> – a<k-2>
r = \(\frac { an }{ a(n – 1) } \)
\(\frac { a2 }{ a1 } \) = \(\frac { 1 }{ 2 } \)
a<n> = a1 . r^(n-1)
Hence, by using the formula, the form of the sequence is a<n> = 24. (\(\frac { 1 }{ 2 } \))^(n-1)
= 24 . \(\frac { 1 }{ 2^(n-1) } \)

Lesson 11.6 Making Inferences from Experiments

Essential Question How can you test a hypothesis about an experiment?

EXPLORATION 1

Resampling Data
Work with a partner. A randomized comparative experiment tests whether water with dissolved calcium affects the yields of yellow squash plants. The table shows the results.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 1
a. Find the mean yield of the control group and the mean yield of the treatment group. Then find the difference of the two means. Record the results.
b. Write each yield measurement from the table on an equal-sized piece of paper. Place the pieces of paper in a bag, shake, and randomly choose 10 pieces of paper. Call this the “control” group, and call the 10 pieces in the bag the “treatment” group. Then repeat part (a) and return the pieces to the bag. Perform this resampling experiment five times.
c. How does the difference in the means of the control and treatment groups compare with the differences resulting from chance?

EXPLORATION 2

Evaluating Results
Work as a class. To conclude that the treatment is responsible for the difference in yield, you need strong evidence to reject the hypothesis:
Water dissolved in calcium has no effect on the yields of yellow squash plants.
To evaluate this hypothesis, compare the experimental difference of means with the resampling differences.
a. Collect all the resampling differences of means found in Exploration 1(b) for the whole class and display these values in a histogram.
b. Draw a vertical line on your class histogram to represent the experimental difference of means found in Exploration 1(a).
c. Where on the histogram should the experimental difference of means lie to give evidence for rejecting the hypothesis?
d. Is your class able to reject the hypothesis? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 2

Communicate Your Answer

Question 3.
How can you test a hypothesis about an experiment?
Answer:
1. Select your research area as a null hypothesis and an alternate hypothesis.
2. Collect the data designed to test the hypothesis.
3. Next perform an appropriate statistical test.
4. Describe whether to reject your null hypothesis.
5. Write the findings in your results and discussion section.

Question 4.
The randomized comparative experiment described in Exploration 1 is replicated and the results are shown in the table. Repeat Explorations 1 and 2 using this data set. Explain any differences in your answers.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 3
Answer:

Monitoring Progress

Question 1.
In Example 1, interpret the meaning of \(\bar{x}_{\text {treatment }}\) \(\bar{x}_{\text {control }}\) control when the difference is (a) negative, (b) zero, and (c) positive.
Answer:
a. Positive means the treatment increases the total yield of cherry tomato plants.
b. Negative means the treatment decreases the total yield of cherry tomato plants.
c. Zero means the treatment made no difference.

Question 2.
In Example 3, what are the consequences of concluding that the hypothesis is false when it is actually true?
Answer:
a. The mean score of the control group = \(\frac { 49 + 45 + 43 + 47 + 46 + 45 + 47 + 46 }{ 8 } \)
= \(\frac { 368 }{ 8 } \)
= 46
b. The mean score of the treatment group = \(\frac { 39 + 40 + 39 + 37 + 41 + 40 + 42 + 43 }{ 8 } \)
= \(\frac { 281 }{ 8 } \)
= 35.13
c. The experimental difference of the means = 35.13 – 46
= -10.87
d. The double data plot of the data is
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 1
e. The data shows that the treatment groups show the lowest scores compared to the control group. We can conclude that music therapy may be effective for college students with depression.

Making Inferences from Experiments 11.6 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
A method in which new samples are repeatedly drawn from the data set is called ____________.
Answer:
A method in which new samples are repeatedly drawn from the data set is called resampling.

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 4

Answer:
The experimental difference of the means between the averages of the experimental (treatment) group and control groups, which can be written as μ<treatment> – μ<control>
So, the different question is the third question which asks to determine the square root of the average of the squared differences from 2.85.
μ<treatment> = \(\frac { 1 }{ 6 } \)(0.4 + 0.6 + 0.5 + 0.6 + 0.7+ 0.5)
= 0.55
μ<control> = \(\frac { 1 }{ 6 } \)(3.3 + 3.2 + 3.7 + 3.5 + 3.3 + 3.4) = 3.4
μ<treatment> – μ<control> = 0.55 – 3.4 = -2.85
The same answer is -2.85
The answer to the different questions is
= √(\(\frac { 1 }{ 12 } \)((3.3 – (-2.85))² + (3.2 – (-2.85))² + . . . +(0.5 – (-2.85))²))
= √(\(\frac { 1 }{ 12 } \)(303.95))
= √(25.32)
= 5.03
The different answer is 5.03

Monitoring Progress and Modeling with Mathematics

Question 3.
PROBLEM SOLVING
A randomized comparative experiment tests whether music therapy affects the depression scores of college students. The depression scores range from 20 to 80, with scores greater than 50 being associated with depression. The control group has eight students and the treatment group, which receives the music therapy, has eight students. The table shows the results.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 5
a. Find the mean score of the control group.
b. Find the mean score of the treatment group.
c. Find the experimental difference of the means.
d. Display the data in a double dot plot.
e. What can you conclude?
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 6
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 3

Question 4.
PROBLEM SOLVING
A randomized comparative experiment tests whether low-level laser therapy affects the waist circumference of adults. The control group has eight adults and the treatment group, which receives the low-level laser therapy, has eight adults. The table shows the results.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 7
a. Find the mean circumference of the control group.
Answer:
By using the given data, let’s find the mean circumference of the control group.
μ<control> = \(\frac { 1 }{ 8 } \)(34.6 + 35.4 + 33 + 34.6 + 35.2 + 35.2 + 36.2 + 35) = \(\frac { 1 }{ 8 } \)(279.2) = 34.9

b. Find the mean circumference of the treatment group.
Answer:
The mean of the treatment group
μ<treatment> = \(\frac { 1 }{ 8 } \)(31.4 + 33 + 32.4 + 32.6 + 33.4 + 33.4 + 33) = \(\frac { 1 }{ 8 } \)(264) = 33

c. Find the experimental difference of the means.
Answer:
The experimental mean is
μ<treatment> – μ<control> = 33 – 34.9 = -1.9

d. Display the data in a double dot plot.
Answer:
The dot plot of the control and treatment groups
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 2

e. What can you conclude?
Answer:
We can conclude that low-level laser therapy has little effect in reducing waist circumference in adults.

Question 5.
ERROR ANALYSIS
In a randomized comparative experiment, the mean score of the treatment group is 11 and the mean score of the control group is 16. Describe and correct the error in interpreting the experimental difference of the means.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 5

Question 6.
REASONING
In Exercise 4, interpret the meaning of \(\bar{x}_{\text {treatment }}-\bar{x}_{\text {control }}\) when the difference is positive, negative, and zero.
Answer:
The mean waist circumference of the treatment group > the mean waist circumference of the control group.
Positive means the treatment made the waist circumference larger.
The mean waist circumference of the treatment group < the mean waist circumference of the control group.
Negative means the treatment made the waist circumference smaller.
The mean waist circumference of the treatment group is the same as the mean waist circumference of the control group.
Zero means the treatment made no difference.

Question 7.
MODELING WITH MATHEMATICS
Resample the data in Exercise 3 using a simulation. Use the means of the new control and treatment groups to calculate the difference of the means.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 7

Question 8.
MODELING WITH MATHEMATICS
Resample the data in Exercise 4 using a simulation. Use the means of the new control and treatment groups to calculate the difference of the means.
Answer:
First, combine the measurements from both groups and assign a number to each value. Let the numbers 1 through 8 represent the data in the original control group, and let the numbers 9 through 16 represent the data in the original treatment group

Original control group 34.6 35.4 33 34.6 35.2 35.2 36.2 35
assigned number 1 2 3 4 5 6 7 8
Original treatment group 31.4 33 32.4 32.6 33.4 33.4 33.8 33
assigned number 9 10 11 12 13 14 15 16

Use a random generator. Randomly generate 16 numbers from 1 through 16 without repeating a number.

8 5 15 7 12 1 13 3
14 4 16 9 11 10 2 6

Use the first 8 numbers to make the new control group, and the next 8 to make the new treatment group.

New control group 35 35.2 34.8 36.2 32.6 34.6 33.4 33
New treatment group 33.4 34.6 33 31.4 32.4 33 35.4 35.2

The mean of the new control and new treatment groups are
μ<new control> = \(\frac { 1 }{ 8 } \)(35 + 35.2 + 34.8 + 36.2 + 32.6 + 34.6 + 33.4 + 33)
= 34.4
μ<new treatment> = \(\frac { 1 }{ 8 } \)(33.4 + 34.6 + 33 + 31.4 + 32.4 + 33 + 35.4 + 35.2)
= 33.6
The experimental difference = μ<new treatment> – μ<new control> = 33.6 – 34.4 = -0.8
This is greater than the experimental difference found in example 4.

Question 9.
DRAWING CONCLUSIONS
To analyze the hypothesis below, use the histogram which shows the results from 200 resamplings of the data in Exercise 3.
Music therapy has no effect on the depression score.
Compare the experimental difference in Exercise 3 with the resampling differences. What can you conclude about the hypothesis? Does music therapy have an effect on the depression score?
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 9

Question 10.
DRAWING CONCLUSIONS
Suppose the experimental difference of the means in Exercise 3 had been −0.75. Compare this experimental difference of means with the resampling differences in the histogram in Exercise 9. What can you conclude about the hypothesis? Does music therapy have an effect on the depression score?
Answer:
The histogram is approximately bell-shaped and fairly symmetric, so the differences have an approximately normal distribution. Note that the hypothesis assumes that the difference in the mean yields is 0.
The experimental difference of -0.75, however, lies close to the left tail.
From the graph, there are about 106 to 113 values out of 200 that are greater than -0.75 which is more than 50% of the values.
Also, the experimental difference falls inside the middle 90% of the resampling differences. It is likely to get a difference this large when you assume that the difference is 0, suggesting the control group and the treatment group is not different.
You can conclude that the hypothesis is most likely true. Thus, music therapy has no effect on depression scores.

Question 11.
WRITING
Compare the histogram in Exercise 9 to the histogram below. Determine which one provides stronger evidence against the hypothesis, Music therapy has no effect on the depression score. Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 11

Question 12.
HOW DO YOU SEE IT?
Without calculating, determine whether the experimental difference, \(\bar{x}_{\text {treatment }}-\bar{x}_{\text {control }}\), is positive, negative, or zero. What can you conclude about the effect of the treatment? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 11

Answer:
Most of the control group values are greater than the treatment group values which implies that the mean of the control group values will be greater than the mean of the treatment group values.
The experimental difference is μ<treatment> – μ<control>
The experimental difference is negative. Also, by using the given graph, since the difference between the control values is treatment values is little.
We have a slight effect.

Question 13.
MAKING AN ARGUMENT
Your friend states that the mean of the resampling differences of the means should be close to 0 as the number of resamplings increase. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 13

Question 14.
THOUGHT PROVOKING
Describe an example of an observation that can be made from an experiment. Then give four possible inferences that could be made from the observation.
Answer:
The 4 possible inferences are as follows,
1. The hair growth shampoo has no effect on hair growth.
2. The hair growth shampoo causes decreases in hair growth.
3. The hair growth shampoo causes a slight increase in the hair growth
4. The hair growth shampoo causes a significant increase in hair growth.

Question 15.
CRITICAL THINKING
In Exercise 4, how many resamplings of the treatment and control groups are theoretically possible? Explain.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 15

Maintaining Mathematical Proficiency

Factor the polynomial completely.
Question 16.
5x3 − 15x2

Answer:
Given,
5x3 − 15x2
Take 5x² as common.
5x3 − 15x = 5. x² . x – 5. x² . 3
= 5x²(x – 3)
Thus, 5x3 − 15x2 = 5x²(x – 3)

Question 17.
y3 − 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 17

Question 18.
z3 + 5z2 − 9z − 45
Answer:
Given,
z3 + 5z2 − 9z − 45
z3 + 5z2 − 9z − 45 = z³ + 5z² – 9z – 45 + 6z – 6z
= z³ + 5z² + 6z – 15z – 45
= z(z² + 5z + 6) – 15(z + 3)
z² + 5z + 6 = 0
Using the quadratic formula
z₁,₂ = \(\frac { -5 ± √(25 – 4 . 6)}{ 2 . 1 } \)
= \(\frac { -5 ± √(25 – 24)}{ 2} \)
= \(\frac { -5 ± 1}{ 2 } \)
The solutions are \(\frac { -5 + 1 }{ 2 } \) = -2, \(\frac { -5 – 1 }{ 2 } \) = -3
The fcators are
z² + 5z + 6 = (z + 2)(z + 3)
z3 + 5z2 − 9z − 45 = z(z² + 5z + 6) – 15(z + 3)
= z(z + 2)(z + 3) – 15(z + 3)
= (z +3)(z² + 2z – 15)
z² + 2z – 15 = 0
Using the quadratic formula
z₁,₂ = \(\frac { -2 ± √(4 – 4 . (-15))}{ 2 . 1 } \)
= \(\frac { -2 ± √(4 + 60)}{ 2 } \)
= \(\frac { -2 ± 8}{ 2} \)
Thus, the solutions are \(\frac { -2 + 8}{ 2} \) = 3, \(\frac { -2 – 8}{ 2} \) = -5
z² + 2z – 15 = (z – 3)(z + 5)
(z +3)(z² + 2z – 15) = (z+3)(z – 3)(z + 5)

Question 19.
81w4 − 16
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 19

Determine whether the inverse of f is a function. Then find the inverse.
Question 20.
f(x) = \(\frac{3}{x+5}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 3
From the graph
y = \(\frac { 3 }{ x + 5 } \)
x = \(\frac { 3 }{ y + 5 } \)
x(y + 5) = 3
y + 5 = \(\frac { 3 }{ x } \)
y = \(\frac { 3 }{ x } \) – 5

Question 21.
f(x) = \(\frac{1}{2x-1}\)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 21

Question 22.
f(x) = \(\frac{2}{x}\) − 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 4
From the graph
y = \(\frac { 2 }{ x } \) – 4
x = \(\frac { 2 }{ y } \) – 4
x + 4 = \(\frac { 2 }{ y } \)
y(x + 4) = 2
y = \(\frac { 2 }{ x + 4 } \)

Question 23.
f(x) = \(\frac{3}{x^{2}}\) + 1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 a 23

Data Analysis and Statistics Performance Task: Curving the Test

11.4–11.6 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 12

Core Concepts
Section 11.4
Randomization in Experiments and Observational Studies, p. 621
Comparative Studies and Causality, p. 621
Analyzing Experimental Designs, p. 622

Section 11.5
Estimating Population Parameters, p. 626
Analyzing Estimated Population Parameters, p. 628

Section 11.6
Experiments with Two Samples, p. 634
Resampling Data Using Simulations, p. 635
Making Inferences About Treatments, p. 636

Mathematical Practices
Question 1.
In Exercise 7 on page 623, find a partner and discuss your answers. What questions should you ask your partner to determine whether an observational study or an experiment is more appropriate?
Answer:

Question 2.
In Exercise 23 on page 632, how did you use the given interval to find the sample size?
Answer:

Performance Task: Curving the Test
Test scores are sometimes curved for different reasons using different techniques. Curving began with the assumption that a good test would result in scores that were normally distributed about a C average. Is this assumption valid? Are test scores in your class normally distributed? If not, how are they distributed? Which curving algorithms preserve the distribution and which algorithms change it?
To explore the answers to these questions and more, go to BigIdeasMath.com.
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 11.6 13

Data Analysis and Statistics Chapter Review

11.1 Using Normal Distributions (pp. 595–602)

Question 1.
A normal distribution has mean μ and standard deviation σ. An x-value is randomly selected from the distribution. Find P(x ≤ μ − 3σ).
Answer:
Find the probability P(x ≤ μ – 3σ)
P(x ≤ μ – 3σ) = \(\frac { 1 – 0.997 }{ 2 } \) = \(\frac { 0.003 }{ 2 } \)
= 0.0015
The probability is 0.0015

Question 2.
The scores received by juniors on the math portion of the PSAT are normally distributed with a mean of 48.6 and a standard deviation of 11.4. What is the probability that a randomly selected score is at least 76?
Answer:
The scores received by juniors on the math portion of the PSAT are normally distributed with μ = 48.6, σ = 11.4
z = \(\frac { x – μ }{ σ } \)
P(x ≥76) = P(\(\frac { x – 48.6 }{ 11.4 } \)≥\(\frac { 76 – 48.6 }{ 11.4 } \))
= P(z≥2.4) = 1 – P(z≤2.4)
= 1 – 0.9918
= 0.0082

11.2 Populations, Samples, and Hypotheses (pp. 603−608)

Question 3.
To estimate the average number of miles driven by U.S. motorists each year, a researcher conducts a survey of 1000 drivers, records the number of miles they drive in a year, and then determines the average. Identify the population and the sample.
Answer:
As the survey does not participate all motorists from the US, we can conclude that in the population of all motorist drivers in the US, a sample of 1000 drivers is surveyed.

Question 4.
A pitcher throws 40 fastballs in a game. A baseball analyst records the speeds of 10 fastballs and finds that the mean speed is 92.4 miles per hour. Is the mean speed a parameter or a statistic? Explain.
Answer:
A pitcher throws 40 fastballs in a game. A baseball analyst records the speeds of 10 fastballs and finds that the mean speed is 92.4 miles per hour
Since we have an average speed of 10 fastballs of some pitchers, the mean value is calculated from a sample, not from the entire population. It represents statistics. Thus, this average represents statistics.

Question 5.
A prize on a game show is placed behind either Door A or Door B. You suspect the prize is more often behind Door A. The show host claims the prize is randomly placed behind either door. What should you conclude when the prize is behind Door A for 32 out of 50 contestants?
Answer:
A prize on a game show is placed behind either Door A or Door B.
Since more than half of the prizes for the customers are behind door A, the host’s claim that the prize is randomly placed behind either door is most likely false.

11.3 Collecting Data (pp. 609−616)

Question 6.
A researcher wants to determine how many people in a city support the construction of a new road connecting the high school to the north side of the city. Fifty residents from each side of the city are surveyed. Identify the type of sample described and determine whether the sample is biased.
Answer:
Fifty residents from each side of the city are surveyed to determine how many people in a city support the construction of a new road connecting the high school to the north side of the city. Thus, it is a stratified sample.

Question 7.
A researcher records the number of people who use a coupon when they dine at a certain restaurant. Identify the method of data collection.
Answer:
It is an observational study because the researcher simply observes the subjects without interfering.
The method of data collection is an observational study.

Question 8.
Explain why the survey question below may be biased or otherwise introduce bias into the survey. Then describe a way to correct the flaw.
“Do you think the city should replace the outdated police cars it is using?”
Answer:
The answer is biased because it is set in a way that encourages the answer yes. The question can be set as “Do you think the police should replace cars?”

11.4 Experimental Design (pp. 619–624)

Question 9.
A restaurant manager wants to know which type of sandwich bread attracts the most repeat customers. Is the topic best investigated through an experiment or an observational study? Describe how you would design the experiment or observational study.
Answer:
We can do an experimental study on this problem. For example, suppose we have 5 types of sandwich bread available.
Then we can make 5 randomly selected groups with the same number of people and sell each group a sandwich with a certain type of bread.
For each group, note how many people came again to buy the same sandwich.

Question 10.
A researcher wants to test the effectiveness of a sleeping pill. Identify a potential problem, if any, with the experimental design below. Then describe how you can improve it.
The researcher asks for 16 volunteers who have insomnia. Eight volunteers are given the sleeping pill and the other 8 volunteers are given a placebo. Results are recorded for 1 month.
Answer:
As the participants in the study are not randomly selected, they cannot represent the whole population. Thus, we must randomly select the members of the population for the study.

Question 11.
Determine whether the study is a randomized comparative experiment. If it is, describe the treatment, the treatment group, and the control group. If it is not, explain why not and discuss whether the conclusions drawn from the study are valid.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics cr 1

Answer:
It is a randomized comparative experiment because customers are randomly assigned to the control group or the treatment group. The treatment is a new design of a car wash.
The treatment group is the group that uses the new design to wash their cars and the control group will be a group that uses the old design to wash their cars.

11.5 Making Inferences from Sample Surveys (pp. 625−632)

Question 12.
In a survey of 1017 U.S. adults, 62% said that they prefer saving money over spending it. Give an interval that is likely to contain the exact percent of all U.S. adults who prefer saving money over spending it.
Answer:
Given,
In a survey of 1017 U.S. adults, 62% said that they prefer saving money over spending it.
n = 1017 U.S adults,
The margin error will be
±\(\frac { 1 }{ √n } \) = ±\(\frac { 1 }{ √1017 } \) = ±0.031
Thus, the interval that is likely to contain the exact percentage of all adults from the U.S who prefer saving money over spending it will be
(62% + 3.1%, 62% – 3.1%) = (65.1%, 58.9%)
Thus, the interval is between 65.1% t0 58.9%

Question 13.
There are two candidates for homecoming king. The table shows the results from four random surveys of the students in the school. The students were asked whether they will vote for Candidate A.Do you think Candidate A will be the homecoming king? Explain.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics cr 2

Answer:
When the sample size increases, it approaches 46.8%. Since that is less than 50% it is not as likely as the probability of candidate B winning homecoming king.
A sample size of 62 has 46.8% of voters saying yes. As the sample size increases, the percentage decreases.

11.6 Making Inferences from Experiments (pp. 633−638)

Question 14.
Describe how to use a simulation to resample the data in the example above. Explain how this allows you to make inferences about the data when the sample size is small.
Answer:
When the sample size is small, we can repeat this process more times and make new control and new treatment groups, and see how often you get differences between the new groups that are at least as large as the one you measured.
Combine the measurements from both groups and assign a number to each value. Let the numbers 1 through 10 represent the data in the original control group
Let the numbers 11 through 20 represent the data in the original treatment group

Control group 4.5 4.5 4.8 4.4 4.4 4.7 4.3 4.5 4.1 4.2
Assigned number 1 2 3 4 5 6 7 8 9 10
Treatment group 4.6 4.8 5.0 4.8 4.7 4.6 4.9 4.9 4.8 4.4
Assigned number 11 12 13 14 15 16 17 18 19 20

Use a random generator. Randomly generate 20 numbers from 1 through 20 without repeating a number.

5 1 15 13 18 17 4 20 10 8
9 2 12 16 6 19 11 3 7 14

Use the first 10 numbers to make the new control group, and the next 10 to make the new treatment group.

New control group 4.4 4.5 4.7 5.0 4.9 4.9 4.4 4.4 4.2 4.5
New treatment group 4.1 4.5 4.8 4.6 4.7 4.8 4.6 4.8 4.3 4.8

Now let’s find the mean yields of the new control and treatment groups
μ<new control> = \(\frac { 1 }{ 10 } \)(4.4 + 4.5 + 4.7 + 5.0 + 4.9 + 4.9 + 4.4 + 4.4 + 4.2 + 4.5)
= 4.59
μ<new treatment> = \(\frac { 1 }{ 10 } \)(4.1 + 4.5 + 4.8 + 4.6 + 4.7 + 4.8 + 4.6 + 4.8 + 4.3 + 4.8)
= 4.6
So, μ<new treatment> – μ<new control> = 4.6 – 4.59 = 0.01

Data Analysis and Statistics Chapter Test

Question 1.
Market researchers want to know whether more men or women buy their product. Explain whether this research topic is best investigated through an experiment or an observational study. Then describe the design of the experiment or observational study.
Answer:
For the researcher, it is best investigated through an observational study because random samples of customers can be selected for the groups being studied. For example, we can record every th customer who buys the product and record if he was women or men.

Question 2.
You want to survey 100 of the 2774 four-year colleges in the United States about their tuition cost. Describe a method for selecting a random sample of colleges to survey.
Answer:
Of the four-year colleges in the United States, we must randomly select colleges. Therefore, we assign one integer to each college. Now, using a random integer generator, we can select numbers from to and make a random sample of 100 colleges that match a random sample of numbers.

Question 3.
The grade point averages of all the students in a high school are normally distributed with a mean of 2.95 and a standard deviation of 0.72. Are these numerical values parameters or statistics? Explain.
Answer:
Since these values represent the mean and standard deviation of all the students in a high school, these values represent paramenters because they are calculated from the entire population.

A normal distribution has a mean of 72 and a standard deviation of 5. Find the probability that a randomly selected x-value from the distribution is in the given interval.
Question 4.
between 67 and 77
Answer:
The normal distribution has a mean of 72 and a standard deviation of 5.
z = \(\frac { x – μ }{ σ } \)
P(67≤x≤77) = P(\(\frac { 67 – 72 }{ 5 } \)≤ \(\frac { x – 72 }{ 5 } \) ≤\(\frac { 77 – 72 }{ 5 } \))
= P(-1≤z≤1) = P(z<1) – P(z<-1)
= 0.8413 – 0.1587 = 0.6826
The probability is 0.68

Question 5.
at least 75
Answer:
The normal distribution has a mean of 72 and a standard deviation of 5.
82 – μ = 82 – 72 = 10 = 2 . 5 = 2σ
10 = μ – 2σ
P(x≤82) = P(x≤μ – 2σ) = 0.5 + \(\frac { 0.95 }{ 2 } \)
= 0.5 + 0.475 = 0.975
The probability is 0.975

Question 6.
at most 82
Answer:
The normal distribution has a mean of 72 and a standard deviation of 5.
z = \(\frac { x – μ }{ σ } \)
P(x≥75) = P(\(\frac { x – 72 }{ 5 } \)≥ \(\frac { 75 – 72 }{ 5 } \))
= P(z≥0.6) = 1 – P(z< 0.6)
= 1 – 0.7257
= 0.2743
The probability is 0.2743

Question 7.
A researcher wants to test the effectiveness of a new medication designed to lower blood pressure. Identify a potential problem, if any, with the experimental design. Then describe how you can improve it.
The researcher identifies 30 people with high blood pressure. Fifteen people with the highest blood pressure are given the medication and the other 15 are given a placebo. After 1 month, the subjects are evaluated.

Answer:
Since 15 people with the highest blood pressure are assigned to one group, we can notice that people are not randomly assigned. Thus, to fix this problem, we can randomly form two groups of 15 people, and then one group will take the medicine and another one will take a placebo.

Question 8.
A randomized comparative experiment tests whether a vitamin supplement increases human bone density (in grams per square centimeter). The control group has eight people and the treatment group, which receives the vitamin supplement, has eight people. The table shows the results.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics ct 1
a. Find the mean yields of the control group, \(\bar{x}_{\text {control }}\), and the treatment group, \(\bar{x}_{\text {treatment }}\).
b. Find the experimental difference of the means, \(\bar{x}_{\text {treatment }}-\bar{x}_{\text {control }}\).
c. Display the data in a double dot plot. What can you conclude?
d. Five hundred resamplings of the data are simulated. Out of the 500 resampling differences, 231 are greater than the experimental difference in part (b). What can you conclude about the hypothesis, The vitamin supplement has no effect on human bone density? Explain your reasoning.
Answer:
a. μ<control> = \(\frac { 1 }{ 8 } \)(0.9 + 1.2 + 1.0 + 0.8 + 1.3 + 1.1 + 0.9 + 1.0)
= 1.025
μ<treatment> = \(\frac { 1 }{ 8 } \)(1.2 + 1.0 + 0.9 + 1.3 + 1.2 + 0.9 + 1.3 + 1.2)
= 1.125
b. μ<treatment> – μ<control> = 1.125 – 1.025 = 0.1
c. The double dot plot is
Big Ideas Math Algebra 2 Answer Key Chapter 11 Data Analysis and Statistics 1
d. The hypothesis which claims that vitamin supplement has no effect on human bone density assumes that the difference of the means is equal to 0. Now, from 500 resampling differences, 231 are greater than the experimental differences in part b, which is 0.1. Now this implies that more than half of the resamplings are less than 0.1, which is less than one SD. So, the hypothesis is probably true.

Question 9.
In a recent survey of 1600 randomly selected U.S. adults, 81% said they have purchased a product online.
Big Ideas Math Algebra 2 Answers Chapter 11 Data Analysis and Statistics ct 2
a. Identify the population and the sample. Describe the sample.
Answer:
In the population of all U.S adults a sample of 1600 randomly selected adults have participated in the survey

b. Find the margin of error for the survey.
Answer:
Since in the survey participate n = 1600 U.S adults, the margin error will be
±\(\frac { 1 }{ √n } \) = ±\(\frac { 1 }{ √1600 } \) = ±0.025
The margin error will be ±2.5%

c. Give an interval that is likely to contain the exact percent of all U.S. adults who have purchased a product online.
Answer:
Thus the interval that is likely to contain the exact percentage of all adults from the U.S who said that they have purchased a product online will be
(81% – 2.5%, 81% + 2.5%) = (78.5%, 83.5%)
The interval is between 78.5% to 83.5%.

d. You survey 75 teachers at your school. The results are shown in the graph. Would you use the recent survey or your survey to estimate the percent of U.S. adults who have purchased a product online? Explain.
Answer:
You survey 75 teachers at your school.
Since we surveyed only the teachers from one school we cannot use this survey because it cannot represent the entire population. So, we will use a recent survey.

Data Analysis and Statistics Cumulative Assessment

Question 1.
Your friend claims any system formed by three of the following equations will have exactly one solution.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics ca 1
a. Write a linear system that would support your friend’s claim.
b. Write a linear system that shows your friend’s claim is incorrect.
Answer:
a. Each of the given equations represents a plane equation. A system of 3 equations will have one solution if these planes intersect at one point.
3x + y + 3z = 6 — (i)
x + y + z = 2 — (ii)
3x + y + 9z = 12 — (iii)
We can write 2nd equation as y = 2 – x – z
Solve 1 & 3,
3x + 2 – x – z + 3z = 6
2x + 2z = 4 — (iv)
3x + 2 – x – z +9z = 12
2x + 8z = 10 — (v)
Subtract (iv) & (v)
2x + 8z – 2x – 2z = 10 – 4
6z = 6
z = 1
Put z = 1 in (iv)
2x + 2 = 4
x = 1
Put x, z values in y = 2 – x – z
y = 2 – 1 – 1
y = 0
Hence, the solution of the system is (x, y, z) = (1, 0, 1) and the friend’s calim is correct. There is only one solution to the system.
b. To prove thae claim is incorrect, take the following equations
3x + y + 3z = 6 — (i)
x + y + z = 2 — (ii)
4x – 2y + 4z = 8 — (iii)
We can write 2nd equation as y = 2 – x – z
Change the first & third equations
3x + 2 – x – z + 3z = 6
2x + 2z = 4
x + z = 2— (iv)
4x – 2(2 – x – z) + 4z = 8
4x – 4 + 2x + 2z + 4z = 8
6x + 6z = 12
x + z = 2 — (v)
As both (iv), v are same, the equation has many solutions
y = 2 – 2
y = 0
Hence, the system has infinately many solutions (x, y , z) = (2 – z, 0, z) and the friend’s claim is incorrect.

Question 2.
Which of the following samples are biased? If the sample is biased, explain why it is biased.
A. A restaurant asks customers to participate in a survey about the food sold at the restaurant. The restaurant uses the surveys that are returned.
Answer:
The sample is biased because only customers with strong opinions will return the survey

B. You want to know the favorite sport of students at your school. You randomly select athletes to survey at the winter sports banquet.
Answer:
The sample is biased because the survey doesn’t participate randomly selected students.

C. The owner of a store wants to know whether the store should stay open 1 hour later each night. Each cashier surveys every fifth customer.
Answer:
The sample is nonbiased because in the survey participants randomly selected customers.

D. The owner of a movie theater wants to know whether the volume of its movies is too loud. Patrons under the age of 18 are randomly surveyed.
Answer:
The sample is biased because only patrons under the age of 18 participate in the survey.

Question 3.
A survey asks adults about their favorite way to eat ice cream. The results of the survey are displayed in the table shown.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics ca 3
a. How many people were surveyed?
b. Why might the conclusion, “Adults generally do not prefer to eat their ice cream in a cone” be inaccurate to draw from this data?
c. You decide to test the results of the poll by surveying adults chosen at random. What is the probability that at least three out of the six people you survey prefer to eat ice cream in a cone?
d. Four of the six respondents in your study said they prefer to eat their ice cream in a cone. You conclude that the other survey is inaccurate. Why might this conclusion be incorrect?
e. What is the margin of error for your survey?
Answer:
a. 2246
b. The conclusion is inaccurate
c. 0.2374
d. Conclusion is wrong
e. ±40.82

Explanation:
a. Given that,
Margin error = ±2.11
±0.0211 = ±\(\frac { 1 }{ √n } \)
(±0.0211)² = (±\(\frac { 1 }{ √n } \))²
0.000445 = \(\frac { 1 }{ n } \)
n = 2246
b. From the result if the survey we see that people prefer different types of ice cream with different percentages and as 2915 people prefer ice cream in a cone, we cannot conclude that in general people do not eat ice cream in a cone.
c. from the result of the survey we see that 29%of people prefer ice cream in acone. So we can compute the probability that a person chooses ice cream in the cone and the probability that doesn’t choose ice cream in the cone
A: “A person prefers to eat ice cream in a cone”
B: “A person doesn’t prefer to eat ice cream in a cone”
P(A) = \(\frac { 29 }{ 100 } \) = 0.291
P(B) = 1 – P(A) = 1 – 0.29 = 0.71
P(X≥3) = 1 – P(X<3)
= 1 – P(X = 2) – P(x = 1) – P(X = 0)
P(X = 2) = 0.3206
P(X = 1) = 0.3139
P(X = 0) = 0.1281
P(X≥3) = 1 – 0.3206 – 0.3139 – 0.1281
= 0.2374
d. The conclusion might be wrong because in your survey, we have only 6 persons that participate and since the sample size is too small, we cannot conclude that the other survey is inaccurate.
e. The sample size of our survey is 6
The margin error = ±\(\frac { 1 }{ √n } \)
= ±\(\frac { 1 }{ √6 } \)
= ±0.4082
The margin error is ±40.82%

Question 4.
You are making a lampshade out of fabric for the lamp shown. The pattern for the lampshade is shown in the diagram on the left.
a. Use the smaller sector to write an equation that relates θ and x.
b. Use the larger sector to write an equation that relates θ and x+ 10.
c. Solve the system of equations from parts (a) and (b) for x and θ.
d. Find the amount of fabric (in square inches) that you will use to make the lampshade.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics ca 4
Answer:
a. θ = \(\frac { 5π }{ x } \)
b. θ = \(\frac { 14π }{ x + 10} \)
c. x = 5.56, θ = 2.82
d. 298.17

Explanation:
a. r = x, s = 5π
Arc length s = rθ
5π = x . θ
θ = \(\frac { 5π }{ x } \)
b. r = x + 10, s = 14π
Arc length s = rθ
14π = (x + 10)θ
θ = \(\frac { 14π }{ x + 10 } \)
c. From a and b
θ = \(\frac { 5π }{ x } \)
θ = \(\frac { 14π }{ x + 10 } \)
\(\frac { 5π }{ x } \) = \(\frac { 14π }{ x + 10 } \)
5(x + 10) = 14
5x + 50 = 14
x = 5.56
θ = \(\frac { 5π }{ 5.56 } \)
= 2.82 radians
d. Area = \(\frac { 1 }{ 2 } \)r²θ
A<G> = \(\frac { 1 }{ 2 } \)(5.56 + 10)² . 2.82
= \(\frac { 1 }{ 2 } \) . 242.11 . 2.82
= 341.76
A<S> = \(\frac { 1 }{ 2 } \)r²θ
= \(\frac { 1 }{ 2 } \) . (5.56)² . 2.82
= \(\frac { 1 }{ 2 } \) . 30.91 . 2.82
= 43.59
A = A<G> – A<S>
= 341.76 – 43.59
= 298.17

Question 5.
For all students taking the Medical College Admission Test over a period of 3 years, the mean score was 25.1. During the same 3 years, a group of 1000 students who took the test had a mean score of 25.3. Classify each mean as a parameter or a statistic. Explain.
Answer:
25.1 Parameter
25.3 statistic

Explanation:
The mean score of 25.1 is computed from the all students from the college. So, it is a parameter and the mean score of 25.3 is computed from a sample of 1000 students. So, it is a statistic.

Question 6.
Complete the table for the four equations. Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics ca 6

Answer:

Question 7.
The normal distribution shown has mean 63 and standard deviation 8. Find the percent of the area under the normal curve that is represented by the shaded region. Then describe another interval underthe normal curve that has the same area.
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics ca 7

Answer:
P(55≤x≤83) = P(\(\frac { 55 – 63 }{ 8 } \) ≤\(\frac { x – 63 }{ 8 } \) ≤\(\frac { 83 – 63 }{ 8 } \))
= P(-1 ≤z≤2.5) = P(z<2.5) – P(z<-1)
= 0.9938 – 0.1587 = 0.84
The probability is 0.84 or 84%

Question 8.
Which of the rational expressions cannot be simplified? (HSA-APR.D.6)
Big Ideas Math Algebra 2 Solutions Chapter 11 Data Analysis and Statistics ca 8

Answer:
A. Factors of 2x² + 5x – 3 = 0
= \(\frac { -5  ± √(25 – 4 . 2 . (-3))}{ 2 . 2 } \)
= \(\frac { -5  ± √(25 + 24)}{ 4 } \)
= \(\frac { -5  ± 7}{ 4 } \)
The solutions are \(\frac { 1 }{ 2 } \), -3
2x² + 5x – 3 = 2(x – \(\frac { 1 }{ 2 } \)) (x + 3)
Factors of x² – 7x + 12 = 0
= \(\frac { -(-7)  ± √(49 – 4 . 1 . 12)}{ 2 . 1 } \)
= \(\frac { 7  ± √(49 – 48)}{ 2 } \)
= \(\frac { 7 ± 1}{ 2 } \)
The solutions are 4, 3
x² – 7x + 12 = (x – 4)(x – 3)
\(\frac { 2x² + 5x – 3 }{ x² – 7x + 12 } \)= \(\frac { 2(x – 1/2)(x + 3) }{ (x – 4)(x – 3) } \)

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions

View Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions with step by step explanation here. Students can find the easy methods to solve polynomial functions from this page. The solutions are prepared by the subject experts in a quick and simple manner that helps the students to score the highest marks in the exams. Our Chapter 4 Polynomial Functions BIM Algebra 2 Textbook Answers help the students to complete homework assignments in time and with accuracy.

Big Ideas Math Book Algebra 2 Answer Key Chapter 4 Polynomial Functions

It is not only important to score marks in the exams it is also necessary to understand the concept of the topic and know how it is applicable in the real life. Keeping this point in mind we have provided Big Ideas Math Textbook Algebra 2 Answers Chapter 4 Polynomial Functions with examples in detail. Math is a fun learning subject if you love to solve problems. Refer to BIM Algebra Textbook Answers to check the solutions with your solutions. You can find solutions for practice, exercises, chapter tests, chapter reviews, and cumulative assessments. Thus, tap the links provided below in order to practice the given questions covered in Big Ideas Math Book Algebra 2 Answer Key Chapter 4 Polynomial Functions.

Polynomial Functions Maintaining Mathematical Proficiency

Simplify the expression.

Question 1.
6x − 4x
Answer:
Given,
6x − 4x
Terms can be combined only if they have the exact same variable portion and combine like terms.
6x – 4x = 2x
So, the simplified expression is 2x.

Question 2.
12m − m − 7m + 3
Answer:
Given,
12m − m − 7m + 3
Terms can be combined only if they have the exact same variable portion and combine like terms.
(12m-m-7m)+3
=m(12 – 1 – 7) + 3
= 4m + 3
So, 12m − m − 7m + 3 = 4m + 3

Question 3.
3(y + 2) − 4y
Answer:
3(y + 2) − 4y
Terms can be combined only if they have the exact same variable portion and combine like terms.
3(y+2)-4y
= 3 × y + 3 × 2 – 4y
= 3y + 6 – 4y
= -y + 6
So, 3(y + 2) − 4y = -y + 6

Question 4.
9x − 4(2x − 1)
Answer:
Given,
9x − 4(2x − 1)
= 9x – 4 × 2x – 4 × -1 (since – × – = +)
Terms can be combined only if they have the exact same variable portion and combine like terms.
= 9x – 8x + 4
= x + 4
So, 9x − 4(2x − 1) = x + 4

Question 5.
−(z + 2) − 2(1 − z)
Answer:
Given,
−(z + 2) − 2(1 − z)
= -z – 2 – 2 + 2z (since – × – = +)
Terms can be combined only if they have the exact same variable portion and combine like terms.
=-z + 2z – 2 – 2
= z -4
So, −(z + 2) − 2(1 − z) = z – 4.

Question 6.
−x2 + 5x + x2
Answer:
Given,
−x2 + 5x + x2
Terms can be combined only if they have the exact same variable portion and combine like terms.
5x−x2 + x2
= 5x
So, −x2 + 5x + x2 = 5x

Find the volume of the solid.

Question 7.
cube with side length 4 inches
Answer:
Given,

s = 4
We know that,
The volume of cube=s³
V =4 × 4 × 4=64 cu. inches
So, the volume of the solid is 64 cu. inches

Question 9.
rectangular prism with length 4 feet, width 2 feet, and height 6 feet
Answer:
Given,
length= 4 feet
width=2 feet
height= 6 feet
We know that,
volume of rectangular prism= length x width x height
Substitute the values in the formula
V = 4 × 2 × 6= 48 cubic feet
Thus the volume of the rectangular prism is 48 cu. feet.

Question 10.
right cylinder with radius 3 centimeters and height 5 centimeters
Answer:
Given that
Radius = 3 centimeters
Height = 5 centimeter
We know that,
Volume of a cylinder = area of base × height
V = π × r2× h
π = 3.14
V = 3.14 × 3 × 3 × 5=141.3 cubic centimeter

Question 11.
ABSTRACT REASONING Does doubling the volume of a cube have the same effect on the side length? Explain your reasoning
Answer:
If the side of the cube is doubled. The volume is 8 times larger.
Example: let’s use a cube that has a side length of 2 as an example.
The volume of that cube would be length x and width x height
V = 2 × 2 ×  2 = 8
If we double the edges to 4.
V = 4 × 4 × 4 = 64.
64/8=8
So, the volume increases by 8 times.

Polynomial Functions Mathematical Practices

Monitoring Progress

Use a graphing calculator to determine whether the function is continuous. Explain your reasoning.

Question 1.
f(x) = \(\frac{x^{2}-x}{x}\)
Answer:
Given,
f(x) = \(\frac{x^{2}-x}{x}\)
f(x) = x(x-1)/x
f(x) = x – 1
x = 0
f(0) = 0 – 1 = -1
f(1) = 1 – 1 = 0
f(2) = 2 – 1 = 1
f(3) = 3 – 1 = 2
Big Ideas Math Answers Polynomial Functions Mathematical Practices_1

Question 2.
f(x) = x3 − 3
Answer:
Given,
f(x) = x3 − 3
f(0) = 0³ – 3 = -3
f(1) = 1³ – 3 = 1 – 3 = -2
f(2) = 2³ – 3 = 8 – 3 = 5
f(-1) = (-1)³ – 3 = -1 – 3 = -4
f(-2) = (-2)³ – 3 = -8 – 3 = -11
Big Ideas Math Answers Polynomial Functions Mathematical Practices_2

Question 3.
f(x) = \(\sqrt{x^{2}+1}\)
Answer:
Given,
f(x) = \(\sqrt{x^{2}+1}\)
f(0) = \(\sqrt{0^{2}+1}\) = 1
f(1) = \(\sqrt{1^{2}+1}\) = \(\sqrt{2}\)
Big Ideas Math Answers Polynomial Functions Mathematical Practices_3

Question 4.
f(x) = | x + 2 |
Answer:
Given,
f(x) = | x + 2 |
f(0) = |0 + 2| = 2
f(1) = |1 + 2| = 3
f(-1) = |-1 + 2| = 1
f(-2) = |-2 + 2| = 0
Big Ideas Math Answers Polynomial Functions Mathematical Practices_4

Question 5.
f(x) = \(\frac{1}{x}\)
Answer:
Given,
f(x) = \(\frac{1}{x}\)
f(0) = \(\frac{1}{x}\) = ∞
f(1) = \(\frac{1}{1}\) = 1
f(2) = \(\frac{1}{2}\) = 0.5
f(-1) = \(\frac{1}{-1}\) = -1
f(-2) = \(\frac{1}{-2}\) = -0.5
Big Ideas Math Answers Polynomial Functions Mathematical Practices_5

Question 6.
f(x) = \(\frac{1}{\sqrt{x^{2}-1}}\)
Answer:
Big Ideas Math Answers Polynomial Functions Mathematical Practices_6

Question 7.
f(x) = x
Answer:
Given function is f(x) = x
f(0) = 0
f(1) = 1
f(2) = 2
f(-1) = -1
f(-2) = -2
Big Ideas Math Answers Polynomial Functions Mathematical Practices_7

Question 8.
f(x) = 2x − 3
Answer:
Given,
f(x) = 2x − 3
f(0) = 2(0) – 3 = -3
f(1) = 2(1) – 3 = -1
f(-1) = 2(-1) – 3 = -5
f(2) = 2(2) – 3 = 1
f(-2) = 2(-2) – 3 = -7
Big Ideas Math Answers Polynomial Functions Mathematical Practices_9

Question 9.
f(x) = \(\frac{x}{x}\)
Answer:
Given,
f(x) = \(\frac{x}{x}\)
f(1) = 1/1 = 1
Big Ideas Math Answers Polynomial Functions Mathematical Practices_8

Lesson 4.1 Graphing Polynomial Functions

Essential Question
What are some common characteristics of the graphs of cubic and quartic polynomial functions?
A polynomial function of the form
f(x) = anxn + an – 1xn– 1 +. . .+ a1x + a0
where an ≠ 0, is cubic when n = 3 and quartic when n = 4.

EXPLORATION 1
Identifying Graphs of Polynomial Functions
Work with a partner.
Match each polynomial function with its graph. Explain your reasoning. Use a graphing calculator to verify your answers.
a. f(x) = x3 − x
b. f(x) = −x3 + x
c. f(x) = −x4 + 1
d. f(x) = x4
e. f(x) = x3
f. f(x) = x4 − x2

EXPLORATION 2
Identifying x-Intercepts of Polynomial Graphs
Work with a partner.
Each of the polynomial graphs in Exploration 1 has x-intercept(s) of −1, 0, or 1. Identify the x-intercept(s) of each graph. Explain how you can verify your answers.

Communicate Your Answer

Question 3.
What are some common characteristics of the graphs of cubic and quartic polynomial functions?
Answer:
The polynomial functions with degree 2 are called quadratic functions and more have graphs that do not have sharp corners.
The characteristics of a graph of polynomial functions are.
A cubic function of form f(x) = ax³ + bx² + cx + d.
The polynomial function of degree 3 is a cubic function.
The cube function has 1 or 3 real numbers and 0 or 2 complex numbers.
The graph of the quadratic function is a plane curve which is a U-shaped curve called a parabola

Question 4.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 1
Determine whether each statement is true or false. Justify your answer.
a. When the graph of a cubic polynomial function rises to the left, it falls to the right.
Answer:
The statement “When the graph of the cubic polynomial function rises to the left, it falls to the right” is true.

b. When the graph of a quartic polynomial function falls to the left, it rises to the right.
Answer:
The statement “When the graph of the quartic polynomial function falls to the left, it rises to the right” is false.

4.1 Lesson

Monitoring Progress

Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.

Question 1.
f(x) = 7 − 1.6x2 − 5x
Answer:
Given,
f(x) = 7 − 1.6x2 − 5x
The standard form of polynomials has the exponents of the terms arranged in descending order.
The leading coefficient is the numerical factor of the terms with the highest degree. The highest degree in the given function is 2.
Hence the given function f(x) = 7 − 1.6x2 − 5x has the  following characteristic
Standard form f (x) = -1.6x2 -5x +7
Degree: 2
Type of function: quadratic
Leading coefficient: -1.6

Question 2.
p(x) =x+ 2x-2 + 9.5
Answer:
Given,
p(x) =x+ 2x-2 + 9.5
The given function is not a polynomial function because the term 2x-2 has an exponent that is not a whole number.

Question 3.
q(x) = x3 − 6x + 3x4
Answer:
The standard form of polynomials has the exponents of the terms arranged in descending order.
The degree of a polynomial is the highest exponent of a term. The leading coefficient is the numerical factor of the terms with the highest degree. The highest degree of the function is 4.
The given function q(x) = x3 − 6x + 3x4  has the  following characteristic
Standard form : q(x) = 3x4 + x3 – 6x
Highest Degree: 4
Type of function: quadratic
Leading coefficient: 3

Evaluate the function for the given value of x.

Question 4.
f(x) = −x3 + 3x2 + 9; x = 4
Answer:
Given,
f(x) = −x3 + 3x2 + 9
substitute for x
f(4) =−x3 + 3x2 + 9
⇒-(4)3+ 3(4)2 +9
⇒-(64)+ 3(16) +9
⇒-64 + 48 + 9 = -7
So, f(x) = −x3 + 3x2 + 9 when x = 4 is -7.

Question 5.
f(x) = 3x5 − x4 − 6x + 10; x = −2
Answer:
Given,
f(x) = 3x5 − x4 − 6x + 10
Substitute x = -2 in the given equation
f(-2) = 3(-2)5-(-2)4-6(-2)+10
Simplify the expression
⇒ 3(-32)-16+12+10
⇒ -96 -16 +12 +10
⇒ -90
So, f(x) = 3x5 − x4 − 6x + 10 when x = -2 is -90.

Question 6.
Describe the end behavior of the graph of f(x) = 0.25x3 − x2 − 1.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.1-Answer-6.jpg

Graph the polynomial Function

Question 7.
f(x) = x4 + x2 − 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.1-Answer7.jpg

Question 8.
f(x) = 4 − x3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.1-Answer-8JPG.jpg

Question 9.
f(x) = x3 − x2 + x − 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.1-Answer-9.jpg

Question 10.
Sketch a graph of the polynomial function f having these characteristics.

  • f is decreasing when x < −1.5 and x > 2.5; f is increasing when −1.5 < x < 2.5.
  • f(x) > 0 when x < −3 and 1 < x < 4; f(x) < 0 when −3 < x < 1 and x > 4.
    Answer:
    https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.1-Answer-10JPG.jpg

Use the graph to describe the degree and leading coefficient of f.

Question 11.
WHAT IF? Repeat Example 6 using the alternative model for electric vehicles of
V(t) = −0.0290900t4 + 0.791260t3 − 7.96583t2 + 36.5561t − 12.025.
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.1-Answer-11.jpg

Graphing Polynomial Functions 4.1 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Explain what is meant by the end behavior of a polynomial function.
Answer: The end behavior of a polynomial function describes the behavior of the graph of the polynomial function as x value approach ∞ or -∞.

Question 2.
WHICH ONE DOESN’T BELONG? Which function does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 2
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-2JPG.jpg

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.

Question 3.
f(x) = −3x + 5x3 − 6x2 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 3

Question 4.
p(x) = \(\frac{1}{2}\)x2 + 3x − 4x3 + 6x4 − 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-4.jpg

Question 5.
f(x) = 9x4 + 8x3 − 6x-2 + 2x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 5

Question 6.
g(x) = \(\sqrt{3}\) − 12x + 13x2
Answer:

The polynomial is a quadratic function
Degree: 2
Type: quadratic
leading coefficient: 13

Question 7.
h(x) = \(\frac{5}{3}\)x2 − \(\sqrt{7}\)x4 + 8x3 − \(\frac{1}{2}\) + x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 7

Question 8.
h(x) = 3x4 + 2x − \(\frac{5}{x}\) + 9x3 − 7
Answer:
The polynomial is a quadratic function.
Standard form: 3x4 + 9x3 + 2x − \(\frac{5}{x}\) − 7
Degree: 4
Type: Quadratic
Coefficient: 4

ERROR ANALYSIS In Exercises 9 and 10, describe and correct the error in analyzing the function.

Question 9.
f(x) = 8x3 − 7x4 − 9x − 3x2 + 11
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 9

Question 10.
f(x) = 2x4 + 4x – 9\(\sqrt{x}\) + 3x2 – 8
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4
Answer:

In Exercises 11–16, evaluate the function for the given value of x.

Question 11.
h(x) = −3x4 + 2x3 − 12x − 6; x = −2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 11

Question 12.
f(x) = 7x4 − 10x2 + 14x − 26; x = −7
Answer:
Given,
f(x) = 7x4 − 10x2 + 14x − 26
Substitute x = -7 in the given function
f(-7) = 7(-7)4– 10(-7)2 + 14(-7) – 26
Simplify the expression
= 7(2401) – 10(49)- 98-26
= 16807- 490- 98- 26
= 16,193
So, f(x) = 7x4 − 10x2 + 14x − 26 when x = −7 is 16,193

Question 13.
g(x) = x6 − 64x4 + x2 − 7x − 51; x = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 13

Question 14.
g(x) = −x3 + 3x2 + 5x + 1; x = −12
Answer:
Given,
g(x) = −x3 + 3x2 + 5x + 1
put x = -12
g(-12) =(-12 )3+ 3(-12) 2+ 5(-12 ) + 1
= 1728 + 3(144) – 60 +1
= 1728 + 432 – 60 +1
= 2101
So, g(x) = −x3 + 3x2 + 5x + 1 when x = −12 is 2101.

Question 15.
p(x) = 2x3 + 4x2 + 6x + 7; x = \(\frac{1}{2}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 15

Question 16.
h(x) = 5x3 − 3x2 + 2x + 4; x = −12
Answer:
Given polynomial
h(x) = 5x3 − 3x2 + 2x + 4
put x = -12
h(-12) = 5(-12)3 – 3(-12)2 + 2(-12) + 4
= 5(-1728) – 3 (144) – 24 + 4
Simplify the expression
= -8640 – 432 – 24 + 4
= -9092
So, h(x) = 5x3 − 3x2 + 2x + 4 when x = −12 is -9092.

In Exercises 17–20, describe the end behavior of the graph of the function.

Question 17.
h(x) = −5x4 + 7x3 − 6x2 + 9x + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 17

Question 18.
g(x) = 7x7 + 12x5 − 6x3 − 2x − 18
Answer:
Given polynomial is g(x) = 7x7 + 12x5 − 6x3 − 2x − 18
The given function has a degree of 7 and a leading coefficient of 7. Because the  degree is odd and leading coefficient  is positive g(x)->-∞ as x ->-∞  and  g(x)->+∞ as x->+∞

Question 19.
f(x) = −2x4 + 12x8 + 17 + 15x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 19

Question 20.
f(x) = 11 − 18x2 − 5x5 − 12x4 − 2x
Answer:
Given polynomial is  f(x)= − 5x5 − 12x4− 18x2− 2x + 11
The highest exponent is 5, then the polynomial is of odd degree.
Hence the end behaviors are the opposite. With the negative leading coefficient
Then   f(x) ─> ∞ as x ─>-∞
f(x)─>-∞ as  x ─>+∞

In Exercises 21 and 22, describe the degree and leading coefficient of the polynomial function using the graph.

Question 21.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 21

Question 22.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 6
Answer:
The degree of the polynomial is even since one side goes up and the other goes up.
The leading coefficient is positive since the left side goes up and the right side goes up.

Question 23.
USING STRUCTURE Determine whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.
f(x) = 5x3x + \(\frac{5}{2}\)x3 − 9x4 + \(\sqrt{2}\)x2 + 4x − 1 − x-5x5 − 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 23

Question 24.
WRITING Let f(x) = 13. State the degree, type, and leading coefficient. Describe the end behavior of the function. Explain your reasoning.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-24.jpg
In Exercises 25–32, graph the polynomial function.

Question 25.
p(x) = 3 − x4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 25

Question 26.
g(x) = x3 + x + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-26.jpg

Question 27.
f(x) = 4x − 9 − x3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 27

Question 28.
p(x) = x5 − 3x3 + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-28.jpg

Question 29.
h(x) = x4 − 2x3 + 3x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 29

Question 30.
h(x) = 5 + 3x2 − x4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-30.jpg

Question 31.
g(x) = x5 − 3x4 + 2x − 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 31

Question 32.
p(x) = x6 − 2x5 − 2x3 + x + 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-32JPG.jpg

ANALYZING RELATIONSHIPS In Exercises 33–36, describe the x-values for which
(a) f is increasing or decreasing,
(b) f(x) > 0, and
(c) f(x) < 0.

Question 33.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 33

Question 34.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 8
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-34.jpg

Question 35.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 35

Question 36.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 10
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-36.jpg

In Exercises 37–40, sketch a graph of the polynomial function f having the given characteristics. Use the graph to describe the degree and leading coefficient of the function f.

Question 37.

  • f is increasing when x > 0.5; f is decreasing when x < 0.5.
  • f(x) > 0 when x < −2 and x > 3; f(x) < 0 when −2 < x < 3.

Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 37

Question 38.

  • f is increasing when −2 < x < 3;f is decreasing when x < −2 and x > 3.
  • f(x) > 0 when x < −4 and 1 < x < 5; f(x) < 0 when −4 < x < 1 and x > 5.

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-38.jpg
Question 39.

  • f is increasing when −2 < x < 0 and x > 2;f is decreasing when x < −2 and 0 < x < 2.
  • f(x) > 0 when x < −3, −1 < x < 1, and x > 3; f(x) < 0 when −3 < x < −1 and 1 < x < 3.

Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 39

Question 40.

  • f is increasing when x < −1 and x > 1;f is decreasing when −1 < x < 1.
  • f(x) > 0 when −1.5 < x < 0 and x > 1.5; f(x) < 0 when x < −1.5 and 0 < x < 1.5.

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-40.jpg
Question 41.
MODELING WITH MATHEMATICS From 1980 to 2007 the number of drive-in theaters in the United States can be modeled by the function
d(t) = −0.141t3 + 9.64t2 − 232.5t + 2421
where d(t) is the number of open theaters and t is the number of years after 1980.
a. Use a graphing calculator to graph the function for the interval 0 ≤ t ≤ 27. Describe the behavior of the graph on this interval.
b. What is the average rate of change in the number of drive-in movie theaters from 1980 to 1995 and from 1995 to 2007? Interpret the average rates of change.
c. Do you think this model can be used for years before 1980 or after 2007? Explain.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 41.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 41.2

Question 42.
PROBLEM SOLVING The weight of an ideal round-cut diamond can be modeled by
w = 0.00583d3 − 0.0125d2 + 0.022d − 0.01
where w is the weight of the diamond (in carats) and d is the diameter (in millimeters). According to the model, what is the weight of a diamond with a diameter of 12 millimeters?
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 12
Answer:
Given,
w = 0.00583d3 − 0.0125d2 + 0.022d − 0.01
put d = 12 in the function.

Question 43.
ABSTRACT REASONING Suppose f(x) → ∞ as x → −∞ and f(x) →−∞ as x → ∞. Describe the end behavior of g(x) = −f(x). Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 43

Question 44.
THOUGHT PROVOKING Write an even degree polynomial function such that the end behavior of f is given by f(x) → −∞ as x → −∞ and f(x) → −∞ as x → ∞. Justify your answer by drawing the graph of your function.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-44.jpg

Question 45.
USING TOOLS When using a graphing calculator to graph a polynomial function, explain how you know when the viewing window is appropriate.
Answer: The viewing window is appropriate when all of the high and low points are visible and the end behavior can be determined.

Question 46.
MAKING AN ARGUMENT Your friend uses the table to speculate that the function f is an even degree polynomial and the function g is an odd degree polynomial. Is your friend correct? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 13
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-46.jpg

Question 47.
DRAWING CONCLUSIONS The graph of a function is symmetric with respect to the y-axis if for each point (a, b) on the graph, (−a, b) is also a point on the graph. The graph of a function is symmetric with respect to the origin if for each point (a, b) on the graph, (−a, −b) is also a point on the graph.
a. Use a graphing calculator to graph the function y = xn when n = 1, 2, 3, 4, 5, and 6. In each case, identify the symmetry of the graph.
b. Predict what symmetry the graphs of y = x10 and y = x11 each have. Explain your reasoning and then confirm your predictions by graphing.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 47.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 47.2

Question 48.
HOW DO YOU SEE IT? The graph of a polynomial function is shown.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 14
a. Describe the degree and leading coefficient of f.
b. Describe the intervals where the function is increasing and decreasing.
c. What is the constant term of the polynomial function?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-48.jpg

Question 49.
REASONING A cubic polynomial function f has a leading coefficient of 2 and a constant term of −5. When f(1) = 0 and f(2) = 3, what is f(−5)? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 49

Question 50.
CRITICAL THINKING The weight y (in pounds) of a rainbow trout can be modeled by y = 0.000304x3, where x is the length (in inches) of the trout.
a. Write a function that relates the weight y and length x of a rainbow trout when y is measured in kilograms and x is measured in centimeters. Use the fact that 1 kilogram ≈ 2.20 pounds and 1 centimeter ≈ 0.394 inch.
b. Graph the original function and the function from part (a) in the same coordinate plane. What type of transformation can you apply to the graph of y = 0.000304x3 to produce the graph from part (a)?
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 15
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.1-Answer-50.jpg

Maintaining Mathematical Proficiency

Simplify the expression. (Skills Review Handbook)

Question 51.
xy + x2 + 2xy + y2 − 3x2
Answer:
Given,
xy + x2 + 2xy + y2 − 3x2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 51

Question 52.
2h3g + 3hg3 + 7h2g2 + 5h3g+ 2hg3
Answer:
Given,
2h3g + 3hg3 + 7h2g2 + 5h3g+ 2hg³
Use the factoring method to simplify the expression.
h³g(2 + 5) + hg³(3 + 2) + 7h²g²
7h³g + 5hg³ + 7h²g²
So, 2h3g + 3hg3 + 7h2g2 + 5h3g+ 2hg³ = 7h³g + 5hg³ + 7h²g²

Question 53.
−wk + 3kz − 2kw + 9zk − kw
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 53

Question 54.
a2(m − 7a3) − m(a2 − 10)
Answer:
Given,
a2(m − 7a3) − m(a2 − 10)
((a2) •(m-(7•(a3))))-m•(a2-10)
((a2) • (m – 7a3)) – m • (a2 – 10)
a2 • (m – 7a3) –  m • (a2 – 10)
= 10m – 7a5
So, a2(m − 7a3) − m(a2 − 10) = 10m – 7a5

Question 55.
3x(xy − 4) + 3(4xy + 3) − xy(x2y − 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.1 Question 55

Question 56.
cv(9 − 3c) + 2c(v − 4c) + 6c
Answer:

Lesson 4.2 Adding, Subtracting, and Multiplying Polynomials

Essential Question
How can you cube a binomial?

EXPLORATION 1
Cubing Binomials
Work with a partner.
Find each product. Show your steps.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 16
Answer:
a. (x + 1)³ = (x + 1)(x + 1)²
= (x + 1)(x² + 1 + 2x)
= x³ + x² + x + 1 + 2x² + 2x
= x³ + 3x² + 3x + 1

b. (a + b)³ = (a + b)(a + b)²
= (a + b)(a² + b² + 2ab)
= a³ + b³ + 2a²b + 2ab²

c. (x – 1)³ = (x – 1)(x – 1)²
= (x – 1)(x² + 1 – 2x)
= x³ + x² + x – 1 – 2x² + 2x
= x³ – x² + 3x – 1

d. (a – b)³ = (a – b)(a – b)²
= (a – b)(a² + b² – 2ab)
= a³ – b³ – 2a²b + 2ab²

EXPLORATION 2
Generalizing Patterns for Cubing a Binomial
Work with a partner.

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 17
a. Use the results of Exploration 1 to describe a pattern for the coefficients of the terms when you expand the cube of a binomial. How is your pattern related to Pascal’s Triangle, shown at the right?
b. Use the results of Exploration 1 to describe a pattern for the exponents of the terms in the expansion of a cube of a binomial.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 18
c. Explain how you can use the patterns you described in parts (a) and (b) to find the product (2x − 3)3. Then find this product.

Communicate Your Answer

Question 3.
How can you cube a binomial?
Answer:
First, write the cube of the binomial in the form of multiplication.
For example (p + q)³ = (p + q) (p + q) (p + q)
Next, multiply the first two binomials and remain the third one as same.
Then multiply the remaining binomial by the third term.

Question 4.
Find each product.
a. (x + 2)3
b. (x − 2)3
c. (2x − 3)3
d. (x − 3)3
e. (−2x + 3)3
f. (3x − 5)3

Answer:
a. (x + 2)3
This is in the form of (a + b)³ = a³ + b³ + 2a²b + 2ab²
(x + 2)³ = (x)³ + (2)³ + 3(x)²(2) + 3(x)(2)²
= x³ + 8 + 6x² + 12x
= x³ + 6x² + 12x + 8
b. (x – 2)³
This is in the form of (a – b)³ = a³ – b³ – 2a²b + 2ab²
(x – 2)³ = (x)³ – (2)³ – 3(x)²(2) + 3(x)(2)²
= x³ – 8 – 6x² + 12x
= x³ – 6x² + 12x – 8
c. (2x − 3)3
This is in the form of (a – b)³ = a³ – b³ – 2a²b + 2ab²
(2x − 3)3 = (2x)³ – (3)³ – 3(2x)²(3) + 3(2x)(3)²
= 8x³ – 27 – 36x² + 54x
= 8x³ – 36x² + 54x – 27
d. (x – 3)³
This is in the form of (a – b)³ = a³ – b³ – 2a²b + 2ab²
(x – 3)³ = (x)³ – (3)³ – 3(x)²(3) + 3(x)(3)²
= x³ – 27 – 9x² + 27x
= x³ – 9x² + 27x – 27
e. (−2x + 3)³
This is in the form of (a + b)³ = a³ + b³ + 2a²b + 2ab²
(−2x + 3)³ = (-2x)³ + (3)³ + 3(-2x)²(3) + 3(-2x)(3)²
= -8x³ + 27 – 36x² – 54x
= -8x³ – 36x² – 54x + 27
f. (3x − 5)³
This is in the form of (a – b)³ = a³ – b³ – 2a²b + 2ab²
(3x − 5)³ = (3x)³ – (5)³ – 3(3x)²(5) + 3(3x)(5)²
= 27x³ – 125 -135x² + 225x
= 27x³ -135x² + 225x – 125

4.2 Lesson

Monitoring Progress

Find the sum or difference.

Question 1.
(2x2 − 6x + 5) + (7x2 − x − 9)
Answer:
Given,
(2x2 − 6x + 5) + (7x2 − x − 9)
Removing the  grouping symbols and combining the like terms, the given expression is
(2x2 − 6x + 5) + (7x2 − x − 9) = 2x2 − 6x + 5 + 7x2 − x − 9
= (2x2 + 7x2) + ( − 6x − x)+( 5-9)
= 9x2 – 7x – 4
So, the sum of (2x2 − 6x + 5) + (7x2 − x − 9) = 9x2 – 7x – 4

Question 2.
(3t3 + 8t2 − t − 4) − (5t3 − t2 + 17)
Answer:
Given,
(3t3 + 8t2 − t − 4) − (5t3 − t2 + 17)
=3t3 + 8t2 -t – 4 – 5t3 + t2 -17
= -2t3+ 9t2 -t -21
Thus (3t3 + 8t2 − t − 4) − (5t3 − t2 + 17) = -2t3+ 9t2 -t -21

Find the product.

Question 3.
(4x2 + x − 5)(2x + 1)
Answer:
Given,
(4x2 + x − 5)(2x + 1)
Multiply both the polynomials
=4x2(2x+1) + x(2x+1) – 5(2x+1)
=8x3+ 4x2+ 2x2+ x- 10x – 5
= 8x3+ 6x2– 9x – 5
So, the product of (4x2 + x − 5)(2x + 1) = 8x3+ 6x2– 9x – 5

Question 4.
(y − 2)(5y2 + 3y − 1)
Answer:
(y − 2)(5y2 + 3y − 1)
Multiply both the polynomials
=(y – 2) 5y2+ (y – 2) 3y – 1(y – 2)
= 5y3-10y2+3y3-6y-y+2
= 5y3-7y2-7y+2
So, the product of (y − 2)(5y2 + 3y − 1) = 5y3-7y2-7y+2

Question 5.
(m − 2)(m − 1)(m + 3)
Answer:
Given,
(m − 2)(m − 1)(m + 3)
find the product of the polynomials
= (m2 – m – 2m + 2)(m + 3)
=(m2 – 3m + 2)(m + 3)
=(m2 – 3m + 2)m+(m2 – 3m + 2)3
=m3 – 3m2 + 2m + 3m2– 9m + 6
=m3 – 7m +6
So, the product of (m − 2)(m − 1)(m + 3) = m3 – 7m +6

Question 6.
(3t − 2)(3t + 2)
Answer:
(3t − 2)(3t + 2)
Use the formula: a2-b2= (a+ b) (a-b)
(3t − 2)(3t + 2) = (3t)2-(2)2
= 9t2-4
So, the product of (3t − 2)(3t + 2) = 9t2 – 4

Question 7.
(5a + 2)2
Answer:
(5a + 2)2
This is in the form of (a + b)² = (a² + b² + 2ab)
=(5a)2+2. 5a .2+22
=25a2+20a+4
So, the product of (5a + 2)2 = 25a2+20a+4

Question 8.
(x y − 3)3
Answer:
(xy-3)3
(a-b)3=a3-3a2 b +3a b 2-b3
= (x y)3-3(x y)2.3+3. xy.32-33
= x3y3– 9x 2y2+27xy-27
So, the product of (x y − 3)3 = x3y3– 9x 2y2+27xy-27

Question 9.
(a) Prove the polynomial identity for the cube of a binomial representing a difference: (a − b)3 = a3 − 3a2b + 3ab2 − b3.
(b) Use the cube of a binomial in part (a) to calculate 93.
Answer:
(a) . (a-b)3 =a3– 3a 2b+3ab2-b3
= (a-b)3 =(a-b)(a-b)(a-b)
= (a2-2ab+b2)(a-b)
= a3-a 2b-2a 2b+2ab2 +ab 2-b3
= a 3-3a 2b+3ab 2– b3

(b). 93=(10-1)3=103-3.102.1+3.10.12-13
= 1000-300+30-1
= 729

Question 10.
Use Pascal’s Triangle to expand
(a) (z + 3)4
Answer:
(z + 3)4 = (z)4 + (3)4 + 6z²(3)² + 4z(3)³
= z4 + 81 + 54z² + 12z³ + 108z

(b) (2t − 1)5
Answer:
1(2t)5 + 5(2t)4(-1) + 10(2t)³ (-1) + 10(2t)²(-1) + 5(2t)(-1)4 + 1(-1)
= (2t)5 – 5(2t)4 + 10(2t)³ – 10(2t)² + 5(2t) – 1
= 32t5 – 80t4 + 80t³ – 40t² + 10t – 1

Adding, Subtracting, and Multiplying Polynomials 4.2 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Describe three different methods to expand (x + 3)3.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 1

Question 2.
WRITING Is (a + b)(a − b) = a2 − b2 an identity? Explain your reasoning.
Answer:
(a + b) (a – b)
a(a – b) + b(a – b)
a² – ab + ab – b²
= a² – b²
This is called standard algebraic identity called a difference of squares.

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, find the sum.

Question 3.
(3x2 + 4x − 1) + (−2x2 − 3x + 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 3

Question 4.
(−5x2 + 4x − 2) + (−8x2 + 2x + 1)
Answer:

Question 5.
(12x5 − 3x4 + 2x − 5) + (8x4 − 3x3 + 4x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 5

Question 6.
(8x4 + 2x2 − 1) + (3x3 − 5x2 + 7x + 1)
Answer:

Question 7.
(7x6 + 2x5 − 3x2 + 9x) + (5x5 + 8x3 − 6x2 + 2x − 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 7

Question 8.
(9x4 − 3x3 + 4x2 + 5x + 7) + (11x4 − 4x2 − 11x − 9)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-8.jpg

In Exercises 9–14, find the difference.

Question 9.
(3x3 − 2x2 + 4x − 8) − (5x3 + 12x2 − 3x − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 9

Question 10.
(7x4 − 9x3 − 4x2 + 5x + 6) − (2x4 + 3x3 − x2 + x − 4)
Answer:

Question 11.
(5x6 − 2x4 + 9x3 + 2x − 4) − (7x5 − 8x4 + 2x− 11)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 11

Question 12.
(4x5 − 7x3 − 9x2 + 18) − (14x5 − 8x4 + 11x2 + x)
Answer:

Question 13.
(8x5 + 6x3 − 2x2 + 10x) − (9x5 − x3 − 13x2 + 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 13

Question 14.
(11x4 − 9x2 + 3x + 11) − (2x4 + 6x3 + 2x − 9)
Answer:

Question 15.
MODELING WITH MATHEMATICS During a recent period of time, the numbers (in thousands) of males Mand females F that attend degree-granting institutions in the United States can be modeled by
M = 19.7t2 + 310.5t + 7539.6
F = 28t2 + 368t + 10127.8
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 19
where t is time in years. Write a polynomial to model the total number of people attending degree-granting institutions. Interpret its constant term.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 15

Question 16.
MODELING WITH MATHEMATICS A farmer plants a garden that contains corn and pumpkins. The total area (in square feet) of the garden is modeled by the expression 2x2 + 5x + 4. The area of the corn is modeled by the expression x2 − 3x + 2. Write an expression that models the area of the pumpkins.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-16.jpg

In Exercises 17–24, find the product.

Question 17.
7x3(5x2 + 3x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 17

Question 18.
−4x5(11x3 + 2x2 + 9x + 1)
Answer:

Question 19.
(5x2 − 4x + 6)(−2x + 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 19

Question 20.
(−x − 3)(2x2 + 5x + 8)
Answer:

Question 21.
(x2 − 2x − 4)(x2 − 3x − 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 21

Question 22.
(3x2 + x − 2)(−4x2 − 2x − 1)
Answer:

Question 23.
(3x3 − 9x + 7)(x2 − 2x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 23

Question 24.
(4x2 − 8x − 2)(x4 + 3x2 + 4x)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-24.jpg

ERROR ANALYSIS In Exercises 25 and 26, describe and correct the error in performing the operation.

Question 25.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 20
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 25

Question 26.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 21
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Questionn-26.jpg

In Exercises 27–32, find the product of the binomials.

Question 27.
(x − 3)(x + 2)(x + 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 27

Question 28.
(x − 5)(x + 2)(x − 6)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-28.jpg

Question 29.
(x − 2)(3x + 1)(4x − 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 29

Question 30.
(2x + 5)(x − 2)(3x + 4)
Answer:

Question 31.
(3x − 4)(5 − 2x)(4x + 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 31

Question 32.
(4 − 5x)(1 − 2x)(3x + 2)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-32.jpg

Question 33.
REASONING Prove the polynomial identity (a + b)(a − b) = a2 − b2. Then give an example of two whole numbers greater than 10 that can be multiplied using mental math and the given identity. Justify your answers.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 33

Question 34.
NUMBER SENSE You have been asked to order textbooks for your class. You need to order 29 textbooks that cost $31 each. Explain how you can use the polynomial identity (a + b)(a − b) = a2 − b2 and mental math to find the total cost of the textbooks.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 22
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-34.jpg

In Exercises 35–42, find the product.

Question 35.
(x − 9)(x + 9)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 35

Question 36.
(m + 6)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-36.jpg

Question 37.
(3c − 5)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 37

Question 38.
(2y − 5)(2y + 5)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-38.jpg

Question 39.
(7h + 4)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 39

Question 40.
(9g − 4)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-40.jpg
Question 41.
(2k + 6)3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 41

Question 42.
(4n − 3)3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-42.jpg

In Exercises 43–48, use Pascal’s Triangle to expand the binomial.

Question 43.
(2t + 4)3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 43

Question 44.
(6m + 2)2
Answer:
Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-44

Question 45.
(2q − 3)4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 45

Question 46.
(g + 2)5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-46.jpg

Question 47.
(yz + 1)5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 47

Question 48.
(np − 1)4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-48.jpg

Question 49.
COMPARING METHODS Find the product of the expression (a2 + 4b2)2(3a2 − b2)2 using two different methods. Which method do you prefer? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 49

Question 50.
THOUGHT PROVOKING Adjoin one or more polygons to the rectangle to form a single new polygon whose perimeter is double that of the rectangle. Find the perimeter of the new polygon.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 23
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/CaptureBig-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-50.jpg

MATHEMATICAL CONNECTIONS In Exercises 51 and 52, write an expression for the volume of the figure as a polynomial in standard form.

Question 51.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 24
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 51

Question 52.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 25
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/CaptureBig-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-52.jpg

Question 53.
MODELING WITH MATHEMATICS Two people make three deposits into their bank accounts earning the same simple interest rate r.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 26
Person A’s account is worth 2000(1 + r)3 + 3000(1 + r)2 + 1000(1 + r) on January 1, 2015.
a. Write a polynomial for the value of Person B’s account on January 1, 2015.
b. Write the total value of the two accounts as a polynomial in standard form. Then interpret the coefficients of the polynomial.
c. Suppose their interest rate is 0.05. What is the total value of the two accounts on January 1, 2015?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 53.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 53.2

Question 54.
PROBLEM SOLVING
The sphere is centered in the cube. Find an expression for the volume of the cube outside the sphere.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 27
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-54.jpg

Question 55.
MAKING AN ARGUMENT Your friend claims the sum of two binomials is always a binomial and the product of two binomials is always a trinomial. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 55

Question 56.
HOW DO YOU SEE IT? You make a tin box by cutting x-inch-by-x-inch pieces of tin off the corners of a rectangle and folding up each side. The plan for your box is shown.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 28
a. What are the dimensions of the original piece of tin?
b. Write a function that represents the volume of the box. Without multiplying, determine its degree.
Answer:

USING TOOLS In Exercises 57–60, use a graphing calculator to make a conjecture about whether the two functions are equivalent. Explain your reasoning.

Question 57.
f(x) = (2x − 3)3; g(x) = 8x3 − 36x2 + 54x − 27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 57

Question 58.
f(x) = (x + 2)5;
g(x) = x5 + 10x4 + 40x3 + 80x2 + 64x
Answer:
Two Functions are not equal.
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Questionn-58.png

Question 59.
f(x) = (−x − 3)4;
g(x) = x4 + 12x3 + 54x2 + 108x + 80
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 59

Question 60.
f(x) = (−x + 5)3; g(x) = −x3 + 15x2 − 75x + 125
Answer:
They are Equal
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-60.png

Question 61.
REASONING Copy Pascal’s Triangle and add rows for n = 6, 7, 8, 9, and 10. Use the new rows to expand (x + 3)7 and (x − 5)9.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 61.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 61.2

Question 62.
ABSTRACT REASONING You are given the function f(x) = (x + a)(x + b)(x + c)(x + d). When f(x) is written in standard form, show that the coefficient of x3 is the sum of a, b, c, and d, and the constant term is the product of a, b, c, and d.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-62.jpg

Question 63.
DRAWING CONCLUSIONS Let g(x) = 12x4 + 8x + 9 and h(x) = 3x5 + 2x3 − 7x + 4.
a. What is the degree of the polynomial g(x) + h(x)?
b. What is the degree of the polynomial g(x) − h(x)?
c. What is the degree of the polynomial g(x) • h(x)?
d. In general, if g(x) and h(x) are polynomials such that g(x) has degree m and h(x) has degree n, and m > n, what are the degrees of g(x) + h(x), g(x) − h(x), and g(x) • h(x)?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 63.1

Question 64.
FINDING A PATTERN In this exercise, you will explore the sequence of square numbers. The first four square numbers are represented below.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 29
a. Find the differences between consecutive square numbers. Explain what you notice.
b. Show how the polynomial identity (n + 1)2 − n2 = 2n + 1 models the differences between square numbers.
c. Prove the polynomial identity in part (b).
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-64.jpg

Question 65.
CRITICAL THINKING Recall that a Pythagorean triple is a set of positive integers a, b, and c such that a2 + b2 = c2. The numbers 3, 4, and 5 form a Pythagorean triple because 32 + 42 = 52. You can use the polynomial identity (x2 − y2)2 + (2xy)2 = (x2 + y2)2 to generate other Pythagorean triples.
a. Prove the polynomial identity is true by showing that the simplified expressions for the left and right sides are the same. b. Use the identity to generate the Pythagorean triple when x = 6 and y = 5.
c. Verify that your answer in part (b) satisfies a2 + b2 = c2.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 65.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 65.2

Maintaining Mathematical Proficiency

Perform the operation. Write the answer in standard form. (Section 3.2)

Question 66.
(3 − 2i) + (5 + 9i)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-66.jpg

Question 67.
(12 + 3i) − (7 − 8i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 67

Question 68.
(7i)(−3i)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-68.jpg

Question 69.
(4 + i)(2 − i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.2 Question 69

Lesson 4.3 Dividing Polynomials

Essential Question
How can you use the factors of a cubic polynomial to solve a division problem involving the polynomial?

EXPLORATION 1
Dividing Polynomials
Work with a partner.
Match each division statement with the graph of the related cubic polynomial f(x). Explain your reasoning. Use a graphing calculator to verify your answers.
a. \(\frac{f(x)}{x}\) = (x − 1)(x + 2)
b. \(\frac{f(x)}{x-1}\) = (x − 1)(x + 2)
c. \(\frac{f(x)}{x+1}\) = (x − 1)(x + 2)
d. \(\frac{f(x)}{x-2}\) = (x− 1)(x+ 2)
e. \(\frac{f(x)}{x+2}\) = (x− 1)(x+ 2)
f. \(\frac{f(x)}{x-3}\) = (x− 1)(x+ 2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 29.1

EXPLORATION 2
Dividing Polynomials

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 30
Work with a partner. Use the results of Exploration 1 to find each quotient. Write your answers in standard form. Check your answers by multiplying.
a. (x3 + x2 − 2x) ÷ x
b. (x3 − 3x + 2) ÷ (x − 1)
c. (x3 + 2x2 − x − 2) ÷ (x + 1)
d. (x3 − x2 − 4x + 4) ÷ (x − 2)
e. (x3 + 3x2 − 4) ÷ (x + 2)
f. (x3 − 2x2 − 5x + 6) ÷ (x − 3)

Communicate Your Answer

Question 3.
How can you use the factors of a cubic polynomial to solve a division problem involving the polynomial?

4.3 Lesson

Monitoring Progress

Divide using polynomial long division.

Question 1.
(x3 − x2 − 2x + 8) ÷ (x − 1)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Question-1.jpg

Question 2.
(x4 + 2x2 − x + 5) ÷ (x2 − x + 1)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Question-2.jpg

Divide using synthetic division.

Question 3.
(x3 − 3x2 − 7x + 6) ÷ (x − 2)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.2-Question-3.jpg

Question 4.
(2x3 − x − 7) ÷ (x + 3)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Question-4.jpg

Use synthetic division to evaluate the function for the indicated value of x.

Question 5.
f(x) = 4x2 − 10x − 21; x = 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Question-5.jpg

Question 6.
f(x) = 5x4 + 2x3 − 20x − 6; x = 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Question-6.jpg

Dividing Polynomials 4.3 Exercises

Monitoring Progress and Modeling with Mathematics

Question 1.
WRITING Explain the Remainder Theorem in your own words. Use an example in your explanation.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 1

Question 2.
VOCABULARY What form must the divisor have to make synthetic division an appropriate method for dividing a polynomial? Provide examples to support your claim.
Answer: In order to use the synthetic division, the divisor must be of the first degree and must have the form ax.

Question 3.
VOCABULARY Write the polynomial divisor, dividend, and quotient functions represented by the synthetic division shown at the right.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 31
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 3

Question 4.
WRITING Explain what the colored numbers represent in the synthetic division in Exercise 3.
Answer:
Quotient: Red colors
Remainder: Blue colors

Vocabulary and Core Concept Check

In Exercises 5–10, divide using polynomial long division.

Question 5.
(x2 + x − 17 ) ÷ (x − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 5

Question 6.
(3x2 − 14x − 5) ÷ (x − 5)
Answer:

Question 7.
(x3 + x2 + x + 2 ) ÷ (x2 − 1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 7

Question 8.
(7x3 + x2 + x ) ÷ (x2 + 1)
Answer:

Question 9.
(5x4 − 2x3 − 7x2 − 39) ÷ (x2 + 2x − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 9

Question 10.
(4x4 + 5x − 4) ÷ (x2 − 3x − 2)
Answer:

In Exercises 11–18, divide using synthetic division.

Question 11.
(x2 + 8x + 1) ÷ (x − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 11

Question 12.
(4x2 − 13x − 5) ÷ (x − 2)
Answer:
Given,
(4x2 − 13x − 5) ÷ (x − 2)
use synthetic division. Because the divisor is x-2, k=2

Question 13.
(2x2 − x + 7) ÷ (x + 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 13

Question 14.
(x3 − 4x + 6) ÷ (x + 3)
Answer:

Question 15.
(x2 + 9) ÷ (x − 3)
Answer:
Given,
(x2 + 9) ÷ (x − 3)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 15

Question 16.
(3x3 − 5x2 − 2) ÷ (x − 1)
Answer:
Given,
(3x3 − 5x2 − 2) ÷ (x − 1)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-16.jpg

Question 17.
(x4 − 5x3 − 8x2 + 13x − 12) ÷ (x − 6)
Answer:
Given,
(x4 − 5x3 − 8x2 + 13x − 12) ÷ (x − 6)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 17

Question 18.
(x4 + 4x3 + 16x − 35) ÷ (x + 5 )
Answer:
Given,
(x4 + 4x3 + 16x − 35) ÷ (x + 5 )
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-18.jpg

ANALYZING RELATIONSHIPS In Exercises 19–22, match the equivalent expressions. Justify your answers.

Question 19.
(x2 + x − 3) ÷ (x − 2)
Answer:
Given,
(x2 + x − 3) ÷ (x − 2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 19

Question 20.
(x2 − x − 3) ÷ (x − 2)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-20.jpg

Question 21.
(x2 − x + 3) ÷ (x − 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 21

Question 22.
(x2 + x + 3) ÷ (x − 2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 32
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-22.jpg

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in using synthetic division to divide x3 – 5x + 3 by x – 2.

Question 23.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 33
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 23

Question 24.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 34
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-24.jpg

In Exercises 25–32, use synthetic division to evaluate the function for the indicated value of x.

Question 25.
f(x) = −x2 − 8x + 30; x = −1
Answer:
Given,
f(x) = −x2 − 8x + 30
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 25

Question 26.
f(x) = 3x2 + 2x − 20; x = 3
Answer:
Given,
f(x) = 3x2 + 2x − 20
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-26.jpg

Question 27.
f(x) = x3 − 2x2 + 4x + 3; x = 2
Answer:
Given,
f(x) = x3 − 2x2 + 4x + 3
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 27

Question 28.
f(x) = x3 + x2 − 3x + 9; x = −4
Answer:
Given,
f(x) = x3 + x2 − 3x + 9
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-28.jpg

Question 29.
f(x) = x3 − 6x + 1; x = 6
Answer:
Given,
f(x) = x3 − 6x + 1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 29

Question 30.
f(x) = x3 − 9x − 7; x = 10
Answer:
Given,
f(x) = x3 − 9x − 7
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-30.jpg

Question 31.
f(x) = x4 + 6x2 − 7x + 1; x = 3
Answer:
Given,
f(x) = x4 + 6x2 − 7x + 1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 31

Question 32.
f(x) = −x4 − x3 − 2; x = 5
Answer:
Given,
f(x) = −x4 − x3 − 2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-32.jpg

Question 33.
MAKING AN ARGUMENT You use synthetic division to divide f(x) by (x − a) and find that the remainder equals 15. Your friend concludes that f(15) = a. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 33

Question 34.
THOUGHT PROVOKING A polygon has an area represented by A = 4x2 + 8x + 4. The figure has at least one dimension equal to 2x + 2. Draw the figure and label its dimensions.
Answer:
A = 4x2 + 8x + 4
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-34.jpg

Question 35.
USING TOOLS The total attendance A (in thousands) at NCAA women’s basketball games and the number T of NCAA women’s basketball teams over a period of time can be modeled by
A = −1.95x3 + 70.1x2 − 188x + 2150
T = 14.8x + 725
where x is in years and 0 < x < 18. Write a function for the average attendance per team over this period of time.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 35
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 35

Question 36.
COMPARING METHODS The profit P (in millions of dollars) for a DVD manufacturer can be modeled by P = −6x3 + 72x, where x is the number (in millions) of DVDs produced. Use synthetic division to show that the company yields a profit of $96 million when 2 million DVDs are produced. Is there an easier method? Explain.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-36.jpg

Question 37.
CRITICAL THINKING What is the value of k such that (x3 − x2 + kx − 30) ÷ (x − 5) has a remainder of zero?
A. −14
B. −2
C. 26
D. 32
Answer:
Given,
(x3 − x2 + kx − 30) ÷ (x − 5)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 37

Question 38.
HOW DO YOU SEE IT? The graph represents the polynomial function f(x) = x3 + 3x2 − x − 3.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 36
a. The expression f(x) ÷ (x − k) has a remainder of −15. What is the value of k?
b. Use the graph to compare the remainders of (x3 + 3x2 − x − 3) ÷ (x + 3) and (x3 + 3x2 − x − 3) ÷ (x + 1).
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-38.jpg

Question 39.
MATHEMATICAL CONNECTIONS The volume V of the rectangular prism is given by V = 2x3 + 17x2 + 46x + 40. Find an expression for the missing dimension.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 37
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 39

Question 40.
USING STRUCTURE You divide two polynomials and obtain the result 5x2 − 13x + 47 − \(\frac{102}{x+2}\). What is the dividend? How did you find it?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.3-Questionn-40.jpg

Maintaining Mathematical Proficiency

Find the zero(s) of the function. (Sections 3.1and 3.2)

Question 41.
f(x) = x2 − 6x + 9
Answer:
Given,
f(x) = x2 − 6x + 9
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 41

Question 42.
g(x) = 3(x + 6)(x − 2)
Answer:
Given,
g(x) = 3(x + 6)(x − 2)
g(x) = 0
3(x + 6)(x − 2) = 0
(3x + 18)(x – 2) = 0
3x + 18 = 0 or x – 2 = 0
3x = -18 or x = 2
x = -18/3
x = -6
So, the solution x = -6.

Question 43.
g(x) = x2 + 14x + 49
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.3 Question 43

Question 44.
h(x) = 4x2 + 36
Answer:
Given,
h(x) = 4x2 + 36
4x2 + 36 = 0
x² + 9 = 0
x² = -9
x = 3i
The solution is x = 3i

Lesson 4.4 Factoring Polynomials

Essential Question
How can you factor a polynomial?

EXPLORATION 1
Factoring Polynomials
Work with a partner.
Match each polynomial equation with the graph of its related polynomial function. Use the x-intercepts of the graph to write each polynomial in factored form. Explain your reasoning.
a. x2 + 5x + 4 = 0
b. x3 − 2x2 − x + 2 = 0
c. x3 + x2 − 2x = 0
d. x3 − x = 0
e. x4 − 5x2 + 4 = 0
f. x4 − 2x3 − x2 + 2x = 0
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 38

EXPLORATION 2
Factoring Polynomials

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 39
Work with a partner. Use the x-intercepts of the graph of the polynomial function to write each polynomial in factored form. Explain your reasoning. Check your answers by multiplying.
a. f(x) = x2 − x − 2
b. f(x) = x3 − x2 − 2x
c. f(x) = x3 − 2x2 − 3x
d. f(x) = x3 − 3x2 − x + 3
e. f(x) = x4 + 2x3 − x2 − 2x
f. f(x) = x4 − 10x2 + 9

Communicate Your Answer

Question 3.
How can you factor a polynomial?
Answer:
There are four ways to find the factors of a polynomial.

  1. Find the greatest common factor.
  2. Grouping the method.
  3. Adding the difference in two squares.
  4. Find the sum or difference in cubes.

Question 4.
What information can you obtain about the graph of a polynomial function written in factored form?

4.4 Lesson

Monitoring Progress

Question 1.
x3 − 7x2 + 10x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-1.jpg

Question 2.
3n7 − 75n5

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-2.jpg

Question 3.
8m5 − 16m4 + 8m3

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-3.jpg

Factor the polynomial completely

Question 4.
a3 + 27
Answer:
Given,
a3 + 27
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-4.jpg

Question 5.
6z5 − 750z2

Answer:
Given,
6z5 − 750z2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-5.jpg
Question 6.
x3 + 4x2 − x − 4
Answer:
Given,
x3 + 4x2 − x − 4
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-6.jpg

Question 7.
3y3 + y2 + 9y + 3
Answer:
Given,
3y3 + y2 + 9y + 3
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-7.jpg

Question 8.
−16n4 + 625
Answer:
Given,
−16n4 + 625
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-8.jpg

Question 9.
5w6 − 25w4 + 30w2

Answer:
Given,
5w6 − 25w4 + 30w2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-9.jpg

Question 10.
Determine whether x− 4 is a factor of f(x) = 2x2 + 5x − 12.
Answer:
Given,
f(x) = 2x2 + 5x − 12
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-10.jpg

Question 11.
Show that x − 6 is a factor of f(x) = x3 − 5x2 − 6x. Then factor f(x) completely.
Answer:
Given,
f(x) = x3 − 5x2 − 6x
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-11.jpg

Question 12.
In Example 7, does your answer change when you first determine whether 2 is a zero and then whether −1 is a zero? Justify your answer.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questionn-12.jpg

Factoring Polynomials 4.4 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE The expression 9x4 − 49 is in _________ form because it can be written as u2 − 49 where u = _____.
Answer:
The expression 9x4 − 49 is in Quadratic form because it can be written as u2 − 49 where u = 3x².

Question 2.
VOCABULARY Explain when you should try factoring a polynomial by grouping.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-2.jpg

Question 3.
WRITING How do you know when a polynomial is factored completely?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 3

Question 4.
WRITING Explain the Factor Theorem and why it is useful.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-4.jpg

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, factor the polynomial completely.

Question 5.
x3 − 2x2 − 24x
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 5

Question 6.
4k5 − 100k3
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-6.jpg

Question 7.
3p5 − 192p3
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 7

Question 8.
2m6 − 24m5 + 64m4
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-8.jpg

Question 9.
2q4 + 9q3 − 18q2
Answer:
Given,
2q4 + 9q3 − 18q2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 9

Question 10.
3r6 − 11r5 − 20r4
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-10.jpg
Question 11.
10w10 − 19w9 + 6w8
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 11

Question 12.
18v9 + 33v8 + 14v7
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-12.jpg

In Exercises 13–20, factor the polynomial completely.

Question 13.
x3 + 64
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 13

Question 14.
y3 + 512
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-14.jpg

Question 15.
g3 − 343
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 15

Question 16.
c3 − 27
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-16.jpg

Question 17.
3h9 − 192h6
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 17

Question 18.
9n6 − 6561n3
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-18.jpg

Question 19.
16t7 + 250t4
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 19

Question 20.
135z11 − 1080z8
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-20.jpg

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in factoring the polynomial.

Question 21.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 40
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 21

Question 22.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 41
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-22.jpg

In Exercises 23–30, factor the polynomial completely.

Question 23.
y3 − 5y2 + 6y − 30
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 23

Question 24.
m3 − m2 + 7m − 72
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-24.jpg

Question 25.
3a3 + 18a2 + 8a + 48
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 25

Question 26.
2k3 − 20k2 + 5k − 50
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-26.jpg

Question 27.
x3 − 8x2 − 4x + 32
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 27

Question 28.
z3 − 5z2 − 9z + 45
Answer:
Given,
z3 − 5z2 − 9z + 45
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-28.jpg

Question 29.
4q3 − 16q7 − 9q + 36
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 29

Question 30.
16n3 + 32n7 − n − 2
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-30.jpg

In Exercises 31–38, factor the polynomial completely.

Question 31.
49k4 − 9
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 31

Question 32.
4m4 − 25
Answer:
Given,
4m4 − 25
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-32.jpg

Question 33.
c4 + 9c2 + 20
Answer:
Given,
c4 + 9c2 + 20
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 33

Question 34.
y4 − 3y2 − 28
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-34.jpg

Question 35.
16z4 − 81
Answer:
Given,
16z4 − 81
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 35

Question 36.
81a4 − 256
Answer:
Given,
81a4 − 256
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-36.jpg

Question 37.
3r8 + 3r5 − 60r2
Answer:
Given,
3r8 + 3r5 − 60r2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 37

Question 38.
4n12 − 32n7 + 48n2
Answer:
Given,
4n12 − 32n7 + 48n2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-38.jpg

In Exercises 39–44, determine whether the binomial is a factor of the polynomial.

Question 39.
f(x) = 2x3 + 5x2 − 37x − 60; x − 4
Answer:
Given,
f(x) = 2x3 + 5x2 − 37x − 60
x – 4 = 0
x = 4
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 39

Question 40.
g(x) = 3x3 − 28x2 + 29x + 140; x + 7
Answer:
Given,
g(x) = 3x3 − 28x2 + 29x + 140
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-40.jpg

Question 41.
h(x) = 6x5 − 15x4 − 9x3; x + 3
Answer:
Given,
h(x) = 6x5 − 15x4 − 9x3
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 41

Question 42.
g(x) = 8x5 − 58x4 + 60x3 + 140; x − 6
Answer:
Given,
g(x) = 8x5 − 58x4 + 60x3 + 140
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-42.jpg

Question 43.
h(x) = 6x4 − 6x3 − 84x2 + 144x; x + 4
Answer:
Given,
h(x) = 6x4 − 6x3 − 84x2 + 144x
x + 4 = 0
x = -4
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 43

Question 44.
t(x) = 48x4 + 36x3 − 138x2 − 36x; x + 2
Answer:
Given,
t(x) = 48x4 + 36x3 − 138x2 − 36x
x + 2 = 0
x = -2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-44.jpg

In Exercises 45–50, show that the binomial is a factor of the polynomial. Then factor the polynomial completely.

Question 45.
g(x) = x3 − x2 − 20x; x + 4
Answer:
Given,
g(x) = x3 − x2 − 20x
x + 4 = 0
x = -4
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 45

Question 46.
t(x) = x3 − 5x2 − 9x + 45; x − 5
Answer:
Given,
t(x) = x3 − 5x2 − 9x + 45
x – 5 = 0
x = 5
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-46.jpg

Question 47.
f(x) = x4 − 6x3 − 8x + 48; x − 6
Answer:
Given,
f(x) = x4 − 6x3 − 8x + 48
x – 6 = 0
x = 6
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 47

Question 48.
s(x) = x4 + 4x3 − 64x − 256; x + 4
Answer:
Given,
s(x) = x4 + 4x3 − 64x − 256
x + 4 = 0
x = -4
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-48.jpg

Question 49.
r(x) = x3 − 37x + 84; x + 7
Answer:
Given,
r(x) = x3 − 37x + 84
x + 7 = 0
x = -7
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 49

Question 50.
h(x) = x3 − x2 − 24x − 36; x + 2
Answer:
Given,
h(x) = x3 − x2 − 24x − 36
x + 2 = 0
x = -2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-50.jpg

ANALYZING RELATIONSHIPS In Exercises 51–54, match the function with the correct graph. Explain your reasoning.

Question 51.
f(x) = (x − 2)(x − 3)(x + 1)
Answer:
Given,
f(x) = (x − 2)(x − 3)(x + 1)
x – 2 = 0
x = 1
x – 3 = 0
x = 3
x + 1 = 0
x = -1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 51

Question 52.
g(x) = x(x + 2)(x + 1)(x − 2)
Answer:
Given,
g(x) = x(x + 2)(x + 1)(x − 2)
x = 0
x + 2 = 0
x = -2
x + 1 = 0
x = -1
x – 2 = 0
x = 2
To easily determine the zeroes of the graph, set each factor equal to zero.
Zeroes: 0, -2, -1,2
Based on the zeroes, match them with the graph that has the same number of zeroes at the same location.

Question 53.
h(x) = (x + 2)(x + 3)(x − 1)
Answer:
Given,
h(x) = (x + 2)(x + 3)(x − 1)
x + 2 = 0
x = -2
x + 3 = 0
x = -3
x – 1 = 0
x = 1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 53

Question 54.
k(x) = x(x − 2)(x − 1)(x + 2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 42
Answer:
Given,
k(x) = x(x − 2)(x − 1)(x + 2)
x = 0
x – 2 = 0
x = 2
x – 1 = 0
x = 1
x + 2 = 0
x = -2
To easily determine the zeroes of the graph, set each factor equal to zero.
Zeroes: 0, 2, 1, -2
Based on the zeroes, match them with the graph that has the same number of zeroes at the same location.

Question 55.
MODELING WITH MATHEMATICS The volume (in cubic inches) of a shipping box is modeled by V = 2x3 − 19x2 + 39x, where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 43
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 55

Question 56.
MODELING WITH MATHEMATICS The volume (in cubic inches) of a rectangular birdcage can be modeled by V = 3x3 − 17x2 + 29x − 15, where x is the length (in inches). Determine the values of x for which the model makes sense. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 44
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-56.jpg

USING STRUCTURE In Exercises 57–64, use the method of your choice to factor the polynomial completely. Explain your reasoning.

Question 57.
a6 + a5 − 30a4
Answer:
Given,
a6 + a5 − 30a4
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 57

Question 58.
8m3 − 343
Answer:
Given,
8m3 − 343
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-58.jpg

Question 59.
z3 − 7z2 − 9z + 63
Answer:
Given,
z3 − 7z2 − 9z + 63
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 59

Question 60.
2p8 − 12p5 + 16p2
Answer:
Given,
2p8 − 12p5 + 16p2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-60.jpg

Question 61.
64r3 + 729
Answer:
Given,
64r3 + 729
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 61

Question 62.
5x5 − 10x4 − 40x3
Answer:
Given,
5x5 − 10x4 − 40x3
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-62.jpg

Question 63.
16n4 − 1
Answer:
Given,
16n4 − 1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 63

Question 64.
9k3 − 24k2 + 3k − 8
Answer:
Given,
9k3 − 24k2 + 3k − 8
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-64.jpg

Question 65.
REASONING Determine whether each polynomial is factored completely. If not, factor completely.
a. 7z4(2z2 − z − 6)
b. (2 − n)(n2 + 6n)(3n − 11)
c. 3(4y − 5)(9y2 − 6y − 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 65

Question 66.
PROBLEM SOLVING The profit P (in millions of dollars) for a T-shirt manufacturer can be modeled by P = −x3 + 4x2 + x, where x is the number (in millions) of T-shirts produced. Currently the company produces 4 million T-shirts and makes a profit of $4 million. What lesser number of T-shirts could the company produce and still make the same profit?
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 44.1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-66.jpg

Question 67.
PROBLEM SOLVING The profit P (in millions of dollars) for a shoe manufacturer can be modeled by P = −21x3 + 46x, where x is the number (in millions) of shoes produced. The company now produces 1 million shoes and makes a profit of $25 million, but it would like to cut back production. What lesser number of shoes could the company produce and still make the same profit?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 67

Question 68.
THOUGHT PROVOKING Find a value of k such that \(\frac{f(x)}{x-k}\) has a remainder of 0. Justify your answer.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-68.jpg

Question 69.
COMPARING METHODS You are taking a test where calculators are not permitted. One question asks you to evaluate g(7) for the function g(x) = x3 − 7x2 − 4x + 28. You use the Factor Theorem and synthetic division and your friend uses direct substitution. Whose method do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 69

Question 70.
MAKING AN ARGUMENT You divide f(x) by (x−a) and find that the remainder does not equal 0. Your friend concludes that f(x) cannot be factored. Is your friend correct? Explain your reasoning.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-70.jpg

Question 71.
CRITICAL THINKING What is the value of k such that x− 7 is a factor of h(x) = 2x3 − 13x2 − kx + 105? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 71

Question 72.
HOW DO YOU SEE IT? Use the graph to write an equation of the cubic function in factored form. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 45
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-72.jpg

Question 73.
ABSTRACT REASONING Factor each polynomial completely.
a. 7ac2 + bc2 − 7ad2 − bd2
b. x2n − 2xn + 1
c. a5b2 − a2b4 + 2a4b − 2ab3 + a3 − b2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 73

Question 74.
REASONING The graph of the function f(x) = x4 + 3x3 + 2x2 + x + 3 is shown. Can you use the Factor Theorem to factor f(x)? Explain.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-74.jpg

Question 75.
MATHEMATICAL CONNECTIONS The standard equation of a circle with radius r and center (h, k) is (x − h)2 + (y − k)2 = r2. Rewrite each equation of a circle in standard form. Identify the center and radius of the circle. Then graph the circle.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 46
a. x2 + 6x + 9 + y2 = 25
b. x2 − 4x + 4 + y2 = 9
c. x2 − 8x + 16 + y2 + 2y + 1 = 36
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 75.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 75.2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 75.3

Question 76.
CRITICAL THINKING Use the diagram to complete parts (a)–(c).
a. Explain why a3 − b3 is equal to the sum of the volumes of the solids I, II, and III.
b. Write an algebraic expression for the volume of each of the three solids. Leave your expressions in factored form.
c. Use the results from part (a) and part (b) to derive the factoring pattern a3 − b3.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 47
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-76.jpg

Maintaining Mathematical Proficiency

Solve the quadratic equation by factoring.(Section 3.1)

Question 77.
x2 − x − 30 = 0
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 77
The solution is x = 6 and x = -5

Question 78.
2x2 − 10x − 72 = 0
Answer:
Given,
2x2 − 10x − 72 = 0
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-78.jpg

Question 79.
3x2 − 11x + 10 = 0
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 79

Question 80.
9x2 − 28x + 3 = 0
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-80.jpg

Solve the quadratic equation by completing the square.(Section 3.3)

Question 81.
x2 − 12x + 36 = 144
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 81

Question 82.
x2 − 8x − 11 = 0
Answer:
Given,
x2 − 8x − 11 = 0
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-82.jpg

Question 83.
3x2 + 30x + 63 = 0
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.4 Question 83
The solution is x = -7 and x = -3.

Question 84.
4x2 + 36x − 4 = 0
Answer:
Given,
4x2 + 36x − 4 = 0
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.4-Questioon-84.jpg

Polynomial Functions Study Skills : Keeping Your Mind Focused

4.1–4.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 48

Core Concepts
Section 4.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 49

Section 4.2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 50

Section 4.3
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 51

Section 4.4
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 52

Mathematical Practices

Question 1.
Describe the entry points you used to analyze the function in Exercise 43 on page 164.

Question 2.
Describe how you maintained oversight in the process of factoring the polynomial in Exercise 49 on page 185.

Study Skills
Keeping Your Mind Focused

  • When you sit down at your desk, review your notes from the last class.
  • Repeat in your mind what you are writing in your notes.
  • When a mathematical concept is particularly difficult, ask your teacher for another example.

Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 52.1

Polynomial Functions 4.1 – 4.4 Quiz

Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.(Section 4.1)

Question 1.
f(x) = 5 + 2x2 − 3x4 − 2x − x3

Answer:
Given,
f(x) = 5 + 2x2 − 3x4 − 2x − x3
The standard form of a polynomial has the exponents of the terms arranged in descending order.
Standard form f (x) = − 3x4− x3 + 2x2− 2x+ 5
The degree of a polynomial is the highest exponent of a term.
Degree of the polynomial: 4
Type of function: quadratic
The leading coefficient is the numerical factor of the terms with the highest degree.
Leading coefficient: -3

Question 2.
g(x) = \(\frac{1}{4}\)x3 + 2x − 3x2 + 1
Answer:
The type of a polynomial depends on the degree.
Given,
g(x) = \(\frac{1}{4}\)x3 + 2x − 3x2 + 1
The standard form of a polynomial has the exponents of the terms arranged in descending order.
Standard form: g(x) = \(\frac{1}{4}\)x3 − 3x2 + 2x  + 1
The degree of a polynomial is the highest exponent of a term.
Degree of the polynomial: 3
Type of function: a quadratic function
The leading coefficient is the numerical factor of the terms with the highest degree.
Leading coefficient: \(\frac{1}{4}\)

Question 3.
h(x) = 3 − 6x3 + 4x − 2 + 6x
Answer:
Given,
h(x) = 3 − 6x3 + 4x − 2 + 6x
The standard form of a polynomial has the exponents of the terms arranged in descending order.
Standard form:
h(x) = 3 − 6x3 + 4x − 2 + 6x
h(x) = − 6x3 + 10x + 1
The degree of a polynomial is the highest exponent of a term.
Degree of the polynomial: 3
The type of a polynomial depends on the degree.
Type of polynomial: quadratic
The leading coefficient is the numerical factor of the terms with the highest degree.
Leading coefficient: -6

Question 4.
Describe the x-values for which
(a) f is increasing or decreasing,
(b) f(x) > 0, and
(c) f(x) < 0. (Section 4.1)
Answer:

 

Question 5.
Write an expression for the area and perimeter for the figure shown. (Section 4.2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 54
Answer:
Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-54
Figure 1:
L = (x + 3)
W = (x + 1)
We know that,
Area of the rectangle = l × w
Area of the rectangle = (x + 3)(x + 1)
= x(x + 1) + 3(x + 1)
x² + x + 3x + 3
x² + 4x + 3
A = x² + 4x + 3
Perimeter = 2((x + 3) + (x + 1))
2(x + 3 + x + 1)
= 2(2x + 4)
= 4x + 8
Figure 2:
L = x
W = (2x + 3)
We know that,
Area of the rectangle = l × w
Area of the rectangle = x(2x + 3)
2x² + 3x
A = 2x² + 3x
P = 2(L + W)
Perimeter = 2(x + 2x + 3)
= 2(3x + 3)
= 6x + 6

Question 6.
(7x2 − 4) − (3x2 − 5x + 1)
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-6.jpg

Question 7.
(x2 − 3x + 2)(3x − 1)
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-7.jpg

Question 8.
(x − 1)(x + 3)(x − 4)
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-8.jpg

Question 9.
Use Pascal’s Triangle to expand (x + 2)5. (Section 4.2)
Answer:
Given,
(x + 2)5.
We need to expand the expression.
This is in the form of (a + b)5 = a5b0 + 5a4b + 10a3b2 + 10a2b3 + 5ab4 + a0b5
(x + 2)5 = x5(20) + 5x4(2) + 10x3(2)2 + 10x2(2)3 + 5x(2)4 + x0(2)5
= x5 + 10x4 + 40x3 + 80x2 + 80x + 32
So, (x + 2)5 = x5 + 10x4 + 40x3 + 80x2 + 80x + 32

Question 10.
Divide 4x4 − 2x3 + x2 − 5x + 8 by x2 − 2x − 1. (Section 4.3)
Answer:

Factor the polynomial completely.(Section 4.4)

Question 11.
a3 − 2a2 − 8a
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-11.jpg

Question 12.
8m3 + 27
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-12.jpg

Question 13.
z3 + z2 − 4z − 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-13.jpg

Question 14.
49b4 − 64
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-14.jpg

Question 15.
Show that x + 5 is a factor of f(x) = x3 − 2x2 − 23x + 60. Then factor f(x) completely. (Section 4.4)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-15.jpg

Question 16.
The estimated price P (in cents) of stamps in the United States can be modeled by the polynomial function P(t)= 0.007t3 − 0.16t2 + 1t + 17, where t represents the number of years since 1990. (Section 4.1)
a. Use a graphing calculator to graph the function for the interval 0 ≤ t ≤ 20. Describe the behavior of the graph on this interval.
b. What was the average rate of change in the price of stamps from 1990 to 2010?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-quiz-Exercise-Answer-16.jpg

Question 17.
The volume V (in cubic feet) of a rectangular wooden crate is modeled by the function V(x) = 2x3 − 11x2 + 12x, where x is the width (in feet) of the crate. Determine the values of x for which the model makes sense. Explain your reasoning. (Section 4.4)
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 55
Answer:

Lesson 4.5 Solving Polynomial Equations

Essential Question
How can you determine whether a polynomial equation has a repeated solution?

EXPLORATION 1
Cubic Equations and Repeated Solutions
Work with a partner.
Some cubic equations have three distinct solutions. Others have repeated solutions. Match each cubic polynomial equation with the graph of its related polynomial function. Then solve each equation. For those equations that have repeated solutions, describe the behavior of the related function near the repeated zero using the graph or a table of values.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 56
a. x3 − 6x2 + 12x − 8 = 0
b. x3 + 3x2 + 3x + 1 = 0
c. x3− 3x + 2 = 0
d. x3 + x2 − 2x = 0
e. x3 − 3x − 2 = 0
f. x3 − 3x2 + 2x = 0
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 57

EXPLORATION 2
Quartic Equations and Repeated Solutions
Work with a partner.
Determine whether each quartic equation has repeated solutions using the graph of the related quartic function or a table of values. Explain your reasoning. Then solve each equation.
a. x4 − 4x3 + 5x2 − 2x = 0
b. x4 − 2x3 − x2 + 2x = 0
c. x4 − 4x3 + 4x2 = 0
d. x4 + 3x3 = 0

Communicate Your Answer

Question 3.
How can you determine whether a polynomial equation has a repeated solution?
Answer:
A solution of a polynomial is a repeated solution if and only if when the polynomial is written in the factored form then it appears more than once.

Question 4.
Write a cubic or quartic polynomial equation that is different from the equations in Explorations 1 and 2 and has a repeated solution.
Answer:

4.5 Lesson

Monitoring Progress

Solve the equation.

Question 1.
4x4 − 40x2 + 36 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.5-Answer-1.jpg
Question 2.
2x5 + 24x = 14x3

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.5-Answer-2.jpg

Find the zeros of the function. Then sketch a graph of the function.

Question 3.
f(x) = 3x4 − 6x2 + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.5-Answer-3.jpg

Question 4.
f(x) = x3 + x2 − 6x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.5-Answer-4.jpg

Question 5.
Find all real solutions of x3 − 5x2 − 2x + 24 = 0.
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.5-Answer-5.jpg
Question 6.
Find all real zeros of f(x) = 3x4 − 2x3 − 37x2 + 24x + 12.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.5-Answer-6.jpg

Question 7.
Write a polynomial function f of least degree that has rational coefficients, aleading coefficient of 1, and the zeros 4 and 1 − \(\sqrt{5}\).
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-Monitoring-progress-exercise-4.5-Answer-7.jpg

Solving Polynomial Equations 4.5 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE If a polynomial function f has integer coefficients, then every rational solution of f(x) = 0 has the form \(\frac{p}{q}\), where p is a factor of the _____________ and q is a factor of the _____________.
Answer:
If a polynomial function f has integer coefficients, then every rational solution of f(x) = 0 has the form \(\frac{p}{q}\), where p is a factor of the constant term and q is a factor of the leading coefficient.

Question 2.
DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 58
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation.

Question 3.
z3 − z2 − 12z = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 3

Question 4.
a3 − 4a2 + 4a = 0
Answer:

Question 5.
2x4 − 4x3 = −2x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 5

Question 6.
v3 − 2v2 − 16v = − 32
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.5-Answer-6.jpg

Question 7.
5w3 = 50w
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 7

Question 8.
9m5 = 27m3
Answer:

Question 9.
2c4 − 6c3 = 12c2 − 36c
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 9

Question 10.
p4 + 40 = 14p2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.5-Answer-10.jpg
Question 11.
12n2 + 48n = −n3 − 64
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 11

Question 12.
y3 − 27 = 9y2 − 27y
Answer:
Given,
y3 − 27 = 9y2 − 27y
Take (y – 3) as a common factor
(y – 3)(y^2 + 3y + 9) = 9y(y – 3)
(y – 3)(y^2 + 3y + 9) – 9y(y – 3) = 0
(y – 3)(y^2 + 3y + 9 – 9y) = 0
(y – 3)(y^2 – 6y + 9) = 0
(y – 3)(y – 3)(y – 3) = 0
(y – 3)^3 = 0
y – 3 = 0
y = 3

In Exercises 13–20, find the zeros of the function. Then sketch a graph of the function.

Question 13.
h(x) = x4 + x3 − 6x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 13

Question 14.
f(x) = x4 − 18x2 + 81
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.5-Answer-14.jpg

Question 15.
p(x) = x6 − 11x5 + 30x4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 15

Question 16.
g(x) = −2x5 + 2x4 + 40x3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.5-Answer-16.jpg

Question 17.
g(x) = −4x4 + 8x3 + 60x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 17

Question 18.
h(x) = −x3 − 2x2 + 15x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.5-Answer-18.jpg

Question 19.
h(x) = −x3 − x2 + 9x + 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 19

Question 20.
p(x) = x3 − 5x2 − 4x + 20
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.5-Answer-20.jpg

Question 21.
USING EQUATIONS According to the Rational Root Theorem, which is not a possible solution of the equation 2x4 − 5x3 + 10x2 − 9 = 0?
A. −9
B. −\([\frac{1}{2}/latex]
C. [latex][\frac{5}{2}/latex]
D. 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-algerbra-2-chapter-4.-Polynomials-exercise-4.5-Answer-.21JPG.jpg
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 21

Question 22.
USING EQUATIONS According to the Rational Root Theorem, which is not a possible zero of the function f(x) = 40x5 − 42x4 − 107x3 + 107x2 + 33x − 36?
A. −[latex][\frac{2}{3}/latex]
B. − [latex][\frac{3}{8}/latex]
C. [latex][\frac{3}{4}/latex]
D. [latex][\frac{4}{5}/latex]
Answer: Option A is the correct answer

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in listing the possible rational zeros of the function.

Question 23.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 59
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 23

Question 24.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 60
Answer:

In Exercises 25–32, find all the real solutions of the equation.

Question 25.
x3 + x2 − 17x + 15 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 25

Question 26.
x3 − 2x2 − 5x + 6 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-26.jpg

Question 27.
x3 − 10x2 + 19x + 30 = 0
Answer:
Given,
x3 − 10x2 + 19x + 30 = 0
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 27

Question 28.
x3 + 4x2 − 11x − 30 = 0
Answer:
Given,
x3 + 4x2 − 11x − 30 = 0
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-28.jpg

Question 29.
x3 − 6x2 − 7x + 60 = 0
Answer:
Given,
x3 − 6x2 − 7x + 60 = 0
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 29

Question 30.
x3 − 16x2 + 55x + 72 = 0
Answer:

Question 31.
2x3 − 3x2 − 50x − 24 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 31

Question 32.
3x3 + x2 − 38x + 24 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-32JPG.jpg

In Exercises 33–38, find all the real zeros of the function.

Question 33.
f(x) = x3 − 2x2 − 23x + 60
Answer:
Given,
f(x) = x3 − 2x2 − 23x + 60
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 33

Question 34.
g(x) = x3 − 28x − 48
Answer:
Given,
g(x) = x3 − 28x − 48
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-34.jpg

Question 35.
h(x) = x3 + 10x2 + 31x + 30
Answer:
Given,
h(x) = x3 + 10x2 + 31x + 30
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 35

Question 36.
f(x) = x3 − 14x2 + 55x − 42
Answer:
Given,
f(x) = x3 − 14x2 + 55x − 42
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-36.jpg

Question 37.
p(x) = 2x3 −x2 − 27x + 36
Answer:
Given,
p(x) = 2x3 −x2 − 27x + 36
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 37

Question 38.
g(x) = 3x3 − 25x2 + 58x − 40
Answer:
Given,
g(x) = 3x3 − 25x2 + 58x − 40

USING TOOLS In Exercises 39 and 40, use the graph to shorten the list of possible rational zeros of the function. Then find all real zeros of the function.

Question 39.
f(x) = 4x3 − 20x + 16
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 61
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 39

Question 40.
f(x) = 4x3 − 49x − 60
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 62
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-40.jpg

In Exercises 41–46, write a polynomial function f of least degree that has a leading coefficient of 1 and the given zeros.

Question 41.
−2, 3, 6
Answer:
Given,
The values of x are −2, 3, 6
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 41

Question 42.
−4, −2, 5
Answer:
Given,
The values of x are −4, −2, 5

Question 43.
−2, 1 + [latex]\sqrt{7}\)
Answer:
Given,
The values of x are −2, 1 + \(\sqrt{7}\)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 43

Question 44.
4, 6 − \(\sqrt{7}\)
Answer:
Given,
The values of x are 4, 6 − \(\sqrt{7}\)

Question 45.
−6, 0, 3 −\(\sqrt{5}\)
Answer:
Given,
The values of x are −6, 0, 3 −\(\sqrt{5}\)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 45

Question 46.
0, 5, −5 + \(\sqrt{8}\)
Answer:
Given,
The values of x are 0, 5, −5 + \(\sqrt{8}\)

Question 47.
COMPARING METHODS Solve the equation x3 − 4x2 − 9x + 36 = 0 using two different methods. Which method do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 47.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 47.2

Question 48.
REASONING Is it possible for a cubic function to have more than three real zeros? Explain.
Answer:
No, it is not possible.
A cubic function is to have at most three real zeros.

Question 49.
PROBLEM SOLVING At a factory, molten glass is poured into molds to make paperweights. Each mold is a rectangular prism with a height 3 centimeters greater than the length of each side of its square base. Each mold holds 112 cubic centimeters of glass. What are the dimensions of the mold?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 49

Question 50.
MATHEMATICAL CONNECTIONS The volume of the cube shown is 8 cubic centimeters.
a. Write a polynomial equation that you can use to find the value of x.
b. Identify the possible rational solutions of the equation in part (a).
c. Use synthetic division to find a rational solution of the equation. Show that no other real solutions exist.
d. What are the dimensions of the cube?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 63
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-50JPG.jpg

Question 51.
PROBLEM SOLVING Archaeologists discovered a huge hydraulic concrete block at the ruins of Caesarea with a volume of 945 cubic meters. The block is x meters high by 12x − 15 meters long by 12x − 21 meters wide. What are the dimensions of the block?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 64
Answer:
Given,
Archaeologists discovered a huge hydraulic concrete block at the ruins of Caesarea with a volume of 945 cubic meters.
The block is x meters high by 12x − 15 meters long by 12x − 21 meters wide.
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 51

Question 52.
MAKING AN ARGUMENT Your friend claims that when a polynomial function has a leading coefficient of 1 and the coefficients are all integers, every possible rational zero is an integer. Is your friend correct? Explain your reasoning.
Answer:

Question 53.
MODELING WITH MATHEMATICS During a 10-year period, the amount (in millions of dollars) of athletic equipment Esold domestically can be modeled by E(t) = −20t3 + 252t2 − 280t + 21,614, where t is in years.
a. Write a polynomial equation to find the year when about $24,014,000,000 of athletic equipment is sold.
b. List the possible whole-number solutions of the equation in part (a). Consider the domain when making your list of possible solutions.
c. Use synthetic division to find when $24,014,000,000 of athletic equipment is sold.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 53

Question 54.
THOUGHT PROVOKING Write a third or fourth degree polynomial function that has zeros at ± \(\frac{3}{4}\). Justify your answer.
Answer:

Question 55.
MODELING WITH MATHEMATICS You are designing a marble basin that will hold a fountain for a city park. The sides and bottom of the basin should be 1 foot thick. Its outer length should be twice its outer width and outer height. What should the outer dimensions of the basin be if it is to hold 36 cubic feet of water?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 65
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 55

Question 56.
HOW DO YOU SEE IT? Use the information in the graph to answer the questions.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 66
a. What are the real zeros of the function f ?
b. Write an equation of the quartic function in factored form.
Answer:

Question 57.
REASONING Determine the value of k for each equation so that the given x-value is a solution.
a. x3 − 6x2 − 7x + k = 0; x = 4
b. 2x3 + 7x2 − kx − 18 = 0; x = −6
c. kx3 − 35x2 + 19x + 30 = 0; x = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 57

Question 58.
WRITING EQUATIONS Write a polynomial function gof least degree that has rational coefficients, a leading coefficient of 1, and the zeros −2 + \(\sqrt{7}\) and 3 + \(\sqrt{2}\).
Answer:

In Exercises 59–62, solve f(x) = g(x) by graphing and algebraic methods.

Question 59.
f(x) = x3 + x2 − x − 1; g(x) = −x + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 59

Question 60.
f(x) = x4 − 5x3 + 2x2 + 8x; g(x) = −x2 + 6x − 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-60.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-60.aJPG_.jpg

Question 61.
f(x) = x3 − 4x2 + 4x; g(x) = −2x + 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 61

Question 62.
f(x) = x4 + 2x3 − 11x2 − 12x + 36;
g(x) = −x2 − 6x − 9
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-62.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-62a.jpg

Question 63.
MODELING WITH MATHEMATICS You are building a pair of ramps for a loading platform. The left ramp is twice as long as the right ramp. If 150 cubic feet of concrete are used to build the ramps, what are the dimensions of each ramp?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 67
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 63

Question 64.
MODELING WITH MATHEMATICS Some ice sculptures are made by filling a mold and then freezing it. You are making an ice mold for a school dance. It is to be shaped like a pyramid with a height 1 foot greater than the length of each side of its square base. The volume of the ice sculpture is 4 cubic feet. What are the dimensions of the mold?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 68
Answer:

Question 65.
ABSTRACT REASONING Let an be the leading coefficient of a polynomial function f and a0 be the constant term. If an has r factors and a0 has s factors, what is the greatest number of possible rational zeros of f that can be generated by the Rational Zero Theorem? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 65

Maintaining Mathematical Proficiency

Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.(Section 4.1)

Question 66.
h(x) = −3x² + 2x − 9 + \(\sqrt{4}\)x³
Answer:
Given,
h(x) = −3x² + 2x − 9 + \(\sqrt{4}\)x³
h(x) =\(\sqrt{4}\)x³ − 3x² + 2x − 9
Degree: 3
Type: Cubic function
Leading Coefficient: \(\sqrt{4}\) or 2

Question 67.
g(x) = 2x3 − 7x2 − 3x − 1 + x
Answer:
Given,
g(x) = 2x3 − 7x2 − 3x − 1 + x
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 67

Question 68.
f(x) = \(\frac{1}{3}\)x2 + 2x3 − 4x4 − \(\sqrt{3}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.4-Polynomials-Exercise-7.5-Answer-68.jpg
Question 69.
p(x) = 2x − 5x3 + 9x2 + \(\sqrt[4]{x}\) + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 69

Find the zeros of the function.(Section 3.2)

Question 70.
f(x) = 7x2 + 42
Answer:

Question 71.
g(x) = 9x2 + 81
Answer:
Given,
g(x) = 9x2 + 81
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 71

Question 72.
h(x) = 5x2 + 40
Answer:
Given,
h(x) = 5x2 + 40

Question 73.
f(x) = 8x2 − 1
Answer:
Given,
f(x) = 8x2 − 1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.5 Question 73

Lesson 4.6 The Fundamental Theorem of Algebra

Essential Question

How can you determine whether a polynomial equation has imaginary solutions?

EXPLORATION 1
Cubic Equations and Imaginary Solutions
Work with a partner.
Match each cubic polynomial equation with the graph of its related polynomial function. Then find all solutions. Make a conjecture about how you can use a graph or table of values to determine the number and types of solutions of a cubic polynomial equation.
a. x3 − 3x2 + x + 5 = 0
b. x3 − 2x2 − x + 2 = 0
c. x3 − x2 − 4x + 4 = 0
d. x3 + 5x2 + 8x + 6 = 0
e. x3 − 3x2 + x − 3 = 0
f. x3 − 3x2 + 2x = 0
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 69

EXPLORATION 2
Quartic Equations and Imaginary Solutions
Work with a partner.
Use the graph of the related quartic function, or a table of values, to determine whether each quartic equation has imaginary solutions. Explain your reasoning. Then find all solutions.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 70
a. x4 − 2x3 − x2 + 2x = 0
b. x4 − 1 = 0
c. x4 + x3 − x − 1 = 0
d. x4 − 3x3 + x2 + 3x − 2 = 0

Communicate Your Answer

Question 3.
How can you determine whether a polynomial equation has imaginary solutions?
Answer:
In order to find the Imaginary roots of the function use the quadratic equation when the discriminant of the quadratic equation — the part under the square root sign (b2 – 4ac) — is negative.

Question 4.
Is it possible for a cubic equation to have three imaginary solutions? Explain your reasoning.
Answer:
The equation has at most three distinct roots, it follows that it cannot have three distinct complex nonreal or imaginary roots. Your argument is almost correct proof of the fact that a cubic equation cannot have three complexes (non-real) roots.

4.6 Lesson

Monitoring Progress

Question 1.
How many solutions does the equation x4 + 7x2 − 144 = 0 have?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-1.jpg

Question 2.
How many zeros does the function f(x) = x3 − 5x2 − 8x + 48 have? Find all zeros of the polynomial function.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-2.jpg

Question 3.
f(x) = x3 + 7x2 + 16x + 12
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-3.jpg

Question 4.
f(x) = x5 − 3x4 + 5x3 − x2 − 6x + 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-4.jpg

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 5.
−1, 4i
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-5.jpg

Question 6.
3, 1 + i\(\sqrt{5}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-6.jpg

Question 7.
\(\sqrt{2}\), 1 − 3i
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-7.jpg

Question 8.
2, 2i, 4 − \(\sqrt{6}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-8.jpg

Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.

Question 9.
f(x) = x3 + 9x − 25
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-9.jpg

Question 10.
f(x) = 3x4 − 7x3 + x2 − 13x + 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-10.jpg

Question 11.
WHAT IF? In Example 5, what is the tachometer reading when the boat travels 20 miles per hour?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Questioon-11.jpg

The Fundamental Theorem of Algebra 4.6 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE The expressions 5 + i and 5 − i are _____________.
Answer:
The expressions 5 + i and 5 − i are complex conjugates.

Question 2.
WRITING How many solutions does the polynomial equation (x + 8)3(x − 1) = 0 have? Explain.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Question-2.jpg

Monitoring Progress and Modeling with Mathematics

Question 3.
x4 + 2x3 − 4x2 + x = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 3

Question 4.
5y3 − 3y2 + 8y = 0
Answer:
Given,
5y3 − 3y2 + 8y = 0
The degree of the polynomial is 3.
It has 3 solutions.

Question 5.
9t6 − 14t3 + 4t − 1 = 0
Answer:
Given,
9t6 − 14t3 + 4t − 1 = 0
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 5

Question 6.
f(z) = −7z4 + z2 − 25
Answer:
Given,
f(z) = −7z4 + z2 − 25
Because −7z4 + z2 − 25 =0 this is a polynomial equation of degree 4.
It has four solutions.

Question 7.
g(s) = 4s5 − s3 + 2s7 − 2
Answer:
Given,
g(s) = 4s5 − s3 + 2s7 − 2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 7

Question 8.
h(x) = 5x4 + 7x8 − x12
Answer:
Given,
h(x) = 5x4 + 7x8 − x12
Because 5x4 + 7x8 − x12 is a polynomial equation of degree 12
It has twelve solutions.

In Exercises 9–16, find all zeros of the polynomial function.

Question 9.
f(x) = x4 − 6x3 + 7x2 + 6x − 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 9

Question 10.
f(x) = x4 + 5x3 − 7x2 − 29x + 30
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Question-10.jpg

Question 11.
g(x) = x4 − 9x2 − 4x + 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 11

Question 12.
h(x) = x3 + 5x2 − 4x − 20
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Question-12.jpg

Question 13.
g(x) = x4 + 4x3 + 7x2 + 16x + 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 13

Question 14.
h(x) = x4 − x3 + 7x2 − 9x − 18
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Question-14.jpg

Question 15.
g(x) = x5 + 3x4 − 4x3 − 2x2 − 12x − 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 15

Question 16.
f(x) = x5 − 20x3 + 20x2 − 21x + 20
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.6-Question-16.jpg

ANALYZING RELATIONSHIPS In Exercises 17–20, determine the number of imaginary zeros for the function with the given degree and graph. Explain your reasoning.

Question 17.
Degree: 4
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 70.1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 17

Question 18.
Degree: 5
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 71
Answer:
The graph has 1 real root, which means there are 4 imaginary roots.

Question 19.
Degree: 2
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 72
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 19

Question 20.
Degree: 3
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 73
Answer:
There are three real roots, which means there are no imaginary roots.

In Exercises 21–28, write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 21.
−5, −1, 2
Answer:
Given,
−5, −1, 2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 21

Question 22.
−2, 1, 3
Answer:
Given,
−2, 1, 3
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.6Answer-22.jpg

Question 23.
3, 4 + i
Answer:
Given,
3, 4 + i
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 23

Question 24.
2, 5 − i
Answer:
Given,
2, 5 − i
f(x) = (x – 2)[x – (5 – i)] [x – (5 + i)]
= (x – 2)[(x – 5) – i] [(x – 5) + i]
= (x – 2)[(x – 5)² – i²]
= (x – 2)[x² + 25 – 10x – 1]
= (x – 2)[x² – 10x + 24]
= x³ – 2x² -10x² + 20x + 24x – 48
= x³ – 12x² + 48x – 48

Question 25.
4, −\(\sqrt{5}\)
Answer:
Given,
4, −\(\sqrt{5}\)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 25

Question 26.
3i, 2 − i
Answer:
Given,
3i, 2 − i

Question 27.
2, 1 + i, 2 −\(\sqrt{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 27

Question 28.
3, 4 + 2i, 1 + \(\sqrt{7}\)
Answer:
Given,
3, 4 + 2i, 1 + \(\sqrt{7}\)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-28.jpg

ERROR ANALYSIS In Exercises 29 and 30, describe and correct the error in writing a polynomial function with rational coefficients and the given zero(s).

Question 29.
Zeros: 2, 1 + i
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 74
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 29

Question 30.
Zero: 2 +i
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 75
Answer:

Question 31.
OPEN-ENDED Write a polynomial function of degree 6 with zeros 1, 2, and −i. Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 31

Question 32.
REASONING Two zeros of f(x) = x3 − 6x2 − 16x + 96 are 4 and −4. Explain why the third zero must also be a real number.
Answer:

In Exercises 33–40, determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.

Question 33.
g(x) = x4 − x2 − 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 33

Question 34.
g(x) = −x3 + 5x2 + 12
Answer:

Question 35.
g(x) = x3 − 4x2 + 8x + 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 35

Question 36.
g(x) = x5 − 2x3 − x2 + 6
Answer:

Question 37.
g(x) = x5 − 3x3 + 8x − 10
Answer:
Given,
g(x) = x5 − 3x3 + 8x − 10
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 37

Question 38.
g(x) = x5 + 7x4 − 4x3 − 3x2 + 9x − 15
Answer:
Given,

Question 39.
g(x) = x6 + x5 − 3x4 + x3 + 5x2 + 9x − 18
Answer:
Given,
g(x) = x6 + x5 − 3x4 + x3 + 5x2 + 9x − 18
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 39

Question 40.
g(x) = x7 + 4x4 − 10x + 25
Answer:

Question 41.
REASONING Which is not a possible classification of zeros for f(x) = x5 − 4x3 + 6x2 + 2x − 6? Explain.
A. three positive real zeros, two negative real zeros, and zero imaginary zeros
B. three positive real zeros, zero negative real zeros, and two imaginary zeros
C. one positive real zero, four negative real zeros, and zero imaginary zeros
D. one positive real zero, two negative real zeros, and two imaginary zeros
Answer:
Given,
f(x) = x5 − 4x3 + 6x2 + 2x − 6
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 41

Question 42.
USING STRUCTURE Use Descartes’s Rule of Signs to determine which function has at least 1 positive real zero.
A. f(x) = x4 + 2x3 − 9x2 − 2x – 8
B. f(x) = x4 + 4x3 + 8x2 + 16x + 16
C. f(x) = −x4 − 5x2 − 4
D. f(x) = x4 + 4x3 + 7x2 + 12x + 12
Answer:
A has sign changes, therefore this function has at least 1 positive real zero.

Question 43.
MODELING WITH MATHEMATICS From 1890 to 2000, the American Indian, Eskimo, and Aleut population P (in thousands) can be modeled by the function P = 0.004t3 − 0.24t2 + 4.9t + 243, where t is the number of years since 1890. In which year did the population first reach 722,000?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 43

Question 44.
MODELING WITH MATHEMATICS Over a period of 14 years, the number N of inland lakes infested with zebra mussels in a certain state can be modeled by
N = −0.0284t4 + 0.5937t3 − 2.464t2 + 8.33t − 2.5
where t is time (in years). In which year did the number of infested inland lakes first reach 120?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 76
Answer:

Question 45.
MODELING WITH MATHEMATICS For the 12 years that a grocery store has been open, its annual revenue R (in millions of dollars) can be modeled by the function
R = 0.0001(−t4 + 12t3 − 77t2 + 600t + 13,650)where t is the number of years since the store opened. In which year(s) was the revenue $1.5 million?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 77
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 45

Question 46.
MAKING AN ARGUMENT Your friend claims that 2 − i is a complex zero of the polynomial function f(x) = x3 − 2x2 + 2x + 5i, but that its conjugate is not a zero. You claim that both 2 − i and its conjugate must be zeros by the Complex Conjugates Theorem. Who is correct? Justify your answer.
Answer:

Question 47.
MATHEMATICAL CONNECTIONS A solid monument with the dimensions shown is to be built using 1000 cubic feet of marble. What is the value of x?
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 78
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 47

Question 48.
THOUGHT PROVOKING Write and graph a polynomial function of degree 5 that has all positive or negative real zeros. Label each x-intercept. Then write the function in standard form.
Answer:

Question 49.
WRITING The graph of the constant polynomial function f(x) = 2 is a line that does not have any x-intercepts. Does the function contradict the Fundamental Theorem of Algebra? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 49

Question 50.
HOW DO YOU SEE IT? The graph represents a polynomial function of degree 6.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 79
a. How many positive real zeros does the function have? negative real zeros? imaginary zeros?
b. Use Descartes’s Rule of Signs and your answers in part (a) to describe the possible sign changes in the coefficients of f(x).
Answer:

Question 51.
FINDING A PATTERN Use a graphing calculator to graph the function f(x) = (x + 3)n for n = 2, 3, 4, 5, 6, and 7.
a. Compare the graphs when n is even and n is odd.
b. Describe the behavior of the graph near the zero x = −3 as n increases.
c. Use your results from parts (a) and (b) to describe the behavior of the graph of g(x) = (x − 4)20 near x = 4.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 51.1
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 51.2

Question 52.
DRAWING CONCLUSIONS Find the zeros of each function.
f(x) = x5 − 5x + 6
g(x) = x3 − 7x + 6
h(x) = x4 + 2x3 + x2 + 8x − 12
k(x) = x5 − 3x4 − 9x3 + 25x2 − 6x
a. Describe the relationship between the sum of the zeros of a polynomial function and the coefficients of the polynomial function.
b. Describe the relationship between the product of the zeros of a polynomial function and the coefficients of the polynomial function.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.6Answer-52.aJPG_.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-52b.jpg

Question 53.
PROBLEM SOLVING You want to save money so you can buy a used car in four years. At the end of each summer, you deposit $1000 earned from summer jobs into your bank account. The table shows the value of your deposits over the four-year period. In the table, g is the growth factor 1 + r, where r is the annual interest rate expressed as a decimal.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 80
a. Copy and complete the table.
b. Write a polynomial function that gives the value v of your account at the end of the fourth summer in terms of g.
c. You want to buy a car that costs about $4300. What growth factor do you need to obtain this amount? What annual interest rate do you need?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 53.1

Maintaining Mathematical Proficiency
Describe the transformation of f(x) = x2 represented by g. Then graph each function. (Section 2.1)

Question 54.
g(x) = −3x2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.6Answer-54.jpg

Question 55.
g(x) = (x − 4)2 + 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 55

Question 56.
g(x) = −(x − 1)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.6Answer-56.jpg

Question 57.
g(x) = 5(x + 4)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 57

Write a function g whose graph represents the indicated transformation of the graph of f.(Sections 1.2and 2.1)

Question 58.
f(x) = x; vertical shrink by a factor of \(\frac{1}{3}\) and a reflection in the y-axis
Answer:

Question 59.
f(x) = | x + 1 ∣− 3; horizontal stretch by a factor of 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.6 Question 59

Question 60.
f(x) = x2; reflection in the x-axis, followed by a translation 2 units right and 7 units up
Answer:

Lesson 4.7 Tranformations of Polynomial Functions

Essential Question
How can you transform the graph of a polynomial function?

EXPLORATION 1
Transforming the Graph of a Cubic Function
Work with a partner.
The graph of the cubic function f(x) = x3 is shown. The graph of each cubic function g represents a transformation of the graph of f. Write a rule for g. Use a graphing calculator to verify your answers.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 81
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 81.1

EXPLORATION 2
Transforming the Graph of a Quartic Function
Work with a partner.
The graph of the quartic function f(x) = x4 is shown. The graph of each quartic function g represents a transformation of the graph of f. Write a rule for g. Use a graphing calculator to verify your answers.
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 82
Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 82.1

Communicate Your Answer

Question 3.
How can you transform the graph of a polynomial function?
Answer: The polynomial function can be transformed through translation along horizontal and vertical axes, Reflection against horizontal and vertical axes, and Stretching or Shrinking on horizontal and vertical axes.

Big Ideas Math Answers Algebra 2 Chapter 4 Polynomial Functions 83
Question 4.
Describe the transformation of f(x) = x4 represented by g(x) = (x + 1)4 + 3. Then graph g.
Answer:
First identify what changed from f(x) to g(x), which is the +1 inside of what’s being raised to the 4th and +3 outside of it.
Inside the parentheses means a horizontal shift to the left if positive and right if negative, while outside the parentheses means a vertical shift upwards if positive and downwards if negative.
In this case a+ 1 and a + 3 respectively.
The differences are just matters of teacher preference, the one on the right is more basic while the one on the left is more advanced.

4.7 Lesson

Monitoring Progress

Question 1.
Describe the transformation of f(x) = x4 represented by g(x) = (x − 3)4 − 1. Then graph each function.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-chapter-4-polynomial-exercise-4.7-Answer-no-1.jpg

Question 2.
Describe the transformation of f(x) = x3 represented by g(x) = 4(x + 2)3. Then graph each function.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-chapter-4-polynomial-exercise-4.7-Answer-no-2-1.jpg

Question 3.
Let f(x) = x5 − 4x + 6 and g(x) = −f(x). Write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f.
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-chapter-4-polynomial-exercise-4.7-Answer-no-3.jpg

Question 4.
Let the graph of g be a horizontal stretch by a factor of 2, followed by a translation 3 units to the right of the graph of f(x) = 8x3 + 3. Write a rule for g.
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-chapter-4-polynomial-exercise-4.7-Answer-no-4.jpg

Question 5.
WHAT IF? In Example 5, the height of the pyramid is 6x, and the volume (in cubic feet) is represented by V(x) = 2x3. Write a rule for W. Find and interpret W(7).
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-chapter-4-polynomial-exercise-4.7-Answer-no-5.jpg

Tranformations of Polynomial Functions 4.7 Exercises

Question 1.
COMPLETE THE SENTENCE The graph of f(x) = (x + 2)3 is a ____________ translation of the graph of f(x) = x3.
Answer:
The graph of f(x) = (x + 2)3 is a horizontal translation of the graph of f(x) = x3.

Question 2.
VOCABULARY Describe how the vertex form of quadratic functions is similar to the form f(x) = a(x − h)3 + k for cubic functions.
Answer:

In Exercises 3–6, describe the transformation of f represented by g. Then graph each function.
Question 3.
f(x) = x4, g(x) = x4 + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 3

Question 4.
f(x) = x4, g(x) = (x − 5)4
Answer:
Given,
f(x) = x4, g(x) = (x − 5)4
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-4.jpg

Question 5.
f(x) = x5, g(x) = (x − 2)5 − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 5

Question 6.
f(x) = x6, g(x) = (x + 1)6 − 4
Answer:

ANALYZING RELATIONSHIPS In Exercises 7–10, match the function with the correct transformation of the graph of f. Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 84

Question 7.
y = f(x − 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 7

Question 8.
y = f(x + 2) + 2
Answer:
Graph C is correct.
The graph of f is 2 units left  and 2 units up

Question 9.
y = f(x − 2) + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 9

Question 10.
y = f(x) − 2
Answer:

Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 85

In Exercises 11–16, describe the transformation of f represented by g. Then graph each function.

Question 11.
f(x) = x4, g(x) = −2x4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 11

Question 12.
f(x) = x6, g(x) = −3x6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-12.jpg

Question 13.
f(x) = x3, g(x) = 5x3 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 13

Question 14.
f(x) = x4, g(x) = \(\frac{1}{2}\)x4 + 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-14.jpg

Question 15.
f(x) = x5, g(x) = \(\frac{3}{4}\)(x + 4)5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 15

Question 16.
f(x) = x4, g(x) = (2x)4− 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-16.jpg

In Exercises 17–20, write a rule for g and then graph each function. Describe the graph of g as a transformation of the graph of f.

Question 17.
f(x) = x4 + 1, g(x) = f(x + 2)
Answer:
Given,
f(x) = x4 + 1,
g(x) = f(x + 2)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 17

Question 18.
f(x) = x5 − 2x + 3, g(x) = 3f(x)
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-18.jpg

Question 19.
f(x) = 2x3 − 2x2 + 6, g(x) = − \(\frac{1}{2}\)f(x)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 19

Question 20.
f(x) = x4 + x3 − 1, g(x) = f(−x) − 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-20.jpg

Question 21.
ERROR ANALYSIS Describe and correct the error in graphing the function g(x) = (x + 2)4 − 6
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 86
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 21

Question 22.
ERROR ANALYSIS Describe and correct the error in describing the transformation of the graph of f(x) = x5 represented by the graph of g(x) = (3x)5− 4.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 87
Answer:
Given,
f(x) = x5
g(x) = (3x)5− 4.
The graph of g is a horizontal shrink by a factor of 1/3, followed by a translation of 4 units down of the graph of f.

Question 23.
f(x) = x3 − 6; translation 3 units left, followed by a reflection in the y-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 23

Question 24.
f(x) = x4 + 2x + 6; vertical stretch by a factor of 2, followed by a translation 4 units right
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-24.jpg

Question 25.
f(x) = x3 + 2x2 − 9; horizontal shrink by a factor of \(\frac{1}{3}\) and a translation 2 units up, followed by a reflection in the x-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 25

Question 26.
f(x) = 2x5 − x3 + x2 + 4; reflection in the y-axis and a vertical stretch by a factor of 3, followed by a translation 1 unit down
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-26.jpg

Question 27.
MODELING WITH MATHEMATICS The volume V(in cubic feet) of the pyramid is given by V(x) = x3 − 4x. The function W(x) = V(3x) gives the volume (in cubic feet) of the pyramid when x is measured in yards. Write a rule for W. Find and interpret W(5).
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 88
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 27

Question 28.
MAKING AN ARGUMENT The volume of a cube with side length x is given by V(x) = x3. Your friend claims that when you divide the volume in half, the volume decreases by a greater amount than when you divide each side length in half. Is your friend correct? Justify your answer.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-28.jpg

Question 29.
OPEN-ENDED Describe two transformations of the graph of f(x) = x5 where the order in which the transformations are performed is important. Then describe two transformations where the order is not important. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 29

Question 30.
THOUGHT PROVOKING Write and graph a transformation of the graph of f(x) = x5 − 3x4 + 2x − 4 that results in a graph with a y-intercept of −2.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-30.jpg

Question 31.
PROBLEM SOLVING A portion of the path that a hummingbird flies while feeding can be modeled by the function
f(x) = −\(\frac{1}{5}\)x(x − 4)2(x − 7), 0 ≤ x ≤ 7 w
here x is the horizontal distance (in meters) and f(x) is the height (in meters). The hummingbird feeds each time it is at ground level.
a. At what distances does the hummingbird feed?
b. A second hummingbird feeds 2 meters farther away than the first hummingbird and flies twice as high. Write a function to model the path of the second hummingbird.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 89
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 31

Question 32.
HOW DO YOU SEE IT? Determine the real zeros of each function. Then describe the transformation of the graph of f that results in the graph of g.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 90
Answer:
Real Zeroes of F is : -2, 0, 2
Real Zeroes of g is : 0, 2, 2
g(x) = f( x- 2).

Question 33.
MATHEMATICAL CONNECTIONS Write a function V for the volume (in cubic yards) of the right circular cone shown. Then write a function W that gives the volume (in cubic yards) of the cone when x is measured in feet. Find and interpret W(3).
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 91
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 33

Maintaining Mathematical Proficiency

Find the minimum value or maximum value of the function. Describe the domain and range of the function, and where the function is increasing and decreasing.(Section 2.2)

Question 34.
h(x) = (x + 5)2 − 7
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-34.jpg

Question 35.
f(x) = 4 − x2
Answer:
Given,
f(x) = 4 − x2
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 35

Question 36.
f(x) = 3(x − 10)(x + 4)
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-36.jpg

Question 37.
g(x) = −(x + 2)(x + 8)
Answer:
Given,
g(x) = −(x + 2)(x + 8)
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 37

Question 38.
h(x) = \(\frac{1}{2}\)(x − 1)2 − 3
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-Ideas-Math-Algebra-2-Answers-Chapter-4-Polynomial-Functions-4.7-Question-38.jpg

Question 39.
f(x) = −2x2 + 4x − 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.7 Question 39

Lesson 4.8 Analyzing Graphs of Polynomial Functions

Essential Question
How many turning points can the graph of a polynomial function have?
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 92
A turning point of the graph of a polynomial function is a point on the graph at which the function changes from

  • increasing to decreasing, or
  • decreasing to increasing.

EXPLORATION 1
Approximating Turning Points
Work with a partner.
Match each polynomial function with its graph. Explain your reasoning. Then use a graphing calculator to approximate the coordinates of the turning points of the graph of the function. Round your answers to the nearest hundredth.
a. f(x) = 2x2 + 3x − 4
b. f(x) = x2 + 3x + 2
c. f(x) = x3 − 2x2 − x + 1
d. f(x) = −x3 + 5x − 2
e. f(x) = x4 − 3x2 + 2x − 1
f. f(x) = −2x5 − x2 + 5x + 3
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 93
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 94

Communicate Your Answer

Question 2.
How many turning points can the graph of a polynomial function have?
Answer:
The turning points are those points where the graph changes direction i.e, where the function changes from growing to decreasing or from decreasing to growing.
x = -2, x = -1, x = 1 and x = 2
It has four turning points.

Question 3.
Is it possible to sketch the graph of a cubic polynomial function that has no turning points? Justify your answer.
Answer:
The maximum number of turning points of a polynomial function is always one less than the degree of the function.
Understand the relationship between degree and turning points | College Algebra

Monitoring Progress

Graph the function.

Question 1.
f(x) = \(\frac{1}{2}\)(x + 1)(x − 4)

Question 2.
f(x) = \(\frac{1}{4}\)(x + 2)(x − 1)(x − 3)

Question 3.
Find all real zeros of f(x) = 18x3 + 21x2 − 13x − 6.

Question 4.
Graph f(x) = 0.5x3 + x2 − x + 2. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.

Determine whether the function is even, odd, or neither.

Question 5.
f(x) = −x2 + 5

Question 6.
f(x) = x4 − 5x3

Question 7.
f(x) = 2x5

Answer:
f(x) = 2x⁵
First find f(-x)
Find f(-x) by substituting -x in place of x in f(x).
f(-x) = 2(-x)⁵
Apply the product rule to -x.
f(-x) = 2((-1)⁵ x⁵)
Raise -1 to the power of 5
f(-x) = -2(-x⁵)
Multiply -1 by 2.
f(-x) = -2x⁵
A function is even if f(-x) = f(x).
f(-xx) = f(x)
-2x⁵ ≠ 2x⁵
So, the function is not even.
A function is odd if f(-x) = -f(x)
Multiply 2 by -1.
-f(x) = -2x⁵
Since -2x⁵ = -2x⁵
Thus the function is odd.

Analyzing Graphs of Polynomial Functions 4.8 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE A local maximum or local minimum of a polynomial function occurs at a ______________ point of the graph of the function.
Answer:
A local maximum or local minimum of a polynomial function occurs at a turning point in the graph of the function.

Question 2.
WRITING Explain what a local maximum of a function is and how it may be different from the maximum value of the function.
Answer: A local maximum is a maximum point in relation to its nearby points. The maximum value of the function, on the other hand, is the maximum point of the entire graph or equation.

ANALYZING RELATIONSHIPS In Exercises 3–6, match the function with its graph.

Question 3.
f(x) = (x − 1)(x − 2)(x + 2)
Answer:
Graph A matches the equation.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 3

Question 4.
h(x) = (x + 2)2(x + 1)
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-4.jpg
Question 5.
g(x) = (x + 1)(x − 1)(x + 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 5

Question 6.
f(x) = (x − 1)2(x + 2)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer6-.jpg
In Exercises 7–14, graph the function.

Question 7.
f(x) = (x − 2)2(x + 1)
Answer:
Given,
f(x) = (x − 2)2(x + 1)
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 7

Question 8.
f(x) = (x + 2)2(x + 4)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer8.jpg

Question 9.
h(x) = (x + 1)2(x − 1)(x − 3)
Answer:
Given,
h(x) = (x + 1)2(x − 1)(x − 3)
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 9

Question 10.
g(x) = 4(x + 1)(x + 2)(x − 1)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-10.jpg

Question 11.
h(x) = \(\frac{1}{3}\)(x − 5)(x + 2)(x − 3)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 11

Question 12.
g(x) = \(\frac{1}{12}\)(x + 4)(x + 8)(x − 1)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-12.jpg

Question 13.
h(x) = (x− 3)(x2 + x + 1)
Answer:
Given,
h(x) = (x− 3)(x2 + x + 1)
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 13

Question 14.
f(x) = (x − 4)(2x2 − 2x + 1)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-14.jpg

ERROR ANALYSIS In Exercises 15 and 16, describe and correct the error in using factors to graph f.

Question 15.
f(x) = (x + 2)(x − 1)2
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 96
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 15

Question 16.
f(x) = x2(x − 3)
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 97
Answer:

In Exercises 17–22, find all real zeros of the function.

Question 17.
f(x) = x3 − 4x2 − x + 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 17

Question 18.
f(x) = x3 − 3x2 − 4x + 12
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-18.jpg

Question 19.
h(x) = 2x3 + 7x2 − 5x − 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 19

Question 20.
h(x) = 4x3 − 2x2 − 24x − 18
Answer:

Question 21.
g(x) = 4x3 + x2 − 51x + 36
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 21

Question 22.
f(x) = 2x3 − 3x2 − 32x − 15
Answer:

In Exercises 23–30, graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.

Question 23.
g(x) = 2x3 + 8x2 − 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 23

Question 24.
g(x) = −x4 + 3x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-24.jpg

Question 25.
h(x) = x4 − 3x2 + x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 25

Question 26.
f(x) = x5 − 4x3 + x2 + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-26.jpg

Question 27.
f(x) = 0.5x3 − 2x + 2.5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 27

Question 28.
f(x) = 0.7x4 − 3x3 + 5x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-28.jpg

Question 29.
h(x) = x5 + 2x2 − 17x − 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 29

Question 30.
g(x) = x4 − 5x3 + 2x2 + x − 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-30.jpg

In Exercises 31–36, estimate the coordinates of each turning point. State whether each corresponds to a local maximum or a local minimum. Then estimate the real zeros and find the least possible degree of the function.

Question 31.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 98
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 31

Question 32.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 99
Answer:

Question 33.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 100
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 33

Question 34.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 101
Answer:

Question 35.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 102
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 35

Question 36.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 103
Answer:

OPEN-ENDED In Exercises 37 and 38, sketch a graph of a polynomial function f having the given characteristics.

Question 37.

  • The graph of f has x-intercepts at x = −4, x = 0, and x = 2.
  • f has a local maximum value when x = 1.
  • f has a local minimum value when x = −2.

Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 37

Question 38.

  • The graph of f has x-intercepts at x = −3, x = 1, and x = 5.
  • f has a local maximum value when x = 1.
  • f has a local minimum value when x = −2 and when x = 4.

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-38.jpg

In Exercises 39–46, determine whether the function is even, odd, or neither.

Question 39.
h(x) = 4x7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 39

Question 40.
g(x) = −2x6 + x2
Answer:

Question 41.
f(x) = x4 + 3x2 − 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 41

Question 42.
f(x) = x5 + 3x3 − x
Answer:

Question 43.
g(x) = x2 + 5x + 1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 43

Question 44.
f(x) = −x3 + 2x − 9
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-44.jpg

Question 45.
f(x) = x4 − 12x2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 45

Question 46.
h(x) = x5 + 3x4
Answer:

Question 47.
USING TOOLS When a swimmer does the breaststroke, the function
S = −241t7 + 1060t6 − 1870t5 + 1650t4 − 737t3 + 144t2 − 2.43t
models the speed S (in meters per second) of the swimmer during one complete stroke, where t is the number of seconds since the start of the stroke and 0 ≤ t ≤ 1.22. Use a graphing calculator to graph the function. At what time during the stroke is the swimmer traveling the fastest?
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 104
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 47

Question 48.
USING TOOLS During a recent period of time, the number S (in thousands) of students enrolled in public schools in a certain country can be modeled by S = 1.64x3 − 102x2 + 1710x + 36,300, where x is time (in years). Use a graphing calculator to graph the function for the interval 0 ≤ x ≤ 41. Then describe how the public school enrollment changes over this period of time.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-48.jpg

Question 49.
WRITING Why is the adjective local, used to describe the maximums and minimums of cubic functions, sometimes not required for quadratic functions?
Answer:
A quadratic function only has one turning point, and it is always the maximum or minimum value of the function.

Question 50.
HOW DO YOU SEE IT? The graph of a polynomial function is shown.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 105
a. Find the zeros, local maximum, and local minimum values of the function.
b. Compare the x-intercepts of the graphs of y = f(x) and y = −f(x).
c. Compare the maximum and minimum values of the functions y = f(x) and y = −f(x).
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-50JPG.jpg

Question 51.
MAKING AN ARGUMENT Your friend claims that the product of two odd functions is an odd function. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 51

Question 52.
MODELING WITH MATHEMATICS You are making a rectangular box out of a 16-inch-by-20-inch piece of cardboard. The box will be formed by making the cuts shown in the diagram and folding up the sides. You want the box to have the greatest volume possible.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 106
a. How long should you make the cuts?
b. What is the maximum volume?
c. What are the dimensions of the finished box?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.8-Answer-52.jpg

Question 53.
PROBLEM SOLVING Quonset huts are temporary, all-purpose structures shaped like half-cylinders. You have 1100 square feet of material to build a quonset hut.
a. The surface area S of a quonset hut is given by S = πr2 + πrℓ. Substitute 1100 for S and then write an expression for ℓ in terms of r.
b. The volume V of a quonset hut is given by V = 1 — 2πr2ℓ. Write an equation that gives V as a function in terms of r only.c. Find the value of r that maximizes the volume of the hut.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 107
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 53

Question 54.
THOUGHT PROVOKING Write and graph a polynomial function that has one real zero in each of the intervals −2 < x < −1, 0 < x < 1, and 4 < x < 5. Is there a maximum degree that such a polynomial function can have? Justify your answer.
Answer:
As it is seen below example, it is possible.

Question 55.
MATHEMATICAL CONNECTIONS A cylinder is inscribed in a sphere of radius 8 inches. Write an equation for the volume of the cylinder as a function of h. Find the value of h that maximizes the volume of the inscribed cylinder. What is the maximum volume of the cylinder?
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 108
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 55.1
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 55.2

Maintaining Mathematical Proficiency

State whether the table displays linear data, quadratic data, or neither. Explain

Question 56.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 109
Answer:

Question 57.
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 110
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 4 Polynomial Functions 4.8 a 57

Lesson 4.9 Modeling with Polynomial Functions

Essential Question
How can you find a polynomial model for real-life data?

EXPLORATION 1
Modeling Real-Life Data
Work with a partner.
The distance a baseball travels after it is hit depends on the angle at which it was hit and the initial speed. The table shows the distances a baseball hit at an angle of 35° travels at various initial speeds.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 111
a. Recall that when data have equally-spaced x-values, you can analyze patterns in the differences of the y-values to determine what type of function can be used to model the data. If the first differences are constant, then the set of data fits a linear model. If the second differences are constant, then the set of data fits a quadratic model. Find the first and second differences of the data. Are the data linear or quadratic? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 112
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 113
b. Use a graphing calculator to draw a scatter plot of the data. Do the data appear linear or quadratic? Use the regression feature of the graphing calculator to find a linear or quadratic model that best fits the data.12019075400
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 113.1
c. Use the model you found in part (b) to find the distance a baseball travels when it is hit at an angle of 35° and travels at an initial speed of 120 miles per hour.
d. According to the Baseball Almanac, “Any drive over400 feet is noteworthy. A blow of 450 feet shows exceptional power, as the majority of major league players are unable to hit a ball that far. Anything in the 500-foot range is genuinely historic.” Estimate the initial speed of a baseball that travels a distance of 500 feet.

Communicate Your Answer

Question 2.
How can you find a polynomial model for real-life data?
Answer:
People use polynomials for modeling various buildings and objects, used in industries, and used in construction. They are even used in marketing, finance, and stocks.

Question 3.
How well does the model you found in Exploration 1(b) fit the data? Do you think the model is valid for any initial speed? Explain your reasoning.

4.9 Lesson

Monitoring Progress

write a cubic function whose graph passes through the given points. 

Question 1.
(−4, 0), (0, 10), (2, 0), (5, 0)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9Answer-1.jpg

Question 2.
Answer:
(−1, 0), (0, −12), (2, 0), (3, 0)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Monitoring-Exercise-4.9Answer-2.jpg

Question 3.
Use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 118
Answer:

Use a graphing calculator to find a polynomial function that fits the data.

Question 4.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 119
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Monitoring-progress-Exercise-4.9-Answer-4.jpg

Question 5.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 120
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Monitoring-progress-Exercise-4.9-Answer-5.jpg

Modeling with polynomial Functions 4.9 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE When the x-values in a set of data are equally spaced, the differences of consecutive y-values are called ________________.
Answer:
When the x-values in a set of data are equally spaced, the differences of consecutive y-values are called first differences.

Question 2.
WRITING Explain how you know when a set of data could be modeled by a cubic function.
Answer:
Utilizing finite differences to decide the degree of the polynomial function that fits the data, you know when a set of data could be demonstrated by a cubic function on the off chance that you get a third difference that is nonzero and constant.

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, write a cubic function whose graph is shown.

Question 3.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 121
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 3

Question 4.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 122
Answer:
https://ccssanswers.com/wp-content/uploads/2021/05/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-16.jpg

Question 5.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 123
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 5

Question 6.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 124
Answer:
Given,
-3, -6, 3 and 0.
y=a(x+6)(x+3)(x-3)
-9=a(0+6)(0+3)(0-3)
-9=-54a
1=6a
a=1/6
y=1/6(x+6)(x+3)(x-3)

In Exercises 7–12, use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.

Question 7.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 125
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 7

Question 8.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 126
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-8.jpg

Question 9.
(−4, −317), (−3, −37), (−2, 21), (−1, 7), (0, −1), (1, 3), (2, −47), (3, −289), (4, −933)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 9

Question 10.
(−6, 744), (−4, 154), (−2, 4), (0, −6), (2, 16), (4, 154), (6, 684), (8, 2074), (10, 4984)
Answer:
Big Ideas Math Answers Modeling with polynomial Functions 4.9 Exercises_10

Question 11.
(−2, 968), (−1, 422), (0, 142), (1, 26), (2, −4), (3, −2), (4, 2), (5, 2), (6, 16)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 11

Question 12.
(1, 0), (2, 6), (3, 2), (4, 6), (5, 12), (6, −10), (7, −114), (8, −378), (9, −904)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-12.jpg

Question 13.
ERROR ANALYSIS Describe and correct the error in writing a cubic function whose graph passes through the given points.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 127
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 13

Question 14.
MODELING WITH MATHEMATICS The dot patterns show pentagonal numbers. The number of dots in the nth pentagonal number is given by f(n) = \(\frac{1}{2}\)n(3n − 1). Show that this function has constant second-order differences.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 128
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-14.jpg

Question 15.
OPEN-ENDED Write three different cubic functions that pass through the points (3, 0), (4, 0), and (2, 6). Justify your answers.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 15

Question 16.
MODELING WITH MATHEMATICS The table shows the ages of cats and their corresponding ages in human years. Find a polynomial model for the data for the first 8 years of a cat’s life. Use the model to estimate the age (in human years) of a cat that is 3 years old.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 129
Answer:
https://ccssanswers.com/wp-content/uploads/2021/05/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-16.jpg

Question 17.
MODELING WITH MATHEMATICS The data in the table show the average speeds y (in miles per hour) of a pontoon boat for several different engine speeds x (in hundreds of revolutions per minute, or RPMs). Find a polynomial model for the data. Estimate the average speed of the pontoon boat when the engine speed is 2800 RPMs.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 130
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 17

Question 18.
HOW DO YOU SEE IT? The graph shows typical speeds y (in feet per second) of a space shuttle x seconds after it is launched.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 131
a. What type of polynomial function models the data? Explain.
b. Which nth-order finite difference should be constant for the function in part (a)? Explain.
Answer:
a) The given graph suggests a cubic model as it has the shape of a cubic polynomial’s graph.

Question 19.
MATHEMATICAL CONNECTIONS The table shows the number of diagonals for polygons with n sides. Find a polynomial function that fits the data. Determine the total number of diagonals in the decagon shown.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 132
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 133
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 19

Question 20.
MAKING AN ARGUMENT Your friend states that it is not possible to determine the degree of a function given the first-order differences. Is your friend correct? Explain your reasoning.
Answer:

Question 21.
WRITING Explain why you cannot always use finite differences to find a model for real-life data sets.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 21

Question 22.
THOUGHT PROVOKING A, B, and C are zeros of a cubic polynomial function. Choose values for A, B, and C such that the distance from A to B is less than or equal to the distance from A to C. Then write the function using the A, B, and C values you chose.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Monitoring-progress-Exercise-4.9-Answer-5.jpg

Question 23.
MULTIPLE REPRESENTATIONS Order the polynomial functions according to their degree, from least to greatest.
A.f(x) = −3x + 2x2 + 1
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 134
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 23

Question 24.
ABSTRACT REASONING Substitute the expressions z, z + 1, z + 2, …….. , z + 5 for x in the function f(x) = ax3 + bx2 + cx + d to generate six equally-spaced ordered pairs. Then show that the third-order differences are constant.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-24.jpg

Maintaining Mathematical Proficiency

Solve the equation using square roots.(Section 3.1)

Question 25.
x2 − 6 = 30
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 25

Question 26.
5x2 − 38 = 187
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-26.jpg

Question 27.
2(x − 3)2 = 24
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 27

Question 28.
\(\frac{4}{3}\)(x + 5)2 = 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-28.jpg

Solve the equation using the Quadratic Formula.(Section 3.4)

Question 29.
2x2 + 3x = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 29

Question 30.
2x2 + \(\frac{1}{2}\) = 2x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-30JPG.jpg

Question 31.
2x2 + 3x =−3x2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 4 Polynomial Functions 4.9 a 31

Question 32.
4x − 20 = x2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Exercise-4.9-Answer-32.jpg

Polynomial Functions Performance Task: For the Birds-Wildlife Management

Core Vocabulary
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 135

Core Concepts
Section 4.5
Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 136

Section 4.6
Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 137

Section 4.7

Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 138

Section 4.8
Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 139

Section 4.9
Big Ideas Math Answer Key Algebra 2 Chapter 4 Polynomial Functions 140

Mathematical Practices

Question 1.
Explain how understanding the Complex Conjugates Theorem allows you to construct your argument in Exercise 46 on page 203.

Question 2.
Describe how you use structure to accurately match each graph with its transformation in Exercises 7–10 on page 209.

Performance Task
For the Birds -Wildlife Management
How does the presence of humans affect the population of sparrows in a park? Do more humans mean fewer sparrows? Or does the presence of humans increase the number of sparrows up to a point? Are there a minimum number of sparrows that can be found in a park, regardless of how many humans there are? What can a mathematical model tell you?
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 204
To explore the answers to these questions and more, go to BigIdeasMath.com.

Polynomial Functions Chapter Review

Decide whether the function is a polynomial function. If so, write it in standard form and state its degree, type, and leading coefficient.

Question 1.
h(x) = −x3 + 2x2 − 15x7

Answer:


Question 2.
p(x) = x3 − 5x0.5 + 13x2 + 8
Answer:

Thus p(x) is not a polynomial function

Graph the polynomial function.

Question 3.
h(x) = x2 + 6x5 − 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-3.jpg

Question 4.
f(x) = 3x4 − 5x2 + 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-4.jpg

Question 5.
g(x) = −x4 + x + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-5.jpg

Find the sum or difference.

Question 6.
(4x3 − 12x2 − 5) − (−8x2 + 4x + 3)
Answer:
Given,
4x3 − 12x2 − 5 + 8x2 – 4x – 3
4x3 − 4x2 − 8
4(x3 − x2 − 2)
So, (4x3 − 12x2 − 5) − (−8x2 + 4x + 3) = 4(x3 − x2 − 2)

Question 7.
(x4 + 3x3 − x2 + 6) + (2x4 − 3x + 9)
Answer:
Given,
(x4 + 3x3 − x2 + 6) + (2x4 − 3x + 9)
x4 + 3x3 − x2 + 6 + 2x4 − 3x + 9
3x4 + 3x3 − x2 -3x + 15
So, (x4 + 3x3 − x2 + 6) + (2x4 − 3x + 9) = 3x4 + 3x3 − x2 -3x + 15

Question 8.
(3x2 + 9x + 13) − (x2 − 2x + 12)
Answer:
Given,
(3x2 + 9x + 13) − (x2 − 2x + 12)
3x2 + 9x + 13 − x2 − 2x + 12
2x2 + 7x + 25
So, (3x2 + 9x + 13) − (x2 − 2x + 12) = 2x2 + 7x + 25

Find the product.

Question 9.
(2y2 + 4y − 7)(y + 3)
Answer:
(2y2 + 4y − 7)(y + 3)
y(2y2 + 4y − 7) + 3(2y2 + 4y − 7)
2y³ + 4y² – 7y + 6y² + 12y – 21
2y³ + 10y² + 5y – 21

Question 10.
(2m + n)3

Answer:
Given,
(2m + n)³
This is in the form of (a + b)³
(2m + n)³ = (2m)³ + (n)³ + 3(2m)²(n) + 3(2m)(n)²
= 8m³ + n³ + 12m²n + 6mn²

Question 11.
(s + 2)(s + 4)(s − 3)
Answer:
Given,
s(s + 4) + 2(s + 4) (s – 3)
(s² + 4s + 2s + 8)(s – 3)
s³ + 4s² + 2s² + 8s – 3s² – 12s – 6s – 24
s³ + 3s² – 10s – 24
So, (s + 2)(s + 4)(s − 3) = s³ + 3s² – 10s – 24

Use Pascal’s Triangle to expand the binomial.

Question 12.
(m + 4)4

Answer:
Given,(m + 4)4
(m + 4)4 = (m)4 + (4)4 + 6m²(4)² + 4m(4)³
= m4 + 256 + 96m² + 256m
So, (m + 4)4 = m4 + 256 + 96m² + 256m

Question 13.
(3s + 2)5

Answer:
Given,
(3s + 2)5
This is in the form of (x + y)5
1(3s)5 + 5(3s)4(2) + 10(3s)³ (2) + 10(3s)²(2) + 5(3s)(2)4 + 1(2)
= (3s)5 – 10(3s)4 + 20(3s)³ – 20(3s)² + 80(3s) – 2
= 243s5 – 810s4 + 540s³ – 180s² + 240s – 2
So, (3s + 2)5 = 243s5 – 810s4 + 540s³ – 180s² + 240s – 2

Question 14.
(z + 1)6

Divide using polynomial long division or synthetic division.

Question 15.
(x3 + x2 + 3x − 4) ÷ (x2 + 2x + 1)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-15.jpg

Question 16.
(x4 + 3x3 − 4x2 + 5x + 3) ÷ (x2 + x + 4)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-16.jpg

Question 17.
(x4 − x2 − 7) ÷ (x + 4)
Answer:

Question 18.
Use synthetic division to evaluate g(x) = 4x3 + 2x2 − 4 when x = 5.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-18.jpg

Factor the polynomial completely.

Question 19.
64x3 − 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-19.jpg

Question 20.
2z5 − 12z3 + 10z
Answer:

Question 21.
2a3 − 7a2 − 8a + 28
Answer:

Question 22.
Show that x + 2 is a factor of f(x) = x4 + 2x3 − 27x − 54. Then factor f(x) completely.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-22JPG.jpg

Find all real solutions of the equation.

Question 23.
x3 + 3x2 − 10x − 24 = 0
Answer:

Question 24.
x3 + 5x2 − 2x − 24 = 0
Answer:

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 25.
1, 2 − \(\sqrt{3}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-25.jpg

Question 26.
2, 3, \(\sqrt{5}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-26.jpg

Question 27.
−2, 5, 3 + \(\sqrt{6}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-27.jpg

Question 28.
You use 240 cubic inches of clay to make a sculpture shaped as a rectangular prism. The width is 4 inches less than the length and the height is 2 inches more than three times the length. What are the dimensions of the sculpture? Justify your answer.
Answer:
Given,
L = x
W = x-4
H = 3x + 2
V = 240
We know that,
V = LWH
240 = x × (x – 4) (3x + 2)
240 = 3x³ – 10x² – 8x
3x³ – 10x² – 8x – 240 = 0
3x³ – 18x² – 8x²- 48x – 40x – 240 = 0
3x²(x – 6) + 8x(x – 6) + 40(x – 6) = 0
(x – 6)(3x² + 8x + 40) = 0
x – 6 = 0
x = 6
3x² + 8x + 40 = 0
Δ = 8² – 4 × 3 × 40 = 64 – 480 = -416
x = L = 6
H = 6 – 2 = 4
W = 3 × 6 + 2 = 20

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 29.
3, 1 + 2i
Answer:

Question 30.
−1, 2, 4i
Answer:

Question 31.
−5, −4, 1 −i\(\sqrt{3}\)
Answer:

Determine the possible numbers of positive real zeros, negative real zeros, and imaginary zeros for the function.

Question 32.
f(x) = x4 − 10x + 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-32.jpg

Question 33.
f(x) = −6x4 − x3 + 3x2 + 2x + 18
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-33.jpg

Describe the transformation of f represented by g. Then graph each function.

Question 34.
f(x) = x3, g(x) = (−x)3 + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-34.jpg

Question 35.
f(x) = x4, g(x) = −(x + 9)4

Answer:

Write a rule for g.

Question 36.
Let the graph of g be a horizontal stretch by a factor of 4, followed by a translation 3 units right and 5 units down of the graph of f(x) = x5 + 3x.
Answer:

Question 37.
Let the graph of g be a translation 5 units up, followed by a reflection in the y-axis of the graph of f(x) = x4 − 2x3 − 12.
Answer:

Graph the function. Identify the x-intercepts and the points where the local maximums and local minimums occur. Determine the intervals for which the function is increasing or decreasing.

Question 38.
f(x) = −2x3 − 3x2 − 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-38.jpg

Question 39.
f(x) = x4 + 3x3 − x2 − 8x + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-review-Answer-39.jpg

Determine whether the function is even, odd, or neither.

Question 40.
f(x) = 2x3 + 3x
Answer:

Question 41.
g(x) = 3x2 − 7
Answer:

Question 42.
h(x) = x6 + 3x5

Answer:

Question 43.
Write a cubic function whose graph passes through the points (−4, 0), (4, 0), (0, 6), and (2, 0).
Answer:
Given,
(-4, 0);
(4, 0);
(0, 6);
(2, 0).
Develop the general equation based on zeros.
f(0) = a(x+4)(0-4)(0-2)
Point (0,6) means x = 0 and y = 6.
6 = a(4)(-4)(-2)
6 = a(32)
6/32 = a
3/16 = a.
f(x) = (3/16)(x+4)(x-4)(x-2).

Question 44.
Use finite differences to determine the degree of the polynomial function that fits the data. Then use technology to find the polynomial function.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 206
Answer:

Polynomial Functions Chapter Test

Write a polynomial function f of least degree that has rational coefficients, a leading coefficient of 1, and the given zeros.

Question 1.
3, 1 − \(\sqrt{2}\)
Answer:

Question 2.
−2, 4, 3i
Answer:

Find the product or quotient.

Question 3.
(x6 − 4)(x2 − 7x + 5)
Answer:
Given,
(x6 − 4)(x2 − 7x + 5)
x6(x2 − 7x + 5) – 4(x2 − 7x + 5)
x8 – 7x7 + 5x6 – 4x2 + 28x – 20
So, (x6 − 4)(x2 − 7x + 5) = x8 – 7x7 + 5x6 – 4x2 + 28x – 20

Question 4.
(3x4 − 2x3 − x − 1) ÷ (x2 − 2x + 1)
Answer:

Question 5.
(2x3 − 3x2 + 5x − 1) ÷ (x + 2)
Answer:

Question 6.
(2x + 3)3

Answer:
Given,
(2x + 3)3
This is in the form of (a + b)³
(2x + 3)3 = (2x)³ + (3)³ + 3(2x)²(3) + 3(2x)(3)²
= 8x³ + 27 + 36x² + 54x

Question 7.
The graphs of f(x) = x4 and g(x) = (x − 3)4 are shown.
a. How many zeros does each function have? Explain.
b. Describe the transformation of f represented by g.
c. Determine the intervals for which the function g is increasing or decreasing.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 207
Answer:

Question 8.
The volume V (in cubic feet) of an aquarium is modeled by the polynomial function V(x) = x3 + 2x2 − 13x + 10, where x is the length of the tank.
a. Explain how you know x = 4 is not a possible rational zero.
b. Show that x − 1 is a factor of V(x). Then factor V(x) completely.
c. Find the dimensions of the aquarium shown.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 208
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-test-Answer-8.jpg

Question 9:
One special product pattern is (a − b)2 = a2 − 2ab + b2. Using Pascal’s Triangle to expand (a − b)2 gives 1a2 + 2a(−b) + 1(−b)2. Are the two expressions equivalent? Explain.
Answer:

Question 10.
Can you use the synthetic division procedure that you learned in this chapter to divide any two polynomials? Explain.
Answer:

Question 11.
Let T be the number (in thousands) of new truck sales. Let C be the number (in thousands) of new car sales. During a 10-year period, T and C can be modeled by the following equations where t is time (in years).
T = 23t4 − 330t3 + 3500t2 − 7500t + 9000
C = 14t4 − 330t3 + 2400t2 − 5900t + 8900
a. Find a new model S for the total number of new vehicle sales.
b. Is the function S even, odd, or neither? Explain your reasoning.
Answer:

Question 12.
Your friend has started a golf caddy business. The table shows the profits p (in dollars) of the business in the first 5 months. Use finite differences to find a polynomial model for the data. Then use the model to predict the profit after 7 months.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 209
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-test-12Answer-.jpg

Polynomial Functions Cumulative Assessment

Question 1.
The synthetic division below represents f(x) ÷ (x− 3). Choose a value for m so that x − 3 is a factor of f(x). Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 210

Question 2.
Analyze the graph of the polynomial function to determine the sign of the leading coefficient, the degree of the function, and the number of real zeros. Explain.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 211
Answer:

Question 3.
Which statement about the graph of the equation 12(x− 6) = −( y + 4)2 is not true?
A. The vertex is (6, −4).
B. The axis of symmetry is y = −4.
C. The focus is (3, −4).
D. The graph represents a function.
Answer:

Question 4.
A parabola passes through the point shown in the graph. The equation of the axis of symmetry is x = −a. Which of the given points could lie on the parabola? If the axis of symmetry was x = a, then which points could lie on the parabola? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 212
Answer:

Question 5.
Select values for the function to model each transformation of the graph of f(x) = x.
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 213
a. The graph is a translation 2 units up and 3 units left.
b. The graph is a translation 2 units right and 3 units down.
c. The graph is a vertical stretch by a factor of 2, followed by a translation 2 units up.
d. The graph is a translation 3 units right and a vertical shrink by a factor of \(\frac{1}{2}\), followed by a translation 4 units down.
Answer:

Question 6.
The diagram shows a circle inscribed in a square. The area of the shaded region is21.5 square meters. To the nearest tenth of a meter, how long is each side of the square?
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 214
A. 4.6 meters
B. 8.7 meters
C. 9.7 meters
D. 10.0 meters
Answer:

Question 7.
Classify each function as even, odd, or neither. Justify your answer.
a. f(x) = 3x5
b. f(x) = 4x3 + 8x
c. f(x) = 3x5 + 12x2+ 1
d. f(x) = 2x4
e. f(x) = x11 − x7
f. f(x) = 2x8 + 4x4 + x2 − 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-assissment-Answer-7.jpg

Question 8.
The volume of the rectangular prism shown is given by V = 2x3 + 7x2 − 18x − 63. Which polynomial represents the area of the base of the prism?
A. 2x2 + x − 21
B. 2x2 + 21 − x
C. 13x + 21 + 2x2
D. 2x2 − 21 − 13x
Big Ideas Math Algebra 2 Solutions Chapter 4 Polynomial Functions 215
Answer:

Question 9.
The number R (in tens of thousands) of retirees receiving Social Security benefits is represented by the function R = 0.286t3 − 4.68t2 + 8.8t + 403, 0 ≤ t ≤ 10
where t represents the number of years since 2000. Identify any turning points on the given interval. What does a turning point represent in this situation?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-Chapter.-4-Polynomials-Chapter-assissment-Answer-9.jpg

Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers

Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers

Download Algebra 2 Big Ideas Math Answers Chapter 3 Quadratic Equations and Complex Numbers free pdf here. Get a brief explanation for all the questions from Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers. So students who are worried about the practice sessions and scores of the upcoming exams can make use of our Big Ideas Math Textbook Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers. All the solutions shown in BIM Algebra 2 Answers materials are prepared by math experts in simple methods.

Big Ideas Math Book Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers

Check out the modules according to the topics from Big Ideas Math Textbook Algebra 2 Ch 3 Quadratic Equations and Complex Numbers Solution Key. With the help of BIM Algebra 2 Solution Key Chapter 3 Quadratic Equations and Complex Numbers guidance, students can learn something new in easy methods and also score better grades on the exams. Hit the links provided below to start practicing the exercises.

Quadratic Equations and Complex Numbers Maintaining Mathematical Proficiency

Simplify the expression.
Question 1.
\(\sqrt{27}\)
Answer:
Given expression is \(\sqrt{27}\)
27 can be written as 3³
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-1.jpg
\(\sqrt{27}\) = 3.

Question 2.
–\(\sqrt{112}\)
Answer:
Given expression is –\(\sqrt{112}\)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-2.jpg
–\(\sqrt{112}\) = -4-\(\sqrt{7}\)

Question 3.
\(\sqrt{\frac{11}{64}}\)
Answer:
Given expression is \(\sqrt{\frac{11}{64}}\)
\(\sqrt{\frac{11}{8²}}\)
\(\sqrt{\frac{11}{64}}\) = \(\sqrt{11}\)/8

Question 4.
\(\sqrt{\frac{147}{100}}\)
Answer:
Given expression is \(\sqrt{\frac{147}{100}}\)
\(\sqrt{\frac{147}{100}}\) = \(\sqrt{\frac{147}{10²}}\)
\(\sqrt{\frac{147}{10²}}\) = \(\sqrt{147}\)/10
So, \(\sqrt{\frac{147}{10²}}\) = \(\sqrt{147}\)/10

Question 5.
\(\sqrt{\frac{18}{49}}\)
Answer:
Given expression is \(\sqrt{\frac{18}{49}}\)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-5.jpg
\(\sqrt{\frac{18}{49}}\) = \(\sqrt{18}\)/7 = 6\(\sqrt{2}\)/10 = 3/5 . \(\sqrt{2}\)

Question 6.
–\(\sqrt{\frac{65}{121}}\)
Answer:
Given expression is –\(\sqrt{\frac{65}{121}}\)
121 = 11²
–\(\sqrt{\frac{65}{121}}\) = –\(\sqrt{65}\)/11

Question 7.
\(\sqrt{80}\)
Answer:
Given expression is \(\sqrt{80}\)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-7.jpg
So, \(\sqrt{80}\) = 4\(\sqrt{5}\)

Question 8.
\(\sqrt{32}\)
Answer:
Given expression is \(\sqrt{32}\)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-8.jpg
So, \(\sqrt{32}\) = 4\(\sqrt{2}\)

Factor the polynomial.
Question 9.
x2 − 36
Answer:
Given,
x² – 36
x2 – 62
(x +6) (x- 6)
So, x2 − 36 = (x +6) (x- 6)

Question 10.
x2 − 9
Answer:
Given,
x2 – 9
x2 – 32
(x – 3) ( x+3)
So, x2 − 9 = (x – 3) (x + 3)

Question 11.
4x2 − 25
Answer:
Given,
4x2 – 25
= (2x)2  –  5²
= (2x +5)( 2x – 5)
So, 4x2 – 25 = (2x +5)(2x – 5)

Question 12.
x2 − 22x + 121
Answer:
Given,
x2– 22x +121
x2 – 11x -11x +121
x(x-11)-11(x-11)
(x – 11)(x – 11)
So, x2 − 22x + 121 = (x – 11)(x – 11)

Question 13.
x2 + 28x + 196
Answer:
Given,
X2 + 28x +196x
x2+14x+14x+196
x(x+14)+14(x+14)
(x+14)(x+14)
So, x2 + 28x + 196 = (x+14)(x+14)

Question 14.
49x2 + 210x + 225
Answer:
Given,
49x2 + 210x + 225
x = (-b ± √b² – 4ac)/2a
x = (-210 ± √210² – 4(49)(225))/2(49)
x = -210/98
b² – 4ac = 0
x = -210/98
x = -15/7 or -2.14
The solution is x = -15/7 or -2.14

Question 15.
ABSTRACT REASONING
Determine the possible integer values of a and c for which the trinomial ax2+ 8x+c is factorable using the Perfect Square Trinomial Pattern. Explain your reasoning.
Answer:
The term ‘a’ is referred to as the leading coefficient, while ‘c’ is the absolute term of f(x). Every quadratic equation has two values of the unknown variable.

Quadratic Equations and Complex Numbers Mathematical Practices

Mathematically proficient students recognize the limitations of technology

Monitoring Progress

Question 1.
Explain why the second viewing window in Example 1 shows gaps between the upper and lower semicircles, but the third viewing window does not show gaps.
Answer:

Use a graphing calculator to draw an accurate graph of the equation. Explain your choice of viewing window.

Question 2.
y = \(\sqrt{x^{2}-1.5}\)
Answer:
We can use the graphing calculator to draw y = \(\sqrt{x^{2}-1.5}\).
Write the equation y = \(\sqrt{x^{2}-1.5}\) in the input.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers Mathematical Practices_2

Question 3.
y = \(\sqrt{x-2.5}\)
Answer:
We can use the graphing calculator to draw y = \(\sqrt{x-2.5}\)
Write the equation y = \(\sqrt{x-2.5}\) in the input.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers Mathematical Practices_3

Question 4.
x2 + y2= 12.25
Answer:
We can use the graphing calculator to draw x2 + y2= 12.25
Write the equation x2 + y2= 12.25 in the input.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers Mathematical Practices_4

Question 5.
x2 + y2 = 20.25
Answer:
We can use the graphing calculator to draw x2 + y2= 12.25
Write the equation x2 + y2= 12.25 in the input.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers Mathematical Practices_5

Question 6.
x2 + 4y2 = 12.25
Answer:
We can use the graphing calculator to draw x2 + 4y2 = 12.25
Write the equation x2 + 4y2 = 12.25 in the input.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers Mathematical Practices_6

Question 7.
4x2 + y2 = 20.25
Answer:
We can use the graphing calculator to draw 4x2 + y2 = 20.25
Write the equation 4x2 + y2 = 20.25 in the input.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers Mathematical Practices_7

Lesson 3.1 Solving Quadratic Equations

Essential Question How can you use the graph of a quadratic equation to determine the number of real solutions of the equation?

EXPLORATION 1

Matching a Quadratic Function with Its Graph
Work with a partner. Match each quadratic function with its graph. Explain your reasoning. Determine the number of x-intercepts of the graph.
a. f(x) = x2 − 2x
b. f(x) = x2 − 2x + 1
c. f(x) = x2 − 2x + 2
d. f(x) = −x2 + 2x
e. f(x) = −x2 + 2x − 1
f. f(x) = −x2 + 2x − 2
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 1

EXPLORATION 2

Solving Quadratic Equations
Work with a partner. Use the results of Exploration 1 to find the real solutions (if any) of each quadratic equation.
a. x2 − 2x = 0
b. x2 − 2x + 1 = 0
c. x2 − 2x + 2 = 0
d. −x2 + 2x = 0
e. −x2 + 2x − 1 = 0
f. −x2 + 2x − 2 = 0

Communicate Your Answer

Question 3.
How can you use the graph of a quadratic equation to determine the number of real solutions of the equation?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 2.1
Answer:
If the graph of a quadratic function crosses the x-axis at two points, then the has two real rational solutions. The number of real solutions of a quadratic equation can be found in the number of x-intercepts on the graph. If the curve passes through the x-axis twice, then the equation has two real solutions. If it passes through once, (the vertex is on the x-axis), then there is one real solution, and if it does not touch the x-axis at all, then it has no real solutions.

Question 4.
How many real solutions does the quadratic equation x2 + 3x + 2 = 0 have? How do you know? What are the solutions?
Answer:
Given equation is x2 + 3x + 2 = 0
x2 + 3x + 2 = 0
x2 + 1x + 2x + 2 = 0
x(x + 1) +2 (x + 1) = 0
(x + 2) (x + 1) = 0
x = -2 or x = -1
There are 2 solutions.
Parabolas have a highest or the lowest point called the Vertex

Monitoring Progress

Solve the equation by graphing.
Question 1.
x2 − 8x + 12 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-1.jpg
Question 2.
4x2 − 12x + 9 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-2.jpg
Question 3.
\(\frac{1}{2}\)x2 = 6x − 20
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-3.jpg

Solve the equation using square roots.
Question 4.
\(\frac{2}{3}\)x2 + 14 = 20
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-4.jpg

Question 5.
−2x2 + 1 = −6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-5.jpg

Question 6.
2(x − 4)2 = −5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-6.jpg

Solve the equation by factoring.
Question 7.
x2 + 12x + 35 = 0
Answer:

Question 8.
3x2 − 5x = 2
Answer:

Find the zero(s) of the function.
Question 9.
f(x) = x2 − 8x
Answer:

Question 10.
f(x) = 4x2 + 28x + 49
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-10.jpg

Question 11.
WHAT IF?
The magazine initially charges $21 per annual subscription. How much should the magazine charge to maximize annual revenue? What is the maximum annual revenue?
Answer:
The magazine initially charges $21 per annual subscription.
Let x be the increase in price and R(x) is annual revenue with initial subscribers of 48,000.
Annual revenue = Number of subscribers + subscription price
Number of subscribers = Initial subscribers – Number of subscribers lost × x
= 48,000 – 2000x
Subscription price = initial price + increase price
= 21 + x
R(x) = (48000 – 2000x)(21 + x)
R(x) = -2000(x – 24)(x + 21)
x – 24 = 0
x + 21 = 0
x = 24, -21
24-21/2 = 3/2
Thus to maximize profit it must cost 21 + 1.5 = 22.5
R(x) = (48,000 – 2000x)(21 + x)
R(x) = -2000 (x – 24)(x + 21)
Substitute x = 1.5 in the equation.
R(1.5) = -2000(1.5 – 2.4) (1.5 + 21)
= $1,012,500
The maximum annual revenue is $1,012,500

Question 12.
WHAT IF?
The egg container is dropped from a height of 80 feet. How does this change your answers in parts (a) and (b)?
Answer:
Given,
h = 80 ft
h = -16t² + h
h = -16t² + 80
0 = -16t² + 80
-16t² + 80 = 0
-16t² = -80
t² = -80/-16
t² = 5
t = ± \(\sqrt{5}\)
For h(1) and h(1.5) we will replace t by 1and 1.5
h(1) = -16(1)² + 80
h(1) = 64
h(1.5) = -16(1.5)² + 80
= -36 + 80  = 44

Solving Quadratic Equations 3.1 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Explain how to use graphing to find the roots of the equation ax2 + bx + c = 0.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 2
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, solve the equation by graphing.
Question 3.
x2 + 3x + 2 = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 3

Question 4.
−x2 + 2x + 3 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-1-4.jpg

Question 5.
y = x2 − 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 5

Question 6.
−8 = −x2 − 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-6-1.jpg

Question 7.
8x = −4 − 4x2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 7

Question 8.
3x2 = 6x − 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-8.jpg

Question 9.
7 = −x2 − 4x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 9

Question 10.
2x = x2 + 2
Answer:
Given,
2x = x2 + 2
x2 – 2x + 2 = 0
Now the graph the related function.

There is no x-intercept. So, there is no solution.

Question 11.
\(\frac{1}{5}\)x2 + 6 = 2x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 11

Question 12.
3x = \(\frac{1}{4}\)x2 + 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-MonitoringProgress-ex-1-12.jpg

In Exercises 13–20, solve the equation using square roots.
Question 13.
s2 = 144
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 13

Question 14.
a2 = 81
Answer:

Question 15.
(z − 6)2 = 25
Answer:
Given,
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 15

Question 16.
(p − 4)2 = 49
Answer:

Question 17.
4(x − 1)2 + 2 = 10
Answer:
Given,
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 17

Question 18.
2(x + 2)2 − 5 = 8
Answer:
Given
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-1-18.jpg

Question 19.
\(\frac{1}{2}\)r2 − 10 = \(\frac{3}{2}\)r2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 19

Question 20.
\(\frac{1}{5}\)x2 + 2 = \(\frac{3}{5}\)x2
Answer:

Question 21.
ANALYZING RELATIONSHIPS
Which equations have roots that are equivalent to the x-intercepts of the graph shown?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 3
A. −x2 − 6x − 8 = 0
B. 0 = (x + 2)(x + 4)
C. 0 = −(x + 2)2 + 4
D. 2x2 − 4x − 6 = 0
E. 4(x + 3)2 − 4 = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 21

Question 22.
ANALYZING RELATIONSHIPS
Which graph has x-intercepts that are equivalent to the roots of the equation (x − \(\frac{3}{2}\))2 = \(\frac{25}{4}\)? Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 4
Answer:

ERROR ANALYSIS In Exercises 23 and 24, describe and correct the error in solving the equation.
Question 23.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 5
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 23

Question 24.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 6
Answer:
-2x² – 8 = 0
x² + 8 = 0
x² = -8
x = -2√2

Question 25.
OPEN-ENDED
Write an equation of the form x2 = d that has (a) two real solutions, (b) one real solution, and (c) no real solution.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 25

Question 26.
ANALYZING EQUATIONS
Which equation has one real solution? Explain.
A. 3x2 + 4 = −2(x2 + 8)
B. 5x2 − 4 = x2 − 4
C. 2(x + 3)2 = 18
D. \(\frac{3}{2}\)x2 − 5 = 19
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-1-26.jpg

In Exercises 27–34, solve the equation by factoring.
Question 27.
0 = x2 + 6x + 9
Answer:
Given,
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 27

Question 28.
0 = z2 − 10z + 25
Answer:

Question 29.
x2 − 8x = −12
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 29

Question 30.
x2 − 11x = −30
Answer:

Question 31.
n2 − 6n = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 31

Question 32.
a2 − 49 = 0
Answer:

Question 33.
2w2 − 16w = 12w − 48
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 33

Question 34.
−y + 28 + y2 = 2y + 2y2
Answer:

MATHEMATICAL CONNECTIONS In Exercises 35–38, find the value of x.
Question 35.
Area of rectangle = 36
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 35

Question 36.
Area of circle = 25π
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 8
Answer:

Question 37.
Area of triangle = 42
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 37

Question 38.
Area of trapezoid = 32
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 10
Answer:

In Exercises 39–46, solve the equation using any method. Explain your reasoning.
Question 39.
u2 = −9u
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 39

Question 40.
\(\frac{t^{2}}{20}\) + 8 = 15
Answer:

Question 41.
−(x + 9)2 = 64
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 41

Question 42.
−2(x + 2)2 = 5
Answer:

Question 43.
7(x − 4)2 − 18 = 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 43

Question 44.
t2 + 8t + 16 = 0
Answer:

Question 45.
x2 + 3x + \(\frac{5}{4}\) = 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 45

Question 46.
x2 − 1.75 = 0.5
Answer:

In Exercises 47–54, find the zero(s) of the function.
Question 47.
g(x) = x2 + 6x + 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 47

Question 48.
f(x) = x2 − 8x + 16
Answer:

Question 49.
h(x) = x2 + 7x − 30
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 49

Question 50.
g(x) = x2 + 11x
Answer:

Question 51.
f(x) = 2x2 − 2x − 12
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 51

Question 52.
f(x) = 4x2 − 12x + 9
Answer:

Question 53.
g(x) = x2 + 22x + 121
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 53

Question 54.
h(x) = x2 + 19x + 84
Answer:

Question 55.
REASONING
Write a quadratic function in the form f(x) = x2 + bx + c that has zeros 8 and 11.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 55

Question 56.
NUMBER SENSE
Write a quadratic equation in standard form that has roots equidistant from 10 on the number line.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-1-56.jpg

Question 57.
PROBLEM SOLVING
A restaurant sells 330 sandwiches each day. For each $0.25 decrease in price, the restaurant sells about 15 more sandwiches. How much should the restaurant charge to maximize daily revenue? What is the maximum daily revenue?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 11
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 57

Question 58.
PROBLEM SOLVING
An athletic store sells about 200 pairs of basketball shoes per month when it charges $120 per pair. For each $2 increase in price, the store sells two fewer pairs of shoes. How much should the store charge to maximize monthly revenue? What is the maximum monthly revenue?
Answer:
Given,
An athletic store sells about 200 pairs of basketball shoes per month when it charges $120 per pair.
For each $2 increase in price, the store sells two fewer pairs of shoes.
Revenue = (120+2x)(200-2x)
Revenue formula
Set 120+2x = 0
and 200-2x = 0
Find the zeros of the equation 3
p = -60
and q = 100
f(x) = (x-p)(x-q) = 0
maximum at x = 20

Charge = $160
Charge = 120+2(20) = $160
Maximum Revenue = $25,600
Revenue = 160(200-2*20)= 160*160 = $25,600

Question 59.
MODELING WITH MATHEMATICS
Niagara Falls is made up of three waterfalls. The height of the Canadian Horseshoe Falls is about 188 feet above the lower Niagara River. A log falls from the top of Horseshoe Falls.
a. Write a function that gives the height h (in feet) of the log after t seconds. How long does the log take to reach the river?
b. Find and interpret h(2) − h(3).
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 59

Question 60.
MODELING WITH MATHEMATICS
According to legend, in 1589, the Italian scientist Galileo Galilei dropped rocks of different weights from the top of the Leaning Tower of Pisa to prove his conjecture that the rocks would hit the ground at the same time. The height h (in feet) of a rock after t seconds can be modeled by h(t) = 196 − 16t2.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 12
a. Find and interpret the zeros of the function. Then use the zeros to sketch the graph.
b. What do the domain and range of the function represent in this situation?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-1-60.jpg

Question 61.
PROBLEM SOLVING
You make a rectangular quilt that is 5 feet by 4 feet. You use the remaining 10 square feet of fabric to add a border of uniform width to the quilt. What is the width of the border?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 13
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 61

Question 62.
MODELING WITH MATHEMATICS
You drop a seashell into the ocean from a height of 40 feet. Write an equation that models the height h (in feet) of the seashell above the water after t seconds. How long is the seashell in the air?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-1-62.jpg

Question 63.
WRITING
The equation h = 0.019s2 models the height h (in feet) of the largest ocean waves when the wind speed is s knots. Compare the wind speeds required to generate 5-foot waves and 20-foot waves.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 14
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 63

Question 64.
CRITICAL THINKING
Write and solve an equation to find two consecutive odd integers whose product is 143.
Answer:

Question 65.
MATHEMATICAL CONNECTIONS
A quadrilateral is divided into two right triangles as shown in the figure. What is the length of each side of the quadrilateral?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 15
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 65

Question 66.
ABSTRACT REASONING
Suppose the equation ax2 + bx + c = 0 has no real solution and a graph of the related function has a vertex that lies in the second quadrant.
a. Is the value of a positive or negative? Explain your reasoning.
b. Suppose the graph is translated so the vertex is in the fourth quadrant. Does the graph have any x-intercepts? Explain.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-1-66.jpg

Question 67.
REASONING
When an object is dropped on any planet, its height h (in feet) after t seconds can be modeled by the function h = −\(\frac{g}{2}\)t2 + h0, where h0 is the object’s initial height and g is the planet’s acceleration due to gravity. Suppose a rock is dropped from the same initial height on the three planets shown. Make a conjecture about which rock will hit the ground first. Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 16
Answer:
Given,
When an object is dropped on any planet, its height h (in feet) after t seconds can be modeled by the function h = −\(\frac{g}{2}\)t2 + h0, where h0 is the object’s initial height and g is the planet’s acceleration due to gravity.
We need to make a conjecture about which rock will hit the ground first.
The rock will hit the ground on Jupiter first. Because the first term is negative, the height of the falling object will decrease faster as ga gets larger.

Question 68.
PROBLEM SOLVING
A café has an outdoor, rectangular patio. The owner wants to add 329 square feet to the area of the patio by expanding the existing patio as shown. Write and solve an equation to find the value of x. By what distance should the patio be extended?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 17
Answer:

Question 69.
PROBLEM SOLVING
A flea can jump very long distances. The path of the jump of a flea can be modeled by the graph of the function y = −0.189x2 + 2.462x, where x is the horizontal distance (in inches) and y is the vertical distance (in inches). Graph the function. Identify the vertex and zeros and interpret their meanings in this situation.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 69.1
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 69.2

Question 70.
HOW DO YOU SEE IT?
An artist is painting a mural and drops a paintbrush. The graph represents the height h (in feet) of the paintbrush after t seconds.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 18
a. What is the initial height of the paintbrush?
b. How long does it take the paintbrush to reach the ground? Explain.
Answer:

Question 71.
MAKING AN ARGUMENT
Your friend claims the equation x2 + 7x =−49 can be solved by factoring and has a solution of x = 7. You solve the equation by graphing the related function and claim there is no solution. Who is correct? Explain.
Answer:
Given,
x2 + 7x =−49
x = 7
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 71

Question 72.
ABSTRACT REASONING
Factor the expressions x2 − 4 and x2 − 9. Recall that an expression in this form is called a difference of two squares. Use your answers to factor the expression x2 − a2. Graph the related function y = x2 − a2. Label the vertex, x-intercepts, and axis of symmetry.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-1-72.jpg

Question 73.
DRAWING CONCLUSIONS
Consider the expression x2 + a2, where a > 0.
a. You want to rewrite the expression as (x + m)(x + n). Write two equations that m and n must satisfy.
b. Use the equations you wrote in part (a) to solve for m and n. What can you conclude?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 73

Question 74.
THOUGHT PROVOKING
You are redesigning a rectangular raft. The raft is 6 feet long and 4 feet wide. You want to double the area of the raft by adding to the existing design. Draw a diagram of the new raft. Write and solve an equation you can use to find the dimensions of the new raft.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 19
Answer:

Question 75.
MODELING WITH MATHEMATICS
A high school wants to double the size of its parking lot by expanding the existing lot as shown. By what distance x should the lot be expanded?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 20
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 75

Maintaining Mathematical Proficiency

Find the sum or difference.
Question 76.
(x2 + 2) + (2x2 − x)
Answer:

Question 77.
(x3 + x2 − 4) + (3x2 + 10)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 77

Question 78.
(−2x + 1) − (−3x2 + x)
Answer:

Question 79.
(−3x3 + x2 − 12x) − (−6x2 + 3x − 9)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 79

Find the product.
Question 80.
(x + 2)(x − 2)
Answer:

Question 81.
2x(3 − x + 5x2)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 81

Question 82.
(7 − x)(x − 1)
Answer:
Given,
(7 − x)(x − 1)
(7x)(x1)=7x+7(−1)+(x)x+(x)(−1)=7x7x²+x=x²+(7x+x)7=x²+8x7Use (1)SimplifyGroup like terms Reduce like terms

Question 83.
11x(−4x2 + 3x + 8)
Answer:

Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.1 a 83

Lesson 3.2 Complex Numbers

Essential Question What are the subsets of the set of complex numbers?
In your study of mathematics, you have probably worked with only real numbers, which can be represented graphically on the real number line. In this lesson, the system of numbers is expanded to include imaginary numbers. The real numbers and imaginary numbers compose the set of complex numbers.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 1

EXPLORATION 1

Classifying Numbers
Work with a partner. Determine which subsets of the set of complex numbers contain each number.
a. \(\sqrt{9}\)
b. \(\sqrt{0}\)
c. −\(\sqrt{4}\)
d. \(\sqrt{\frac{4}{9}}\)
e. \(\sqrt{2}\)
f. \(\sqrt{-1}\)

EXPLORATION 2

Complex Solutions of Quadratic Equations
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 2
Work with a partner. Use the definition of the imaginary unit i to match each quadratic equation with its complex solution. Justify your answers.
a. x2 − 4 = 0
b. x2 + 1 = 0
c. x2 − 1 = 0
d. x2 + 4 = 0
e. x2 − 9 = 0
f. x2 + 9 = 0
A. i
B. 3i
C. 3
D. 2i
E. 1
F. 2

Communicate Your Answer

Question 3.
What are the subsets of the set of complex numbers? Give an example of a number in each subset.
Answer:
Real numbers are the subset of complex numbers. A combination of a real and an imaginary number in the form a + bi, where a and b are real, and i is imaginary.
The values a and b can be zero, so the set of real numbers and the set of imaginary numbers are subsets of the set of complex numbers.
Example:
-3.5i is an example of the subset of the imaginary number.
9 is an example of real numbers subset.
2 + i, 3 – 6i combination of real and imaginary numbers

Question 4.
Is it possible for a number to be both whole and natural? natural and rational? rational and irrational? real and imaginary? Explain your reasoning.
Answer:

  • A natural number is a subset of a whole number all natural numbers are whole numbers. But all the whole numbers are not natural numbers except zero.
  • A number cannot be rational or irrational. All rational numbers can be written as a fraction with an integer as the numerator and a non-zero integer as the denominator.
  • All natural numbers are not rational numbers.
  • A square of an imaginary number is an imaginary number but not an actual number. The 0 is considered both real and imaginary.

Monitoring Progress

Find the square root of the number.
Question 1.
\(\sqrt{-4}\)
Answer:

Question 2.
\(\sqrt{-12}\)
Answer:

Question 3.
−\(\sqrt{-36}\)
Answer:

Question 4.
2\(\sqrt{-54}\)
Answer:

Find the values of x and y that satisfy the equation.
Question 5.
x + 3i = 9 − yi
Answer:
Given,
x + 3i = 9 − yi
Equate the equation to find the values of x and y.
x = 9 and 3i = -yi
3i = -yi
y = -3
So, x = 9 and y = -3

Question 6.
9 + 4yi = −2x + 3i
Answer:
9 + 4yi = −2x + 3i
Equate the equation to find the values of x and y.
-2x = 9 and 4yi = 3i
x = -9/2 and 4y = 3
y = 3/4
So, x = -9/2 and y = 3/4

Question 7.
WHAT IF?
In Example 4, what is the impedance of the circuit when the capacitor is replaced with one having a reactance of 7 ohms?
Answer:

Perform the operation. Write the answer in standard form.
Question 8.
(9 − i ) + (−6 + 7i )
Answer:
Given,
(9 − i ) + (−6 + 7i )
9 – i – 6 – 7i
Combine the like terms to write the given expression in the standard form.
3 – 8i
So, (9 − i ) + (−6 + 7i ) = 3 – 8i

Question 9.
(3 + 7i ) − (8 − 2i )
Answer:
Given,
(3 + 7i ) − (8 − 2i )
Combine the like terms to write the given expression in the standard form.
3 + 7i – 8 + 2i
-5 + 9i
So, (3 + 7i ) − (8 − 2i ) = -5 + 9i

Question 10.
−4 − (1 + i) − (5 + 9i)
Answer:
Given,
−4 − (1 + i) − (5 + 9i)
-4 – 1 – i – 5 – 9i
-5 – i – 5 – 9i
Combine the like terms to write the given expression in the standard form.
-10 – 10i
Take -10 as common.
-10(1 + i)
So, −4 − (1 + i) − (5 + 9i) = -10(1 + i)

Question 11.
(−3i)(10i)
Answer:
Given,
(−3i)(10i)
-30i²
We know that
i² = -1
-30(-1)
= 30
So, (−3i)(10i) = 30

Question 12.
i(8 − i)
Answer:
Given,
i(8 − i)
8i – i²
We know that
i² = -1
8i – (-1)
8i + 1
i(8 − i) = 8i + 1

Question 13.
(3 + i)(5 −i)
Answer:
Given,
(3 + i)(5 −i)
3(5 – i) + i(5 – i)
15 – 3i + 5i – i²
We know that
i² = -1
15 + 2i + 1
16 + 2i
Thus, (3 + i)(5 −i) = 16 + 2i

Solve the equation.
Question 14.
x2 = −13
Answer:
Given,
x2 = −13
Applying square root on both sides
x = \(\sqrt{-13}\)

Question 15.
x2= −38
Answer:
Given,
x2= −38
Applying square root on both sides
x = \(\sqrt{-38}\)

Question 16.
x2 + 11 = 3
Answer:
Given,
x2 + 11 = 3
x2 = 3 – 11
x2 = -8
Applying square root on both sides
x = 2√-2

Question 17.
x2 − 8 = −36
Answer:
Given,
x2 − 8 = −36
x2 = -36 + 8
x2 = -28
Applying square root on both sides
x = 2√-7
So, x = 2√-7

Question 18.
3x2 − 7 = −31
Answer:
Given,
3x2 − 7 = −31
3x2 = −31 + 7
3x2 = -24
x2 = -24/3
x2 = -8
Applying square root on both sides
x = 2√-2

Question 19.
5x2 + 33 = 3
Answer:
Given,
5x2 + 33 = 3
5x2 = 3 – 33
5x2 = -30
x2 = -30/5
x2 = -6
Applying square root on both sides
x = √-6

Find the zeros of the function.
Question 20.
f(x) = x2 + 7
Answer:
Given,
f(x) = x2 + 7
f(x) = 0
Equate the function to zero.
x2 + 7 = 0
x(x + 7) = 0
x = 0 or x + 7 = 0
x = 0 or x = -7

Question 21.
f(x) = −x2 − 4
Answer:
Given,
f(x) = −x2 − 4
f(x) = 0
Equate the function to zero.
−x2 − 4 = 0
x2 + 4 = 0
(x + 2) (x + 2) = 0
x + 2 = 0 or x + 2 = 0
x = -2 or x = -2

Question 22.
f(x) = 9x2 + 1
Answer:
f(x) = 9x2 + 1
f(x) = 0
Equate the function to zero.
9x2 + 1 = 0
(3x + 1) (3x + 1) = 0
3x + 1 = 0 or 3x + 1 = 0
3x = -1 or 3x = -1
x = -1/3 or x = -1/3

Complex Numbers 3.2 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What is the imaginary unit i defined as and how can you use i?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 1

Question 2.
COMPLETE THE SENTENCE
For the complex number 5 + 2i, the imaginary part is ____ and the real part is ____.
Answer: For the complex number 5 + 2i, the imaginary part is 2i and the real part is 5.

Question 3.
WRITING
Describe how to add complex numbers.
Answer: To add two complex numbers, add the real parts and the imaginary parts separately.

Question 4.
WHICH ONE DOESN’T BELONG?
Which number does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 3
Answer:
3 + 0i = 3 real number
2 + 5i is a complex number
\(\sqrt{3}\) + 6i is a complex number
0 – 7i is a complex number
3 + 0i does not belong to the other three because it is a real number.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–12, find the square root of the number.
Question 5.
\(\sqrt{-36}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 5

Question 6.
\(\sqrt{-64}\)
Answer:
Given,
\(\sqrt{-64}\)
= \(\sqrt{64}\) . \(\sqrt{-1}\)
= 8i
So, \(\sqrt{-64}\) = 8i

Question 7.
\(\sqrt{-18}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 7

Question 8.
\(\sqrt{-24}\)
Answer:
Given,
\(\sqrt{-24}\)
= \(\sqrt{24}\) . \(\sqrt{-1}\)
= \(\sqrt{4}\) . \(\sqrt{6}\) . \(\sqrt{-1}\)
= 2\(\sqrt{6}\) . i
= 2i\(\sqrt{6}\)
So, \(\sqrt{-24}\) = 2i\(\sqrt{6}\)

Question 9.
2\(\sqrt{-16}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 9

Question 10.
−3\(\sqrt{-49}\)
Answer:
Given,
−3\(\sqrt{-49}\)
= -3\(\sqrt{49}\) . \(\sqrt{-1}\)
-3 × 7i
= -21i
So, −3\(\sqrt{-49}\) = -21i

Question 11.
−4\(\sqrt{-32}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 11

Question 12.
6\(\sqrt{-63}\)
Answer:
Given,
6\(\sqrt{-63}\)
= 6 \(\sqrt{9}\) . \(\sqrt{7}\) . \(\sqrt{-1}\)
= 6 × 3 \(\sqrt{7}\)i
= 18 × \(\sqrt{7}\)i
= 18i . \(\sqrt{7}\)
So, 6\(\sqrt{-63}\) = 18i . \(\sqrt{7}\)

In Exercises 13–20, find the values of x and y that satisfy the equation.
Question 13.
4x + 2i = 8 + yi
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 13

Question 14.
3x + 6i = 27 + yi
Answer:
Given
3x + 6i = 27 + yi
3x = 27 and 6i = yi
x = 27/3 and 6 = y
x = 9 and y = 6

Question 15.
−10x + 12i = 20 + 3yi
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 15

Question 16.
9x − 18i = −36 + 6yi
Answer:
Given,
9x − 18i = −36 + 6yi
9x = -36 and -18i = 6yi
x = -36/9 and y = -18/6
x = -4 and -3 = y
So, x = -4 and y = -3

Question 17.
2x − yi = 14 + 12i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 17

Question 18.
−12x + yi = 60 − 13i
Answer:
Given,
−12x + yi = 60 − 13i
-12x = 60 and yi = -13i
x = 60/-12 and y = -13
x = -5 and y = -13i
So, the real part x = -5 and imaginary part y = -13i

Question 19.
54 − \(\frac{1}{7}\)yi = 9x− 4i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 19

Question 20.
15 − 3yi = \(\frac{1}{2}\)x + 2i
Answer:
Given,
15 − 3yi = \(\frac{1}{2}\)x + 2i
15 = \(\frac{1}{2}\)x and -3yi = 2i
x = 15 × 2 and y = -2/3
30 = x and y = -2/3
So, x = 30 and y = -2/3
So, the real part x = 30 and imaginary part y = -2/3

In Exercises 21–30, add or subtract. Write the answer in standard form.
Question 21.
(6 − i) + (7 + 3i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 21

Question 22.
(9 + 5i) + (11 + 2i )
Answer:
Given,
(9 + 5i) + (11 + 2i )
Combine the real parts and imaginary parts.
9 + 5i + 11 + 2i
20 + 7i
So, (9 + 5i) + (11 + 2i ) = 20 + 7i

Question 23.
(12 + 4i) − (3 − 7i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 23

Question 24.
(2 − 15i) − (4 + 5i)
Answer:
Given,
(2 − 15i) − (4 + 5i)
Combine the real parts and imaginary parts.
2 – 15i – 4 – 5i
-2 – 20i
-2(1 + 10i)

Question 25.
(12 − 3i) + (7 + 3i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 25

Question 26.
(16 − 9i) − (2 − 9i)
Answer:
Given,
(16 − 9i) − (2 − 9i)
Combine the real parts and imaginary parts.
16 – 9i – 2 + 9i
16 – 2 = 14
So, (16 − 9i) − (2 − 9i) = 14

Question 27.
7 − (3 + 4i) + 6i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 27

Question 28.
16 − (2 − 3i) − i
Answer:
Given,
16 − (2 − 3i) − i
Combine the real parts and imaginary parts.
16 – 2 + 3i  – i
= 14 + 2i
= 2(7 + i)
So, 16 − (2 − 3i) − i = 2(7 + i)

Question 29.
−10 + (6 − 5i) − 9i
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 29

Question 30.
−3 + (8 + 2i) + 7i
Answer:
Given,
−3 + (8 + 2i) + 7i
Combine the real parts and imaginary parts.
-3 + 8 + 2i + 7i
5 + 9i
So, −3 + (8 + 2i) + 7i = 5 + 9i

Question 31.
USING STRUCTURE
Write each expression as a complex number in standard form.
a. \(\sqrt{-9}+\sqrt{-4}-\sqrt{16}\)
b. \(\sqrt{-16}+\sqrt{8}+\sqrt{-36}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 31

Question 32.
REASONING
The additive inverse of a complex number z is a complex number za such that z + za = 0. Find the additive inverse of each complex number.
a. z = 1 + i
b. z = 3 − i
c. z = −2 + 8i
Answer:
a. z = 1 + i
The additive inverse of a complex number is changing the sign of the number
-z = -1 – i
b. z = 3 − i
The additive inverse of a complex number is changing the sign of the number
z = 3 − i = -z = -3 + i
c. z = −2 + 8i
The additive inverse of a complex number is changing the sign of the number
z = −2 + 8i
-z = 2 – 8i

In Exercises 33–36, find the impedance of the series circuit.
Question 33.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 33

Question 35.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 35

In Exercises 37–44, multiply. Write the answer in standard form.
Question 37.
3i(−5 + i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 37

Question 38.
2i(7 − i)
Answer:
Given,
2i(7 − i) = 14i – 2i²
We know that,
i² = -1
= 14i – 2(-1)
= 14i + 2
So, 2i(7 − i) = 14i + 2

Question 39.
(3 − 2i)(4 + i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 39

Question 40.
(7 + 5i)(8 − 6i)
Answer:
Given,
(7 + 5i)(8 − 6i)
7(8 – 6i) + 5i(8 – 6i)
56 – 42i + 40i – 30i²
We know that,
i² = -1
56 – 2i – 30(-1)
56 – 2i + 30
26 – 2i
So, (7 + 5i)(8 − 6i) = 26 – 2i

Question 41.
(4 − 2i)(4 + 2i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 41

Question 42.
(9 + 5i)(9 − 5i)
Answer:
Given,
(9 + 5i)(9 − 5i)
9² – (5i)²
81 – 25i²
We know that,
i² = -1
81 – 25(-1)
81 + 25 = 106
So, (9 + 5i)(9 − 5i) = 106

Question 43.
(3 − 6i)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 43

Question 44.
(8 + 3i)2
Answer:
Given,
(8 + 3i)2
8² + 2 (8)(3i) + (3i)²
64 + 48i + 9 i²
We know that,
i² = -1
64 + 48i + 9(-1)
= 64 + 48i – 9
= 55 + 48i
So, (8 + 3i)2 = 55 + 48i

JUSTIFYING STEPS In Exercises 45 and 46, justify each step in performing the operation.
Question 45.
11 − (4 + 3i) + 5i
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 45

Question 46.
(3 + 2i)(7 − 4i)
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 7
Answer:
Big-Ideas-Math-Algebra-2-Answers-Chapter-3-Quadratic-Equations-and-Complex-Numbers-3.2-7

REASONING In Exercises 47 and 48, place the tiles in the expression to make a true statement.
Question 47.
(____ − ____i) – (____ − ____i ) = 2 − 4i
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 47

Question 48.
____i(____ + ____i ) = −18 − 10i
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 9
Answer:
2i(-5 + 9i)
= -10i + 18i²
We know that,
i² = -1
= -10i + 18(-1)
= -10i – 18
So, 2i(-5 + 9i) = -10i – 18

In Exercises 49–54, solve the equation. Check your solution(s).
Question 49.
x2 + 9 = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 49

Question 50.
x2 + 49 = 0
Answer:
Given,
x2 + 49 = 0
x² + 7² = 0
x² = -7²
Applying square root on both sides
x = -7
x = 7i
So, the value of x is ±7i

Question 51.
x2 − 4 = −11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 51

Question 52.
x2 − 9 = −15
Answer:
Given,
x2 − 9 = −15
x2 = −15 + 9
x2 = −6
x = √-6
x = i√6
So, the value of x is i√6.

Question 53.
2x2 + 6 = −34
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 53

Question 54.
x2 + 7 = −47
Answer:
Given,
x2 + 7 = −47
x² = -47 – 7
x² = -54
x = √-54
x = 3i√6
So, the value of x is 3i√6

In Exercises 55–62, find the zeros of the function.
Question 55.
f(x) = 3x2 + 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 55

Question 56.
g(x) = 7x2 + 21
Answer:
Given,
g(x) = 7x2 + 21
7x2 + 21 = 0
7x² = -21
x² = -21/7
x² = -3
x = i√3
So, the zeros of g is i√3

Question 57.
h(x) = 2x2 + 72
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 57

Question 58.
k(x) = −5x2 − 125
Answer:
Given,
k(x) = −5x2 − 125
−5x2 − 125 = 0
-5(x² – 25) =0
x² – 25 = 0
x² = 25
x = 5
So, the zeros of k are 5 or -5.

Question 59.
m(x) = −x2 − 27
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 59

Question 60.
p(x) = x2 + 98
Answer:
Given,
p(x) = x2 + 98
x2 + 98 = 0
x² = -98
x =7 i√14
So, the zeros of p is 7 i√14

Question 61.
r(x) = − \(\frac{1}{2}\)x2 − 24
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 61

Question 62.
f(x) = −\(\frac{1}{5}\)x2 − 10
Answer:
Given,
f(x) = −\(\frac{1}{5}\)x2 − 10
f(x) = 0
−\(\frac{1}{5}\)x2 − 10 = 0
\(\frac{1}{5}\)x² + 10 = 0
\(\frac{1}{5}\)x² = -10
x² = -50
x = √-50
x = 5i√2
So, the zeros of f is 5i√2

ERROR ANALYSIS In Exercises 63 and 64, describe and correct the error in performing the operation and writing the answer in standard form.
Question 63.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 63

Question 64.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 11
Answer:
Given,
(4 + 6i)² = (4)² + 2(4)(6i) + (6i)²
= 16 + 48i + 36(-1)
= 16 + 48i – 36
= -20 + 48i

Question 65.
NUMBER SENSE
Simplify each expression. Then classify your results in the table below.
a. (−4 + 7i) + (−4 − 7i)
b. (2 − 6i) − (−10 + 4i)
c. (25 + 15i) − (25 − 6i)
d. (5 + i)(8 − i)
e. (17 − 3i) + (−17 − 6i)
f. (−1 + 2i)(11 − i)
g. (7 + 5i) + (7 − 5i)
h. (−3 + 6i) − (−3 − 8i)
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 65

Question 66.
MAKING AN ARGUMENT
The Product Property ofSquare Roots states \(\sqrt{a}\) • \(\sqrt{b}\) = \(\sqrt{ab}\) . Your friend concludes \(\sqrt{-4}\) • \(\sqrt{-9}\) = \(\sqrt{36}\) = 6. Is your friend correct? Explain.
Answer:
Given,
The Product Property of Square Roots states \(\sqrt{a}\) • \(\sqrt{b}\) = \(\sqrt{ab}\) .
\(\sqrt{-4}\) • \(\sqrt{-9}\)
2i × 3i = 6i²
We know that,
i² = -1
= 6(-1) = -6
So, your friend is not correct.

Question 67.
FINDING A PATTERN
Make a table that shows the powers of i from i1 to i8 in the first row and the simplified forms of these powers in the second row. Describe the pattern you observe in the table. Verify the pattern continues by evaluating the next four powers of i.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 67

Question 68.
HOW DO YOU SEE IT?
The graphs of three functions are shown. Which function(s) has real zeros? imaginary zeros? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 13
Answer:
The graphs of three quadratic functions.
From the above graph, we can see only f and g make intercepts on the x-axis.
So, only f and g have real zeros.
The function h makes no intercepts.
From this, we can conclude that h has only imaginary roots.
Also, the function f only intersects the axis once so it has only one real zero.

In Exercises 69–74, write the expression as a complex number in standard form.
Question 69.
(3 + 4i) − (7 − 5i) + 2i(9 + 12i)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 69

Question 70.
3i(2 + 5i) + (6 − 7i) − (9 + i)
Answer:
Given,
3i(2 + 5i) + (6 − 7i) − (9 + i)
Combine the real parts and imaginary parts.
= 6i + 30i² + 6 – 7i – 9 – i
= 6i + 30(-1) + 6 – 7i – 9 – i
= 6i – 30 + 6 – 7i – 9 – i
= -2i -33
= -33 – 2i
So, 3i(2 + 5i) + (6 − 7i) − (9 + i) = -33 – 2i

Question 71.
(3 + 5i)(2 − 7i4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 71

Question 72.
2i3(5 − 12i )
Answer:
Given
2i3(5 − 12i )
= 10i³ – 24i4
= 10i(-1) – 24(1)
= -10i – 24
= -2(5i + 12)
So, 2i3(5 − 12i ) = -2(5i + 12)

Question 73.
(2 + 4i5) + (1 − 9i6) − (3 +i7)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 73

Question 74.
(8 − 2i4) + (3 − 7i8) − (4 + i9)
Answer:
Given,
(8 − 2i4) + (3 − 7i8) − (4 + i9)
8 + 3 – 4 – 2i4 – 7i8 – i9
7 – i4(2 + 7i4 + i5)
i4 = 1
7 – 1(2 + 7 + 1i)
7 – 2 – 7 – i
-2 – i
So, (8 − 2i4) + (3 − 7i8) − (4 + i9) = -2 – i

Question 75.
OPEN-ENDED
Find two imaginary numbers whose sum and product are real numbers. How are the imaginary numbers related?
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 75

Question 76.
COMPARING METHODS
Describe the two different methods shown for writing the complex expression in standard form. Which method do you prefer? Explain.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 14
Answer: I prefer method 1 because it is somewhat easy for writing the expression in the standard form.

Question 77.
CRITICAL THINKING
Determine whether each statement is true or false. If it is true, give an example. If it is false, give a counterexample.
a. The sum of two imaginary numbers is an imaginary number.
b. The product of two pure imaginary numbers is a real number.
c. A pure imaginary number is an imaginary number.
d. A complex number is a real number.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 77

Question 78.
THOUGHT PROVOKING
Create a circuit that has an impedance of 14 − 3i.
Answer:
A resistor has a resistance of 14 ohms.
impedance = 14 ohms
An inductor having a reactance of 5i in ohms. So, its impedance is 5i ohms
A capacitor of having a reactance of 8i in ohms.
So its impedance is -8i ohms.
= 14 + 5i + (-8i)
= 14 + 5i – 8i
= 14 – 3i

Maintaining Mathematical Proficiency

Determine whether the given value of x is a solution to the equation.
Question 79.
3(x − 2) + 4x − 1 = x − 1; x = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 79

Question 80.
x3 − 6 = 2x2 + 9 − 3x; x = −5
Answer:
Given,
x3 − 6 = 2x2 + 9 − 3x
substitute x = -5 in the expression
(-5)3 − 6 = 2(-5)2 + 9 − 3(-5)
-125 – 6 = 50 + 9 + 15
-131 = 74
-131 ≠ 74
So, x = -5 is not a solution to the equation

Question 81.
−x2 + 4x = 19 — 3x2; x = −\(\frac{3}{4}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 81

Write a quadratic function in vertex form whose graph is shown.
Question 82.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 15
Answer:
Given points (0, 3) and (1, 2)
(h, k) = (1, 2)
y = a(x – h)² + k
3 = a(0 – 1)² + 2
3 = a + 2
a = 3 – 2
a = 1
So, the equation of the parabola is y = 1(0 – 1)² + 2

Question 83.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 16
Answer:
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 a 83

Question 84.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers 3.2 17
Answer:
The points are (3, -2) and (2, -1)
(h, k) = (3, -2)
x = 2 and y = -1
y = a(x – h)² + k
-1 = a(2 – 3)² + (-2)
-1 = -1a – 2
-a = 1
a = -1
So, the equation of the parabola is y = -1(2 – 3)² + (-2)

Lesson 3.3 Completing the Square

Essential Question How can you complete the square for a quadratic expression?

EXPLORATION 1

Using Algebra Tiles to Complete the Square
Work with a partner. Use algebra tiles to complete the square for the expression x2 + 6x.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 1
a. You can model x2 + 6x using one x2-tile and six x-tiles. Arrange the tiles in a square. Your arrangement will be incomplete in one of the corners.
b. How many 1-tiles do you need to complete the square?
c. Find the value of c so that the expression x2 + 6x + c is a perfect square trinomial.
d. Write the expression in part (c) as the square of a binomial.

EXPLORATION 2

Drawing Conclusions
Work with a partner.
a. Use the method outlined in Exploration 1 to complete the table.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 2
b. Look for patterns in the last column of the table. Consider the general statement x2 + bx + c = (x + d)2. How are d and b related in each case? How are c and d related in each case?
c. How can you obtain the values in the second column directly from the coefficients of x in the first column?

Communicate Your Answer

Question 3.
How can you complete the square for a quadratic expression?
Answer:
i. Divide all the terms of the coefficient of x².
ii. Move the number c/a to the right side of the equation.
iii. Complete the square on the left side of the equation and balance this by adding the same value to the right side of the equation.

Question 4.
Describe how you can solve the quadratic equation x2 + 6x = 1 by completing the square.
Answer:
Given,
x2 + 6x = 1
x2 + 6x + 9 = 1 + 9
(x + 3)² = 10

Monitoring Progress

Solve the equation using square roots. Check your solution(s).
Question 1.
x2 + 4x + 4 = 36
Answer:
Given,
x2 + 4x + 4 = 36
The expression is in the form of a² + 2ab + b² = (a + b)²
(x + 2)² = 36
Applying square root on both sides
x + 2 = 36
x = 36 – 2
x = 34
The solution is x = 34.

Question 2.
x2 − 6x + 9 = 1
Answer:
Given,
x2 − 6x + 9 = 1
The expression is in the form of a² – 2ab + b² = (a – b)²
x2 − 6x + 9 = 1
(x – 3)² = 1
Applying square root on both sides
x – 3 = 1
x = 1 + 3
x = 4
The solution is x = 4.

Question 3.
x2 − 22x + 121 = 81
Answer:
Given,
x2 − 22x + 121 = 81
This expression is in the form of a² – 2ab + b² = (a – b)²
(x – 11)² = 81
(x – 11)² = 9²
Applying square root on both sides
x – 11 = 9
x = 9 + 11
x = 20
The solution is x = 20.

Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.
Question 4.
x2 + 8x + c
Answer:
Given,
x2 + 8x + c
c = b/2a
b = 8 and a = 1
8/2 = 4
c = 4²
c = 16
x2 + 8x + 16 = (x + 4)²

Question 5.
x2 − 2x + c
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-5-1.jpg

Question 6.
x2 − 9x + c
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-5-1.jpg

Solve the equation by completing the square.
Question 7.
x2 − 4x + 8 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-7-1.jpg

Question 8.
x2 + 8x − 5 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-8.jpg

Question 9.
−3x2 − 18x − 6 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-9.jpg

Question 10.
4x2 + 32x = −68
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-10.jpg

Question 11.
6x(x + 2) = −42
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-11.jpg

Question 12.
2x(x − 2) = 200
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-12.jpg

Write the quadratic function in vertex form. Then identify the vertex.
Question 13.
y = x2 − 8x + 18
Answer:

Question 14.
y = x2 + 6x + 4
Answer:

Question 15.
y = x2 − 2x − 6
Answer:

Question 16.
WHAT IF?
The height of the baseball can be modeled by y = −16t2 + 80t + 2. Find the maximum height of the baseball. How long does the ball take to hit the ground?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-Monitoring-progress-ex-3.3-16.jpg

Completing the Square 3.3 Exercises

Vocabulary and Core Concept Check
Question 1.
VOCABULARY
What must you add to the expression x2 + bx to complete the square?
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 1

Question 2.
COMPLETE THE SENTENCE
The trinomial x2 − 6x + 9 is a ____ because it equals ____.
Answer:
The trinomial x2 − 6x + 9 is a perfect square trinomial because it equals (x – 3)².

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, solve the equation using square roots. Check your solution(s).
Question 3.
x2 − 8x + 16 = 25
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 3

Question 4.
r2 − 10r + 25 = 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-1.jpg

Question 5.
x2 − 18x + 81 = 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 5

Question 6.
m2 + 8m + 16 = 45
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-6.jpg

Question 7.
y2 − 24y + 144 = −100
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 7

Question 8.
x2 − 26x + 169 = −13
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-8.jpg

Question 9.
4w2 + 4w + 1 = 75
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 9

Question 10.
4x2 − 8x + 4 = 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-10.jpg

In Exercises 11–20, find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial.
Question 11.
x2 + 10x + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 11

Question 12.
x2 + 20x + c
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-10-1.jpg

Question 13.
y2 − 12y + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 13

Question 14.
t2 − 22t + c
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-14.jpg

Question 15.
x2 − 6x + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 15

Question 16.
x2 + 24x + c
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-16.jpg

Question 17.
z2 − 5z + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 17

Question 18.
x2 + 9x + c
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-18.jpg

Question 19.
w2 + 13w + c
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 19

Question 20.
s2 − 26s + c
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-20.jpg

In Exercises 21–24, find the value of c. Then write an expression represented by the diagram.
Question 21.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 3
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 21

Question 22.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-22.jpg

Question 23.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 5
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 23

Question 24.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-24.jpg

In Exercises 25–36, solve the equation by completing the square.
Question 25.
x2 + 6x + 3 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 25

Question 26.
s2 + 2s − 6 = 0
Answer:

Question 27.
x2 + 4x − 2 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 27

Question 28.
t2 − 8t − 5 = 0
Answer:

Question 29.
z(z + 9) = 1
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 29

Question 30.
x(x + 8) = −20
Answer:

Question 31.
7t2 + 28t + 56 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 31

Question 32.
6r2 + 6r + 12 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-32.jpg

Question 33.
5x(x + 6) = −50
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 33

Question 34.
4w(w − 3) = 24
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-34.jpg

Question 35.
4x2 − 30x = 12 + 10x
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 35

Question 36.
3s2 + 8s = 2s − 9
Answer:

Question 37.
ERROR ANALYSIS
Describe and correct the error in solving the equation.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 7
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 37

Question 38.
ERROR ANALYSIS
Describe and correct the error in finding the value of c that makes the expression a perfect square trinomial.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 8
Answer:

Question 39.
WRITING
Can you solve an equation by completing the square when the equation has two imaginary solutions? Explain.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 39

Question 40.
ABSTRACT REASONING
Which of the following are solutions of the equation x2 − 2ax + a2 = b2? Justify your answers.
A. ab
B. −a − b
C. b
D. a
E. a − b
F. a + b
Answer:

USING STRUCTURE In Exercises 41–50, determine whether you would use factoring, square roots, or completing the square to solve the equation. Explain your reasoning. Then solve the equation.
Question 41.
x2 − 4x − 21 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 41

Question 42.
x2 + 13x + 22 = 0
Answer:
Given,
x2 + 13x + 22 = 0

Question 43.
(x + 4)2 = 16
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 43

Question 44.
(x − 7)2 = 9
Answer:

Question 45.
x2 + 12x + 36 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 45

Question 46.
x2 − 16x + 64 = 0
Answer:

Question 47.
2x2 + 4x − 3 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 47

Question 48.
3x2 + 12x + 1 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-48.jpg

Question 49.
x2 − 100 = 0
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 49

Question 50.
4x2 − 20 = 0
Answer:
Given,
4x2 − 20 = 0
4x² = 20
x² = 20/4
x² = 5
x = \(\sqrt{5}\)
The solution is x = \(\sqrt{5}\)

MATHEMATICAL CONNECTIONS In Exercises 51–54, find the value of x.
Question 51.
Area of rectangle = 50
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 9
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 51

Question 52.
Area of parallelogram = 48
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 10
Answer:

Question 53.
Area of triangle = 40
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 11
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 53

Question 54.
Area of trapezoid = 20
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 12
Answer:

In Exercises 55–62, write the quadratic function in vertex form. Then identify the vertex.
Question 55.
f(x) = x2 − 8x + 19
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 55

Question 56.
g(x) = x2 − 4x − 1
Answer:

Question 57.
g(x) = x2 + 12x + 37
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 57

Question 58.
h(x) = x2 + 20x + 90
Answer:

Question 59.
h(x) = x2 + 2x − 48
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 59

Question 60.
f(x) = x2 + 6x − 16
Answer:

Question 61.
f(x) = x2 − 3x + 4
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 61

Question 62.
g(x) = x2 + 7x + 2
Answer:

Question 63.
MODELING WITH MATHEMATICS
While marching, a drum major tosses a baton into the air and catches it. The height h (in feet) of the baton t seconds after it is thrown can be modeled by the function h = −16t2 + 32t + 6.
a. Find the maximum height of the baton.
b. The drum major catches the baton when it is 4 feet above the ground. How long is the baton in the air?
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 63

Question 64.
MODELING WITH MATHEMATICS
A firework explodes when it reaches its maximum height. The height h (in feet) of the firework t seconds after it is launched can be modeled by h = \(-\frac{500}{9} t^{2}+\frac{1000}{3} t\) + 10. What is the maximum height of the firework? How long is the firework in the air before it explodes?
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 13
Answer:

Question 65.
COMPARING METHODS
A skateboard shop sells about 50 skateboards per week when the advertised price is charged. For each $1 decrease in price, one additional skateboard per week is sold. The shop’s revenue can be modeled by y = (70 − x)(50 + x).
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 14
a. Use the intercept form of the function to find the maximum weekly revenue.
b. Write the function in vertex form to find the maximum weekly revenue.
c. Which way do you prefer? Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 65

Question 66.
HOW DO YOU SEE IT?
The graph of the function f(x) = (x − h)2 is shown. What is the x-intercept? Explain your reasoning.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 15
Answer:

Question 67.
WRITING
At Buckingham Fountain in Chicago, the height h (in feet) of the water above the main nozzle can be modeled by h = −162 + 89.6t, where t is the time (in seconds) since the water has left the nozzle. Describe three different ways you could find the maximum height the water reaches. Then choose a method and find the maximum height of the water.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 67

Question 68.
PROBLEM SOLVING
A farmer is building a rectangular pen along the side of a barn for animals. The barn will serve as one side of the pen. The farmer has 120 feet of fence to enclose an area of 1512 square feet and wants each side of the pen to be at least 20 feet long.
a. Write an equation that represents the area of the pen.
b. Solve the equation in part (a) to find the dimensions of the pen.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 16
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-68.jpg

Question 69.
MAKING AN ARGUMENT
Your friend says the equation x2 + 10x = −20 can be solved by either completing the square or factoring. Is your friend correct? Explain.
Answer:
x2 + 10x = −20
x2 + 10x + 20 = 0
Your friend is incorrect because x2 + 10x = −20 does not factor into rational numbers.

Question 70.
THOUGHT PROVOKING
Write a function g in standard form whose graph has the same x-intercepts as the graph of f(x) = 2x2 + 8x + 2. Find the zeros of each function by completing the square. Graph each function.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-70.jpg

Question 71.
CRITICAL THINKING
Solve x2 + bx + c = 0 by completing the square. Your answer will be an expression for x in terms of b and c.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 71

Question 72.
DRAWING CONCLUSIONS
In this exercise, you will investigate the graphical effect of completing the square.
a. Graph each pair of functions in the same coordinate plane.
y = x2 + 2x y = x2 − 6x
y = (x + 1)2 y = (x − 3)2
b. Compare the graphs of y = x2 + bx and y = (x + \(\frac{b}{2}\))2. Describe what happens to the graph of y = x2 + bx when you complete the square.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-72.jpg

Question 73.
MODELING WITH MATHEMATICS
In your pottery class, you are given a lump of clay with a volume of 200 cubic centimeters and are asked to make a cylindrical pencil holder. The pencil holder should be 9 centimeters high and have an inner radius of 3 centimeters. What thickness x should your pencil holder have if you want to use all of the clay?
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 17
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 73

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution.
Question 74.
2x − 3 < 5
Answer:

Question 75.
4 − 8y ≥ 12
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 75

Question 76.
\(\frac{n}{3}\) + 6 > 1
Answer:

Question 77.
−\(\frac{2s}{5}\) ≤ 8
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 77

Graph the function. Label the vertex, axis of symmetry, and x-intercepts.
Question 78.
g(x) = 6(x − 4)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-78.jpg

Question 79.
h(x) = 2x(x − 3)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 79

Question 80.
f(x) = x2 + 2x + 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-3-chapter-Quadratic-Equations-and-Complex-Numbers-ex-3.3-80.jpg

Question 81.
f(x) = 2(x + 10)(x − 12)
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 81

Quadratic Equations and Complex Numbers Study Skills: Creating a Positive Study Environment

3.1–3.3 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 18

Core Concepts
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 19

Mathematical Practices
Question 1.
Analyze the givens, constraints, relationships, and goals in Exercise 61 on page 101.
Answer:

Question 2.
Determine whether it would be easier to find the zeros of the function in Exercise 63 on page 117 or Exercise 67 on page 118.
Answer:

Study Skills: Creating a Positive Study Environment

  • Set aside an appropriate amount of time for reviewing your notes and the textbook, reworking your notes, and completing homework.
  • Set up a place for studying at home that is comfortable, but not too comfortable. The place needs to be away from all potential distractions.
  • Form a study group. Choose students who study well together, help out when someone misses school, and encourage positive attitudes.
    Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 20

Quadratic Equations and Complex Numbers 3.1–3.3 Quiz

Solve the equation by using the graph. Check your solution(s).
Question 1.
x2 − 10x + 25 = 0
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 1
Answer:
Given,
f(x) = x2 − 10x + 25
x2 − 10x + 25 = 0
(x – 5)² = 0
Applying square root on both sides
x – 5 = 0
x = 5
The solution is x = 5.

Question 2.
2x2 + 16 = 12x
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 2
Answer:
Given,
g(x) = 2x2 -12x + 16
2x2 + 16 = 12x
2x2 -12x + 16 = 0
2(x² – 6x + 8) = 0
x² – 6x + 8 = 0
x = -b ± √b² – 4ac/2a
x = -(-6) ± √(-6)² – 4(1)(8)/2
x = 6 ± √36-32/2
x = 6 ± √4/2
x = (6+2)/2 = 8/2 = 4
x = (6-2)/2 = 4/2 = 2
The solution is x = 2

Question 3.
x2 = −2x + 8
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 3
Answer:
Given,
h(x) = -x² – 2x + 8
x2 = −2x + 8
x² + 2x – 8 = 0
x = -b ± √b² – 4ac/2a
x = -(2) ± √(2)² – 4(1)(-8)/2
x = -2 ± √4+32/2
x = 4/2 or x = -8/2
x = 2 or -4
The solutions are x = 2 and x = -4.

Solve the equation using square roots or by factoring. Explain the reason for your choice.
Question 4.
2x2 − 15 = 0
Answer:
Given,
2x2 − 15 = 0
2x² = 15
x² = 15/2
x² = 7.5
Applying square root on both sides.
x = √7.5
The solution x = √7.5

Question 5.
3x2 − x − 2 = 0
Answer:
Given,
3x2 − x − 2 = 0
x = (-b ± √b² – 4ac)/2a
x = -(-1) ± √(-1)² – 4(3)(-2)/2(3)
x = 1 ± √1+24/6
x = 1 ± √1+24/6
x = (1 ± 5)/6
x = 6/6 or -4/6
x = 1 or -2/3
The solution is x = 1 or -2/3

Question 6.
(x + 3)2 = 8
Answer:
Given,
x² + 6x + 9 = 8
x² + 6x + 1 = 0
x = (-b ± √b² – 4ac)/2a
b = 6, a = 1 and c = 1
x = -(6) ± √(6)² – 4(1)(1)/2(1)
x = -6 ± √36-4/2
x = -6 ± √32/2
x = -0.17
x = -2.82
The solution is x = -2.82 and -0.17

Question 7.
Find the values of x and y that satisfy the equation 7x − 6i = 14 + yi.
Answer:
Given,
7x − 6i = 14 + yi.
7x = 14 and -6i = yi
x = 14/7 and y = -6
x = 2 and y = -6

Perform the operation. Write your answer in standard form
Question 8.
(2 + 5i) + (−4 + 3i)
Answer:
Given,
(2 + 5i) + (−4 + 3i)
= 2 + 5i – 4 + 3i
= -2 + 8i
So, (2 + 5i) + (−4 + 3i) = -2 + 8i

Question 9.
(3 + 9i) − (1 − 7i)
Answer:
Given,
(3 + 9i) − (1 − 7i)
= 3 + 9i – 1 + 7i
= 2 + 16i
So, (3 + 9i) − (1 − 7i) = 2 + 16i

Question 10.
(2 + 4i)(−3 − 5i)
Answer:
Given,
(2 + 4i)(−3 − 5i)
= 2(-3 – 5i) + 4i(-3 – 5i)
= -6 – 15i – 12i – 20i²
= -6 – 27i + 20
= 14 – 27i
So, (2 + 4i)(−3 − 5i) = 14 – 27i

Question 11.
Find the zeros of the function f(x) = 9x2 + 2. Does the graph of the function intersect the x-axis? Explain your reasoning.
Answer:
Given,
f(x) = 9x2 + 2
f(x) = 0
9x2 + 2 = 0
9x² = -2
x = √(-2/9)
x = i/3 √2

Solve the equation by completing the square.
Question 12.
x2 − 6x + 10 = 0
Answer:
Given,
x² – 6x + 10 = 0
x = (-b ± √b² – 4ac)/2a
b = -6, a = 1 and c = 10
x = -(6) ± √(6)² – 4(1)(10)/2(1)
x = -6 ± √-4/2
x = -6 ± √32/2
x = -3 ± i
So, x = -3 + i or -3 – i

Question 13.
x2 + 12x + 4 = 0
Answer:
Given,
x2 + 12x + 4 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = 12 and c = 4
x = -(12) ± √(12)² – 4(1)(4)/2(1)
x = (-12 ± √144 – 16)/2
x = (-12±8√2)/2
x = -6 ± 4√2
So, x = -6 + 4√2 and x = -6 – 4√2

Question 14.
4x(x + 6) = −40
Answer:
Given,
4x(x + 6) = −40
4x² + 24x + 40 = 0
4(x² + 6x + 10) = 0
x² + 6x + 10 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = 6 and c = 10
x = -(6) ± √(6)² – 4(1)(10)/2(1)
x = -6 ± √-4/2
x = -6 ± √32/2
x = -3 ± i
So, x = -3 + i or -3 – i

Question 15.
Write y = x2 − 10x + 4 in vertex form. Then identify the vertex.
Answer:
Given
y = x2 − 10x + 4
(b/2)² = (-10/2)²
y + 25 = (x² – 10x + 25) + 4
y + 25 = (x – 5)² + 4
y = (x – 5)² – 21
The vertex is (5, -21)

Question 16.
A museum has a café with a rectangular patio. The museum wants to add 464 square feet to the area of the patio by expanding the existing patio as shown.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 4
a. Find the area of the existing patio.
b. Write an equation to model the area of the new patio.
c. By what distance x should the length of the patio be expanded?
Answer:
a. Area of the rectangle = length × breadth
length = 30 ft
breadth = 20 ft
= 30 × 20
= 600 sq. ft
So, the area of the existing patio = 600 sq. ft
b. 600 + 464 = (30 + x) (20 + x)
1064 = 600 + 30x + 20x + x²
x² + 50x + 600 – 1064 = 0
x² + 50x – 464 = 0
c. x² + 50x – 464 = 0
(x + m)(x + n)
mn = -464 and m + n = 50
x² + 58x – 8x – 464 = 0
x(x + 58) – 8(x + 58) = 0
(x + 58) (x – 8) = 0
x + 58 = 0
x = -58
The dimensions cannot be negative.
x – 8 = 0
x = 8
So, x = 8

Question 17.
Find the impedance of the series circuit.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers q 5
Answer:
R = 7Ω
Li = 5i
C = 2
impedance = R + Li + (-Ci)
= 7 + 5i + (-2i)
= 7 + 3i

Question 18.
The height h (in feet) of a badminton birdie t seconds after it is hit can be modeled by the function h = −16t2 + 32t + 4.
a. Find the maximum height of the birdie.
Answer:
h = −16t2 + 32t + 4
-16 + h = -16(t² -2t + 1) + 4
h = -16(t – 1)² + 4 + 16
h = -16(t – 1)² + 20
The maximum height of the birdie is 20 feet

b. How long is the birdie in the air?
Answer:
0 = -16(t – 1)² + 20
-20 = -16(t – 1)²
1.25 = (t – 1)²
1.1 = t – 1
t = 2.1 sec

Lesson 3.4 Using the Quadratic Formula

Essential Question How can you derive a general formula for solving a quadratic equation?

EXPLORATION 1

Deriving the Quadratic Formula
Work with a partner. Analyze and describe what is done in each step in the development of the Quadratic Formula.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 1

EXPLORATION 2

Using the Quadratic Formula
Work with a partner. Use the Quadratic Formula to solve each equation.
a. x2 − 4x + 3 = 0
b. x2 − 2x + 2 = 0
c. x2 + 2x − 3 = 0
d. x2 + 4x + 4 = 0
e. x2 − 6x + 10 = 0
f. x2 + 4x + 6 = 0

Communicate Your Answer

Question 3.
How can you derive a general formula for solving a quadratic equation?
Answer:
Start with an equation of the form ax² + bx + c = 0.
Rewrite the equation so that ax² + bx is isolated on one side.
Complete the square by adding b²/4a² to both sides.
Rewrite the perfect square trinomial as a square of a binomial.

Question 4.
Summarize the following methods you have learned for solving quadratic equations: graphing, using square roots, factoring, completing the square, and using the Quadratic Formula.
Answer:
The methods for solving quadratic equations are summarized as:
Method 1 is graphing
Method 2 is factoring
Method 3 is Completing the square
Graphing: The x-intercepts serve as the solution for the equation.
Using square roots: The properties of square roots are used to solve the problem.
Factoring: The factors are framed for the equation and the zero product property is used to solve the problem.
Completing the square: The square is completed so that the properties of the square root may be applied as illustrated in the solution by using square roots.
Quadratic formula: Start with an equation of the form ax² + bx + c = 0
x = (-b ± √b² – 4ac)/2a

Monitoring Progress

Solve the equation using the Quadratic Formula.
Question 1.
x2 − 6x + 4 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.4-1.jpg

Question 2.
2x2 + 4 = −7x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.4-2.jpg

Question 3.
5x2 = x + 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.4-3.jpg

Solve the equation using the Quadratic Formula.
Question 4.
x2 + 41 = −8x
Answer:
Given equation,
x2 + 41 = −8x
x2 + 41 + 8x = 0
x = -b ± √b² – 4ac/2a
a = 1, b = 8 and c = 41
x = -8 ± √8² – 4(1)(41)/2(1)
x = (-8 ± √64 – 164)/2
x = (-8 ± √-100)/2
x = -8/2 ± 10i/2
x = -4 ± 5i
The solution is x = -4 + 5i and x = -4 – 5i

Question 5.
−9x2 = 30x + 25
Answer:
Given equation,
−9x2 = 30x + 25
-9x² – 30x – 25 = 0
9x² + 30x + 25 = 0
a = 9, b = 30 and c = 25
Substitute the values of a, b and c in the formula.
x = -b ± √b² – 4ac/2a
x = -30 ± √30² – 4(9)(25)/2(9)
x = (-30 ± √900- 900)/2
x = (-30 ± √0)/18
x = -30/18
x = -5/3
The solution x = -5/3.

Question 6.
5x − 7x2 = 3x + 4
Answer:
Given,
5x − 7x2 = 3x + 4
-7x² + 5x – 3x – 4 = 0
-7x² + 2x – 4 = 0
7x² – 2x + 4 = 0
a = 7, b = -2 and c = 4
x = -b ± √b² – 4ac/2a
x = 2 ± √(-2)² – 4(7)(4)/2(7)
x = (2 ± √4- 112)/14
x = 1/7 ± 3i√3/7
The solution is x = 1/7 + 3i√3/7 and x = 1/7 – 3i√3/7

Find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 7.
4x2 + 8x + 4 = 0
Answer:
Given,
4x2 + 8x + 4 = 0
The discriminant is b² – 4ac
d = b² – 4ac
b² – 4ac = 0
8² – 4(4)(4)
a = 4, b = 8 and c = 4
64 – 64 = 0
So, the discriminant is 0.

Question 8.
\(\frac{1}{2}\)x2 + x − 1 = 0
Answer:
\(\frac{1}{2}\)x2 + x − 1 = 0
The discriminant is b² – 4ac
d = b² – 4ac
b² – 4ac = 0
a = \(\frac{1}{2}\), b = 1, c = -1
1² – 4(\(\frac{1}{2}\))(-1)
1 + 2 = 3
The discriminant is 3.

Question 9.
5x2 = 8x − 13
Answer:
Given,
5x2 = 8x − 13
5x2 – 8x + 13 = 0
The discriminant is b² – 4ac
b² – 4ac = 0
(-8)² – 4(5)(-13)
a = 5, b = -8 and c = -13
64 + 260 = 324
The discriminant is 324

Question 10.
7x2 − 3x = 6
Answer:
Given,
7x2 − 3x = 6
The discriminant is b² – 4ac
b² – 4ac = 0
7x2 − 3x – 6 = 0
a = 7, b = -3 and c = -6
(3)² – 4(7)(-6)
9 + 168 = 177
The discriminant is 177.

Question 11.
4x2 + 6x = −9
Answer:
Given,
4x2 + 6x = −9
4x2 + 6x + 9 = 0
The discriminant is b² – 4ac
a = 4, b = 6 and c = 9
6² – 4(4)(-9)
36 + 144 = 180
The discriminant is 180.

Question 12.
−5x2 + 1x = 6 − 10x
Answer:
Given,
−5x2 + 1 = 6 − 10x
-5x² + x – 6 + 10x
-5x² + 11x – 6 = 0
The discriminant is b² – 4ac
a = -5, b = 11 and c = -6
(11)² – 4(-5)(-6)
121 – 120 = 1
The discriminant is 1.

Question 13.
Find a possible pair of integer values for a and c so that the equation ax2 + 3x + c = 0 has two real solutions. Then write the equation.
Answer:
Given,
ax2 + 3x + c = 0
If b² – 4ac > 0 that means there are 2 real solutions
b² – 4ac = 0 there is 1 real solution
b² – 4ac < 0 that means there are imaginary solutions
ax2 + 3x + c = 0
b = 3, a = a and c = c.
3² – 4ac = 0
9 – 4ac = 0
-4ac = -9
ac = 9/4
a = 9/4 c
So, there are 2 real solutions

Question 14.
WHAT IF?
The ball leaves the juggler’s hand with an initial vertical velocity of 40 feet per second. How long is the ball in the air?
Answer:

Using the Quadratic Formula 3.4 Exercises

Vocabulary and Core Concept Check
Question 1.
COMPLETE THE SENTENCE
When a, b, and c are real numbers such that a ≠ 0, the solutions of the quadratic equation ax2 + bx + c = 0 are x= ____________.
Answer:
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers 3.3 a 81

Question 2.
COMPLETE THE SENTENCE
You can use the ____________ of a quadratic equation to determine the number and type of solutions of the equation.
Answer: You can use the discriminant of a quadratic equation to determine the number and type of solutions of the equation.

Question 3.
WRITING
Describe the number and type of solutions when the value of the discriminant is negative.
Answer: If the discriminant of a quadratic equation is negative, then the equation has two imaginary solutions.

Question 4.
WRITING
Which two methods can you use to solve any quadratic equation? Explain when you might prefer to use one method over the other.
Answer: There are three basic methods for solving quadratic equations:

  1. Factoring,
  2. Using the quadratic formula, and
  3. Completing the square.
    The quadratic formula can be used to solve any quadratic equation and it is easy to just plug in the numbers. Graphing would be a little bit more complicated but if you have a graphing calculator, solving this equation would be easy.

Monitoring Progress and Modeling with Mathematics

In Exercises 5–18, solve the equation using the Quadratic Formula. Use a graphing calculator to check your solution(s).
Question 5.
x2 − 4x + 3 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 5

Question 6.
3x2 + 6x + 3 = 0
Answer:
3x2 + 6x + 3 = 0
3(x2 + 2x + 1) = 0
x2 + 2x + 1 = 0
x² + x + x + 1 = 0
x(x + 1) + 1 (x + 1) = 0
(x + 1) (x + 1) = 0
x = -1
The solution is x = -1
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_6
From the above graph, we observe that the solution is -1.

Question 7.
x2 + 6x + 15 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 7

Question 8.
6x2 − 2x + 1 = 0
Answer:
Given equation
6x2 − 2x + 1 = 0
x = -b ± √b² – 4ac/2a
a = 6, b = -2 and c = 1
x = -(-2) ± √(-2)² – 4(6)(1)/2(6)
x = (2 ± √4 – 24)/12
x = (2 ± √-20)/12
x = 2/12 ± 2√5i/12
x = 1/6 ± √5/6i
x = 0.16 + 0.37i
x = 0.16 – 0.37i
The solution is x = 0.16 + 0.37i and x = 0.16 – 0.37i
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_8

Question 9.
x2 − 14x = −49
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 9

Question 10.
2x2 + 4x = 30
Answer:
Given,
2x2 + 4x = 30
2(x2 + 2x – 15) = 0
x2 + 2x – 15 = 0
x = -b ± √b² – 4ac/2a
a = 2, b = 4 and c = -30
x = -(2) ± √(2)² – 4(1)(15)/2(1)
x = (-2 ± √4 – 60)/2
x = (-2 ± √-56)/2
x = -2/2 ± 2√14i/2
x = -1± √14i
x = -1 + 3.74i
x = -1 – 3.74i
The solution is x = -1 + 3.74i and x = -1 – 3.74i
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_10

Question 11.
3x2 + 5 = −2x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 11

Question 12.
−3x = 2x2 − 4
Answer:
Given,
−3x = 2x2 − 4
2x2 − 4 + 3x = 0
x = -b ± √b² – 4ac/2a
a = 2, b = 3 and c = -4
x = -(3) ± √(3)² – 4(2)(-4)/2(2)
x = (-3 ± √9+32)/4
x = (-3 ± √41)/4
x = -3/4 ± √41/4
x = 0.85
x = -2.35
The solution is x = 0.85 and x = -2.35
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_12

Question 13.
−10x = −25 − x2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 13

Question 14.
−5x2 − 6 = −4x
Answer:
Given,
−5x2 − 6 = −4x
5x² – 4x + 6 = 0
x = -b ± √b² – 4ac/2a
x = -(-4) ± √(-4)² – 4(5)(6)/2(5)
x = (4 ± √16-120)/10
x = (4 ± √-104)/10
x = (4 ± 2√26i)/10
x = 2/5 ± √26i/5
x = 0.4 + 1.0198i
x = 0.4 – 1.0198i
The solution is x = 0.4 + 1.0198i and x = 0.4 – 1.0198i
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_14

Question 15.
−4x2 + 3x = −5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 15

Question 16.
x2 + 121 = −22x
Answer:
Given,
x2 + 121 = −22x
x2 + 22x + 121 = 0
b = a × c
x² + 11x + 11x + 121 = 0
x(x + 11) + 11(x + 11) = 0
(x + 11)(x + 11) = 0
x = -11
The solution is x = -11.
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_16

Question 17.
−z2 = −12z + 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 17

Question 18.
−7w + 6 = −4w2
Answer:
−7w + 6 = −4w2
-4w² + 7w – 6 = 0
-(4w² – 7w + 6 = 0)
4w² – 7w + 6 = 0
x = -b ± √b² – 4ac/2a
x = -(-7) ± √(-7)² – 4(4)(6)/2(4)
x = (7 ± √49-96)/8
x = (7 ± √-47)/8
x = (7 ± √47i)/8
x = 7/8 ± √47i/8
x = 0.875 + 0.85i
x = 0.875 – 0.85i
The solution is x = 0.875 + 0.85i and x = 0.875 – 0.85i
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_18

In Exercises 19–26, find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 19.
x2 + 12x + 36 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 19

Question 20.
x2 − x + 6 = 0
Answer:
Given equation
x2 − x + 6 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = -1 and c = 6
x = -(-1) ± √(-1)² – 4(1)(6)/2(1)
x = (1 ± √1-24)/2
x = (1 ± √-23)/2
x = (1 ± √23i)/2
x = 1/2 ± √23i/2
x = 0.5 + 2.39i
x = 0.5 – 2.39i
The solution is x = 0.5 + 2.39i and x = 0.5 – 2.39i
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_20

Question 21.
4n2 − 4n − 24 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 21

Question 22.
−x2 + 2x + 12 = 0
Answer:
Given,
x2 – 2x – 12 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = -2 and c = -12
x = -(-2) ± √(-2)² – 4(1)(12)/2(1)
x = (2 ± √4-48)/2
x = (2 ± √-44)/2
x = (2 ± √44i)/2
x = 1 ± √11i
x = 1 + 3.3i
x = 1 – 3.3i
The solution is x = 1 + 3.3i and x = 1 – 3.3i
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_22

Question 23.
4x2 = 5x − 10
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 23

Question 24.
−18p = p2 + 81
Answer:
Given,
p2 + 18p + 81 = 0
p2 + 9p + 9p + 81 = 0
p(p + 9) + 9(p + p)
(p + 9) (p + 9) = 0
p = -9
The solution is p = -9.
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_24

Question 25.
24x = −48 − 3x2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 25

Question 26.
−2x2 − 6 = x2
Answer:
Given,
−2x2 − 6 = x²
x² + 2x² + 6 = 0
3x² + 6 = 0
3(x² + 2) = 0
x² + 2 = 0
x² = -2
x = √2i
The solution is x = √2i
Check the solution with the graph.
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_26

Question 27.
USING EQUATIONS
What are the complex solutions of the equation 2x2− 16x+ 50 = 0?
A. 4 + 3i, 4 − 3i
B. 4 + 12i, 4 − 12i
C. 16 + 3i, 16 − 3i
D. 16 + 12i, 16 − 12i
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 27

Question 28.
USING EQUATIONS
Determine the number and type of solutions to the equation x2 + 7x = −11.
A. two real solutions
B. one real solution
C. two imaginary solutions
D. one imaginary solution
Answer:
x2 + 7x = −11
x2 + 7x + 11 = 0
It has two imaginary solutions

ANALYZING EQUATIONS In Exercises 29–32, use the discriminant to match each quadratic equation with the correct graph of the related function. Explain your reasoning.
Question 29.
x2 − 6x + 25 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 29

Question 30.
2x2 − 20x + 50 = 0
Answer:
Given,
2x2 − 20x + 50 = 0
x² – 10x + 25 = 0
x² – 5x – 5x + 25 = 0
x(x – 5) -5(x – 5) = 0
(x – 5) (x – 5) = 0
x = 5
The discriminant is b² – 4ac
d = b² – 4ac
a = 1, b = 10 and c = 25
(10)² – 4(1)(25)
100 – 100 = 0

Question 31.
3x2 + 6x − 9 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 31

Question 32.
5x2 − 10x − 35 = 0
Answer:
Given
5x2 − 10x − 35 = 0
Take 5 as the common factor.
5(x² – 2x – 7) = 0
It becomes
x² – 2x – 7 = 0
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_32

Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 2

ERROR ANALYSIS In Exercises 33 and 34, describe and correct the error in solving the equation.
Question 33.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 33

Question 34.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 4
Answer:
Given
x² + 6x + 8 = 2
x² + 6x + 8 – 2 = 0
x² + 6x + 6 = 0
x = -b ± √b² – 4ac/2a
a = 1, b = 6 and c = 6
x = -(6) ± √(6)² – 4(1)(6)/2(1)
x = (-6 ± √36-24)/2
x = (-6 ± √12)/2
x = (-6 ± 2√3)/2
x = -3 ± √3
x = -1.26
x = -4.73
The solution is x = -1.26 and x = -4.73

OPEN-ENDED In Exercises 35–40, find a possible pair of integer values for a and c so that the quadratic equation has the given solution(s). Then write the equation.
Question 35.
ax2 + 4x + c = 0; two imaginary solutions
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 35

Question 36.
ax2 + 6x + c = 0; two real solutions
Answer:
Given,
For a quadratic equation to have two imaginary solutions, the discriminant must be negative. So, the set the discriminant equal to any negative number.
ax2 + 6x + c = 0
1. substitute a and c
2. check for discriminant
3. b² – 4ac ≥ 0
b = 6
4. 36 – 4ac ≥ 0
The ≥ sign will give us the result for two real unequal solutions and two real equal solutions. If we only need Real unequal solutions we only > sign instead of ≥.
36 – 4ac = 0
36 = 4ac
ac = 36/4
ac = 9

Question 37.
ax2 − 8x + c = 0; two real solutions
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 37

Question 38.
ax2 − 6x + c = 0; one real solution
Answer:
Given,
ax2 – 6x + c = 0
For a quadratic equation to have two imaginary solutions, the discriminant must be negative. So, they set the discriminant equal to any negative number.
1. substitute a and c
2. check for discriminant
3. b² – 4ac ≥ 0
4. 36 – 4ac ≥ 0
The ≥ sign will give us the result for two real unequal solutions and two real equal solutions. If we only need Real unequal solutions we only > sign instead of ≥.

Question 39.
ax2 + 10x = c; one real solution
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 39

Question 40.
−4x + c = −ax2; two imaginary solutions
Answer:
-4x + c + ax² = 0
ax² – 4x + c = 0
For a quadratic equation to have two imaginary solutions, the discriminant must be negative. So, they set the discriminant equal to any negative number.
d = b² – 4ac
b² – 4ac = 0
(-4)² – 4ac = 0
16 – 4ac = 0
4 – ac = 0
ac = 4

USING STRUCTURE In Exercises 41–46, use the Quadratic Formula to write a quadratic equation that has the given solutions.
Question 41.
x = \(\frac{-8 \pm \sqrt{-176}}{-10}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 41

Question 42.
x = \(\frac{15 \pm \sqrt{-215}}{22}\)
Answer:
Given,
The solution x = \(\frac{15 \pm \sqrt{-215}}{22}\)
22x = 15 ± √-215
22x – 15 = √-215
Apply Squaring on both sides
(22x – 15)² = (√-215)²
484x² + 225x – 660 = -215
484x² + 225x – 660 + 215 = 0
484x² + 225x – 445 = 0
Thus, the equation is 484x² + 225x – 445 = 0

Question 43.
x = \(\frac{-4 \pm \sqrt{-124}}{-14}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 43

Question 44.
x = \(\frac{-9 \pm \sqrt{137}}{4}\)
Answer:
Given
The solution is x = \(\frac{-9 \pm \sqrt{137}}{4}\)
4x = -9 ± √137
4x + 9 = √137
Apply Squaring on both sides to find the equation
(4x + 9)² = (√137)²
36x² + 81 + 72x = 137
36x² – 72x + 81 – 137 = 0
36x² – 72x – 56 = 0
Thus, the equation is 36x² – 72x – 56 = 0

Question 45.
x = \(\frac{-4 \pm 2}{6}\)
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 45

Question 46.
x = \(\frac{2 \pm 4}{-2}\)
Answer:
Given the solution is
x = \(\frac{2 \pm 4}{-2}\)
-2x = 2 ± 4
-2x – 2 = 4
-x – 1 = 4
x + 1 = -4
Apply to square on both sides
(x + 1)² = 16
x² + 2x + 1 = 16
x² + 2x – 15 = 0
Thus, the equation is x² + 2x – 15 = 0

COMPARING METHODS In Exercises 47–58, solve the quadratic equation using the Quadratic Formula. Then solve the equation using another method. Which method do you prefer? Explain.
Question 47.
3x2 − 21 = 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 47

Question 48.
5x2 + 38 = 3
Answer:
Given,
Use the factoring method to find the solution.
5x2 + 38 = 3
5x2 + 38 -3 = 0
5x2 + 35 = 0
5(x² + 7) = 0
x² + 7 = 0
x² = -7
x = i√7
The solution is x = i√7

Question 49.
2x2 − 54 = 12x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 49

Question 50.
x2 = 3x + 15
Answer:
Given,
x2 = 3x + 15
x2 – 3x – 15 = 0
Use the quadratic equation formula to find the solution.
x = -b ± √b² – 4ac/2a
x = -(-3) ± √(-3)² – 4(1)(-5)/2(1)
x = (3 ± √9+20)/2
x = (3 ± √29)/2
x = 3/2 ± √29/2
x = 4.19
x = -1.19
The solution is x = 4.19 and x = -1.19

Question 51.
x2 − 7x + 12 = 0
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 51

Question 52.
x2 + 8x − 13 = 0
Answer:
Given,
x2 + 8x − 13 = 0
Use the quadratic equation formula to find the solution.
x = -b ± √b² – 4ac/2a
x = -(8) ± √(8)² – 4(1)(-13)/2(1)
a = 1, b = 8 and c = -13
x = (-8 ± √64+54)/2
x = (-8 ± √116)/2
x = -8/2 ± √116/2
x = -4 ± √29
x = 1.38
x = -9.38
The solution is x = 1.38 and -9.38

Question 53.
5x2 − 50x = −135
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 53

Question 54.
8x2 + 4x + 5 = 0
Answer:
Given,
8x2 + 4x + 5 = 0
Use the quadratic equation formula to find the solution.
x = -b ± √b² – 4ac/2a
a = 8, b = 4 and c = 5
x = -(4) ± √(4)² – 4(8)(-5)/2(8)
x = (-4 ± √16+160)/16
x = (-4 ± √176)/16
x = -4/16 ± √176/16
x = -1/4 ± √11/4
x = 0.57
x = -1.07
The solution is x = 0.57 and x = -1.07

Question 55.
−3 = 4x2 + 9x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 55.1
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 55.2

Question 56.
−31x + 56 = −x2
Answer:
Given,
−31x + 56 = −x2
x² – 31x + 56 = 0
Use the quadratic equation formula to find the solution.
x = -b ± √b² – 4ac/2a
a = 1, b = -31 and c = 56
x = -(-31) ± √(-31)² – 4(1)(56)/2(1)
x = (31 ± √961-224)/2
x = (31 ± √737)/2
x = 31/2 ± √737/2
x = 29.07
x = 1.92
The solution is x = 29.07 and x = 1.92

Question 57.
x2 = 1 − x
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 57

Question 58.
9x2 + 36x + 72 = 0
Answer:
Given,
9x2 + 36x + 72 = 0
Take 9 as a common factor.
9(x² + 4x + 8) = 0
x² + 4x + 8 = 0
x = -b ± √b² – 4ac/2a
Use the quadratic equation formula to find the solution.
x = -(4) ± √(4)² – 4(1)(8)/2(1)
x = (-4 ± √16-32)/2
x = (-4 ± √-16)/2
x = -4/2 ± 4i/2
x = -2 + 2i
x = -2 – 2i
The solution is x = -2 + 2i and x = -2 – 2i

MATHEMATICAL CONNECTIONS In Exercises 59 and 60, find the value for x.
Question 59.
Area of the rectangle = 24 m2
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 59

Question 60.
Area of the triangle = 8ft2
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 6
Answer:
Given,
Area of the triangle = 8ft2
We know that,
Area of the triangle = 1/2 × bh
b = (3x – 7)
h = (x + 1)
8 = 1/2 × (3x – 7) (x + 1)
16 = 3x(x + 1) -7(x + 1)
16 = 3x² + 3x – 7x – 7
16 + 7 = 3x² – 4x
3x² – 4x – 23 = 0
x = -b ± √b² – 4ac/2a
x = -(-4) ± √(-4)² – 4(3)(23)/2(3)
x = (4 ± √-260)/6
x = (4 ± √260i)/6
x = 4/6 ± 2√65i/6
x = 2/3 ± √65i/3
x = 0.66 + 2.68i
x = 0.66 – 2.68i
The solution is x = 0.66 + 2.68i and x = 0.66 – 2.68i

Question 61.
MODELING WITH MATHEMATICS
A lacrosse player throws a ball in the air from an initial height of 7 feet. The ball has an initial vertical velocity of 90 feet per second. Another player catches the ball when it is 3 feet above the ground. How long is the ball in the air?
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 7
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 61

Question 62.
NUMBER SENSE
Suppose the quadratic equation ax2 + 5x + c = 0 has one real solution. Is it possible for a and c to be integers? rational numbers? Explain your reasoning. Then describe the possible values of a and c.
Answer:

Question 63.
MODELING WITH MATHEMATICS
In a volleyball game, a player on one team spikes the ball over the net when the ball is 10 feet above the court. The spike drives the ball downward with an initial vertical velocity of 55 feet per second. How much time does the opposing team have to return the ball before it touches the court?
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 63

Question 64.
MODELING WITH MATHEMATICS
An archer is shooting at targets. The height of the arrow is 5 feet above the ground. Due to safety rules, the archer must aim the arrow parallel to the ground.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 8
a. How long does it take for the arrow to hit a target that is 3 feet above the ground?
b. What method did you use to solve the quadratic equation? Explain.
Answer:
An archer is shooting at targets.
The height of the arrow is 5 feet above the ground.
h = 1/2 gt²
h = 5 – 3 = 2
h = 2 ft
g = 32.2 ft/s²
2 = 32.2 × t²
4 – 32.2 t²
0.124 = t²
t = 0.35s
So, the time to hit the target is 0.35 seconds

Question 65.
PROBLEM SOLVING
A rocketry club is launching model rockets. The launching pad is 30 feet above the ground. Your model rocket has an initial vertical velocity of 105 feet per second. Your friend’s model rocket has an initial vertical velocity of 100 feet per second.
a. Use a graphing calculator to graph the equations of both model rockets. Compare the paths.
b. After how many seconds is your rocket 119 feet above the ground? Explain the reasonableness of your answer(s).
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 65

Question 66.
PROBLEM SOLVING
The number A of tablet computers sold (in millions) can be modeled by the function A = 4.5t2 + 43.5t + 17, where t represents the year after 2010.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 9
a. In what year did the tablet computer sales reach 65 million?
b. Find the average rate of change from 2010 to 2012 and interpret the meaning in the context of the situation.
c. Do you think this model will be accurate after a new, innovative computer is developed? Explain.
Answer:
The number A of tablet computers sold (in millions) can be modeled by the function A = 4.5t2 + 43.5t + 17, where t represents the year after 2010.
A = 4.5t2 + 43.5t + 17
where,
A = 65
65 = 4.5t2 + 43.5t + 17
t = 1 and t = -10.67
Therefore, the answer is 2011.

Question 67.
MODELING WITH MATHEMATICS
A gannet is a bird that feeds on fish by diving into the water. A gannet spots a fish on the surface of the water and dives 100 feet to catch it. The bird plunges toward the water with an initial vertical velocity of −88 feet per second.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 10
a. How much time does the fish have to swim away?
b. Another gannet spots the same fish, and it is only 84 feet above the water and has an initial vertical velocity of −70 feet per second. Which bird will reach the fish first? Justify your answer.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 67.1
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 67.2

Question 68.
USING TOOLS
You are asked to find a possible pair of integer values for a and c so that the equation ax2 − 3x + c = 0 has two real solutions. When you solve the inequality for the discriminant, you obtain ac < 2.25. So, you choose the values a = 2 and c = 1. Your graphing calculator displays the graph of your equation in a standard viewing window. Is your solution correct? Explain.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 11
Answer:

Question 69.
PROBLEM SOLVING
Your family has a rectangular pool that measures 18 feet by 9 feet. Your family wants to put a deck around the pool but is not sure how wide to make the deck. Determine how wide the deck should be when the total area of the pool and deck is 400 square feet. What is the width of the deck?
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 12
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 69

Question 70.
HOW DO YOU SEE IT?
The graph of a quadratic function y = ax2 + bx + c is shown. Determine whether each discriminant of ax2 + bx + c = 0 is positive, negative, or zero. Then state the number and type of solutions for each graph. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 13
Answer:
The graph shows the parabola with branches going down in a negative y-direction.
This gives you a < 0.
Since the parabola has no x-intercepts, then the equation ax2 + bx + c = 0 has no solutions and the discriminant is less than 0.
d < 0

Question 71.
CRITICAL THINKING
Solve each absolute value equation.
a. |x2 – 3x – 14| = 4
b. x2 = |x| + 6
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 71.1
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 71.2

Question 72.
MAKING AN ARGUMENT
The class is asked to solve the equation 4x2 + 14x + 11 = 0. You decide to solve the equation by completing the square. Your friend decides to use the Quadratic Formula. Whose method is more efficient? Explain your reasoning.
Answer:
Given,
The class is asked to solve the equation 4x2 + 14x + 11 = 0.
4x2 + 14x + 11 = 0
x = -b ± √b² – 4ac/2a
x = -(14) ± √(14)² – 4(4)(11)/2(4)
a = 4, b = 14 and c = 11
x = (-14 ± √196 – 176)/8
x = (-14 ± √20)/8
x = -14/8 ± 2√5/8
x = -7/4 ± √5/4
x = -1.19
x = -2.3
The solution is x = -1.19 and x = -2.3

Question 73.
ABSTRACT REASONING
For a quadratic equation ax2 + bx + c = 0 with two real solutions, show that the mean of the solutions is \(\frac{b}{2a}\). How is this fact related to the symmetry of the graph of y = ax2 + bx + c?
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 73

Question 74.
THOUGHT PROVOKING
Describe a real-life story that could be modeled by h = −16t2 + v0t + h0 . Write the height model for your story and determine how long your object is in the air.
Answer:
h = −16t2 + v0t + h0 .
Consider the real-life example as, if a stone is launched straight up from the top of a 48 meter tall building and it has an initial speed of 32 meter per second.
h = −16t2 + 32t + 48
let h = 0
−16t2 + 32t + 48 = 0
Take -16 as a common factor.
-16(t² + 2t – 3) = 0
t² + 2t – 3 = 0
t² – 3t + 1t – 3 = 0
t(t – 3) + 1(t – 3) = 0
(t-3) (t + 1) = 0
t – 3 = 0
t = 3
t + 1 = 0
t = -1
Thus the time the stone stays in the air is 3 seconds.

Question 75.
REASONING
Show there is no quadratic equation ax2+bx+c= 0 such that a, b, and c are real numbers and 3i and −2i are solutions.
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 75

Question 76.
MODELING WITH MATHEMATICS
The Stratosphere Tower in Las Vegas is 921 feet tall and has a “needle” at its top that extends even higher into the air. A thrill ride called Big Shot catapults riders 160 feet up the needle and then lets them fall back to the launching pad.
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 14
a. The height h (in feet) of a rider on the Big Shot can be modeled by h = −16t2 + v0 t + 921, where t is the elapsed time (in seconds) after launch and v0 is the initial vertical velocity (in feet per second). Find v0 using the fact that the maximum value of h is 921 + 160 = 1081 feet.
b. A brochure for the Big Shot states that the ride up the needle takes 2 seconds. Compare this time to the time given by the model h = −16t2 + v0 t + 921, where v0 is the value you found in part (a). Discuss the accuracy of the model.
Answer:

Maintaining Mathematical Proficiency

Solve the system of linear equations by graphing.
Question 77.
−x + 2y = 6
x + 4y = 24
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 77

Question 78.
y = 2x − 1
y = x + 1
Answer:
Given two equations
y = 2x − 1
y = x + 1
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_80
The equation of the line meet at 2. It has one solution.
The graph intersects at point (2,3)

Question 79.
3x + y = 4
6x + 2y = −4
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 79

Question 80.
y = −x + 2
−5x + 5y = 10
Answer:
Given two linear equations
y = −x + 2
−5x + 5y = 10
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_80
The graph intersects at point 2, so the solution is (0,2)

Graph the quadratic equation. Label the vertex and axis of symmetry.
Question 81.
y = −x2 + 2x + 1
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 81

Question 82.
y = 2x2 − x + 3
Answer:
Given quadratic equation is y = 2x2 − x + 3
2x2 − x + 3
2x2 + 2x – 3x + 3 = 0
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_82

Question 83.
y = 0.5x2 + 2x + 5
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.4 a 83

Question 84.
y = −3x2 − 2
Answer:
Given equation is y = −3x2 − 2
Big Ideas Math Answers Chapter 3 Using the Quadratic Formula 3.4 Exercises_84

Lesson 3.5 Solving Nonlinear Systems

Essential Question How can you solve a nonlinear system of equations?

EXPLORATION 1

Solving Nonlinear Systems of Equations
Work with a partner. Match each system with its graph. Explain your reasoning. Then solve each system using the graph.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 1

EXPLORATION 2

Solving Nonlinear Systems of Equations
Work with a partner. Look back at the nonlinear system in Exploration 1(f). Suppose you want a more accurate way to solve the system than using a graphical approach.
a. Show how you could use a numerical approach by creating a table. For instance, you might use a spreadsheet to solve the system.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 2
b. Show how you could use an analytical approach. For instance, you might try solving the system by substitution or elimination.

Communicate Your Answer

Question 3.
How can you solve a nonlinear system of equations?
Answer: We can solve a nonlinear system of equations with two variables by graphing, substitution, and elimination.

  • Identify the graph of each equation.
  • Graph the first equation.
  • Solve one of the equations for either variable.
  • Substitute the expression from Step 2 into the other equation.
  • Check whether the graph intersects or not.
  • Identify the point of intersection.
  • Solve the resulting equation.

Question 4.
Would you prefer to use a graphical, numerical, or analytical approach to solve the given nonlinear system of equations? Explain your reasoning.
Answer:
The numerical methods approximate solutions to equations when exact solutions can’t be determined by algebraic methods. Graphical methods are used for solving non-linear equations by seeing graphs we have an idea for solving systems. We can use the analytical method to find the nonlinear system of equations.

Solve the system using any method. Explain your choice of method.
Question 1.
y = −x2 + 4
y = −4x + 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.5-1.jpg

Question 2.
x2 + 3x + y = 0
2x + y = 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.5-2.jpg

Question 3.
2x2 + 4x − y =−2
x2 + y = 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.5-3.jpg

Solve the system.
Question 4.
x2 + y2 = 16
y = −x + 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.5-4.jpg

Question 5.
x2 + y2 = 4
y = x + 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.5-5.jpg

Question 6.
x2 + y2 = 1
y = \(\frac{1}{2}\)x + \(\frac{1}{2}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.5-6.jpg

Solve the equation by graphing.
Question 7.
x2 − 6x + 15 = −(x − 3)2 + 6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.5-7.jpg

Question 8.
(x + 4)(x − 1) = −x2 + 3x + 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.5-8.jpg

Solving Nonlinear Systems 3.5 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Describe the possible solutions of a system consisting of two quadratic equations.
Answer: The possible number of solutions of a system of two quadratic equations is two, when they intersect in two points, one when they intersect in one point, or zero when they do not intersect.

Question 2.
WHICH ONE DOESN’T BELONG?
Which system does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-2.jpg

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, solve the system by graphing. Check your solution(s).
Question 3.
y = x + 2
y = 0.5(x + 2)2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 3

Question 4.
y = (x − 3)2 + 5
y = 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-4.jpg

Question 5.
y = \(\frac{1}{3}\)x + 2
y = −3x2 − 5x − 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 5

Question 6.
y = −3x2 − 30x − 71
y = −3x − 17
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-6.jpg

Question 7.
y = x2 + 8x + 18
y = −2x2 − 16x − 30
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 7

Question 8.
y = −2x2 − 9
y = −4x − 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-8.jpg

Question 9.
y = (x − 2)2
y = −x2 + 4x − 2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 9

Question 10.
y = \(\frac{1}{2}\)(x + 2)2
y = −\(\frac{1}{2}\)x2 + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-10.jpg

In Exercises 11–14, solve the system of nonlinear equations using the graph.
Question 11.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 11

Question 12.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 5
Answer:
Observe the two functions carefully.
The two graphs represent two parabolas: one opens up and the other opens down.
The functions do not intersect.
So the system has no solution.

Question 13.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 6
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 13

Question 14.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 7
Answer:

In Exercises 15–24, solve the system by substitution.
Question 15.
y = x + 5
y = x2 − x + 2
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 15

Question 16.
x2 + y2 = 49
y = 7 − x
Answer:

Question 17.
x2 + y2 = 64
y = −8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 17

Question 18.
x = 3
−3x2 + 4x − y = 8
Answer:

Question 19.
2x2 + 4x − y = −3
−2x + y = −4
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 19

Question 20.
2x − 3 = y + 5x2
y = −3x − 3
Answer:

Question 21.
y = x2 − 1
−7 = −x2 − y
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 21

Question 22.
y + 16x − 22 = 4x2
4x2 − 24x + 26 + y = 0
Answer:

Question 23.
x2 + y2 = 7
x + 3y = 21
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 23

Question 24.
x2 + y2 = 5
−x + y = −1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-24.jpg

Question 25.
USING EQUATIONS
Which ordered pairs are solutions of the nonlinear system?
y = \(\frac{1}{2}\)x2 − 5x + \(\frac{21}{2}\)
y = −\(\frac{1}{2}\)x + \(\frac{13}{2}\)
A. (1, 6)
B. (3, 0)
C. (8, 2.5)
D. (7, 0)
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 25

Question 26.
USING EQUATIONS
How many solutions does the system have? Explain your reasoning.
y = 7x2 − 11x + 9
y = −7x2 + 5x − 3
A. 0
B. 1
C. 2
D. 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-26.jpg

In Exercises 27–34, solve the system by elimination.
Question 27.
2x2 − 3x −y =−5
−x + y = 5
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 27

Question 28.
−3x2 + 2x − 5 = y
−x + 2 = −y
Answer:

Question 29.
−3x2 + y = −18x + 29
−3x2 − y = 18x − 25
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 29

Question 30.
y = −x2 − 6x 10
y = 3x2 + 18x + 22
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-30.jpg

Question 31.
y + 2x = −14
−x2 − y − 6x = 11
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 31

Question 32.
y = x2 + 4x + 7
−y = 4x + 7
Answer:
Given,
y = x2 + 4x + 7
−y = 4x + 7
Add both the equations
y = x2 + 4x + 7
−y = 4x + 7
0 = x² + 8x + 14
x² + 8x + 14 = 0
Use the quadratic function to find the solution of x.
x = -b ± √b² – 4ac/2a
a = 1, b = 8 and c = 14.
x = -(8) ± √(8)² – 4(1)(14)/2(1)
x = (-8 ± √64 – 56)/2
x = (-8 ± √8)/2
x = -8/2 ± √8/2
x = -4 ± √2
x = -2.58
x = -5.41
So, the solution is -2.58 and -5.41

Question 33.
y = −3x2 − 30x − 76
y = 2x2 + 20x + 44
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 33

Question 34.
−10x2 + y = −80x + 155
5x2 + y = 40x − 85
Answer:
Given,
−10x2 + y = −80x + 155
5x2 + y = 40x − 85
–     –       –       +
-15x² = -120x + 240
-x² = -8x + 16
x² – 8x + 16 = 0
Use the quadratic function to find the solution of x.
x = -b ± √b² – 4ac/2a
a = 1, b = -8 and c = 16
x = -(8) ± √(8)² – 4(1)(16)/2(1)
x = (-8 ± √64 – 64)/2
x = (-8)/2
x = -4
So, the solution is x = -4

Question 35.
ERROR ANALYSIS
Describe and correct the error in using elimination to solve a system.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 35

Question 36.
NUMBER SENSE
The table shows the inputs and outputs of two quadratic equations. Identify the solution(s) of the system. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 9
Answer:

In Exercises 37–42, solve the system using any method. Explain your choice of method.
Question 37.
y = x2 − 1
−y = 2x2 + 1
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 37

Question 38.
y = −4x2 − 16x − 13
−3x2 + y + 12x = 17
Answer:

Question 39.
−2x + 10 + y = \(\frac{1}{3}\)x2
y = 10
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 39

Question 40.
y = 0.5x2 − 10
y = −x2 + 14
Answer:

Question 41.
y = −3(x − 4)2 + 6
(x − 4)2 + 2 − y = 0
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 41

Question 42.
−x2 + y2 = 100
y = −x + 14
Answer:
Given,
−x2 + y2 = 100
y = −x + 14
x² + y² – 28x = 196

−x2 + y2 = 100
(-)x² + y² = 28x + 196
-2x² = 28x – 96
-x² = 14x – 48
x² + 14x – 48 = 0
Use the quadratic function to find the solution of x.
x = -b ± √b² – 4ac/2a
a = 1, b = 14 and c = -48
x = -(14) ± √(14)² – 4(1)(-48)/2(1)
x = (-14 ± √196 + 192)/2
x = (-14 ± √388)/2
x = -14/2 ± 2√97/2
x = -7 ± √97
The solution is x = -7 + √97 and x = -7 – √97

USING TOOLS In Exercises 43–48, solve the equation by graphing.
Question 43.
x2 + 2x = −\(\frac{1}{2}\)x2 + 2x
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 43

Question 44.
2x2 − 12x − 16 = −6x2 + 60x − 144
Answer:

Question 45.
(x + 2)(x − 2) = −x2 + 6x − 7
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 45

Question 46.
−2x2 − 16x − 25 = 6x2 + 48x + 95
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-46.jpg

Question 47.
(x − 2)2 − 3 = (x + 3)(−x + 9) − 38
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 47

Question 48.
(−x + 4)(x + 8) − 42 = (x + 3)(x + 1) − 1
Answer:

Question 49.
REASONING
A nonlinear system contains the equations of a constant function and a quadratic function. The system has one solution. Describe the relationship between the graphs.
Answer: There is only one solution to the system, the graph of the constant function intersects the graph of the quadratic function at the vertex.

Question 50.
PROBLEM SOLVING
The range (in miles) of a broadcast signal from a radio tower is bounded by a circle given by the equation x2 + y2= 1620.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 10
A straight highway can be modeled by the equation y = −\(\frac{1}{3}\)x + 30.
For what lengths of the highway are cars able to receive the broadcast signal?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-50.jpg

Question 51.
PROBLEM SOLVING
A car passes a parked police car and continues at a constant speed r. The police car begins accelerating at a constant rate when it is passed. The diagram indicates the distance d (in miles) the police car travels as a function of time t (in minutes) after being passed. Write and solve a system of equations to find how long it takes the police car to catch up to the other car.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 11
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 51

Question 52.
THOUGHT PROVOKING
Write a nonlinear system that has two different solutions with the same y-coordinate. Sketch a graph of your system. Then solve the system.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-52.jpg
Question 53.
OPEN-ENDED
Find three values for m so the system has no solution, one solution, and two solutions. Justify your answer using a graph.
3y = −x2 + 8x − 7
y = mx + 3
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 53

Question 54.
MAKING AN ARGUMENT
You and a friend solve the system shown and determine that x = 3 and x = −3. You use Equation 1 to obtain the solutions (3, -3), (3, −3), (−3, 3), and (−3, −3). Your friend uses Equation 2 to obtain the solutions (3, 3) and (−3, −3). Who is correct? Explain your reasoning.
x2 + y2 = 18 Equation 1
x − y = 0 Equation 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-54.jpg

Question 55.
COMPARING METHODS
Describe two different ways you could solve the quadratic equation. Which way do you prefer? Explain your reasoning.
−2x2 + 12x − 17 = 2x2 − 16x + 31
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 55

Question 56.
ANALYZING RELATIONSHIPS
Suppose the graph of a line that passes through the origin intersects the graph of a circle with its center at the origin. When you know one of the points of intersection, explain how you can find the other point of intersection without performing any calculations.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-56.jpg

Question 57.
WRITING
Describe the possible solutions of a system that contains (a) one quadratic equation and one equation of a circle, and (b) two equations of circles. Sketch graphs to justify your answers.
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 57.1
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 57.2
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 57.3

Question 58.
HOW DO YOU SEE IT?
The graph of a nonlinear system is shown. Estimate the solution(s). Then describe the transformation of the graph of the linear function that results in a system with no solution.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 12
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-58.jpg

Question 59.
MODELING WITH MATHEMATICS
To be eligible for a parking pass on a college campus, a student must live at least 1 mile from the campus center.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 13
a. Write equations that represent the circle and Oak Lane.
b. Solve the system that consists of the equations in part (a).
c. For what length of Oak Lane are students not eligible for a parking pass?
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 59.1
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 59.2

Question 60.
CRITICAL THINKING
Solve the system of three equations shown.
x2 + y2 = 4
2y = x2 − 2x + 4
y = −x + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-60.jpg

Maintaining Mathematical Proficiency

Solve the inequality. Graph the solution on a number line.
Question 61.
4x − 4 > 8
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 61

Question 62.
−x + 7 ≤ 4 − 2x
Answer:

Question 63.
−3(x − 4) ≥ 24
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 63

Write an inequality that represents the graph.
Question 64.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 14
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-64.jpg

Question 65.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 15
Answer:
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 a 65

Question 66.
Big Ideas Math Answers Algebra 2 Chapter 3 Quadratic Equations and Complex Numbers 3.5 16
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.5-66.jpg

Lesson 3.6 Quadratic Inequalities

Essential Question How can you solve a quadratic inequality?

EXPLORATION 1

Solving a Quadratic Inequality
Work with a partner. The graphing calculator screen shows the graph of f(x) = x2 + 2x − 3.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 1
Explain how you can use the graph to solve the inequality x2 + 2x − 3 ≤ 0.
Then solve the inequality.

EXPLORATION 2

Solving Quadratic Inequalities
Work with a partner. Match each inequality with the graph of its related quadratic function. Then use the graph to solve the inequality.
a. x2 − 3x + 2 > 0
b. x2 − 4x + 3 ≤ 0
c. x2 − 2x − 3 < 0
d. x2 + x − 2 ≥ 0
e. x2 − x − 2 < 0
f. x2 − 4 > 0
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 2

Communicate Your Answer

Question 3.
How can you solve a quadratic inequality?
Answer:
In order to solve a quadratic inequality we have to try to find all possible values of the variables which will make the equality true.
Suppose we have an equation
x² – x – 6 > 0.
x² – 3x + 2x – 6 > 0
x(x – 3) + 2(x – 3) > 0
x = -2 and x = 3
Here you can try to find all the values of x for which the quadratic is greater than zero or positive.
The internals of the equation is (-∞, -2) or (3, ∞).

Question 4.
Explain how you can use the graph in Exploration 1 to solve each inequality. Then solve each inequality.
Answer:
In exploration 1 the equation of the graph is
x² + 2x – 3 ≤ 0.
Use the quadratic function to find the solution of the graph.
x = -b ± √b² – 4ac/2a
a = 1, b = 2, c = -3.
x = -2 ± √(2² – 4 × 1 × (-3))/2.
x = -2 ± √(6)/2
x = -2 ± 4/2
x = -3 or x = 1.
So, the inequality can be simplified into (x+3) (x-1) ≤ 0.
They’re one of the factors is positive and the other is negative, so the solution will be x = [-3, 1].

Monitoring Progress

Graph the inequality.
Question 1.
y ≥ x2 + 2x − 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.6-1.jpg

Question 2.
y ≤ 2x2 −x − 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.6-2.jpg

Question 3.
y > −x2 + 2x + 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.6-3.jpg

Question 4.
Graph the system of inequalities consisting of y ≤ −x2 and y > x2 − 3.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.6-4.jpg

Solve the inequality.
Question 5.
2x2 + 3x ≤ 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.6-5.jpg

Question 6.
−3x2 − 4x + 1 < 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-Monitoring-3.6-6.jpg

Question 7.
2x2 + 2 > −5x
Answer:

Question 8.
WHAT IF?
In Example 6, the area must be at least 8500 square feet. Describe the possible lengths of the parking lot.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-8.jpg

Quadratic Inequalities 3.6 Exercises

Vocabulary and Core Concept Check
Question 1.
WRITING
Compare the graph of a quadratic inequality in one variable to the graph of a quadratic inequality in two variables.
Answer: The graph of a quadratic inequality in one variable is an interval on the real number line, whereas the graph of a quadratic inequality in two variables is a region in the coordinate plane.

Question 2.
WRITING
Explain how to solve x2 + 6x − 8 < 0 using algebraic methods and using graphs.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, match the inequality with its graph. Explain your reasoning.
Question 3.
y ≤ x2 + 4x + 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 3

Question 4.
y > −x2 + 4x − 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-4.jpg

Question 5.
y < x2 − 4x + 3
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 5

Question 6.
y ≥ x2 + 4x + 3
Answer:

Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 3

In Exercises 7–14, graph the inequality.
Question 7.
y < −x2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 7

Question 8.
y ≥ 4x2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-8-1.jpg

Question 9.
y > x2 − 9
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 9

Question 10.
y < x2 + 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-10.jpg

Question 11.
y ≤ x2 + 5x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 11

Question 12.
y ≥ −2x2 + 9x − 4
Answer:
graph y≥-2x²+9x-4
X= – b/2a
X= – 9/-4 = 2.25
y = -2(2.25)² + 9(2.25) – 4
y = -2(5.0625)+20.25 – 4
y = -10.125 +20.25 – 4
y = 6.125 or 49/8
V(2.25,6.125)

Question 13.

y > 2(x + 3)2 − 1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 13

Question 14.
y ≤ (x − \(\frac{1}{2}\))2 + \(\frac{5}{2}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-14.jpg

ANALYZING RELATIONSHIPS In Exercises 15 and 16, use the graph to write an inequality in terms of f (x) so point P is a solution.
Question 15.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 15

Question 16.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 5
Answer:

ERROR ANALYSIS In Exercises 17 and 18, describe and correct the error in graphing y ≥ x2 + 2.
Question 17.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 6
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 17

Question 18.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 7
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-18.jpg

Question 19.
MODELING WITH MATHEMATICS
A hardwood shelf in a wooden bookcase can safely support a weight W (in pounds) provided W ≤ 115x2, where x is the thickness (in inches) of the shelf. Graph the inequality and interpret the solution.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 19

Question 20.
MODELING WITH MATHEMATICS
A wire rope can safely support a weight W (in pounds) provided W ≤ 8000d2, where d is the diameter (in inches) of the rope. Graph the inequality and interpret the solution.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-20.jpg

In Exercises 21–26, graph the system of quadratic inequalities.
Question 21.
y ≥ 2x2
y < −x2 + 1
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 21

Question 22.
y > −5x2
y > 3x2 − 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-22.jpg

Question 23.
y ≤ −x2 + 4x − 4
y < x2 + 2x − 8
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 23

Question 24.
y ≥ x2 − 4
y ≤ −2x2 + 7x + 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-24.jpg

Question 25.
y ≥ 2x2 + x − 5
y < −x2 + 5x + 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 25

Question 26.
y ≥ x2 − 3x − 6
y ≥ x2 + 7x + 6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-26.jpg

In Exercises 27–34, solve the inequality algebraically.
Question 27.
4x2 < 25
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 27

Question 28.
x2 + 10x + 9 < 0
Answer:
Given,
x2 + 10x + 9 < 0
Use the factoring method to find the solution of the equation.
x2 + 1x + 9x + 9 < 0
x(x + 1) + 9(x + 1) < 0
(x + 9) (x + 1) < 0
x = -9 or x = -1
The solution is x = -9 and x = -1.
-9 < 0 and -1 < 0

Question 29.
x2 − 11x ≥ −28
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 29

Question 30.
3x2 − 13x > −10
Answer:
Given,
3x2 − 13x > −10
3x2 − 13x + 10 > 0
Use the factoring method to find the solution of the equation.
3x² – 3x – 10x + 10 >0
3x(x – 1) -1(x – 1) > 0
(3x – 1) >0 or (x – 1) >0
3x – 1 > 0
x > 1/3 or x > 1
The solution is x = 1/3 and x = 1.
1/3 > x > 1

Question 31.
2x2 − 5x − 3 ≤ 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 31

Question 32.
4x2 + 8x − 21 ≥ 0
Answer:
Given,
4x2 + 8x − 21 ≥ 0
Use the factoring method to find the solution of the equation.
2x(2x + 3) + 1(2x + 3)
(2x + 1)(2x + 3) ≥ 0
x = -1/2 or x = -3/2
-1/2 < 0 and -3/2 < 0

Question 33.
\(\frac{1}{2}\)x2 − x > 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 33.1
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 33.2

Question 34.
−\(\frac{1}{2}\)x2 + 4x ≤ 1
Answer:
Given,
−\(\frac{1}{2}\)x2 + 4x ≤ 1
−\(\frac{1}{2}\)x2 + 4x – 1 ≤ 0
Use the quadratic function method to find the solution.
x = -b ± √b² – 4ac/2a
a = -1/2, b = 4 and c = -1
x = -(4) ± √(4)² – 4(-0.5)(-1)/2(-0.5)
x = (-14 ± √16 – 2)/-1
x = (-4 ± √14)/-1
x = 4 ± √14
The solution is x = 4 + √14 and x = 4 – √14

In Exercises 35–42, solve the inequality by graphing.
Question 35.
x2 − 3x + 1 < 0
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 35

Question 36.
x2 − 4x + 2 > 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-36.jpg

Question 37.
x2 + 8x > −7
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 37

Question 38.
x2 + 6x < −3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-38.jpg

Question 39.
3x2 − 8 ≤ − 2x
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 39

Question 40.
3x2 + 5x − 3 < 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-40.jpg

Question 41.
\(\frac{1}{3}\)x2 + 2x ≥ 2
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 41

Question 42.
\(\frac{3}{4}\)x2 + 4x ≥ 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-42.jpg

Question 43.
DRAWING CONCLUSIONS
Consider the graph of the function f(x) = ax2 + bx + c.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 8
a. What are the solutions of ax2 + bx + c < 0?
b. What are the solutions of ax2 + bx + c > 0?
c. The graph of g represents a reflection in the x-axis of the graph of f. For which values of x is g(x) positive?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 43

Question 44.
MODELING WITH MATHEMATICS
A rectangular fountain display has a perimeter of 400 feet and an area of at least 9100 feet. Describe the possible widths of the fountain.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 9
Answer:

Question 45.
MODELING WITH MATHEMATICS
The arch of the Sydney Harbor Bridge in Sydney, Australia, can be modeled by y = −0.00211x2 + 1.06x, where x is the distance (in meters) from the left pylons and y is the height (in meters) of the arch above the water. For what distances x is the arch above the road?
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 45

Question 46.
PROBLEM SOLVING
The number T of teams that have participated in a robot-building competition for high-school students over a recent period of time x(in years) can be modeled by T(x) = 17.155x2 + 193.68x + 235.81, 0 ≤ x ≤ 6.
After how many years is the number of teams greater than 1000? Justify your answer.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 11
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-46.jpg

Question 47.
PROBLEM SOLVING
A study found that a driver’s reaction time A(x) to audio stimuli and his or her reaction time V(x) to visual stimuli (both in milliseconds) can be modeled by
A(x) = 0.0051x2 − 0.319x + 15, 16 ≤ x ≤ 70
V(x) = 0.005x2 − 0.23x + 22, 16 ≤ x ≤ 70
where x is the age (in years) of the driver.
a. Write an inequality that you can use to find the x-values for which A(x) is less than V(x).
b. Use a graphing calculator to solve the inequality A(x) < V(x). Describe how you used the domain 16 ≤ x ≤ 70 to determine a reasonable solution. c. Based on your results from parts (a) and (b), do you think a driver would react more quickly to a traffic light changing from green to yellow or to the siren of an approaching ambulance? Explain.
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 47

Question 48.
HOW DO YOU SEE IT?
The graph shows a system of quadratic inequalities.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 12
a. Identify two solutions of the system.
b. Are the points (1, −2) and (5, 6) solutions of the system? Explain.
c. Is it possible to change the inequality symbol(s) so that one, but not both of the points in part (b), is a solution of the system? Explain.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-3.6-48.jpg

Question 49.
MODELING WITH MATHEMATICS
The length L (in millimeters) of the larvae of the black porgy fish can be modeled by L(x) = 0.00170x2 + 0.145x + 2.35, 0 ≤ x ≤ 40 where x is the age (in days) of the larvae. Write and solve an inequality to find at what ages a larva’s length tends to be greater than 10 millimeters. Explain how the given domain affects the solution.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 13
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 49

Question 50.
MAKING AN ARGUMENT
You claim the system of inequalities below, where a and b are real numbers, has no solution. Your friend claims the system will always have at least one solution. Who is correct? Explain.
y < (x + a)2
y < (x + b)2
Answer:

Question 51.
MATHEMATICAL CONNECTIONS
The area A of the region bounded by a parabola and a horizontal line can be modeled by A= \(\frac{2}{3}\)bh, where b and h are as defined in the diagram. Find the area of the region determined by each pair of inequalities.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 14
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 51.1
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 51.2

Question 52.
THOUGHT PROVOKING
Draw a company logo that is created by the intersection of two quadratic inequalities. Justify your answer.
Answer:

Question 53.
REASONING
A truck that is 11 feet tall and 7 feet wide is traveling under an arch. The arch can be modeled by y = −0.0625x2 + 1.25x + 5.75, where x and y are measured in feet.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 15
a. Will the truck fit under the arch? Explain.
b. What is the maximum width that a truck 11 feet tall can have and still make it under the arch?
c. What is the maximum height that a truck 7 feet wide can have and still make it under the arch?
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 53.1
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 53.2

Maintaining Mathematical Proficiency

Graph the function. Label the x-intercept(s) and the y-intercept.
Question 54.
f(x) = (x + 7)(x − 9)
Answer:
Given
f(x) = (x + 7)(x − 9)
f(x) = 0
x + 7 = 0 or x – 9 = 0
x = -7 or x = 9

Question 55.
g(x) = (x − 2)2 − 4
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 55

Question 56.
h(x) = −x2 + 5x − 6
Answer:
Given,
h(x) = −x2 + 5x − 6
h(x) = 0
−x2 + 5x − 6 = 0
x = -b ± √b² – 4ac/2a
a = -1, b = 5 and c = -6
x = -5 ± √5² – 4(-1)(-6)/2(-1)
x = -5 ± √25 – 24/-2
x = -5 ±-(1/2)
x = -5 – 1/2 = -5 1/2
x = -5 + 1/2 = -4 1/2
Big Ideas Math Answers Chapter 3 Quadratic Inequalities 3.6 Exercises_56

Find the minimum value or maximum value of the function. Then describe where the function is increasing and decreasing.
Question 57.
f(x) = −x2 − 6x − 10
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 57

Question 58.
h(x) = \(\frac{1}{2}\)(x + 2)2 − 1
Answer:

Question 59.
f(x) = −(x − 3)(x + 7)
Answer:
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 a 59

Question 60.
h(x) = x2 + 3x − 18
Answer:

Quadratic Equations and Complex Numbers Performance Task: Algebra in Genetics: The Hardy-Weinberg Law

3.4–3.6 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 16

Core Concepts
Section 3.4
Solving Equations Using the Quadratic Formula, p. 122
Analyzing the Discriminant of ax2+bx+c= 0, p. 124
Methods for Solving Quadratic Equations, p. 125
Modeling Launched Objects, p. 126

Section 3.5
Solving Systems of Nonlinear Equations, p. 132
Solving Equations by Graphing, p. 135

Section 3.6
Graphing a Quadratic Inequality in Two Variables, p. 140
Solving Quadratic Inequalities in One Variable, p. 142

Mathematical Practices
Question 1.
How can you use technology to determine whose rocket lands first in part (b) of Exercise 65 on page 129?
Answer:

Question 2.
What question can you ask to help the person avoid making the error in Exercise 54 on page 138?
Answer:

Question 3.
Explain your plan to find the possible widths of the fountain in Exercise 44 on page 145.
Answer:

Performance Task: Algebra in Genetics: The Hardy-Weinberg Law
Some people have attached earlobes, the recessive trait. Some people have free earlobes, the dominant trait. What percent of people carry both traits?
To explore the answers to this question and more, go to BigIdeasMath.com.
Big Ideas Math Algebra 2 Answer Key Chapter 3 Quadratic Equations and Complex Numbers 3.6 17

Quadratic Equations and Complex Numbers Chapter Review

3.1 Solving Quadratic Equations (pp. 93–102)

Question 1.
Solve x2 − 2x − 8 = 0 by graphing.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-1.jpg

Solve the equation using square roots or by factoring.
Question 2.
3x2 − 4 = 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-2.jpg

Question 3.
x2 + 6x − 16 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-.3JPG.jpg

Question 4.
2x2 − 17x = −30
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-4.jpg

Question 5.
A rectangular enclosure at the zoo is 35 feet long by 18 feet wide. The zoo wants to double the area of the enclosure by adding the same distance x to the length and width. Write and solve an equation to find the value of x. What are the dimensions of the enclosure?
Answer:

3.2 Complex Numbers (pp. 103–110)

Question 6.
Find the values x and y that satisfy the equation 36 − yi = 4x + 3i.
Answer:

Perform the operation. Write the answer in standard form.
Question 7.
(−2 + 3i ) + (7 − 6i )
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-7.jpg

Question 8.
(9 + 3i ) − (−2 − 7i )
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-8.jpg

Question 9.
(5 + 6i )(−4 + 7i )
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-9.jpg

Question 10.
Solve 7x2 + 21 = 0.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-10.jpg

Question 11.
Find the zeros of f(x) = 2x2 + 32.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-11.jpg

3.3 Completing the Square (pp. 111–118)

Question 12.
An employee at a local stadium is launching T-shirts from a T-shirt cannon into the crowd during an intermission of a football game. The height h (in feet) of the T-shirt after t seconds can be modeled by h = −16t2 + 96t + 4. Find the maximum height of the T-shirt.
Answer:
Given,
h = −16t2 + 96t + 4.
t = 3
h(t) = −16t2 + 96t + 4
h(t) = 0
−16t2 + 96t + 4 = 0
Put t = 3 in the equation.
h = -16(3)² + 96 (3) + 4
h = 148 ft
Thus the height of the T-shirt is 148 ft.

Solve the equation by completing the square.
Question 13.
x2 + 16x + 17 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-13.jpg

Question 14.
4x2 + 16x + 25 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-14.jpg

Question 15.
9x(x − 6) = 81
Answer:

Question 16.
Write y = x2 − 2x + 20 in vertex form. Then identify the vertex.
Answer:

3.4 Using the Quadratic Formula (pp. 121–130)

Solve the equation using the Quadratic Formula.
Question 17.
−x2 + 5x = 2
Answer:
Given,
−x2 + 5x = 2
x² – 5x + 2 = 0
Use the quadratic formula to solve the equation.
x = -b ± √b² – 4ac/2a
x = -(-5) ± √(-5)² – 4(1)(2)/2(1)
x = (5 ± √25 – 8)/2
x = (5± √17)/2
x = 5/2 ± √17/2
x = 4.56
x = 0.43
The solution is x = 4.56 and 0.43

Question 18.
2x2 + 5x = 3
Answer:
Given,
2x2 + 5x = 3
2x2 + 5x – 3 = 0
Use the quadratic formula to solve the equation.
x = -b ± √b² – 4ac/2a
x = -(5) ± √(5)² – 4(2)(-3)/2(2)
x = (-5 ± √25 +24)/4
x = (-5± √49)/4
x = 2/4 ± 12/4
x = 1/2 ± 3
The solution is x = 1/2 + 3 = 3 1/2 and x = 1/2 – 3 = -2 1/2

Question 19.
3x2 − 12x + 13 = 0
Answer:

Find the discriminant of the quadratic equation and describe the number and type of solutions of the equation.
Question 20.
−x2 − 6x − 9 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-20.jpg

Question 21.
x2 − 2x − 9 = 0
Answer:
Given,
x2 − 2x − 9 = 0
Use the quadratic formula to solve the equation.
x = -b ± √b² – 4ac/2a
a = 1, b = -2 and c = -9
x = -(-2) ± √(-2)² – 4(1)(-9)/2(1)
x = (2 ± √4 +36)/2
x = (2 ± √40)/2
x = 1 ± √10
The solution is x = 1 + √10 and x = 1 – √10

Question 22.
x2 + 6x + 5 = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-22.jpg

3.5 Solving Nonlinear Systems (pp. 131–138)

Solve the system by any method. Explain your choice of method.
Question 23.
2x2 − 2 = y
−2x + 2 = y
Answer:

Question 24.
x2 − 6x + 13 = y
−y = −2x + 3
Answer:

Question 25.
x2 + y2 = 4
−15x + 5 = 5y
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-25.jpg

Question 26.
Solve −3x2 + 5x − 1 = 5x2 − 8x − 3 by graphing.
Answer:

3.6 Quadratic Inequalities (pp. 139–146)

Graph the inequality.
Question 27.
y > x2 + 8x + 16
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-27.jpg

Question 28.
y ≥ x2 + 6x + 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-28.jpg

Question 29.
x2 + y ≤ 7x − 12
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-chapter-reviw-29.jpg

Graph the system of quadratic inequalities.
Question 30.
x2 − 4x + 8 > y
−x2 + 4x + 2 ≤ y
Answer:

Question 31.
2x2 − x ≥ y − 5
0.5x2> y − 2x− 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-31.jpg

Question 32.
−3x2 − 2x ≤ y + 1
−2x2 + x − 5 > −y
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-32.jpg

Solve the inequality.
Question 33.
3x2 + 3x − 60 ≥0
Answer:

Question 34.
−x2 − 10x < 21
Answer:

Question 35.
3x2 + 2 ≤ 5x
Answer:

Quadratic Equations and Complex Numbers Chapter Test

Solve the equation using any method. Provide a reason for your choice.
Question 1.
0 = x2 + 2x + 3
Answer:
Given equation is
0 = x2 + 2x + 3
x2 + 2x + 3 = 0
Use the quadratic function method to solve the equation
x = -b ± √b² – 4ac/2a
a = 1, b = 2 and c = 3
x = -(2) ± √(2)² – 4(1)(3)/2(1)
x = (-2 ± √4 – 12)/2
x = (-2± √-8)/2
x = (-2± 2√2i)/2
x = -2/2 ± 2√2i/2
x = -1 ± √2i
The solution is x = -1 + √2i or x = -1 – √2i

Question 2.
6x = x2 + 7
Answer:
Given equation is
6x = x2 + 7
x² – 6x + 7 = 0
Use the quadratic function method to solve the equation
x = -b ± √b² – 4ac/2a
a = 1, b = -6 and c = 7
x = -(-6) ± √(-6)² – 4(1)(7)/2(1)
x = (6 ± √36 – 28)/2
x = (6± √8)/2
x = (6± 2√2)/2
x = 6/2 ± 2√2/2
x = 3 ± √2
The solution is x = 3 + √2 or 3 – √2

Question 3.
x2 + 49 = 85
Answer:
Given equation is
x2 + 49 = 85
x² + 49 – 85 = 0
x² – 36 = 0
x² = 36
x = 6
The solution is x = 6

Question 4.
(x + 4)(x − 1) = −x2 + 3x + 4
Answer:

Explain how to use the graph to find the number and type of solutions of the quadratic equation. Justify your answer by using the discriminant.
Question 5.
\(\frac{1}{2}\)x2 + 3x + \(\frac{9}{2}\) = 0
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-5-1.jpg

Question 6.
4x2 + 16x + 18 = 0
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-6-1.jpg

Question 7.
−x2 + \(\frac{1}{2}\)x + \(\frac{3}{2}\) = 0
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 7
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-7-1.jpg

Solve the system of equations or inequalities.
Question 8.
x2 + 66 = 16x − y
2x − y = 18
Answer:

Question 9.
y ≥ \(\frac{1}{4}\)x2 − 2
y < −(x + 3)2x − y = 18 + 4
Answer:

Question 10.
0 = x2 + y2 − 40
y = x + 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-10.jpg

Question 11.
Write (3 + 4i )(4 − 6i ) as a complex number in standard form.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-11JPG.jpg

Question 12.
The aspect ratio of a widescreen TV is the ratio of the screen’s width to its height, or 16 : 9. What are the width and the height of a 32-inch widescreen TV? Justify your answer. (Hint: Use the Pythagorean Theorem and the fact that TV sizes refer to the diagonal length of the screen.)
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 12
Answer:

Question 13.
The shape of the Gateway Arch in St. Louis, Missouri, can be modeled by y = −0.0063x2 + 4x, where x is the distance (in feet) from the left foot of the arch and y is the height (in feet) of the arch above the ground. For what distances x is the arch more than 200 feet above the ground? Justify your answer.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-14.jpg

Question 14.
You are playing a game of horseshoes. One of your tosses is modeled in the diagram, where x is the horseshoe’s horizontal position (in feet) and y is the corresponding height (in feet). Find the maximum height of the horseshoe. Then find the distance the horseshoe travels. Justify your answer.
Big Ideas Math Algebra 2 Answers Chapter 3 Quadratic Equations and Complex Numbers ct 14
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-14a.jpg

Quadratic Equations and Complex Numbers Cumulative Assessment

Question 1.
The graph of which inequality is shown?
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 1
A. y2 > x2 + x – 6
B. y ≥ x2 + x – 6
C. y > x2 – x – 6
D. y ≥ x2 – x – 6
Answer:

Question 2.
Classify each function by its function family. Then describe the transformation of the parent function.
a. g(x) = x + 4
b. h(x) = 5
c. h(x) = x2 − 7
d. g(x) = −∣x + 3∣− 9
e. g(x) = \(\frac{1}{4}\)(x − 2)2 − 1
f. h(x) = 6x+ 11
Answer:

Question 3.
Two baseball players hit back-to-back home runs. The path of each home run is modeled by the parabolas below, where x is the horizontal distance (in feet) from home plate and y is the height (in feet) above the ground. Choose the correct symbol for each inequality to model the possible locations of the top of the outfield wall.(HSA-CED.A.3)
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 3
Answer:

Question 4.
You claim it is possible to make a function from the given values that has an axis of symmetry at x = 2. Your friend claims it is possible to make a function that has an axis of symmetry at x = −2. What values can you use to support your claim? What values support your friend’s claim?(HSF-IF.B.4)
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 4
Answer:

Question 5.
Which of the following values are x-coordinates of the solutions of the system?
y = x2 – 6x + 14
y = 2x + 7
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 5
Answer:

Question 6.
The table shows the altitudes of a hang glider that descends at a constant rate. How long will it take for the hang glider to descend to an altitude of 100 feet? Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 6
A. 25 seconds
B. 35 seconds
C. 45 seconds
D. 55 seconds
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-6a-.jpg

Question 7.
Use the numbers and symbols to write the expression x2 + 16 as the product of two binomials. Some may be used more than once. Justify your answer.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 7
Answer:

Question 8.
Choose values for the constants h and k in the equation x = \(\frac{1}{4}\)( y − k)2 + h so that each statement is true.(HSA-CED.A.2)
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-8-1.jpg

Question 9.
Which of the following graphs represents a perfect square trinomial? Write each function in vertex form by completing the square.
Big Ideas Math Algebra 2 Solutions Chapter 3 Quadratic Equations and Complex Numbers ca 9
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-3-Quadratic-Equations-and-Complex-Numbers-9.jpg

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions

Make solving all homework & assignments questions with Big Ideas Math Algebra 2 Answers guide. Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions guide is the best option for all students. In this article, the students can find topic-wise Big Ideas Math Algebra 2 Ch 2 Quadratic Functions. Get BIM Algebra 2 Solution Key for Exercise questions & answers, practices, quizzes, chapter reviews, chapter tests, and cumulative assessments. Do practice with the Algebra 2 Big Ideas Math Answers Chapter 2 Quadratic Functions Guide and kickstart your preparation for the exam.

Big Ideas Math Book Algebra 2 Answer Key Chapter 2 Quadratic Functions

Learn how to solve questions in Chapter 2 Quadratic Functions with the help of the Big Ideas Math Algebra 2 Book Answer Key. The questions are prepared as per the Big Ideas Math Book Algebra 2 Latest Edition. Thus, make use of our BIM Book Algebra 2 Solution Key Chapter 2 Quadratic Functions direct links and prepare well to score good results in any examinations. Below you can find the quick links to download Big Ideas Math Book Algebra 2 Answer Key Chapter 2 Quadratic Functions pdf.

Quadratic Functions Maintaining Mathematical Proficiency

Find the x-intercept of the graph of the linear equation.

Question 1.
y = 2x + 7
Answer:
Given linear equation is
y = 2x + 7
To find the x-intercept set y = 0
0 = 2x + 7
2x = -7
x = -7/2
So, the x-intercept of the graph of the linear equation is -7/2

Question 2.
y = -6x + 8
Answer:
Given linear equation is
y = -6x + 8
To find the x-intercept set y = 0
0 = -6x + 8
-6x = -8
6x = 8
x = 8/6
x = 4/3
So, the x-intercept of the graph of the linear equation is 4/3.

Question 3.
y = -10x – 36
Answer:
Given linear equation is
y = -10x – 36
To find the x-intercept set y = 0
0 = -10x – 36
-10x = 36
x = -36/10
x = -3.6
So, the x-intercept of the graph of the linear equation is -3.6

Question 4.
y = 3(x – 5)
Answer:
Given linear equation is
y = 3(x – 5)
To find the x-intercept set y = 0
0 = 3(x – 5)
x – 5 = 0
x = 5
So, the x-intercept of the graph of the linear equation is 5.

Question 5.
y = -4(x + 10)
Answer:
Given linear equation is
y = -4(x + 10)
To find the x-intercept set y = 0
0 = -4(x + 10)
0 = x + 10
x = -10
So, the x-intercept of the graph of the linear equation is -10.

Question 6.
3x + 6y = 24
Answer:
Given linear equation is
3x + 6y = 24
To find the x-intercept set y = 0
3x + 6(0) = 24
3x = 24
x = 24/3
x = 8
So, the x-intercept of the graph of the linear equation is 8.

Find the distance between the two points.

Question 7.
(2, 5), (-4, 7)
Answer:
Given two points (x1, y1) = (2, 5) and (x2, y2) = (-4, 7)

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-2.1-7.jpg

Question 8.
(-1, 0), (-8, 4)
Answer:
Given two points (x1, y1) = (-1, 0) and (x2, y2) = (-8, 4)

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-2.1-8.jpg

Question 9.
(3, 10), (5, 9)
Answer:
Given two points (x1, y1) = (3, 10) and (x2, y2) = (5, 9)

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-2.1-9.jpg

Question 10.
(7, -4), (-5, 0)
Answer:
Given two points (x1, y1) = (7, -4) and (x2, y2) = (-5, 0)

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-2.1-10.jpg

Question 11.
(4, -8), (4, 2)
Answer:
Given two points (x1, y1) = (4, -8) and (x2, y2) = (4, 2)

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-2.1-11.jpg

Question 12.
(0, 9), (-3, -6)
Answer:
Given two points (x1, y1) = (0, 9) and (x2, y2) = (-3, -6)

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-2.1-12.jpg

Question 13.
ABSTRACT REASONING Use the Distance Formula to write an expression for the distance between the two points (a, c) and (b, c). Is there an easier way to find the distance when the x-coordinates are equal? Explain your reasoning
Answer:
Given two points (a, c) and (b, c)
The distance between the two points are √(x2 – x1)² + (y2 – y1)²
d = √(b – a)² + (c – c)²
d = √(b – a)²
The y-coordinates are equal in the given points,
If the x-coordinates are equal then the distance will be 0.

Quadratic Functions Mathematical Practices

Monitoring Progress

Decide whether the syllogism represents correct or flawed reasoning. If flawed, explain why the conclusion is not valid.

Question 1.
All mammals are warm-blooded.
All dogs are mammals.
Therefore, all dogs are warm-blooded.
Answer: The conclusion is not valid

Question 2.
All mammals are warm-blooded.
My pet is warm-blooded.
Therefore, my pet is a mammal.
Answer:
The given syllogism represents correct reasoning. So, the conclusion is valid.

Question 3.
If I am sick, then I will miss school.
I missed school.
Therefore, I am sick.
Answer:
The given syllogism represents correct reasoning. So, the conclusion is valid.

Question 4.
If I am sick, then I will miss school.
I did not miss school.
Therefore, I am not sick.
Answer:
The given syllogism represents the flawed reasoning that there might be an incident when the person attended school when he was sick.

Lesson 2.1 Transformations of Quadratic Functions

Essential Question

How do the constants a, h, and k affect the graph of the quadratic function g(x) = a(x – h)2 + k?
The parent function of the quadratic family is f(x) = x2. A transformation of the graph of the parent function is represented by the function g(x) = a(x – h)2 + k, where a ≠ 0.

EXPLORATION 1
Identifying Graphs of Quadratic Functions
Work with a partner.
Match each quadratic function with its graph. Explain your reasoning. Then use a graphing calculator to verify that your answer is correct.
a. g(x) = -(x – 2)2
b. g(x) = (x – 2)2 + 2
c. g(x) = -(x + 2)2 – 2
d. g(x) = 0.5(x – 2)2 + 2
e. g(x) = 2(x – 2)2
f. g(x) = -(x + 2)2 + 2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 1

Communicate Your Answer

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2
Question 2.
How do the constants a, h, and k affect the graph of the quadratic function g(x) =a(x – h)2 + k?
Graphing Quadratic Functions

Question 3.
Write the equation of the quadratic function whose graph is shown at the right. Explain your reasoning. Then use a graphing calculator to verify that your equation is correct.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 3

2.1 Lesson

Monitoring Progress

Describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 1.
g(x) = (x – 3)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-1.jpg

Question 2.
g(x) = (x + 2)2 – 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-2.jpg

Question 3.
g(x) = (x + 5)2 + 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-3.jpg

Describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 4.
g(x) = (\(\frac{1}{3} x\))2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-4.jpg
Question 5.
g(x) = 3(x – 1)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-5.jpg

Question 6.
g(x) = -(x + 3)2 + 2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-6.jpg

Question 7.
Let the graph of g be a vertical shrink by a factor of \(\frac{1}{2}\) followed by a translation 2 units up of the graph of f(x) = x2. Write a rule for g and identify the vertex.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-7.jpg

Question 8.
Let the graph of g be a translation 4 units left followed by a horizontal shrink by a factor of \(\frac{1}{3}\) of the graph of f(x) = x2 + x. Write a rule for g.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-8.jpg

Question 9.
WHAT IF? In Example 5, the water hits the ground 10 feet closer to the fire truck after lowering the ladder. Write a function that models the new path of the water.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-9.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.1-9a.jpg

Transformations of Quadratic Functions 2.1 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE The graph of a quadratic function is called a(n) ________.
Answer:
The graph of a quadratic function is called a parabola.

Question 2.
VOCABULARY Identify the vertex of the parabola given by f(x) = (x + 2)2 – 4.
Answer:

Monitoring Progress and Modeling with Mathematics

In Exercises 3–12, describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 3.
g(x) = x2 – 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 3

Question 4.
g(x) = x2 + 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-4.jpg

Question 5.
g(x) = (x + 2)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 5

Question 6.
g(x) = (x – 4)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-6.jpg

Question 7.
g(x) = (x – 1)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 7

Question 8.
g(x) = (x + 3)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-8.jpg

Question 9.
g(x) = (x + 6)2 – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 9

Question 10.
g(x) = (x – 9)2 + 5
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-10.jpg

Question 11.
g(x) = (x – 7)2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 11

Question 12.
g(x) = (x + 10)2 – 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-12.jpg

ANALYZING RELATIONSHIPS In Exercises 13–16, match the function with the correct transformation of the graph of f. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 4

Question 13.
y = f(x – 1)
Answer:
Given,
y = f(x – 1)
The answer is A.
The graph is the original graph translated 1 unit right.

Question 14.
y = f(x) + 1
Answer:
Given,

Question 15.
y = f(x – 1) + 1
Answer:
Given,
y = f(x – 1) + 1
The graph is the original graph translated 1 unit up and 1 unit right.
The correct answer is C.

Question 16.
y = f(x + 1)
Answer:
The correct answer is Graph B.

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 5

In Exercises 17–24, describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 17.
g(x) = -x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 17

Question 18.
g(x) = (-x)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-18.jpg

Question 19.
g(x) = 3x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 19

Question 20.
g(x) = \(\frac{1}{3}\)x2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-20.jpg

Question 21.
g(x) = (2x)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 21

Question 22.
g(x) = -(2x)2
Answer:

Question 23.
g(x) = \(\frac{1}{5}\)x2 – 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 23

Question 24.
g(x) = \(\frac{1}{2}\)(x – 1)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-24.jpg

ERROR ANALYSIS In Exercises 25 and 26, describe and correct the error in analyzing the graph of f(x) = −6x2 + 4.

Question 25.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 6
Answer:
The graph is a reflection in the x-axis, not y-axis.
The graph is a reflection in the x-axis and a vertical stretch by a factor of 6, followed by a translation 4 units up of the graph of the parent quadratic function.

Question 26.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 7
Answer:
The order of transformations is wrong. The first transformation is either a vertical stretch by a factor of 6 or a reflection in the x-axis.
The second transformation is a vertical translation of the graph by 4 units up. This is the second error.
The graph is a vertical stretch by a factor of 6 and a reflection in the x-axis followed by a vertical translation up by 4 units.

USING STRUCTURE In Exercises 27–30, describe the transformation of the graph of the parent quadratic function. Then identify the vertex.

Question 27.
f(x) = 3(x + 2)2 + 1
Answer:
Given quadratic function is f(x) = 3(x + 2)2 + 1
The transformation is a horizontal translation to the left 2 units, followed by a vertical stretch by a factor of 3, then a vertical translation 1 unit up.
The vertex is (-2, 1).

Question 28.
f(x) = -4(x + 1)2 – 5
Answer:

Question 29.
f(x) = -2x2 + 5
Answer:
Given function is f(x) = -2x2 + 5
The transformation is a vertical stretch by a factor of 2 and a reflection in the x-axis, then a vertical translation 5 units up.
The vertex is (0, 5).

Question 30.
f(x) = \(\frac{1}{2}\)(x – 1)2
Answer:

In Exercises 31–34, write a rule for g described by the transformations of the graph of f. Then identify the vertex.

Question 31.
f(x) = x2 vertical stretch by a factor of 4 and a reflection in the x-axis, followed by a translation 2 units up
Answer:
Given,
f(x) = x2 vertical stretch by a factor of 4 and a reflection in the x-axis, followed by a translation 2 units up
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 31

Question 32.
f(x) = x2; vertical shrink by a factor of \(\frac{1}{3}\) and a reflection in the y-axis, followed by a translation 3 units right
Answer:
Given f(x) = x2

Question 33.
f(x) = 8x2 – 6; horizontal stretch by a factor of 2 and a translation 2 units up, followed by a reflection in the y-axis
Answer:
Given function is f(x) = 8x2 – 6
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 33

Question 34.
f(x) = (x + 6)2 + 3; horizontal shrink by a factor of \(\frac{1}{2}\) and a translation 1 unit down, followed by a reflection in the x-axis
Answer:
Given,
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-34.jpg

USING TOOLS In Exercises 35–40, match the function with its graph. Explain your reasoning.

Question 35.
g(x) = 2(x – 1)2 – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 35

Question 36.
g(x) = \(\frac{1}{2}\)(x + 1)2 – 2
Answer:
Given,
g(x) = \(\frac{1}{2}\)(x + 1)2 – 2
a = \(\frac{1}{2}\), h = -1 and k = -2
(h, k) = (-1, -2)
Graphs B and F has vertex (-1, -2).
a = \(\frac{1}{2}\), there is no reflection in the x-axis.
So, the answer is B.

Question 37.
g(x) = -2(x – 1)2 + 2
Answer:
Given,
g(x) = -2(x – 1)2 + 2
a = -2, h = 1 and k = 2
(h, k) = (1, 2)
This is a vertical stretch by a factor of 2 with a reflection in the x-axis and a translation 1 unit right and 2 units up.
So, the answer is graph D.

Question 38.
g(x) = 2(x + 1)2 + 2
Answer:
Given,
g(x) = 2(x + 1)2 + 2
Where a = 2, h = -1 and k = 2
(h, k) = (-1, 2)
Graph E is the only one with vertex (-1, 2).

Question 39.
g(x) = -2(x + 1)2 – 2
Answer:
Given,
g(x) = -2(x + 1)2 – 2
a = -2, h = -1 and k = -2
(h, k) = (-1, -2)
This is a vertical stretch by a factor of 2 with a reflection in the x-axis and a translation 1 unit left and 2 units down.
The answer is F.

Question 40.
g(x) = 2(x – 1)2 + 2
Answer:
Given,
g(x) = 2(x – 1)2 + 2
a = 2, h = 1 and k = 2
(h, k) = (1, 2)

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 8

JUSTIFYING STEPS In Exercises 41 and 42, justify eachstep in writing a function g based on the transformationsof f(x) = 2x2 + 6x.

Question 41.
translation 6 units down followed by a reflection in the x-axis
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 41

Question 42.
reflection in the y-axis followed by a translation 4 units right
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 10
Answer:

Question 43.
MODELING WITH MATHEMATICS The function h(x) = -0.03(x – 14)2 + 6 models the jump of a red kangaroo, where x is the horizontal distance traveled (in feet) and h(x) is the height (in feet). When the kangaroo jumps from a higher location, it lands 5 feet farther away. Write a function that models the second jump.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 43

Question 44.
MODELING WITH MATHEMATICS The function f(t) = -16t2 + 10 models the height (in feet) of an object t seconds after it is dropped from a height of 10 feet on Earth. The same object dropped from the same height on the moon is modeled by g(t) = –\(\frac{8}{3}\)t2 + 10. Describe the transformation of the graph of f to obtain g. From what height must the object be dropped on the moon so it hits the ground at the same time as on Earth?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.1-44.jpg

Question 45.
MODELING WITH MATHEMATICS Flying fish use their pectoral fins like airplane wings to glide through the air.
a. Write an equation of the form y = a(x – h)2 + k with vertex (33, 5) that models the flight path, assuming the fish leaves the water at (0, 0).
b. What are the domain and range of the function? What do they represent in this situation?
c. Does the value of a change when the flight path has vertex (30, 4)? Justify your answer.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 12
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 45.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 45.2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 45.3

Question 46.
HOW DO YOU SEE IT? Describe the graph of g as a transformation of the graph of f(x) = x2.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 13
Answer:
The graph of g(x) is made of the graph of f(x) using horizontal translation for 4 left followed by vertical translation for 2 down.

Question 47.
COMPARING METHODS Let the graph of g be a translation 3 units up and 1 unit right followed by a vertical stretch by a factor of 2 of the graph of f(x) = x2.
a. Identify the values of a, h, and k and use vertex form to write the transformed function.
b. Use function notation to write the transformed function. Compare this function with your function in part (a).
c. Suppose the vertical stretch was performed first, followed by the translations. Repeat parts (a) and (b).
d. Which method do you prefer when writing a transformed function? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 47.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 47.2

Question 48.
THOUGHT PROVOKING A jump on a pogo stick with a conventional spring can be modeled by f(x) = -0.5(x – 6)2 + 18, where x is the horizontal distance (in inches) and f(x) is the vertical distance (in inches). Write at least one transformation of the function and provide a possible reason for your transformation.
Answer:

Question 49.
MATHEMATICAL CONNECTIONS The area of a circle depends on the radius, as shown in the graph. A circular earring with a radius of r millimeters has a circular hole with a radius of \(\frac{3 r}{4}\) millimeters. Describe a transformation of the graph below that models the area of the blue portion of the earring.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 13.1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 49

Maintaining Mathematical Proficiency
A line of symmetry for the figure is shown in red. Find the coordinates of point A. (Skills Review Handbook)

Question 50.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 14
Answer:

Question 51.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 15
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.1 Question 51

Question 52.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 16
Answer:
Flip it over the line of symmetry making it (-2, 2).

Lesson 2.2 Characteristics of Quadratic Functions

Essential Question
What type of symmetry does the graph of f(x) = a(x – h)2 + k have and how can you describe this symmetry?

EXPLORATION 1
Parabolas and Symmetry
Work with a partner.

a. Complete the table. Then use the values in the table to sketch the graph of the function
f(x) = \(\frac{1}{2}\)x2 – 2x – 2 on graph paper.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 17
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 18
b. Use the results in part (a) to identify the vertex of the parabola.
c. Find a vertical line on your graph paper so that when you fold the paper, the left portion of the graph coincides with the right portion of the graph. What is the equation of this line? How does it relate to the vertex?
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 19
d. Show that the vertex form f(x) = \(\frac{1}{2}\)(x – 2)2 – 4 is equivalent to the function given in part (a).

EXPLORATION 2
Parabolas and Symmetry
Work with a partner.
Repeat Exploration 1 for the function given by f(x) = –\(\frac{1}{3}\)x2 + 2x + 3 = –\(\frac{1}{3}\)(x – 3),sup>2 + 6.

Communicate Your Answer

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 20
What type of symmetry does the graph of f(x) = a(x – h)2 + k have and how can you describe this symmetry?

Question 4.
Describe the symmetry of each graph. Then use a graphing calculator to verify your answer.
a. f(x) = -(x – 1)2 + 4
b. f(x) = (x + 1)2 – 2
c. f(x) = 2(x – 3)2 + 1
d. f(x) = \(\frac{1}{2}\)(x + 2)2
e. f(x) = -2x2 + 3
f. f(x) = 3(x – 5)2 + 2

2.2 Lesson

Monitoring Progress

Graph the function. Label the vertex and axis of symmetry.

Question 1.
f(x) = -3(x + 1)2

Answer:

Question 2.
g(x) = 2(x – 2)2 + 5
Answer:

Question 3.
h(x) = x2 + 2x – 1
Answer:

Question 4.
p(x) = -2x2 – 8x + 1
Answer:

Question 5.
Find the minimum value or maximum value of
(a) f(x) = 4x2 + 16x – 3 and
(b) h(x) = -x2 + 5x + 9. Describe the domain and range of each function, and where each function is increasing and decreasing.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.2-5.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.2-5b.jpg

Graph the function. Label the x-intercepts, vertex, and axis of symmetry.

Question 6.
f(x) = -(x + 1)(x + 5)
Answer:
Given the function is f(x) = -(x + 1)(x + 5)
Big Ideas Math Answers Algebra 2 Lesson 2.2 Characteristics of Quadratic Functions_6

Question 7.
g(x) = \(\frac{1}{4}\)(x – 6)(x – 2)
Answer:
Given equation is g(x) = \(\frac{1}{4}\)(x – 6)(x – 2)
Big Ideas Math Answers Algebra 2 Lesson 2.2 Characteristics of Quadratic Functions_7

Question 8.
WHAT IF? The graph of your third shot is a parabola through the origin that reaches a maximum height of 28 yards when x = 45. Compare the distance it travels before it hits the ground with the distances of the first two shots.
Answer:
Given that,
The graph of your third shot is a parabola through the origin that reaches a maximum height of 28 yards when x = 45.
The vertex of the parabola is (45, 28)
where x is the horizontal distance and y is the corresponding height
As per the given problem parabola pass through the origin.
Thus x-intercepts are 0 and 90.
So, the ball during the third shot travels 90 (twice of the maximum 45) yards before hitting the ground.
The first shot travels 100 yards before hitting the ground, the second shot travels 80 yards before hitting the ground and the third shot travels 90 yards before hitting the ground.

Characteristics of Quadratic Functions 2.2 Exercises

Vocabulary and Core Concept and Check

Question 1.
WRITING Explain how to determine whether a quadratic function will have a minimum value or a maximum value.
Answer:
To determine whether the function is quadratic and will have a minimum value or maximum value, consider the leading coefficient a.
If a is negative, then the function has a maximum value and if a is positive, then the function has a minimum value.

Question 2.
WHICH ONE DOESN’T BELONG? The graph of which function does not belong with the other three? Explain.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 21
Answer:

Question 3.
f(x) = (x – 3)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 3

Question 4.
h(x) = (x + 4)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-4.jpg

Question 5.
g(x) = (x + 3)2 + 5
Answer:
Given function is g(x) = (x + 3)2 + 5
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 5

Question 6.
y = (x – 7)2 – 1
Answer:
Given function is y = (x – 7)2 – 1
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-6.jpg

Question 7.
y = -4(x – 2)2 + 4
Answer:
Given function is y = -4(x – 2)2 + 4
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 7

Question 8.
g(x) = 2(x + 1)2 – 3
Answer:
Given function is g(x) = 2(x + 1)2 – 3
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-8.jpg

Question 9.
f(x) = -2(x – 1)2 – 5
Answer:
Given function is f(x) = -2(x – 1)2 – 5
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 9

Question 10.
h(x) = 4(x + 4)2 + 6
Answer:
Given function is h(x) = 4(x + 4)2 + 6
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-10.jpg

Question 11.
y = –\(\frac{1}{4}\)(x + 2)2 + 1
Answer:
Given function is y = –\(\frac{1}{4}\)(x + 2)2 + 1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 11

Question 12.
y = \(\frac{1}{2}\)(x – 3)2 + 2
Answer:
Given function is y = \(\frac{1}{2}\)(x – 3)2 + 2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-12.jpg

Question 13.
f(x) = 0.4(x – 1)2
Answer:
Given function is f(x) = 0.4(x – 1)2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 13

Question 14.
g(x) = 0.75x2 – 5
Answer:
Given function is g(x) = 0.75x2 – 5
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-14.jpg

ANALYZING RELATIONSHIPS In Exercises 15–18, use the axis of symmetry to match the equation with its graph.

Question 15.
y = 2(x – 3)2 + 1
Answer:
Given function is y = 2(x – 3)2 + 1
a = 2, h = 3 and k = 1
Graph C is the correct answer.
The axis of symmetry is x = 3.

Question 16.
y = (x + 4)2 – 2
Answer:
Given function is y = (x + 4)2 – 2

Question 17.
y = \(\frac{1}{2}\)(x + 1)2 + 3
Answer:
Given function is y = \(\frac{1}{2}\)(x + 1)2 + 3
a = \(\frac{1}{2}\), h = -1 and k = 3
Graph B is the correct answer.
The axis of symmetry is x = -1.

Question 18.
y = (x – 2)2 – 1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 22
Answer:
y = (x – 2)2 – 1
a = 1
h = 2 and k = -1
The axis of the symmetry is x = 2
Graph A is the correct answer.

REASONING In Exercises 19 and 20, use the axis of symmetry to plot the reflection of each point and complete the parabola.

Question 19.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 23
Answer:
The coordinates of the points are (0, -1), (2, 3), (1, 2), (3, 2), (4, -1).
The axis of the symmetry is x = 2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 19

Question 20.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 24
Answer:
The points are (-1, 1), (-2, -2), (-3, -3)
The axis of symmetry is x = -3

In Exercises 21–30, graph the function. Label the vertex and axis of symmetry.

Question 21.
y = x2 + 2x + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 21.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 21.2

Question 22.
y = 3x2 – 6x + 4
Answer:
Given function is y = 3x2 – 6x + 4
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-22.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-22a-1.jpg

Question 23.
y = -4x2 + 8x + 2
Answer:
Given function is y = -4x2 + 8x + 2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 23.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 23.2

Question 24.
f(x) = -x2 – 6x + 3
Answer:
Given function is f(x) = -x2 – 6x + 3
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-24.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-24a.jpg

Question 25.
g(x) = -x2 – 1
Answer:
Given function is g(x) = -x2 – 1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 25.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 25.2

Question 26.
f(x) = 6x2 – 5
Answer:
Given function is f(x) = 6x2 – 5
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-26.jpg

Question 27.
g(x) = -1.5x2 + 3x + 2
Answer:
Given function is g(x) = -1.5x2 + 3x + 2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 27.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 27.2

Question 28.
f(x) = 0.5x2 + x – 3
Answer:
Given function is f(x) = 0.5x2 + x – 3
a = 0.5, b = 1
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-28.jpg

Question 29.
y = \(\frac{3}{2}\)x2 – 3x + 6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 29.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 29.2

Question 30.
y = –\(\frac{5}{2}\)x2 – 4x – 1
Answer:
Given function is y = –\(\frac{5}{2}\)x2 – 4x – 1
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-30.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-30a.jpg

Question 31.
WRITING Two quadratic functions have graphs with vertices (2, 4) and (2, -3). Explain why you can not use the axes of symmetry to distinguish between the two functions.
Answer:
The axis of symmetry, x = 2 is the same for both of the parabolas.
So, axes of symmetry will not make a distinction between the two quadratic functions.

Question 32.
WRITING A quadratic function is increasing to the left of x = 2 and decreasing to the right of x = 2. Will the vertex be the highest or lowest point on the graph of the parabola? Explain.
Answer:
If the slope m goes from positive to negative as x increases, the slope is negative.
Big Ideas Math Answers Algebra 2 Lesson 2.2 Characteristics of Quadratic Functions_7
A parabola opens downward and has a vertex that is the highest point on the graph.

ERROR ANALYSIS In Exercises 33 and 34, describe and correct the error in analyzing the graph of y = 4x2 + 24x − 7.

Question 33.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 25
Answer:
x = -b/2a = -24/2(4) = -24/8 = -3
The error is the formula is missing a negative sign.

Question 34.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 26
Answer:
The student incorrectly labeled the y-intercept c = -7 because the sign is negative, which means c is negative.

MODELING WITH MATHEMATICS In Exercises 35 and 36, x is the horizontal distance (in feet) and y is the vertical distance (in feet). Find and interpret the coordinates of the vertex.

Question 35.
The path of a basketball thrown at an angle of 45° can be modeled by y = -0.02x2 + x + 6.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 35

Question 36.
The path of a shot put released at an angle of 35° can be modeled by y = -0.01x2 + 0.7x + 6.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 27
Answer:

Question 37.
ANALYZING EQUATIONS The graph of which function has the same axis of symmetry as the graph of y = x2 + 2x + 2?
A. y = 2x2 + 2x + 2
B. y = -3x2 – 6x + 2
C. y = x2 – 2x + 2
D. y = -5x2 + 10x + 23
Answer:
Given the line of symmetry for the graph of y = x2 + 2x + 2
x = -b/2a = -2/2(1) = -1.
The only other graph that has a line of symmetry of x = -1.
So, the answer is B.

Question 38.
USING STRUCTURE Which function represents the widest parabola? Explain your reasoning.
A. y = 2(x + 3)2
B. y = x2 – 5
C. y = 0.5(x – 1)2 + 1
D. y = -x2 + 6
Answer:

In Exercises 39–48, find the minimum or maximum value of the function. Describe the domain and range of the function, and where the function is increasing and decreasing.

Question 39.
y = 6x2 – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 39

Question 40.
y = 9x2 + 7
Answer:

Question 41.
y = -x2 – 4x – 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 41

Question 42.
g(x) = -3x2 – 6x + 5
Answer:

Question 43.
f(x) = -2x2 + 8x + 7
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 43

Question 44.
g(x) = 3x2 + 18x – 5
Answer:

Question 45.
h(x) = 2x2 – 12x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 45

Question 46.
h(x) = x2 – 4x
Answer:

Question 47.
y = \(\frac{1}{4}\)x2 – 3x + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 47

Question 48.
f(x) = \(\frac{3}{2}\)x2 + 6x + 4
Answer:

Question 49.
PROBLEM SOLVING The path of a diver is modeled by the function f(x) = -9x2 + 9x + 1, where f(x) is the height of the diver (in meters) above the water and x is the horizontal distance (in meters) from the end of the diving board.
a. What is the height of the diving board?
b. What is the maximum height of the diver?
c. Describe where the diver is ascending and where the diver is descending.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 28
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 49

Question 50.
PROBLEM SOLVING The engine torque y (in foot-pounds) of one model of car is given by y = -3.75x2 + 23.2x + 38.8, where x is the speed (in thousands of revolutions per minute) of the engine.
a. Find the engine speed that maximizes torque. What is the maximum torque?
b. Explain what happens to the engine torque as the speed of the engine increases.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-50.jpg

MATHEMATICAL CONNECTIONS In Exercises 51 and 52, write an equation for the area of the figure. Then determine the maximum possible area of the figure.

Question 51.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 29
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 51

Question 52.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 30
Answer:

In Exercises 53–60, graph the function. Label the x-intercept(s), vertex, and axis of symmetry.

Question 53.
y = (x + 3)(x – 3)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 53.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 53.2

Question 54.
y = (x + 1)(x – 3)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-54.jpg

Question 55.
y = 3(x + 2)(x + 6)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 55.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 55.2

Question 56.
f(x) = 2(x – 5)(x – 1)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-56.jpg

Question 57.
g(x) = -x(x + 6)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 57.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 57.2

Question 58.
y = -4x(x + 7)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-58.jpg

Question 59.
f(x) = -2(x – 3)2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 59.1

Question 60.
y = 4(x – 7)2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-60.jpg

USING TOOLS In Exercises 61–64, identify the x-intercepts of the function and describe where the graph is increasing and decreasing. Use a graphing calculator to verify your answer.

Question 61.
f(x) = \(\frac{1}{2}\)(x – 2)(x + 6)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 61

Question 62.
y = \(\frac{3}{4}\)(x + 1)(x – 3)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-62.jpg

Question 63.
g(x) = -4(x – 4)(x – 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 63.1

Question 64.
h(x) = -5(x + 5)(x + 1)
Answer:
Given,
h(x) = -5(x + 5)(x + 1)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-64.jpg

Question 65.
MODELING WITH MATHEMATICS A soccer player kicks a ball downfield. The height of the ball increases until it reaches a maximum height of 8 yards, 20 yards away from the player. A second kick is modeled by y = x(0.4 – 0.008x). Which kick travels farther before hitting the ground? Which kick travels higher?
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 31
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 65.1

Question 66.
MODELING WITH MATHEMATICS Although a football field appears to be flat, some are actually shaped like a parabola so that rain runs off to both sides. The cross section of a field can be modeled by y = -0.000234x(x – 160), where x and y are measured in feet. What is the width of the field? What is the maximum height of the surface of the field?
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 32
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-66.jpg

Question 67.
REASONING The points (2, 3) and (-4, 2) lie on the graph of a quadratic function. Determine whether you can use these points to find the axis of symmetry. If not, explain. If so, write the equation of the axis of symmetry.
Answer:
Given,
The points (2, 3) and (-4, 2) lie on the graph of a quadratic function.
You cannot determine the axis of symmetry based on the two points that are given. Either of the points could be on the axis of symmetry, or neither of the points could be on the axis of symmetry. You can only determine the axis of symmetry if the y-coordinates of the two points are the same because the axis of symmetry would lie halfway between the two points.

Question 68.
OPEN-ENDED Write two different quadratic functions in intercept form whose graphs have the axis of symmetry x= 3.
Answer:
The axis of the symmetry is x = 3.

Question 69.
PROBLEM SOLVING An online music store sells about 4000 songs each day when it charges $1 per song. For each $0.05 increase in price, about 80 fewer songs per day are sold. Use the verbal model and quadratic function to determine how much the store should charge per song to maximize daily revenue.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 33
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 69

Question 70.
PROBLEM SOLVING An electronics store sells 70 digital cameras per month at a price of $320 each. For each $20 decrease in price, about 5 more cameras per month are sold. Use the verbal model and quadratic function to determine how much the store should charge per camera to maximize monthly revenue.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 34
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-70.jpg

Question 71.
DRAWING CONCLUSIONS Compare the graphs of the three quadratic functions. What do you notice? Rewrite the functions f and g in standard form to justify your answer.
f(x) = (x + 3)(x + 1)
g(x) = (x + 2)2 – 1
h(x) = x2 + 4x + 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 71

Question 72.
USING STRUCTURE Write the quadratic function f(x) = x2 + x – 12 in intercept form. Graph the function. Label the x-intercepts, y-intercept, vertex, and axis of symmetry.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-72a.jpg

Question 73.
PROBLEM SOLVING A woodland jumping mouse hops along a parabolic path given by y = -0.2x2 + 1.3x, where x is the mouse’s horizontal distance traveled (in feet) and y is the corresponding height (in feet). Can the mouse jump over a fence that is 3 feet high? Justify your answer.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 35
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 73

Question 74.
HOW DO YOU SEE IT? Consider the graph of the function f(x) = a(x – p)(x – q).
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 36
a. What does f(\(\frac{p+q}{2}\)) represent in the graph?
b. If a < 0, how does your answer in part (a) change? Explain.
Answer:

Question 75.
MODELING WITH MATHEMATICS The Gateshead Millennium Bridge spans the River Tyne. The arch of the bridge can be modeled by a parabola. The arch reaches a maximum height of 50 meters at a point roughly 63 meters across the river. Graph the curve of the arch. What are the domain and range? What do they represent in this situation?
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 37
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 75

Quadratic 76.
THOUGHT PROVOKING
You have 100 feet of fencing to enclose a rectangular garden. Draw three possible designs for the garden. Of these, which has the greatest area? Make a conjecture about the dimensions of the rectangular garden with the greatest possible area. Explain your reasoning.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-76.jpg

Question 77.
MAKING AN ARGUMENT The point (1, 5) lies on the graph of a quadratic function with axis of symmetry x = -1. Your friend says the vertex could be the point (0, 5). Is your friend correct? Explain.
Answer:
Given that,
The point (1, 5) lies on the graph of a quadratic function with axis of symmetry x = -1.
The x-coordinate of the vertex corresponds to the axis of symmetry.
So, the x-coordinate of the vertex would be -1.
Thus your friend is incorrect.

Question 78.
CRITICAL THINKING Find the y-intercept in terms of a, p, and q for the quadratic function f(x) = a(x – p)(x – q).
Answer:
f(x) = a(x – p)(x – q).

Question 79.
MODELING WITH MATHEMATICS A kernel of popcorn contains water that expands when the kernel is heated, causing it to pop. The equations below represent the “popping volume” y (in cubic centimeters per gram) of popcorn with moisture content x (as a percent of the popcorn’s weight).
Hot-air popping: y = -0.761(x – 5.52)(x – 22.6)
Hot-oil popping:y = -0.652(x – 5.35)(x – 21.8)
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 38
a. For hot-air popping, what moisture content maximizes popping volume? What is the maximum volume?
b. For hot-oil popping, what moisture content maximizes popping volume? What is the maximum volume?
c. Use a graphing calculator to graph both functions in the same coordinate plane. What are the domain and range of each function in this situation? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 79.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 79.2

Question 80.
ABSTRACT REASONING A function is written in intercept form with a > 0. What happens to the vertex of the graph as a increases? as a approaches 0?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-80.jpg

Maintaining Mathematical Proficiency

Solve the equation. Check for extraneous solutions. (Skills Review Handbook)

Question 81.
3\(\sqrt{x}\) – 6 = 0
Answer:
Given,
3\(\sqrt{x}\) – 6 = 0
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 81

Question 82.
2\(\sqrt{x-4}\) – 2 = 2
Answer:
Given,
2\(\sqrt{x-4}\) – 2 = 2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-82.jpg

Question 83.
\(\sqrt{5x}\) + 5 = 0
Answer:
Given,
\(\sqrt{5x}\) + 5 = 0
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 83

Question 84.
\(\sqrt{3x+8}\) = \(\sqrt{x+4}\)
Answer:
Given,
\(\sqrt{3x+8}\) = \(\sqrt{x+4}\)
Squaring on both sides
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-84.jpg

Solve the proportion. (Skills Review Handbook)

Question 85.
\(\frac{1}{2}\) = \(\frac{x}{4}\)
Answer:
Given,
\(\frac{1}{2}\) = \(\frac{x}{4}\)
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 85

Question 86.
\(\frac{2}{3}\) = \(\frac{x}{9}\)
Answer:
Given,
\(\frac{2}{3}\) = \(\frac{x}{9}\)
2 × 9 = 3 × x
18 = 3x
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-86.jpg

Question 87.
\(\frac{-1}{4}\) = \(\frac{3}{x}\)
Answer:
Given,
\(\frac{-1}{4}\) = \(\frac{3}{x}\)
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.2 Question 87

Question 88.
\(\frac{5}{2}\) =-\(\frac{20}{x}\)
Answer:
Given,
\(\frac{5}{2}\) =-\(\frac{20}{x}\)
5 × x = -20 × 2
5x = -40
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.2-88.jpg

Quadratic Functions Study Skills Using the Features of Your Textbook to Prepare for Quizzes and Tests

Core Vocabulary
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 39

Core Concepts

Section 2.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 40
Section 2.2
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 40.1

Mathematical Practices

Question 1.
Why does the height you found in Exercise 44 on page 53 make sense in the context of the situation?

Question 2.
How can you effectively communicate your preference in methods to others in Exercise 47 on page 54?

Question 3.
How can you use technology to deepen your understanding of the concepts in Exercise 79 on page 64?

Study Skills
Using the Features of Your Textbook to Prepare for Quizzes and Tests

  • Read and understand the core vocabulary and the contents of the Core Concept boxes.
  • Review the Examples and the Monitoring Progress questions. Use the tutorials at BigIdeasMath.com for additional help.
  • Review previously completed homework assignments.

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 41

Quadratic Functions 2.1 – 2.2 Quiz

2.1 – 2.2 Quiz

Describe the transformation of f(x) = x2 represented by g. (Section 2.1)

Question 1.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 42
Answer:
The graph g is Translation 2 units up followed by a vertical shrink of a factor 1/2.

Question 2.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 43
Answer: The graph g is a horizontal translation 4 units left and a vertical translation 1 unit down of the graph f.

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 44
Answer:
Graph g is a horizontal translation of 1 unit right and a reflection in the x-axis of the graph f.

Write a rule for g and identify the vertex. (Section 2.1)

Question 4.
Let g be a translation 2 units up followed by a reflection in the x-axis and a vertical stretch by a factor of 6 of the graph of f(x) = x2.
Answer:
Given,
f(x) = x2.

Question 5.
Let g be a translation 1 unit left and 6 units down, followed by a vertical shrink by a factor of \(\frac{1}{2}\) of the graph of f(x) = 3(x + 2)2.
Answer:
Given,

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-5.jpg

Question 6.
Let g be a horizontal shrink by a factor of \(\frac{1}{4}\), followed by a translation 1 unit up and 3 units right of the graph of f(x) = (2x + 1)2 – 11.
Answer:
Given,
f(x) = (2x + 1)2 – 11.

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-6.jpg

Graph the function. Label the vertex and axis of symmetry. (Section 2.2)

Question 7.
f(x) = 2(x – 1)2 – 5
Answer:
Given,
f(x) = 2(x – 1)2 – 5
x = 0
f(0) = 2(0 – 1)2 – 5 = 2(1) – 5 = 2 – 5 = -3
x = 1
f(1) = 2(1 – 1)2 – 5 = -5
x = -1
f(-1) = 2(-1 – 1)2 – 5 = 4 – 5 = -1
f(x) = 2(x – 1)2 – 5
a = 2, h = 1 and k = -5
(h, k) = (1, -5)

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-7.jpg

Question 8.
h(x) = 3x2 + 6x – 2
Answer:
Given function is h(x) = 3x2 + 6x – 2

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-8.jpg

Question 9.
f(x) = 7 – 8x – x2

Answer:
Given function is f(x) = 7 – 8x – x2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-9.jpg

Find the x-intercepts of the graph of the function. Then describe where the function is increasing and decreasing.(Section 2.2)

Question 10.
g(x) = -3(x + 2)(x + 4)
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-10.jpg

Question 11.
g(x) = \(\frac{1}{2}\)(x – 5)(x + 1)
Answer:
Given,

Question 12.
f(x) = 0.4x(x – 6)
Answer:
Given,

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-12.jpg

Question 13.
A grasshopper can jump incredible distances, up to 20 times its length. The height (in inches) of the jump above the ground of a 1-inch-long grasshopper is given by h(x) = –\(\frac{1}{20}\)x2 + x, where x is the horizontal distance (in inches) of the jump. When the grasshopper jumps off a rock, it lands on the ground 2 inches farther. Write a function that models the new path of the jump. (Section 2.1)
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 45
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-13.jpg

Question 14.
A passenger on a stranded lifeboat shoots a distress flare into the air. The height (in feet) of the flare above the water is given by f(t) = -16t(t – 8), where t is time (in seconds) since the flare was shot. The passenger shoots a second flare, whose path is modeled in the graph. Which flare travels higher? Which remains in the air longer? Justify your answer. (Section 2.2)
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 46
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-quiz-2.1-2.2-14.jpg

Lesson 2.3 Focus of a Parabola

Essential Question
What is the focus of a parabola?
EXPLORATION 1
Analyzing Satellite Dishes
Work with a partner
. Vertical rays enter a satellite dish whose cross section is a parabola. When the rays hit the parabola, they reflect at the same angle at which they entered. (See Ray 1 in the figure.)
a. Draw the reflected rays so that they intersect the y-axis.
b. What do the reflected rays have in common?
c. The optimal location for the receiver of the satellite dish is at a point called the focus of the parabola. Determine the location of the focus. Explain why this makes sense in this situation.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 47

EXPLORATION 2
Analyzing Spotlights

Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 48
Work with a partner. Beams of light are coming from the bulb in a spotlight, located at the focus of the parabola. When the beams hit the parabola, they reflect at the same angle at which they hit. (See Beam 1 in the figure.) Draw the reflected beams. What do they have in common? Would you consider this to be the optimal result? Explain.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 49

Communicate Your Answer

Question 3.
What is the focus of a parabola?
Answer:
A parabola is a set of all points in a plane that are an equal distance away from a given point and given line. The point is called the focus of the parabola and the line is called the directrix. The focus lies on the axis of symmetry of the parabola.

Question 4.
Describe some of the properties of the focus of a parabola.
Answer:

  • The parabola is symmetric about its axis.
  • The axis is perpendicular to the directrix.
  • The axis passes through the vertex and the focus.
  • The tangent at the vertex is parallel to the directrix.

2.3 Lesson

Monitoring Progress

Question 1.
Use the Distance Formula to write an equation of the parabola with focus F(0, -3) and directrix y = 3.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 50
Answer:

Identify the focus, directrix, and axis of symmetry of the parabola. Then graph the equation.

Question 2.
y = 0.5x2

Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-2.jpg

Question 3.
-y = x2

Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-3.jpg

Question 4.
y2 = 6x
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-4.jpg

Write an equation of the parabola with vertex at (0, 0) and the given directrix or focus.

Question 5.
directrix: x = -3
Answer:


Question 6.
focus: (-2, 0)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-6.jpg

Question 7.
focus: (0, \(\frac{3}{2}\))
Answer:

Monitoring Progress

Question 8.
Write an equation of a parabola with vertex (-1, 4) and focus (-1, 2).
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-8.jpg

Question 9.
A parabolic microwave antenna is 16 feet in diameter. Write an equation that represents the cross section of the antenna with its vertex at (0, 0) and its focus 10 feet to the right of the vertex. What is the depth of the antenna?
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-9.jpg

Focus of a Parabola 2.3 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE A parabola is the set of all points in a plane equidistant from a fixed point called the ______ and a fixed line called the __________ .
Answer:
A parabola is the set of all points in a plane equidistant from a fixed point called the focus and a fixed line called the directrix.

Question 2.
WRITING Explain how to find the coordinates of the focus of a parabola with vertex (0, 0)and directrix y = 5.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-2.jpg

Monitoring Progress and Modeling with Mathematics

In Exercises 3–10, use the Distance Formula to write an equation of the parabola.

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 51
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 3

Question 4.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 52
Answer:

Question 5.
focus: (0, -2)
directrix: y = 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 5

Question 6.
directrix: y = 7
focus: (0, -7)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-6-1.jpg

Question 7.
vertex: (0, 0)
directrix: y = -6
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 7

Question 8.
vertex: (0, 0)
focus: (0, 5)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-8-1.jpg

Question 9.
vertex: (0, 0)
focus: (0, -10)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 9

Question 10.
vertex: (0, 0)
directrix: y = -9
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Monitoring-progress-2.3-10.jpg

Question 11.
ANALYZING RELATIONSHIPS Which of the given characteristics describe parabolas that open down? Explain your reasoning.
A. focus: (0, -6)
directrix: y = 6
B. focus: (0, -2)
directrix: y = 2
C.focus: (0, 6)
directrix: y = -6
D. focus: (0, -1)
directrix: y = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 11

Question 12.
REASONING Which of the following are possible coordinates of the point P in the graph shown? Explain.
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 53
A. (-6, -1)
B. (3, –\(\frac{1}{4}\))
C. (4, –\(\frac{4}{9}\))
D. (1, –\(\frac{1}{36}\))
E. (6, -1)
F. (2, –\(\frac{1}{18}\))
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-12.jpg

In Exercises 13–20, identify the focus, directrix, and axis of symmetry of the parabola. Graph the equation.

Question 13.
y = \(\frac{1}{8}\)x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 13

Question 14.
y = –\(\frac{1}{12}\)x2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-14.jpg

Question 15.
x = –\(\frac{1}{20}\)y2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 15

Question 16.
x= \(\frac{1}{24}\)y2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-16.jpg

Question 17.
y2 = 16x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 17.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 17.2

Question 18.
-x2 = 48y
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-18.jpg

Question 19.
6x2 + 3y = 0
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 19.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 19.2

Question 20.
8x2 – y = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-20.jpg

ERROR ANALYSIS In Exercises 21 and 22, describe and correct the error in graphing the parabola.

Question 21.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 54
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 21

Question 22.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 55
Answer:

Question 23.
ANALYZING EQUATIONS The cross section (with units in inches) of a parabolic satellite dish can be modeled by the equation y = \(\frac{1}{38}\)x2. How far is the receiver from the vertex of the cross section? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 23

Question 24.
ANALYZING EQUATIONS The cross section (with units in inches) of a parabolic spotlight can be modeled by the equation x = \(\frac{1}{20}\)y2. How far is the bulb from the vertex of the cross section? Explain.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 56
Answer:

In Exercises 25–28, write an equation of the parabola shown.

Question 25.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 57
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 25

Question 26.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 58
Answer:

Question 27.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 59
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 27

Question 28.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 60
Answer:

In Exercises 29–36, write an equation of the parabola with the given characteristics.

Question 29.
focus: (3, 0)
directrix: x = -3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 29

Questio
30.
focus: (\(\frac{2}{3}\), 0)
directrix: x = –\(\frac{2}{3}\)
Answer:

Question 31.
directrix: x = -10
vertex: (0, 0)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 31

Question 32.
directrix: y = \(\frac{8}{3}\)
vertex: (0, 0)
Answer:
Given,
directrix: y = \(\frac{8}{3}\)
vertex: (0, 0)
Because the vertex is at the origin and the axis of symmetry is vertical, the equation has the form y = \(\frac{1}{49}\) x².
The directrix is x = -p, p = –\(\frac{8}{3}\).
x = 1/4(\(\frac{8}{3}\)) × y²
Thus, the equation is 1/4(\(\frac{8}{3}\)) y²

Question 33.
focus: (0, –\(\frac{5}{3}\))
directrix: y = \(\frac{5}{3}\)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 33

Question 34.
focus: (0, \(\frac{5}{4}\))
directrix: y = –\(\frac{5}{4}\)
Answer:

Question 35.
focus: (0, \(\frac{6}{7}\))
vertex: (0, 0)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 35

Question 36.
focus: (-\(\frac{4}{5}\), 0)
vertex: (0, 0)
Answer:

In Exercises 37–40, write an equation of the parabola shown.

Question 37.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 61
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 37

Question 38.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 62
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-38.jpg

Question 39.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 63
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 39

Question 40.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 64
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-40.jpg

In Exercises 41–46, identify the vertex, focus, directrix, and axis of symmetry of the parabola. Describe the transformations of the graph of the standard equation with p = 1 and vertex (0, 0).

Question 41.
y = \(\frac{1}{8}\)(x – 3)2 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 41

Question 42.
y = –\(\frac{1}{4}\)(x + 2)2 + 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-42.jpg

Question 43.
x = \(\frac{1}{16}\)(y – 3)2 + 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 43

Question 44.
y = (x + 3)2 – 5
Answer:

Question 45.
x = -3(y + 4)2 + 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 45

Question 46.
x = 4(y + 5)2 – 1
Answer:

Question 47.
MODELING WITH MATHEMATICS Scientists studying dolphin echolocation simulate the projection of a bottlenose dolphin’s clicking sounds using computer models. The models originate the sounds at the focus of a parabolic reflector. The parabola in the graph shows the cross section of the reflector with focal length of 1.3 inches and aperture width of 8 inches. Write an equation to represent the cross section of the reflector. What is the depth of the reflector?
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 65
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 47

Question 48.
MODELING WITH MATHEMATICS Solar energy can be concentrated using long troughs that have a parabolic cross section as shown in the figure. Write an equation to represent the cross section of the trough. What are the domain and range in this situation? What do they represent?
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 66
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-48.jpg

Question 49.
ABSTRACT REASONING As | p | increases, how does the width of the graph of the equation y = \(\frac{1}{4 p}\)x2 change? Explain your reasoning.
Answer:
As | p | increases, the constant in the function decreases, which results in a vertical shrink, making the graph wider.

Question 50.
HOW DO YOU SEE IT? The graph shows the path of a volleyball served from an initial height of 6 feet as it travels over a net.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 67
a. Label the vertex, focus, and a point on the directrix.
b. An underhand serve follows the same parabolic path but is hit from a height of 3 feet. How does this affect the focus? the directrix?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-50.jpg

Question 51.
CRITICAL THINKING The distance from point P to the directrix is 2 units. Write an equation of the parabola.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 68
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 51

Question 52.
THOUGHT PROVOKING Two parabolas have the same focus (a, b) and focal length of 2 units. Write an equation of each parabola. Identify the directrix of each parabola.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.3-52.jpg

Question 53.
REPEATED REASONING Use the Distance Formula to derive the equation of a parabola that opens to the right with vertex (0, 0), focus (p, 0), and directrix x = -p.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 69
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 53

Question 54.
PROBLEM SOLVING The latus rectum of a parabola is the line segment that is parallel to the directrix, passes through the focus, and has endpoints that lie on the parabola. Find the length of the latus rectum of the parabola shown.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 70
Answer:

Maintaining Mathematical Proficiency

Write an equation of the line that passes through the points.(Section 1.3)

Question 55.
(1, -4), (2, -1)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 55

Question 56.
(-3, 12), (0, 6)
Answer:

Question 57.
(3, 1), (-5, 5)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.3 Question 57

Question 58.
(2, -1), (0, 1)
Answer:

Use a graphing calculator to find an equation for the line of best fit.

Question 59.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 71
Answer:
Given points are (0, 4), (3, 9), (7, 29), (11, 46).
Use the graphing calculator and plot the points on the graph to find the line of best fit.
The line of best fit is y = 3.98x + 0.92

Question 60.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 72
Answer:
Given points are (0, 18), (5, 15), (10, 9), (12, 7), (16, 2)
Use the graphing calculator and plot the points on the graph to find the line of best fit.
The line of best fit is y = -0.92x + 18

Lesson 2.4 Modeling with Quadratic Functions

Essential Question
How can you use a quadratic function to model a real-life situation?

EXPLORATION 1
Modeling with a Quadratic Function

Work with a partner. The graph shows a quadratic function of the form
P(t) = at2 + bt + c
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 73
which approximates the yearly profits for a company, where P(t) is the profit in year t.
a. Is the value of a positive, negative, or zero? Explain.
b. Write an expression in terms of a and b that represents the year t when the company made the least profit.
c. The company made the same yearly profits in 2004 and 2012. Estimate the year in which the company made the least profit.
d. Assume that the model is still valid today. Are the yearly profits currently increasing, decreasing, or constant? Explain.

EXPLORATION 2
Modeling with a Graphing Calculator
Work with a partner.
The table shows the heights h (in feet) of a wrench t seconds after it has been dropped from a building under construction.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 74
a. Use a graphing calculator to create a scatter plot of the data, as shown at the right. Explain why the data appear to fit a quadratic model.
b. Use the quadratic regression feature to find a quadratic model for the data.
c. Graph the quadratic function on the same screen as the scatter plot to verify that it fits the data.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 75
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 76
d. When does the wrench hit the ground? Explain.

Communicate Your Answer

Question 3.
How can you use a quadratic function to model a real-life situation?
Answer: Quadratic equations are used in real life when we calculate areas and determine a product’s profit or formulate the speed of an object.
Example: Throwing a ball, shooting a cannon, diving from a platform and hitting a golf ball are the examples of quadratic functions.

Question 4.
Use the Internet or some other reference to find examples of real-life situations that can be modeled by quadratic functions.
When you throw a ball (or shoot an arrow, fire a missile or throw a stone) it goes up into the air, slowing as it travels, then comes down again faster and faster.ball throw
Answer:
When a ball is thrown up in the sky its trajectory is parabolic, so using the quadratic function highest point which will be the vertex can be calculated and the maximum height attained by the ball from the ground can also be calculated.
We a cannonball is fired it also follows the parabolic path so using the quadratic function we can calculate where it will land from the launching point of the cannonball.

2.4 Lesson

Monitoring Progress

Question 1.
WHAT IF? The vertex of the parabola is (50, 37.5). What is the height of the net?
Answer:
Given,
The vertex of the parabola is (50, 37.5).
h = 50
k = 37.5
y = a(x – h)² + k
y = a(x – 50)² + 37.5

Question 2.
Write an equation of the parabola that passes through the point (-1, 2) and has vertex (4, -9).
Answer:
Given,
vertex (4, -9)
y = a(x – h)² + k
y = a(x – 4)² – 9
2 = a(-1 – 4)² – 9
2 + 9 = a(-1 – 4)²
11 = 25a
a = 11/25
y = 11/25 (x – 4)² – 9

Question 3.
WHAT IF? The y-intercept is 4.8. How does this change your answers in parts (a) and (b)?
Answer:
Given,
The y-intercept is 4.8
y = a(x – p)(x – q)
p = 4 and q = 24
4.8 = a(0 – 4)(0 – 24)
4.8 = 96a
a = 0.05
4.8 = 0.05(x – 4)(x – 24) where 0 ≤ x ≤ 24
x = (4 + 24)/2
x = 14
f(x) = 0.05(14 – 4)(14 – 24)
f(14) = 0.05(10)(-10)
= – 5
Hence the coldest temperature is -5°C at the same point of time when the parabola passes through the point (0, 9.6)
0 < x < 14
The average rate of change f(14) – f(0)/14 – 0
f(14) = -5
f(0) = 4.8
f(14) – f(0)/14 – 0 = -5 – 4.8/14
= -9.8/14
= -0.7
14 < x < 14
The average rate of change f(24) – f(14)/24 – 14
f(24) = 0
f(14) = -5
f(24) – f(14)/24 – 14 = 5/10
= 1/2
= 0.5
|-0.7| > |0.5|, the average rate at which temperature decreases from midnight to 2 p.m. is greater than the average rate at which it increases from 2 p.m. to midnight. When the parabola passes through (0, 9.6)

Question 4.
Write an equation of the parabola that passes through the point (2, 5) and has x-intercepts -2 and 4.
Answer:
Given points (2, 5)
x-intercept = -2 and 4 and the values of x and y as 2 and 5 as the parabola passes through the point (2, 5) in the intercept form
y = a(x – p)(x – q)
5 = a(2 – (-2))(2 – 4)
5 = a(2 + 2)(2 – 4)
5 = a(4)(-2)
5 = -8a
a = -5/8
y = -5/8(x + 2)(x – 4)
The equation of the parabola is y = -5/8(x + 2)(x – 4)

Question 5.
Write an equation of the parabola that passes through the points (-1, 4), (0, 1), and (2, 7).
Answer:
m = y2-y1/x2-x1
m = 1-4/0+1
m = -3/1
m = -3
We can see that one side is decreasing and 1 side is ascending.
The start of the parabola is 0,1 as x = 0 and crosses the line
m = (y2-y1)/(x2-x1)
= 7-1/2-0
= 6/2
y = 3
4 = a(-1)² + b(-1) + c
a + b + c = 4
1 = a(0)² + b(0) + c
c = 1
7 = a(2)² + b(2) + c
4a + 2b = 7
a = -1
b = -1
-a + -b = 3
4a – 0 = 4a
2b – 0 = 2b
7 – 1 = 6
-a + -b = 3
2a + 2b = -6
4a + 2b = 6
2a = -12
a = -6
4(-6) + 2b = 6
-24 + 2b = 6
2b = 30
b = 15
-6 + 15 + c = 4
c = -5
So, a = -6, b = 15 and c = -5.

Question 6.
The table shows the estimated profits y (in dollars) for a concert when the charge is x dollars per ticket. Write and evaluate a function to determine what the charge per ticket should be to maximize the profit.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 77

Answer:
y = ax² + bx + c
(2, 2600) = 4a + 2b + c = 2600
(5, 6500) = 25a + 5b + c = 6500
(8, 8600) = 64a + 8b + c = 8600
Solving eq. 1 & 2
25a + 5b + c = 6500
4a + 2b + c = 2600
21a + 3b = 3900
Solving eq. 2 & 3
64a + 8b + c = 8600
25a + 5b + c = 6500
39a + 3b = 2100

39a + 3b = 2100
21a + 3b = 3900
18a = -1800
a = -100
21a + 3b = 3900
21(-100) + 3b = 3900
b = 2000
4a + 2b + c = 2600
4(-100) + 2(2000) + c = 2600
c = -3000
y(x) = -100x² + 2000x – 3000

Question 7.
The table shows the results of an experiment testing the maximum weights y (in tons) supported by ice x inches thick. Write a function that models the data. How much weight can be supported by ice that is 22 inches thick?
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 78
Answer:
Given points (12, 3.4), (14, 7.6), (15, 10.0), (18, 18.3), (20, 25.0), (24, 40.6), (27, 54.3)
Big Ideas Math Answers Algebra 2 Chapter 2 Lesson 2.4 Modeling with Quadratic Functions_7

Modeling with Quadratic Functions 2.4 Exercises

Vocabulary and Core Concept Check

Question 1.
WRITING Explain when it is appropriate to use a quadratic model for a set of data.
Answer: You can use a quadratic model to model a set of data when the scatter plot of the set appears quadratic.

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 79
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-2.jpg

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, write an equation of the parabola in vertex form.

Question 3.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 80
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 3

Question 4.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 81
Answer:

Question 5.
passes through (13, 8) and has vertex (3, 2)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 5

Question 6.
passes through (-7, -15) and has vertex (-5, 9)
Answer:

Question 7.
passes through (0, -24) and has vertex (-6, -12)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 7

Question 8.
passes through (6, 35) and has vertex (-1, 14)
Answer:

In Exercises 9–14, write an equation of the parabola in intercept form.

Question 9.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 82
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 9.1

Question 10.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 83
Answer:

Question 11.
x-intercepts of 12 and -6; passes through (14, 4)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 11

Question 12.
x-intercepts of 9 and 1; passes through (0, -18)
Answer:

Question 13.
x-intercepts of -16 and -2; passes through (-18, 72)
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 13

Question 14.
x-intercepts of -7 and -3; passes through (-2, 0.05)
Answer:

Question 15.
WRITING Explain when to use intercept form and when to use vertex form when writing an equation of a parabola.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 15

Question 16.
ANALYZING EQUATIONS Which of the following equations represent the parabola?
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 84
A. y = 2(x – 2)(x + 1)
B. y = 2(x + 0.5)2 – 4.5
C. y = 2(x – 0.5)2 – 4.5
D. y = 2(x + 2)(x – 1)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-16.jpg

In Exercises 17–20, write an equation of the parabola in vertex form or intercept form.

Question 17.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 85
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 17

Question 18.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 86
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-18.jpg

Question 19.
Big Ideas Math Answer Key Algebra 2 Chapter 2 Quadratic Functions 87
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 19

Question 20.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 88
Answer:

Question 21.
ERROR ANALYSIS Describe and correct the error in writing an equation of the parabola.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 89
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 21

Question 22.
MATHEMATICAL CONNECTIONS The area of a rectangle is modeled by the graph where y is the area (in square meters) and x is the width (in meters). Write an equation of the parabola. Find the dimensions and corresponding area of one possible rectangle. What dimensions result in the maximum area?
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 90
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-22.jpg

Question 23.
MODELING WITH MATHEMATICS Every rope has a safe working load. A rope should not be used to lift a weight greater than its safe working load. The table shows the safe working loads S (in pounds) for ropes with circumference C (in inches). Write an equation for the safe working load for a rope. Find the safe working load for a rope that has a circumference of 10 inches.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 91
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 23.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 23.2

Question 24.
MODELING WITH MATHEMATICS A baseball is thrown up in the air. The table shows the heights y (in feet) of the baseball after x seconds. Write an equation for the path of the baseball. Find the height of the baseball after 1.7 seconds.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 92
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-24.jpg

Question 25.
COMPARING METHODS You use a system with three variables to find the equation of a parabola that passes through the points (−8, 0), (2, −20), and (1, 0). Your friend uses intercept form to find the equation. Whose method is easier? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 25

Question 26.
MODELING WITH MATHEMATICS The table shows the distances y a motorcyclist is from home after x hours.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 93
a. Determine what type of function you can use to model the data. Explain your reasoning.
b. Write and evaluate a function to determine the distance the motorcyclist is from home after 6 hours.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-26.jpg

Question 27.
USING TOOLS The table shows the heights h (in feet) of a sponge t seconds after it was dropped by a window cleaner on top of a skyscraper.
Big Ideas Math Algebra 2 Solutions Chapter 2 Quadratic Functions 94
a. Use a graphing calculator to create a scatter plot. Which better represents the data, a line or a parabola? Explain.
b. Use the regression feature of your calculator to find the model that best fits the data.
c. Use the model in part (b) to predict when the sponge will hit the ground.
d. Identify and interpret the domain and range in this situation.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 27

Question 28.
MAKING AN ARGUMENT Your friend states that quadratic functions with the same x-intercepts have the same equations, vertex, and axis of symmetry. Is your friend correct? Explain your reasoning.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-28.jpg

In Exercises 29–32, analyze the differences in the outputs to determine whether the data are linear, quadratic, or neither. Explain. If linear or quadratic, write an equation that fits the data.

Question 29.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 95
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 29.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 29.2

Question 30.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 96
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-30.jpg

Question 31.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 97
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 31

Question 32.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 98
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-32.jpg

Question 33.
PROBLEM SOLVING The graph shows the number y of students absent from school due to the flu each day x.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 99
a. Interpret the meaning of the vertex in this situation.
b. Write an equation for the parabola to predict the number of students absent on day 10.
c. Compare the average rates of change in the students with the flu from 0 to 6 days and 6 to 11 days.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 33.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 33.2

Question 34.
THOUGHT PROVOKING Describe a real-life situation that can be modeled by a quadratic equation. Justify your answer.
Answer:

Question 35.
PROBLEM SOLVING The table shows the heights y of a competitive water-skier x seconds after jumping off a ramp. Write a function that models the height of the water-skier over time. When is the water-skier 5 feet above the water? How long is the skier in the air?
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 100
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 35.1

Question 36.
HOW DO YOU SEE IT? Use the graph to determine whether the average rate of change over each interval is positive, negative, or zero.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 101
a. 0 ≤ x ≤ 2
b. 2 ≤ x ≤ 5
c. 2 ≤ x ≤ 4
d. 0 ≤ x ≤ 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-Exercise-2.4-36.jpg

Question 37.
REPEATED REASONING The table shows the number of tiles in each figure. Verify that the data show a quadratic relationship. Predict the number of tiles in the 12th figure.
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 102
Answer:
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 37.1
Big Ideas Math Algebra 2 Answers Chapter 2 Quadratic Functions 2.4 Question 37.2

Maintaining Mathematical Proficiency

Factor the trinomial. (Skills Review Handbook)

Question 38.
x2 + 4x + 3
Answer:

Question 39.
x2 – 3x + 2
Answer:
x2 – 3x + 2 = x2 – 2x – 1x + 2
= x(x – 2) -1(x – 2)
= (x – 1)(x – 2)

Question 40.
3x2 – 15x + 12
Answer:

Question 41.
5x2 + 5x – 30
Answer:
Given,
5x2 + 5x – 30
= 5(x² + x – 6)
= 5(x – 2)(x + 3)

Quadratic Functions Performance Task: Accident Reconstruction

2.3–2.4 What Did You Learn?

Core Vocabulary
focus, p. 68
directrix, p. 68

Core Concepts

Section 2.3
Standard Equations of a Parabola with Vertex at the Origin, p. 69
Standard Equations of a Parabola with Vertex at (h, k), p. 70

Section 2.4
Writing Quadratic Equations, p. 76
Writing Quadratic Equations to Model Data, p. 78

Mathematical Practices

Question 1.
Explain the solution pathway you used to solve Exercise 47 on page 73.

Question 2.
Explain how you used definitions to derive the equation in Exercise 53 on page 74.

Question 3.
Explain the shortcut you found to write the equation in Exercise 25 on page 81.

Question 4.
Describe how you were able to construct a viable argument in Exercise 28 on page 81.

Performance Task

Accident Reconstruction

Was the driver of a car speeding when the brakes were applied? What do skid marks at the scene of an accident reveal about the moments before the collision?
Big Ideas Math Algebra 2 Answer Key Chapter 2 Quadratic Functions 103
To explore the answers to these questions and more, go to BigIdeasMath.com.

Quadratic Functions Chapter Review

Describe the transformation of f(x) = x2 represented by g. Then graph each function.

Question 1.
g(x) = (x + 4)2

Answer:

Question 2.
g(x) = (x – 7)2 + 2

Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-2.jpg

Question 3.
g(x) = -3(x + 2)2 – 1
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-3.jpg

Question 4.
Let the graph of g be a horizontal shrink by a factor of \(\frac{2}{3}\), followed by a translation 5 units left and 2 units down of the graph of f(x) = x2.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-4.jpg

Question 5.
Let the graph of g be a translation 2 units left and 3 units up, followed by a reflection in the y-axis of the graph of f(x) = x2 – 2x.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-5.jpg

Graph the function. Label the vertex and axis of symmetry. Find the minimum or maximum value of f. Describe where the function is increasing and decreasing.

Question 6.
f(x) = 3(x – 1)2 – 4
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-6.jpg

Question 7.
g(x) = -2x2 + 16x + 3
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-7.jpg

Question 8.
h(x) = (x – 3)(x + 7)
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-8.jpg

Question 9.
You can make a solar hot-dog cooker by shaping foil-lined cardboard into a parabolic trough and passing a wire through the focus of each end piece. For the trough shown, how far from the bottom should the wire be placed?
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 105

Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-9.jpg

Question 10.
Graph the equation 36y = x2. Identify the focus, directrix, and axis of symmetry.
Answer:

Write an equation of the parabola with the given characteristics.

Question 11.
vertex: (0, 0)
directrix: x = 2
Answer:
Given,
vertex: (h, k) = (0, 0)
directrix: x = 2
y² = 4px
p = -2
y² = 4 × -2x
y² = -8x

Question 12.
focus: (2, 2)
vertex: (2, 6)
Answer:

Write an equation for the parabola with the given characteristics.

Question 13.
passes through (1, 12) and has vertex (10, -4)
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-13.jpg

Question 14.
passes through (4, 3) and has x-intercepts of -1 and 5
Answer:

Question 15.
passes through (-2, 7), (1, 10), and (2, 27)
Answer:

Question 16.
The table shows the heights y of a dropped object after x seconds. Verify that the data show a quadratic relationship. Write a function that models the data. How long is the object in the air?
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 106
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-2-quadratic-functions-chapter-review-Exercise-16.jpg

Quadratic Functions Chapter Test

Question 1.
A parabola has an axis of symmetry y= 3 and passes through the point (2, 1). Find another point that lies on the graph of the parabola. Explain your reasoning.
Answer:
Given that,
A parabola has an axis of symmetry y= 3 and passes through the point (2, 1).
The x-intercept will remain the same and we only need to find the y-coordinate of the reflected point.
y-coordinate of the given point lies 2 units below the axis of symmetry y = 3
3 + 2 = 5
So, the point will be (2, 5)

Question 2.
Let the graph of g be a translation 2 units left and 1 unit down, followed by a reflection in the y-axis of the graph of f(x) = (2x + 1)2 – 4. Write a rule for g.
Answer:
Let the graph of g is a translation of 2 units left and 1 unit down of the graph of f.
A horizontal translation to the left indicates a negative sign, and so, does a vertical translation down.
To shift the graph of 2 units left means Substitute the value of x in f(x) then you get x-(-2).
And to shift the graph of f 1 unit down add -1 to f(x) then you get x-(1).

Question 3.
Identify the focus, directrix, and axis of symmetry of x = 2y2. Graph the equation.
Answer:

Question 4.
Explain why a quadratic function models the data. Then use a linear system to find the model.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 107
Answer:

Write an equation of the parabola. Justify your answer.

Question 5.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 108

Question 6.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 109
Answer:
Given that,
(h, k) = (5, 3)
(x, y) = (-1, 0)
The formula for the parabola equation is y = a (x – h)² + k.
0 = a(-1-5)² + 3
0 = a(-6)² + 3
0 = 36a + 3
-3 = 36a
a = -3/36
a = 0.083.
The equation is y = 0.083(x – 5)² + 3.

Question 7.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 110
Answer:

Question 8.
A surfboard shop sells 40 surfboards per month when it charges $500 per surfboard. Each time the shop decreases the price by $10, it sells 1 additional surfboard per month. How much should the shop charge per surfboard to maximize the amount of money earned? What is the maximum amount the shop can earn per month? Explain.
Answer:
Given,
A surfboard shop sells 40 surfboards per month when it charges $500 per surfboard.
Each time the shop decreases the price by $10, it sells 1 additional surfboard per month.
P = (40 + x) (500 – 10x)
P = -10x² + 100x + 20000
x = -b/2a
= -100/2(-10) = -100/-20 = 5
P = -10(5)² + 100(5) + 20000
P = 20,250
The shop should charge 500 – 10(5) = $450 per surfboard.
Thus the maximum amount the shop can earn per month is $450.

Question 9.
Graph f(x) = 8x2 – 4x+ 3. Label the vertex and axis of symmetry. Describe where the function is increasing and decreasing.
Answer:
Given,
f(x) = 8x2 – 4x+ 3
(h, k)
vertex = (0.25, 2.5)
x = h
x = 0.25
The axis of symmetry is x = 0.25
The function increasing (0.25, ∞) and decreasing (-∞, 0.25)
BIM Algebra 2 Answer Key Chapter 2 Quadratic Functions Chapter Test_9

Question 10.
Sunfire is a machine with a parabolic cross section used to collect solar energy. The Sun’s rays are reflected from the mirrors toward two boilers located at the focus of the parabola. The boilers produce steam that powers an alternator to produce electricity.
a. Write an equation that represents the cross section of the dish shown with its vertex at (0, 0).
b. What is the depth of Sunfire? Justify your answer.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 111
Answer:
vertex at (0, 0)
focus is (0, 10)
h = 0
k = 0
p = 10
y = 1/4p (x – h)² + k
y = 1/40 x²
x = 37/2
y = 1/40 (37/2)² = 8.56
Thus the depth of Sunfire is 8.56 ft

Question 11.
In 2011, the price of gold reached an all-time high. The table shows the prices (in dollars per troy ounce) of gold each year since 2006 (t = 0 represents 2006). Find a quadratic function that best models the data. Use the model to predict the price of gold in the year 2016.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 112
Answer:
Given,
In 2011, the price of gold reached an all-time high.
The table shows the prices (in dollars per troy ounce) of gold each year since 2006 (t = 0 represents 2006).
y = ax² + bx + c
y(0) = c = 603.46
y(1) = a + b + c = 695.39
a + b = 695.39 – 603.46 = 91.93
y(2) = 4b + 2b + c =871.96
4a + 2b = 871.96 – 603.46 = 268.5
solving 1 & 2
a = 42.32, b = 49.61

Quadratic Functions Cumulative Assessment

Question 1.
You and your friend are throwing a football. The parabola shows the path of your friend’s throw, where x is the horizontal distance (in feet) and y is the corresponding height (in feet). The path of your throw can be modeled by h(x) = −16x2 + 65x + 5. Choose the correct inequality symbol to indicate whose throw travels higher. Explain your reasoning.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 113
Answer:

Question 2.
The function g(x) = \(\frac{1}{2}\)∣x − 4 ∣ + 4 is a combination of transformations of f(x) = | x|. Which combinations describe the transformation from the graph of f to the graph of g?
A. translation 4 units right and vertical shrink by a factor of \(\frac{1}{2}\), followed by a translation 4 units up
B. translation 4 units right and 4 units up, followed by a vertical shrink by a factor of \(\frac{1}{2}\)
C. vertical shrink by a factor of \(\frac{1}{2}\) , followed by a translation 4 units up and 4 units right
D. translation 4 units right and 8 units up, followed by a vertical shrink by a factor of \(\frac{1}{2}\)
Answer:

Question 3.
Your school decides to sell tickets to a dance in the school cafeteria to raise money. There is no fee to use the cafeteria, but the DJ charges a fee of $750. The table shows the profits (in dollars) when x students attend the dance.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 114
a. What is the cost of a ticket?
Answer:
y = zx – 750
250 = 200z – 750
200z = 1000
z = 1000/200
z = 5
Thus the price of the ticket is $5.

b. Your school expects 400 students to attend and finds another DJ who only charges $650. How much should your school charge per ticket to still make the same profit?
Answer:
y = xz – 650
1250 = 400x – 650
400z = 1950
z = 1950/400
z = 4.875
The price of the ticket should be $4.85

c. Your school decides to charge the amount in part (a) and use the less expensive DJ. How much more money will the school raise?
Answer:
c. y = xz – 650
y = 5(400) – 650
y = 2000 – 650
y = 1350
1350 – 1250 = 100
Thus, the school will raise $100 more than the previous value.

Question 4.
Order the following parabolas from widest to narrowest.
A. focus: (0, −3); directrix: y = 3
B. y = \(\frac{1}{16}\)x2 + 4
C. x = \(\frac{1}{8}\)y2
D. y = \(\frac{1}{4}\)(x − 2)2 + 3
Answer:

Question 5.
Your friend claims that for g(x) = b, where b is a real number, there is a transformation in the graph that is impossible to notice. Is your friend correct? Explain your reasoning.
Answer:
Big Ideas Math Answers Algebra 2 Quadratic Functions Cumulative Assessment_5

Question 6.
Let the graph of g represent a vertical stretch and a reflection in the x-axis, followed by a translation left and down of the graph of f(x) = x2. Use the tiles to write a rule for g.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 115
Answer:

Question 7.
Two balls are thrown in the air. The path of the first ball is represented in the graph. The second ball is released 1.5 feet higher than the first ball and after 3 seconds reaches its maximum height 5 feet lower than the first ball.
Big Ideas Math Answers Algebra 2 Chapter 2 Quadratic Functions 116
a. Write an equation for the path of the second ball.
b. Do the balls hit the ground at the same time? If so, how long are the balls in the air? If not, which ball hits the ground first? Explain your reasoning.
Answer:
y = a(x – 3)² + 51.5
6.5 = a(0 – 3)² + 51.5
-45 = a(-3)²
-45 = 9a
a = -45/9
a = -5
y = -5(x – 3)² + 51.5

Question 8.
Let the graph of g be a translation 3 units right of the graph of f. The points (−1, 6), (3, 14), and (6, 41) lie on the graph of f. Which points lie on the graph of g?
A. (2, 6)
B. (2, 11)
C. (6, 14)
D. (6, 19)
E. (9, 41)
F. (9, 46)
Answer:

Question 9.
Gym A charges $10 per month plus an initiation fee of $100. Gym B charges $30 per month, but due to a special promotion, is not currently charging an initiation fee.
a. Write an equation for each gym modeling the total cost y for a membership lasting x months.
b. When is it more economical for a person to choose Gym A over Gym B?
c. Gym A lowers its initiation fee to $25. Describe the transformation this change represents and how it affects your decision in part (b).
Answer:
Gym A is $120
Gym B is $30
12 × 20 = 240
100 + 240 = $340
30 × 12 = $360
Whole year in Gym B is 360
360 – 340 = $20
So, you can save $20 with Gym A in a year.

Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions

Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions

Do you have any doubts regarding solving the linear functions? If our guess is correct, then this is the right page for you to complete your assignments in time. We provide Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions that help you to enhance your problem-solving skills in depth. Build self-confidence by Answering all your questions and clearing your complex doubts regarding Alegra 2 BIM Answer Key Chapter 1 Linear Functions topics. Improve your preparation by referring our Big Ideas Math Algebra 2 Answers that help the students to learn concepts in different methods.

Big Ideas Math Book Algebra 2 Answer Key Chapter 1 Linear Functions

The solutions seen in Big Ideas Math Book Algebra 2 Answer Key is prepared by math professionals in a very simple manner with explanations. With the help of BIM Algebra 2 Answer Key students can score good grades in any of their exams and can make you achieve what you are looking for but improving their subject knowledge. Before that we suggest you to check the list of the chapters covered in BIM Algebra 2 Answer Key Ch 1 Linear Functions. Big Ideas Math Book Algebra 2 Chapter 1 Solutions is the best guide for you to score well in exams. Just tap the links provided below and start solving the questions.

Linear Functions Maintaining Mathematical Proficiency

Evaluate.

Question 1.
5 • 23 + 7
Answer: 47

Explanation:
Given expression,
5 • 23 + 7
= 5.8+7
= 40+7 = 47
So, the expression 5 • 23 + 7 = 47.

Question 2.
4 – 2(3 + 2)²
Answer: -46

Explanation:
Given expression,
4 – 2(3 + 2)2
= 4 – 2 (25)
= 4  – 2 (25)
= 4 – 50
= – 46
So, the expression = 4 – 2(3 + 2)² = -46

Question 3.
48 ÷ 42 + \(\frac{3}{5}\)
Answer: 3.6

Explanation:
Given expression,
48 ÷ 42 + \(\frac{3}{5}\)
= 48 ÷ 16 + \(\frac{3}{5}\)
= 3 + \(\frac{3}{5}\) = 3.6
So, 48 ÷ 42 + \(\frac{3}{5}\) = 3.6

Question 4.
50 ÷ 52 • 2
Answer: 4

Explanation:
50 ÷ 52 • 2
= 50 ÷ 25 . 2
= 2 .2
= 4
So, 50 ÷ 52 • 2 = 4

Question 5.
\(\frac{1}{2}\)(22+ 22)
Answer: 13

Explanation:
Given,
\(\frac{1}{2}\)(22+ 22)
\(\frac{1}{2}\)(4+ 22)
\(\frac{1}{2}\)(26) = 13
So, \(\frac{1}{2}\)(22+ 22) is 13.

Question 6.
\(\frac{1}{6}\)(6 + 18) – 2²
Answer: 0

Explanation:
Given,
\(\frac{1}{6}\)(6 + 18) – 22
\(\frac{1}{6}\)(24) – 4
4 – 4 = 0
So, \(\frac{1}{6}\)(6 + 18) – 2² = 0

Graph the transformation of the figure.

Question 7.
Translate the rectangle 1 unit right and 4 units up.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1
Answer:
Big-Ideas-Math-Algebra-2-Answers-Chapter-1-Linear-Functions-1
We have to draw a rectangle 1 unit right and 4 units up. So, start drawing the rectangle from the origin 0 to 3.

Question 8.
Reflect the triangle in the y-axis. Then translate 2 units left.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 2
Answer:
Big-Ideas-Math-Algebra-2-Answers-Chapter-1-Linear-Functions-2
Start drawing the triangle 2 units left.

Question 9.
Translate the trapezoid 3 units down. Then reflect in the x-axis.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 3
Answer:
Big-Ideas-Math-Algebra-2-Answers-Chapter-1-Linear-Functions-3

Question 10.
ABSTRACT REASONING Give an example to show why the order of operations is important when evaluating a numerical expression. Is the order of transformations of figures important? Justify your answer.
Answer:
The order of operations says the order to solve steps in expressions with more than one operation. First, we solve any operations inside parentheses or brackets.

Linear Functions Maintaining Mathematical Practices

Monitoring Progress

Use a graphing calculator to graph the equation using the standard viewing window and a square viewing window. Describe any differences in the graphs.

Question 1.
y = 2x – 3
Answer:
Given equation is y = 2x – 3
When x = 0
y = 2(0) – 3 = -3
y = -3
x = 1
y = 2(1) – 3 = 2 – 3 = -1
x = 2
y = 2(2) – 3 = 1
x = 3
y = 2(3) – 3 = 6 – 3 = 3
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions Maintaining Mathematical Practices_1

Question 2.
y = | x + 2 |
Answer:
Given equation is y = | x + 2 |
x = -2
y = |-2 + 2| = 0
x = -1
y = |-1 + 2| = 1
x = 0
y = |0 + 2| = 2
x = 1
y = |1 + 2| = 3
x = 2
y = |2 + 2| = 4
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions Maintaining Mathematical Practices_2

Question 3.
y = -x2 + 1
Answer:
Given equation is y = -x2 + 1
x = -1
y = -x2 + 1
y = -1 + 1 = 0
x = 0
y = 0 + 1 = 1
x = 1
y = -1 + 1 = 0
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions Maintaining Mathematical Practices_3

Question 4.
y = \(\sqrt{x-1}\)
Answer:
Given equation is y = \(\sqrt{x-1}\)
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions Maintaining Mathematical Practices_4

Question 5.
y = x3 – 2
Answer:
Given equation is y = x3 – 2
x = 0
y = 0 – 2 = -2
(0, -2)
x = 1
y = 1 – 2 = -1
(1, -1)
x = -1
y = -1 – 2 = -3
(-1, -3)
x = 2
y = 2³ – 2
y = 8 – 2 = 6
(2, 6)
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions Maintaining Mathematical Practices_5

Question 6.
y = 0.25x³
Answer:
Given equation is y = 0.25x³
x = -1
y = 0.25(-1)³
y = -0.25
x = 0
y = 0.25(0)
y = 0
x = 1
y = 0.25(1)³
y = 0.25
x = 2
y = 0.25(2)³
y = 0.25(8)
y = 2
(2, 2)
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions Maintaining Mathematical Practices_6

Determine whether the viewing window is square. Explain.

Question 7.
-8 ≤ x ≤ 8, -2 ≤ y ≤ 8
Answer: Square

Explanation:
Given,
-8 ≤ x ≤ 8, -2 ≤ y ≤ 8
The total range of the X-axis is 16 units and the total range of the Y-axis is 10 units
The ratio of the height to width of the viewing screen is \(\frac{10}{16}\) = \(\frac{5}{8}\)
So, the ratio is 5:8.
Hence the viewing window is square.

Question 8.
-7 ≤ x ≤ 8, -2 ≤ y ≤ 8
Answer: Square

Explanation:
Given,
-7 ≤ x ≤ 8, -2 ≤ y ≤ 8
The total range of the X-axis is 15 units and the total range of the Y-axis is 10 units
The ratio of the height to width of the viewing screen is \(\frac{10}{15}\) = \(\frac{2}{3}\)
So, the ratio is 2:3.
Hence the viewing window is square.

Question 9.
-6 ≤ x ≤ 9, -2 ≤ y ≤ 8
Answer: Square

Explanation:
Given,
-6 ≤ x ≤ 9, -2 ≤ y ≤ 8
The total range of the X-axis is 15 units and the total range of the Y-axis is 10 units
The ratio of the height to width of the viewing screen is \(\frac{10}{15}\) = \(\frac{2}{3}\)
So, the ratio is 2:3.
Thus the viewing window is square.

Question 10.
-2 ≤ x≤ 2, -3 ≤ y ≤ 3
Answer: not a square

Explanation:
Given,
-2 ≤ x≤ 2, -3 ≤ y ≤ 3
The total range of the X-axis is 4 units and the total range of the Y-axis is 6 units
The ratio of the height to width of the viewing screen is \(\frac{6}{4}\) = \(\frac{3}{2}\)
So, the ratio is 3:2.
The viewing window is not a square.

Question 11.
-4 ≤ x ≤ 5, -3 ≤ y ≤ 3
Answer: square

Explanation:
Given,
-4 ≤ x ≤ 5, -3 ≤ y ≤ 3
The total range of the X-axis is 9 units and the total range of the Y-axis is 6 units
The ratio of the height to width of the viewing screen is \(\frac{6}{9}\) = \(\frac{2}{3}\)
So, the ratio is 2:3.
The viewing window is a square.

Question 12.
-4 ≤ x ≤ 4, -3 ≤ y ≤ 3
Answer: square

Explanation:
Given,
-4 ≤ x ≤ 4, -3 ≤ y ≤ 3
The total range of the X-axis is 8 units and the total range of the Y-axis is 6 units
The ratio of the height to width of the viewing screen is \(\frac{6}{8}\) = \(\frac{2}{3}\)
So, the ratio is 2:3.
The viewing window is a square.

Lesson 1.1 Parent Functions and Transformations

Essential Question

What are the characteristics of some of the basic parent functions?
Answer:
Odd. End behavior goes in different directions.
If a function is positive, the left side of the graph will point down and the right side will point up i.e., increasing from left to right.

EXPLORATION 1
Identifying Basic Parent Functions
Work with a partner.
Graphs of eight basic parent functions are shown below. Classify each function as constant, linear, absolute value, quadratic, square root, cubic, reciprocal, or exponential. Justify your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 4

Communicate Your Answer

Question 2.
What are the characteristics of some of the basic parent functions?
Answer: The key common points of linear parent functions include the fact that the: Equation is y = x. Domain and range are real numbers. The slope, or rate of change, is constant.

Question 3.
Write an equation for each function whose graph is shown in Exploration 1. Then use a graphing calculator to verify that your equations are correct.
Answer:
a. The equation for the given graph of the absolute value function in exploration 1 is y = |x|
Big Ideas Math Answers Algebra 2 Lesson 1.1 Parent Functions and Transformations_3
b. y = √x
Big Ideas Math Answers Algebra 2 Lesson 1.1 Parent Functions and Transformations_3b
c. y = c
Big Ideas Math Answers Algebra 2 Lesson 1.1 Parent Functions and Transformations_3c
y = e^x
Big Ideas Math Answers Algebra 2 Lesson 1.1 Parent Functions and Transformations_3d
y = x³
Big Ideas Math Answers Algebra 2 Lesson 1.1 Parent Functions and Transformations_3e
y = x
Big Ideas Math Answers Algebra 2 Lesson 1.1 Parent Functions and Transformations_3f
y = 1/x
Big Ideas Math Answers Algebra 2 Lesson 1.1 Parent Functions and Transformations_3g

y = x²
Big Ideas Math Answers Algebra 2 Lesson 1.1 Parent Functions and Transformations_3h

Use graphing calculator to find the graph of the equation.

1.1 Lesson

Monitoring Progress

Question 1.
Identify the function family to which g belongs. Compare the graph of g to the graph of its parent function.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 5
Answer:

Graph the function and its parent function. Then describe the transformation.

Question 2.
g(x) = x + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-2.jpg

Question 3.
h(x) = (x – 2)2

Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-3.jpg

Question 4.
n(x) = – | x |
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-4.jpg

Graph the function and its parent function. Then describe the transformation.

Question 5.
g(x) = 3x
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-5.jpg

Question 6.
h(x) = \(\frac{3}{2}\)x2

Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-6.jpg

Question 7.
c(x) = 0.2|x|
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-7JPG.jpg

Use a graphing calculator to graph the function and its parent function. Then describe the transformations

Question 8.
h(x) = –\(\frac{1}{4}\)x + 5
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-8.jpg

Question 9.
d(x) = 3(x – 5)2 – 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-9.jpg

Question 10.
The table shows the amount of fuel in a chainsaw over time. What type of function can you use to model the data? When will the tank be empty?
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 6
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.1-10.jpg

Parent Functions and Transformations 1.1 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The function f(x) = x2 is the ______ of f(x) = 2x2 – 3.
Answer:
The function f(x) = x2 is the parent function of f(x) = 2x2 – 3.

Question 2.
DIFFERENT WORDS, SAME QUESTION Which is different? Find “both” answers.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 7
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-1.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-1a.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-1b.jpg

Monitoring Progress and Modeling with Mathematics

In Exercises 3–6, identify the function family to which f belongs. Compare the graph of f to the graph of its parent function.

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 3

Question 4.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 9
Answer:

Question 5.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 5

Question 6.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 11
Answer:

Question 7.
MODELING WITH MATHEMATICS
At 8:00 A.M., the temperature is 43°F. The temperature increases 2°F each hour for the next 7 hours. Graph the temperatures over time t (t = 0 represents 8:00 A.M.). What type of function can you use to model the data? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 7

Question 8.
MODELING WITH MATHEMATICS
You purchase a car from a dealership for $10,000. The trade-in value of the car each year after the purchase is given by the function f(x) = 10,000 – 250x2. What type of function models the trade-in value?
Answer:
Given,
You purchase a car from a dealership for $10,000.
The trade-in value of the car each year after the purchase is given by the function f(x) = 10,000 – 250x2.
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-8JPG.jpg

In Exercises 9–18, graph the function and its parent function. Then describe the transformation.

Question 9.
g(x) = x + 4
Answer:
Given equation is g(x) = x + 4
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 9

Question 10.
f(x) = x – 6
Answer:
f(x) = x – 6
y = x – 6
x = 0
y = 0 – 6 = -6
x = 1
y = 1 – 6 = -5
x = -1
y = -1 – 6 = -7
x = 2
y = 2 – 6 = -4
x = 3
y = 3 – 6 = -3
x = 4
y = 4 – 6 = -2
x = 5
y = 5 – 6 = -1
x = 6
y = 6 – 6 = 0
x = 7
y = 7 – 6 = 1
x = 8
y = 8 – 6 = 2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-10.jpg

The graph is a linear function.

Question 11.
f(x) = x2 – 1
Answer:
Given equation is f(x) = x2 – 1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 11

Question 12.
h(x) = (x+ 4)2
Answer:
Given equation is h(x) = (x+ 4)2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-12.jpg

Question 13.
g(x) = | x – 5 |
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 13

Question 14.
f(x) = 4 + | x |
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-14.jpg

Question 15.
h(x) = -x2

Answer:
Given equation is h(x) = -x2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 15

Question 16.
g(x) = -x
Answer:
Given equation is g(x) = -x
y = -x
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-16.jpg

Question 17.
f(x) = 3
Answer:
Given equation is f(x) = 3
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 17

Question 18.
f(x) = -2
Answer:
Given equation is f(x) = -2
The parent function is f(x) = 1
Find the graph of the given function y = -2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-18.jpg

In Exercises 19–26, graph the function and its parent function. Then describe the transformation.

Question 19.
f(x) = \(\frac{1}{3}\)x
Answer:
Given equation is f(x) = \(\frac{1}{3}\)x
The parent function y = x
Find the graph of the given function f(x) = \(\frac{1}{3}\)x
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 19

Question 20.
g(x) = 4x
Answer:
Given equation is g(x) = 4x
The parent function if f(x) = x
Find the graph of the given function g(x) = 4x
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-20.jpg

Question 21.
f(x) = 2x2
Answer:
Given equation is f(x) = 2x2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 21

Question 22.
h(x) = \(\frac{1}{3}\)x2
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-22.jpg

Question 23.
h(x) = \(\frac{3}{4}\)x
Answer:
Given equation is h(x) = \(\frac{3}{4}\)x
The parent function y = x
Find the graph of the given function h(x) = \(\frac{3}{4}\)x
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 23

Question 24.
g(x) = \(\frac{4}{3}\)x
Answer:
Given equation is g(x) = \(\frac{4}{3}\)x
The parent function y = x
Find the graph of the given function g(x) = \(\frac{4}{3}\)x
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-24.jpg

Question 25.
h(x) = 3 | x |
Answer:
Given equation is h(x) = 3 | x |
The parent function y =|x|
Find the graph of the given function h(x) = 3 | x |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 25

Question 26.
f(x) = \(\frac{1}{2}\) | x |
Answer:
Given equation is f(x) = \(\frac{1}{2}\) | x |
The parent function y =|x|
Find the graph of the given function f(x) = \(\frac{1}{2}\) | x |
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-26.jpg

In Exercises 27–34, use a graphing calculator to graph the function and its parent function. Then describe the transformations.

Question 27.
f(x) = 3x + 2
Answer:
Given equation,
f(x) = 3x + 2
x = 0
y = 3(0) + 2 = 2
x = 1
y = 3(1) + 2 = 5
x = 2
y = 3(2) + 2 = 7
x = -1
y = 3(-1) + 2 = -3 + 2 = -1
x = -2
y = 3(-2) + 2 = -6 + 2 = -4
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 27

Question 28.
h(x) = -x + 5
Answer:
Given equation is h(x) = -x + 5
x = -5
y = -(-5) + 5 = 10
x = 0
y = 0 + 5 = 5
x = 5
y = -5 + 5 = 0
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-28.jpg

Question 29.
h(x) = -3 | x | – 1
Answer:
Given equation is h(x) = -3 | x | – 1
x = -3
y = -3|-3| – 1 = -9 – 1 = -10
x = -2
y = -3|-3| – 1 = -6 – 1 = -7
x = 0
y = -3|0| – 1 = -1
x = 3
y = -3|3| – 1 = -9 – 1 = -10
x = 2
y = -3|2| – 1 = -6 – 1 = -7
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 29

Question 30.
f(x) = \(\frac{3}{4}\) | x | + 1
Answer:
Given equation is f(x) = \(\frac{3}{4}\) | x | + 1
x = -5
y = \(\frac{3}{4}\) | x | + 1
y = \(\frac{3}{4}\) | -5 | + 1 = \(\frac{15}{4}\) + 1 = \(\frac{19}{4}\)
x = 0
y = \(\frac{3}{4}\) | 0 | + 1 = 1
x = -5
y = \(\frac{3}{4}\) | x | + 1
y = \(\frac{3}{4}\) | 5 | + 1 = \(\frac{15}{4}\) + 1 = \(\frac{19}{4}\)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-30.jpg

Question 31.
g(x) = \(\frac{1}{2}\)x2 – 6
Answer:
Given equation is g(x) = \(\frac{1}{2}\)x2 – 6
The parent function is y = x2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 31

Question 32.
f(x) = 4x2 – 3
Answer:
Given equation is f(x) = 4x2 – 3
The parent function is f(x) = x2
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-32.jpg

Question 33.
f(x) = -(x + 3)2 + \(\frac{1}{4}\)
Answer:
Given equation is f(x) = -(x + 3)2 + \(\frac{1}{4}\)
The parent function is f(x) = x2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 33

Question 34.
g(x) = – | x – 1 | – \(\frac{1}{2}\)
Answer:
Given equation is g(x) = – | x – 1 | – \(\frac{1}{2}\)
The parent function is f(x) = |x|
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-34.jpg

ERROR ANALYSIS In Exercises 35 and 36, identify and correct the error in describing the transformation of the parent function.

Question 35.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 12
Answer: The error is there is no vertical shrink of the parent quadratic function. The graph is a reflection in the x-axis followed by a vertical stretch of the parent quadratic function.

Question 36.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 13
Answer: The graph is translated horizontally 3 units to the right.
So, the equation should be f(x) = |x – 3| not f(x) = |x + 3|

MATHEMATICAL CONNECTIONS In Exercises 37 and 38, find the coordinates of the figure after the transformation.

Question 37.
Translate 2 units down.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 13.1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 37

Question 38.
Reflect in the x-axis.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 15
Answer:

USING TOOLS In Exercises 39–44, identify the function family and describe the domain and range. Use a graphing calculator to verify your answer.

Question 39.
g(x) = | x + 2 | – 1
Answer: The function g is in the family of absolute value functions. The domain of the function is all real numbers and the range of the function is y ≥ -1.

Question 40.
h(x) = | x – 3 | + 2
Answer:
Given equation is h(x) = | x – 3 | + 2
x = -1
y = |-1 – 3| + 2 = 4 + 2 = 6
x = 0
y = |0 – 3| + 2 = 3 + 2 = 5
x = 1
y = |1 – 3| + 2 = 2 + 2 = 4
Big Ideas Math Answers Parent Functions and Transformations 1.1 Exercises_40

Question 41.
g(x) = 3x + 4
Answer:
Function g is in the family of linear functions.
Domain: All Real Numbers
Range:  All Real Numbers

Question 42.
f(x) = -4x + 11
Answer:
Given,
f(x) = -4x + 11
x = 0
y = -4(0) + 11 = 11
(0, 11)
x = 1
y = -4(1) + 11 = 7
(1, 7)
x = 2
y = -4(2) + 11 = -8 + 11 = -3
(2, -3)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-42.jpg

Question 43.
f(x) = 5x2 – 2
Answer:
Given function is f(x) = 5x2 – 2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 43

Question 44.
f(x) = -2x2 + 6
Answer:
Given function is f(x) = -2x2 + 6
x = -1
y = -2(-1)² + 6 = -2 + 6 = 4
(-1, 4)
x = 0
y = -2(0)² + 6 = 6
(0, 6)
x = 1
y = -2(1)² + 6 = -2 + 6 = 4
(1, 4)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-44.jpg

Question 45.
MODELING WITH MATHEMATICS The table shows the speeds of a car as it travels through an intersection with a stop sign. What type of function can you use to model the data? Estimate the speed of the car when it is 20 yards past the intersection.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 16
Answer: The type of function that can model the data is an absolute value function because the data are linear and there are positive speeds for the positive and negative displacements.
The speed of the car 20 yards past the intersection is estimated to be 8 miles per hour.

Question 46.
THOUGHT PROVOKING In the same coordinate plane, sketch the graph of the parent quadratic function and the graph of a quadratic function that has no x-intercepts. Describe the transformation(s) of the parent function.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-46.jpg

Question 47.
USING STRUCTURE Graph the functions f(x) = | x – 4 | and g(x) = | x | – 4. Are they equivalent? Explain.
Answer:
Given,
Graph the functions f(x) = | x – 4 | and g(x) = | x | – 4.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 47

Question 48.
HOW DO YOU SEE IT? Consider the graphs of f, g, and h.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 17
a. Does the graph of g represent a vertical stretch or a vertical shrink of the graph of f? Explain your reasoning.
Answer:
g represents a vertical shrink of f because the y-values of g are smaller than the y-values of f at the same x-values.
b. Describe how to transform the graph of f to obtain the graph of h.
Answer:
The graph of f would have to be reflected over the x-axis and then vertically stretched to obtain the graph of h.

Question 49.
MAKING AN ARGUMENT Your friend says two different translations of the graph of the parent linear function can result in the graph of f(x) = x – 2. Is your friend correct? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 49

Question 50.
DRAWING CONCLUSIONS A person swims at a constant speed of 1 meter per second. What type of function can be used to model the distance the swimmer travels? If the person has a 10-meter head start, what type of transformation does this represent? Explain.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 17.1
Answer:
The distance traveled by the swimmer can be modeled by a linear function. The 10-meter head start is modeled by a vertical translation up to 10 units.

Question 51.
PROBLEM SOLVING You are playing basketball with your friends. The height (in feet) of the ball above the ground t seconds after a shot is released from your hand is modeled by the function f(t) = -16t2 + 32t + 5.2.
a. Without graphing, identify the type of function that models the height of the basketball.
b. What is the value of t when the ball is released from your hand? Explain your reasoning.
c. How many feet above the ground is the ball when it is released from your hand? Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 51

Question 52.
MODELING WITH MATHEMATICS The table shows the battery lives of a computer over time. What type of function can you use to model the data? Interpret the meaning of the x-intercept in this situation.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 18
Answer:
The data can be modeled by an absolute value function. The x-intercept in this situation is when the battery of the computer dies.

Question 53.
REASONING Compare each function with its parent function. State whether it contains a horizontal translation, vertical translation, both, or neither. Explain your reasoning.
a. f(x) = 2 | x | – 3
b. f(x) = (x – 8)2
c. f(x) = | x + 2 | + 4
d. f(x) = 4x2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 53

Question 54.
CRITICAL THINKING
Use the values -1, 0, 1, and 2 in the correct box so the graph of each function intersects the x-axis. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 19
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-54-.jpg

Maintaining Mathematical Proficiency

Determine whether the ordered pair is a solution of the equation. (Skills Review Handbook)

Question 55.
f(x) = | x + 2 |; (1, -3)
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 55

Question 56.
f(x) = | x | – 3; (-2, -5)
Answer:

Question 57.
f(x) = x – 3; (5, 2)
Answer:
Given,
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 57

Question 58.
f(x) = x – 4; (12, 8)
Answer:
Given,
f(x) = x – 4

Find the x-intercept and the y-intercept of the graph of the equation. (Skills Review Handbook)

Question 59.
y = x
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 59

Question 60.
y = x + 2
Answer:
To find the x-intercept let y = 0, then solve for x.
y = x + 2
0 = x + 2
x + 2 = 0
x = 0 – 2
x = -2
To find the y-intercept let x = 0, then solve for y.
y = x+ 2
y = 0 + 2
y = 2
Therefore, the intercept is (0, 0) and the y-intercept is (-2, 2)

Question 61.
3x + y = 1
Answer:
Given,
3x + y = 1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.1 Question 61

Question 62.
x – 2y = 8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.1-62.jpg

Lesson 1.2 Transformations of Linear and Absolute Value Functions

Essential Question

How do the graphs of y = f(x) + k, y = f(x – h), and y = -f(x) compare to the graph of the parent function f?

EXPLORATION 1
Transformations of the Parent Absolute Value Function
Work with a partner.
Compare the graph of the function
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 20
y = | x | + k Transformation
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 20.1
to the graph of the parent function
f(x) = | x |.

EXPLORATION 2
Transformations of the Parent Absolute Value Function
Work with a partner.
Compare the graph of the function
y = | x – h | Transformation
to the graph of the parent function
f(x) = | x |. Parent function
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 22

EXPLORATION 3
Transformation of the Parent Absolute Value Function
Work with a partner.
Compare the graph of the function
y = – | x | Transformation
to the graph of the parent function
f(x) = | x | Parent function
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 23

Communicate Your Answer

Question 4.
Transformation How do the graphs of y = f (x) + k, y = f(x – h), and y = -f(x) compare to the graph of the parent function f?

Answer:
The graphs of y = f (x) + k, y = f(x – h), and y = -f(x) are compared to the graph of the parent function by vertical shifts, horizontal shifts and reflections.
Vertical shifts: Let f(x) be the parent function and k be a positive number. To graph the function y = f (x) + k, we shift the graph of y = f(x) up k units by adding k to the y-coordinates of the points on the graph of f.
Horizontal shifts: Let f(x) be the parent function and h be a positive number. To graph the function y = f(x – h), we shift the graph of y = f(x) to the right h units by adding h to the x-coordinates of the points on the graph of f.
Reflections: Let f(x) be the parent function.
To graph the function y = -f(x), we reflect the graph of y = f(x) about the x-axis by multiplying the y-coordinates of the points on the graph of f by -1.

Question 5.
Compare the graph of each function to the graph of its parent function f. Use a graphing calculator to verify your answers are correct.
a. y = \([\sqrt{x}/latex] – 4
b. y = [latex][\sqrt{x + 4}/latex]
c. y = –[latex][\sqrt{x}/latex]
d. y = x2 + 1
e. y = (x – 1)2
f. y = -x2

1.2 Lesson

Monitoring Progress

Write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 1.
f(x) = 3x; translation 5 units up
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.2-1.jpg

Question 2.
f(x) = | x | – 3; translation 4 units to the right
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.2-2.jpg

Question 3.
f(x) = – | x + 2 | – 1; reflection in the x-axis
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.2-3.jpg

Question 4.
f(x) = [latex]\frac{1}{2}\)x+ 1; reflection in the y-axis
Answer:
Given function is f(x) = \(\frac{1}{2}\)x+ 1
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.2-4.jpg

Write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 5.
f(x) = 4x+ 2; horizontal stretch by a factor of 2
Answer:
Given function is 4x + 2

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.2-5.jpg

Question 6.
f(x) = | x | – 3; vertical shrink by a factor of \(\frac{1}{3}\)
Answer:
Given function is f(x) = | x | – 3
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.2-6.jpg

Question 7.
Let the graph of g be a translation 6 units down followed by a reflection in the x-axis of the graph of f(x) = | x |. Write a rule for g. Use a graphing calculator to check your answer.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.2-7.jpg

Question 8.
WHAT IF? In Example 5, your revenue function is f(x) = 3x. How does this affect your profit for 100 downloads?
Answer:

Transformations of Linear and Absolute Value Functions 1.2 Exercises

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
The function g(x) = | 5x |- 4 is a horizontal ___________ of the function f(x) = | x | – 4.
Answer:
The function g(x) = | 5x |- 4 is a horizontal shrink of the function f(x) = | x | – 4.

Question 2.
WHICH ONE DOESN’T BELONG? Which transformation does not belong with the other three? Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 24
Answer:
Horizontal shrink for a factor 1/5.

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 3.
f(x) = x – 5; translation 4 units to the left
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 3

Question 4.
f(x) = x + 2; translation 2 units to the right
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-4.jpg

Question 5.
f(x) = | 4x + 3 | + 2; translation 2 units down
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 5

Question 6.
f(x) = 2x – 9; translation 6 units up
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-6.jpg

Question 7.
f(x) = 4 – | x + 1 |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 25
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 7

Question 8.
f(x) = | 4x | + 5
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 26
Answer:

Question 9.
WRITING Describe two different translations of the graph of f that result in the graph of g.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 27
Answer:
A horizontal translation of 3 units right or a vertical translation of 3 units up will produce the function g from the function f.

Question 10.
PROBLEM SOLVING You open a café. The function f(x) = 4000x represents your expected net income (in dollars) after being open x weeks. Before you open, you incur an extra expense of $12,000. What transformation of f is necessary to model this situation? How many weeks will it take to pay off the extra expense?
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 28
Answer:

In Exercises 11–16, write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 11.
f(x) = -5x+ 2; reflection in the x-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 11

Question 12.
f(x) = \(\frac{1}{2}\)x – 3; reflection in the x-axis
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-1.2-12.jpg

Question 13.
f(x) = | 6x | – 2; reflection in the y-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 13

Question 14.
f(x) = | 2x – 1 | + 3; reflection in the y-axis
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-14.jpg

Question 15.
f(x) = -3 + | x – 11 |; reflection in the y-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 15

Question 16.
f(x) = -x+ 1; reflection in the y-axis
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-16.jpg

In Exercises 17–22, write a function g whose graph represents the indicated transformation of the graph of f. Use a graphing calculator to check your answer.

Question 17.
f(x) = x + 2; vertical stretch by a factor of 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 17

Question 18.
f(x) = 2x+ 6; vertical shrink by a factor of \(\frac{1}{2}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-18.jpg

Question 19.
f(x) = | 2x | + 4; horizontal shrink by a factor of \(\frac{1}{2}\)
Answer:
Given function is f(x) = | 2x | + 4
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 19

Question 20.
f(x) = | x+ 3 | ; horizontal stretch by a factor of 4
Answer:
Given function is f(x) = | x+ 3 |
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-20.jpg

Question 21.
f(x) = -2 | x – 4 | + 2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 29
Answer:
Given function is f(x) = -2 | x – 4 | + 2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 21

Question 22.
f(x) = 6 – x
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 30
Answer:
Given function is f(x) = 6 – x

ANALYZING RELATIONSHIPS In Exercises 23–26, match the graph of the transformation of f with the correct equation shown. Explain your reasoning.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 31

Question 23.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 32
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 23

Question 24.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 33
Answer:

Question 25.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 34
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 25

Question 26.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 35
Answer:
A. y = 2f(x)
B. y = f(2x)
C. y = f(x + 2)
D. y = f(x) + 2

In Exercises 27–32, write a function g whose graph represents the indicated transformations of the graph of f.

Question 27.
f(x) = x; vertical stretch by a factor of 2 followed by a translation 1 unit up
Answer:
Given function is f(x) = x
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 27

Question 28.
f(x) = x; translation 3 units down followed by a vertical shrink by a factor of \(\frac{1}{3}\)
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-28.jpg

Question 29.
f(x) = | x | ; translation 2 units to the right followed by a horizontal stretch by a factor of 2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 29

Question 30.
f(x) = | x |; reflection in the y-axis followed by a translation 3 units to the right
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-30.jpg

Question 31.
f(x) = | x |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 36
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 31

Question 32.
f(x) = | x |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 37
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-32.jpg

ERROR ANALYSIS In Exercises 33 and 34, identify and correct the error in writing the function g whose graph represents the indicated transformations of the graph of f.

Question 33.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 38
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 33

Question 34.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 39
Answer:

Question 35.
MAKING AN ARGUMENT Your friend claims that when writing a function whose graph represents a combination of transformations, the order is not important. Is your friend correct? Justify your answer.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 35

Question 36.
MODELING WITH MATHEMATICS During a recent period of time, bookstore sales have been declining. The sales (in billions of dollars) can be modeled by the function f(t) = –\(\frac{7}{5}\)t + 17.2, where t is the number of years since 2006. Suppose sales decreased at twice the rate. How can you transform the graph of f to model the sales? Explain how the sales in 2010 are affected by this change.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-36.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-36.aJPG_.jpg

MATHEMATICAL CONNECTIONS For Exercises 37–40, describe the transformation of the graph of f to the graph of g. Then find the area of the shaded triangle.

Question 37.
f(x) = | x – 3 |
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 40
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 37

Question 38.
f(x) = – | x | – 2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 41
Answer:

Question 39.
f(x) = -x + 4
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 42
Answer:
The transformation is a reflection in the x-axis.
We know that the area of the triangle is A = 1/2 × bh
A = 1/2 × 8 × 4 = 16 sq. units

Question 40.
f(x) = x – 5
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 43
Answer:

Question 41.
ABSTRACT REASONING The functions f(x) = mx + b and g(x) = mx + c represent two parallel lines.
a. Write an expression for the vertical translation of the graph of f to the graph of g.
b. Use the definition of slope to write an expression for the horizontal translation of the graph of f to the graph of g.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 41

Question 42.
HOW DO YOU SEE IT? Consider the graph of f(x) = mx + b. Describe the effect each transformation has on the slope of the line and the intercepts of the graph.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 44
a. Reflect the graph of f in the y-axis.
b. Shrink the graph of f vertically by a factor of \(\frac{1}{3}\).
c. Stretch the graph of f horizontally by a factor of 2.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-42.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-42a.jpg

Question 43.
REASONING The graph of g(x) = -4 |x | + 2 is a reflection in the x-axis, vertical stretch by a factor of 4, and a translation 2 units down of the graph of its parent function. Choose the correct order for the transformations of the graph of the parent function to obtain the graph of g. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 43

Question 44.
THOUGHT PROVOKING You are planning a cross-country bicycle trip of 4320 miles. Your distance d (in miles) from the halfway point can be modeled by d = 72 |x – 30 |, where x is the time (in days) and x = 0 represents June 1. Your plans are altered so that the model is now a right shift of the original model. Give an example of how this can happen. Sketch both the original model and the shifted model.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-44.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.2-44a.jpg

Question 45.
CRITICAL THINKING Use the correct value 0, -2, or 1 with a, b, and c so the graph of g(x) = a|x – b | + c is a reflection in the x-axis followed by a translation one unit to the left and one unit up of the graph of f(x) = 2 |x – 2 | + 1. Explain your reasoning.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 45

Maintaining Mathematical Proficiency

Evaluate the function for the given value of x. (Skills Review Handbook)

Question 46.
f(x) = x + 4; x = 3
Answer:
Given function is f(x) = x + 4
when x = 4
f(x) = x + 4
f(x) = 3 + 4 = 7
So, f(x) = 7

Question 47.
f(x) = 4x – 1; x = -1
Answer:
f(x) = 4x – 1
x = -1
f(-1) = 4(-1) – 1
= -4 – 1 = -5
So, f(-1) = -5

Question 48.
f(x) = -x + 3; x = 5
Answer:
Given function is f(x) = -x + 3
when x = 5
f(-5) = -(5) + 3
= -5 + 3 = -2

Question 49.
f(x) = -2x – 2; x = -1
Answer:
f(x) = -2x – 2
x = -1
f(-1) = -2(-1) – 2
= 2 – 2 = 0

Create a scatter plot of the data. (Skills Review Handbook)

Question 50.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 45
Answer:
The coordinates are (8, 4), (10, 9), (11, 10), (12, 12) and (15, 12). Plot the points on the graph.

Question 51.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 46
Answer:
The coordinates are (2, 22), (5, 13), (6, 15), (10, 12) and (13, 6).
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.2 Question 51

Linear Functions Study Skills Taking Control of Your Class Time

1.1 – 1.2 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 47

Core Concepts

Section 1.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 48

Section 1.2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 49

Mathematical Practices

Question 1.
How can you analyze the values given in the table in Exercise 45 on page 9 to help you determine what type of function models the data?

Question 2.
Explain how you would round your answer in Exercise 10 on page 16 if the extra expense is $13,500.

Study Skills

Taking Control of Your Class Time

Question 1.
Sit where you can easily see and hear the teacher, and the teacher can see you.

Question 2.
Pay attention to what the teacher says about math, not just what is written on the board.

Question 3.
Ask a question if the teacher is moving through the material too fast.

Question 4.
Try to memorize new information while learning it.

Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 50

Question 5.
Ask for clarification if you do not understand something.

Question 6.
Think as intensely as if you were going to take a quiz on the material at the end of class.

Question 7.
Volunteer when the teacher asks for someone to go up to the board.

Question 8.
At the end of class, identify concepts or problems for which you still need clarification.

Question 9.
Use the tutorials at BigIdeasMath.com for additional help.

Linear Functions 1.1-1.2 Quiz

Identify the function family to which g belongs. Compare the graph of the function to the graph of its parent function. (Section 1.1)

Question 1.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 51
Answer:
The given function is linear.
The function g is shifted down and the slope of g is smaller than the slope of the parent linear function.
The domain of g and its parent function is all real numbers and the range of g and its parent function is all real numbers.

Question 2.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 52
Answer:
The function g is a quadratic function. Function g is shifted left and g is narrower than the parent quadratic function.
The domain of g and its parent function is all real numbers and the range of g and its parent function is y > 0.

Question 3.
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 53
Answer:
Function g is an absolute value function. Function g is shifted left and up.
The domain of g and its parent function is all real numbers but the range of g is y > -2 and the range of its parent function is y > 0.

Graph the function and its parent function. Then describe the transformation. (Section 1.1)

Question 4.
f(x) = \(\frac{3}{2}\)
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-quiz-Exercise-1.11.2-4JPG.jpg

Question 5.
f(x) = 3x
Answer:
Given,
f(x) = 3x
x = 0
y = 3(0) = 0
x = 1
y = 3(1) = 3
x = 2
y = 3(2) = 6
x = -1
y = 3(-1) = -3
x = -2
y = 3(-2) = -6
(0, 0), (1, 3), (2, 6), (-1, -3), (-2, -6).
Big Ideas Math Answers Linear Functions 1.1-1.2 Quiz_5

Question 6.
f(x) = 2(x – 1)²
Answer:
f(x) = 2(x – 1)²
x = 0
f(x) = 2(0 – 1)² = 2
(0, 2)
x = 1
f(x) = 2(1 – 1)² = 0
x = 2
f(x) = 2(2 – 1)² = 2
x = -1
f(x) = 2(-1 – 1)² = 8
(0, 2), (1, 0), (2, 2) and (-1, 8).
Big Ideas Math Answers Linear Functions 1.1-1.2 Quiz_6

Question 7.
f(x) = – | x + 2 | – 7
Answer:
Big Ideas Math Answers Linear Functions 1.1-1.2 Quiz_7

Question 8.
f(x) = \(\frac{1}{4}\)x2 + 1
Answer:
Big Ideas Math Answers Linear Functions 1.1-1.2 Quiz_8

Question 9.
f(x) = –\(\frac{1}{2}\)x – 4
Answer:
Given,
f(x) = –\(\frac{1}{2}\)x – 4
x = 0
y = –\(\frac{1}{2}\)(0) – 4 = -4
x = 1
y = –\(\frac{1}{2}\)(1) – 4 = -4\(\frac{1}{2}\)
x = 2
y = –\(\frac{1}{2}\)(2) – 4 = -4 – 1 = -5
x = -1
y = –\(\frac{1}{2}\)(-1) – 4 = -3\(\frac{1}{2}\)
x = -2
y = –\(\frac{1}{2}\)(-2) – 4 = -4 + 1 = -3

Big Ideas Math Answers Linear Functions 1.1-1.2 Quiz_9

Write a function g whose graph represents the indicated transformation of the graph of f. (Section 1.2)

Question 10.
f(x) = 2x + 1; translation 3 units up
Answer:
Given function is f(x) = 2x + 1
The vertex is (0, 1) of the original graph to move the vertex up by 3 units just add 3 to the y-intercept.
g(x) = 2x + 4

Question 11.
f(x) = -3 | x – 4 | ; vertical shrink by a factor of \(\frac{1}{2}\)
Answer:
Given the function is f(x) = -3 | x – 4 | to vertically shrink a function by a factor by c, multiply the whole function by c f(x) vertically shrunk by a factor of c would be cf(x)
so f(x) = -3 | x – 4 | vertically shrunk by a factor of 1/2 would be f(x) = (-3/2) |x – 4|

Question 12.
f(x) = 3 | x + 5 |; reflection in the x-axis
Answer:
Given,
f(x) = 3 | x + 5 |
The points reflected in the x-axis have opposite y-coordinates
f(x) = -y
-y = -3 |x + 5|
f(x) = -3 |x + 5|

Question 13.
f(x) = \(\frac{1}{3}\)x – \(\frac{2}{3}\) ; translation 4 units left
Answer:
Given,
f(x) = \(\frac{1}{3}\)x – \(\frac{2}{3}\)
= \(\frac{1}{3}\) (x – 2)
= \(\frac{1}{3}\) (x – 2 + 4)
= \(\frac{1}{3}\) (x + 2)
= \(\frac{1}{3}\)x + \(\frac{2}{3}\)
So, the translation 4 units left is f(x) = \(\frac{1}{3}\)x + \(\frac{2}{3}\)

Write a function g whose graph represents the indicated transformations of the graph of f. (Section 1.2)

Question 14.
Let g be a translation 2 units down and a horizontal shrink by a factor of \(\frac{2}{3}\) of the graph of f(x) =x.
Answer:
Given,
f(x) =x
horizontal shrink by a factor of \(\frac{2}{3}\)
f(x) = \(\frac{3}{2}\)x
f(x) = \(\frac{3}{2}\)x – 2
g(x) =\(\frac{3}{2}\)x – 2
So, the translation 2 units down and a horizontal shrink by a factor of \(\frac{2}{3}\) is g(x) =\(\frac{3}{2}\)x – 2

Question 15.
Let g be a translation 9 units down followed by a reflection in the y-axis of the graph of f(x) = x.
Answer:
Given,
f(x) =x
g(x) = f(x) – 9
It is also reflected about the y-axis.
g(x) = f(-x) – 9
So, the translation 9 units down followed by a reflection in the y-axis is g(x) = f(-x) – 9

Question 16.
Let g be a reflection in the x-axis and a vertical stretch by a factor of 4 followed by a translation 7 units down and 1 unit right of the graph of f(x) = | x |.
Answer:
Given,
f(x) = |x|
g(x) = bf(x)
Reflecting function over the x-axis
g(x) = -f(x)
Original function is f(x) = -|x|
Stretching by a factor of 4 means we have to multiply by 4.
g(x) = -4|x|
g(x) = -4x
translation 7 units down and 1 unit right of the graph of f(x) = | x |
g(x) = -4x – 7 -1
g(x) = -4x – 8
g(x) = -4(x + 2)
So, reflection in the x-axis and a vertical stretch by a factor of 4 followed by a translation 7 units down and 1 unit right is g(x) = -4(x + 2).

Question 17.
Let g be a translation 1 unit down and 2 units left followed by a vertical shrink by a factor of \(\frac{1}{2}\) of the graph of f(x) = | x |.
Answer:
Given,
f(x) = |x|
Multiply output with 1/2 to vertically shrink function
g(x) = 1/2 |x|
Subtract 1 to the output of the function to translate 1 unit down and 2 units left.
g(x) = 1/2 |x| – 1 -2
g(x) = 1/2 |x| – 3
So, translation 1 unit down and 2 units left followed by a vertical shrink by a factor of \(\frac{1}{2}\) is g(x) = 1/2 |x| – 3.

Question 18.
The table shows the total distance a new car travels each month after it is purchased. What type of function can you use to model the data? Estimate the mileage after 1 year. (Section 1.1)
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 54
Answer:
From the given data in the above table,
2 months = 2300
5 months = 5750
6 months = 6900
9 months = 10,350
1 month = 2300/2 = 1150
12 months = 1150 × 12 = 13800
The mileage after 1 year is 13800.
The estimated mileage after 1 year is 14,000 miles.

Question 19.
The total cost of an annual pass plus camping for x days in a National Park can be modeled by the function f(x) = 20x+ 80. Senior citizens pay half of this price and receive an additional $30 discount. Describe how to transform the graph of f to model the total cost for a senior citizen. What is the total cost for a senior citizen to go camping for three days? (Section 1.2)
Answer:
Given,
The total cost of an annual pass plus camping for x days in a National Park can be modeled by the function f(x) = 20x+ 80.
Senior citizens pay half of this price and receive an additional $30 discount.
For senior citizen cost
= 1/2 (20x + 80) – 30
= 10x + 40 – 30
= 10x + 10
= 10(x + 1)
x = 3
= 10(3 + 1)
= 10(4)
= $40
Thus the total cost for a senior citizen to go camping for three days is $40.

Lesson 1.3 Modeling with Linear Functions

Essential Question
How can you use a linear function to model and analyze a real-life situation?

EXPLORATION 1
Modeling with a Linear Function
Work with a partner.
A company purchases a copier for $12,000. The spreadsheet shows how the copier depreciates over an 8-year period.
a. Write a linear function to represent the value V of the copier as a function of the number t of years.
b. Sketch a graph of the function. Explain why this type of depreciation is called straight line depreciation.
c. Interpret the slope of the graph in the context of the problem.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 55
Answer:
Given,
A company purchases a copier for $12,000.
The spreadsheet shows how the copier depreciates over an 8-year period.
m = 12,000 – 10,750 = $1,250
b = 12,000
V = -1250t + 12,000
The slope shows that for every year that passes, the value depreciates by $1250.

EXPLORATION 2
Modeling with Linear Functions
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 56
Work with a partner. Match each description of the situation with its corresponding graph. Explain your reasoning.
a. A person gives $20 per week to a friend to repay a $200 loan.
b. An employee receives $12.50 per hour plus $2 for each unit produced per hour.
c. A sales representative receives $30 per day for food plus $0.565 for each mile driven.
d. A computer that was purchased for $750 depreciates $100 per year.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 56.1

Communicate Your Answer

Question 3.
How can you use a linear function to model and analyze a real-life situation?
Answer:
One of the real-life situations is finding Variable costs.
Imagine that you are taking a taxi while on vacation. You know that the taxi service charges 9 rupees to pick your family up from your hotel and another 0.15 rupees per mile for the trip. Without knowing how many miles it will be to each destination, you can set up a linear equation that can be used to find the cost of any taxi trip you take on your trip. By using′ ′x′′ to represent the number of miles to your destination and ”y′′ to represent the cost of that taxi ride, the linear equation would be: y = 0.15x+9

Question 4.
Use the Internet or some other reference to find a real-life example of straight line depreciation.
a. Use a spreadsheet to show the depreciation.
Answer:
The real-life example of straight-line depreciation is the decrease of the speed of the car by ten meters per second which was moving with an initial speed of a hundred meter per second till the speed reaches thirty meters per second.

Speed Time (in second)
100 0
90 1
80 2
70 3
60 4
50 5
40 6
30 7

b. Write a function that models the depreciation.
Answer:
The real-life example of straight-line depreciation is the decrease of the speed of the car by ten meters per second which was moving with an initial speed of a hundred meter per second till the speed reaches thirty meters per second.
The function that represents the statement is y = 100 – 10x, where y is speed and x is time.

c. Sketch a graph of the function.
Answer:
y = 100 – 10x
x = 7
y = 100 – 10(7)
y = 100 – 70
y = 30
when x = 0
y = 100
when x = 4
y = 60
Big Ideas Math Answers Algebra 2 chapter 1 Lesson 1.3 Modeling with Linear Functions_4

1.3 Lesson

Monitoring Progress

Question 1.
The graph shows the remaining balance y on a car loan after making x monthly payments. Write an equation of the line and interpret the slope and y-intercept. What is the remaining balance after 36 payments?
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 57

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-Progress-Exercise-1.3-1.jpg
Question 2.
WHAT IF? Maple Ridge charges a rental fee plus a $10 fee per student. The total cost is $1900 for 140 students. Describe the number of students that must attend for the total cost at Maple Ridge to be less than the total costs at the other two venues. Use a graph to justify your answer.
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-Progress-Exercise-1.3-2.jpg

Question 3.
The table shows the humerus lengths (in centimeters) and heights (in centimeters) of several females.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 58
a. Do the data show a linear relationship? If so, write an equation of a line of fit and use it to estimate the height of a female whose humerus is 40 centimeters long.
b. Use the linear regression feature on a graphing calculator to find an equation of the line of best fit for the data. Estimate the height of a female whose humerus is 40 centimeters long. Compare this height to your estimate in part (a).
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-Progress-Exercise-1.3-3-1.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-Progress-Exercise-1.3-3a.jpg

Modeling with Linear Functions 1.3 Exercises

Question 1.
COMPLETE THE SENTENCE The linear equation y = \(\frac{1}{2}\)x + 3 is written in ____________ form.
Answer:
The linear equation y = \(\frac{1}{2}\)x + 3 is written in slope-intercept form.

Question 2.
VOCABULARY A line of best fit has a correlation coefficient of -0.98. What can you conclude about the slope of the line?
Answer:
The slope of the graph with a correlation coefficient of -0.98 is negative.

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, use the graph to write an equation of the line and interpret the slope.

Question 3.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 59
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 3

Question 4.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 60
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-4.jpg

Question 5.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 61
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 5

Question 6.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 62
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-6.jpg

Question 7.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 63
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 7

Question 8.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 64
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-8.jpg

Question 9.
MODELING WITH MATHEMATICS Two newspapers charge a fee for placing an advertisement in their paper plus a fee based on the number of lines in the advertisement. The table shows the total costs for different length advertisements at the Daily Times. The total cost y (in dollars) for an advertisement that is x lines long at the Greenville Journal is represented by the equation y = 2x + 20. Which newspaper charges less per line? How many lines must be in an advertisement for the total costs to be the same?
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 65
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 9.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 9.2

Question 10.
PROBLEM SOLVING While on vacation in Canada, you notice that temperatures are reported in degrees Celsius. You know there is a linear relationship between Fahrenheit and Celsius, but you forget the formula. From science class, you remember the freezing point of water is 0°C or 32°F, and its boiling point is 100°C or 212°F.
a. Write an equation that represents degrees Fahrenheit in terms of degrees Celsius.
b. The temperature outside is 22°C. What is this temperature in degrees Fahrenheit?
c. Rewrite your equation in part (a) to represent degrees Celsius in terms of degrees Fahrenheit.
d. The temperature of the hotel pool water is 83°F. What is this temperature in degrees Celsius?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-10.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-10.jpg

ERROR ANALYSIS In Exercises 11 and 12, describe and correct the error in interpreting the slope in the context of the situation.

Question 11.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 66
Answer:
Given that the slope of the line is 10, after 7 years the balance is $70.
The slope was correctly used in the situation, however, the intercept was not used correctly.
In this situation, the starting balance is $100, so after 7 years the balance is 100 + 70 = $170.

Question 12.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 67
Answer:
Here the slope m = 3
The income is $3 per hour.
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-12.jpg

In Exercises 13–16, determine whether the data show a linear relationship. If so, write an equation of a line of fit. Estimate y when x = 15 and explain its meaning in the context of the situation.

Question 13.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 68
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 13.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 13.2

Question 14.

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-14.jpg

Question 15.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 70
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 15.1

Question 16.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 71
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-16.jpg

Question 17.
MODELING WITH MATHEMATICS The data pairs (x, y) represent the average annual tuition y (in dollars) for public colleges in the United States x years after 2005. Use the linear regression feature on a graphing calculator to find an equation of the line of best fit. Estimate the average annual tuition in 2020. Interpret the slope and y-intercept in this situation.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 72
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 17

Question 18.
MODELING WITH MATHEMATICS The table shows the numbers of tickets sold for a concert when different prices are charged. Write an equation of a line of fit for the data. Does it seem reasonable to use your model to predict the number of tickets sold when the ticket price is $85? Explain.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 73
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-18a.jpg

USING TOOLS In Exercises 19–24, use the linear regression feature on a graphing calculator to find an equation of the line of best fit for the data. Find and interpret the correlation coefficient.

Question 19.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 74
Answer:
Use the graphing calculator to find the equation of the line of best fit for the data.
Enter the data into two lists.
The line of best fit is y = 0.42x + 1.44
The correlation coefficient is r ≈ 0.61.
This represents a weak positive correlation.

Question 20.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 75
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-20.jpg

Question 21.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 76
Answer:
Use the graphing calculator to find the equation of the line of best fit for the data.
Enter the data into two lists.
The line of best fit is y = -0.45x + 4.26
The correlation coefficient is r ≈ -0.67
This represents a weak negative correlation.

Question 22.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 77
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-22.jpg

Question 23.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 78
Answer:
Use the graphing calculator to find the equation of the line of best fit for the data.
Enter the data into two lists.
The line of best fit is y = 0.61x + 0.10
The correlation coefficient is r ≈ 0.95.
This represents a strong positive correlation.

Question 24.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 79
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-24.jpg

Question 25.
OPEN-ENDED Give two real-life quantities that have
(a) a positive correlation,
(b) a negative correlation, and
(c) approximately no correlation. Explain.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 25

Question 26.
HOW DO YOU SEE IT? You secure an interest-free loan to purchase a boat. You agree to make equal monthly payments for the next two years. The graph shows the amount of money you still owe.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 80
a. What is the slope of the line? What does the slope represent?
Answer:
The points are (0, 30), (24, 0) are on the line.
We can assume the slope is k.
k = (30-0)/(0-24) = -5/4
And the slope represents the amount of money to be paid monthly

b. What is the domain and range of the function? What does each represent?
Answer:
Because x and y are linear functions, and
k = -5/4, (0, 30) is on the function.
y = -5/4 x + 30
And the domain is [0, 24] and the range is [0, 30] and the range represents the total money still to be paid.

c. How much do you still owe after making payments for 12 months?
Answer:
y = -5/4 x + 30
when x = 12 and y = 15
So, you still owe 15 hundred of dollars.

Question 27.
MAKING AN ARGUMENT A set of data pairs has a correlation coefficient r = 0.3. Your friend says that because the correlation coefficient is positive, it is logical to use the line of best fit to make predictions. Is your friend correct? Explain your reasoning.
Answer:
The correlation coefficient is near to 0 and because of that, there line cannot help with the prediction.
Your friend is incorrect.
r = 0.3 is close to 0 than 1, the line of best fit will not make good predictions.

Question 28.
THOUGHT PROVOKING Points A and B lie on the line y = -x + 4. Choose coordinates for points A, B, and C where point C is the same distance from point A as it is from point B. Write equations for the lines connecting points A and C and points B and C.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-28.jpg

Question 29.
ABSTRACT REASONING If x and y have a positive correlation, and y and z have a negative correlation, then what can you conclude about the correlation between x and z? Explain.
Answer:
As x value increases, y increases the value of z also decreases.
The correlation between x and z is negative.

Question 30.
MATHEMATICAL CONNECTIONS Which equation has a graph that is a line passing through the point (8, -5) and is perpendicular to the graph of y = -4x + 1?
A. y = \(\frac{1}{4}\)x – 5
B. y = -4x + 27
C. y = –\(\frac{1}{4}\)x – 7
D. y = \(\frac{1}{4}\)x – 7
Answer:

Question 31.
PROBLEM SOLVING You are participating in an orienteering competition. The diagram shows the position of a river that cuts through the woods. You are currently 2 miles east and 1 mile north of your starting point, the origin. What is the shortest distance you must travel to reach the river?
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 81
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 31.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 31.2

Question 32.
ANALYZING RELATIONSHIPS Data from North American countries show a positive correlation between the number of personal computers per capita and the average life expectancy in the country.
a. Does a positive correlation make sense in this situation? Explain.
Answer:
It makes sense. There is a positive correlation between the average life expectancy and the country’s level of advancement and also between a country’s level of advancement and the number of personal computers per person.
Example: When compared to the USA, a lot of African countries don’t have nearly as many personal computers per person.

b. Is it reasonable to conclude that giving residents of a country personal computers will lengthen their lives? Explain.
Big Ideas Math Answers Algebra 2 Chapter 1 Linear Functions 82
Answer:
No, the number of personal computers per capita is only an indicator that a country is more advanced, not a way to prolong someone’s life expectancy.

Maintaining Mathematical Proficiency

Solve the system of linear equations in two variables by elimination or substitution. (Skills Review Handbook)

Question 33.
3x + y = 7
-2x – y = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 33

Question 34.
4x + 3y = 2
2x – 3y = 1
Answer:

Question 35.
2x + 2y = 3
x = 4y – 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 35

Question 36.
y = 1 + x
2x + y = -2
Answer:

Question 37.
\(\frac{1}{2}\)x + 4y = 4
2x – y = 1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 37.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.3 Question 37.2

Question 38.
y = x – 4
4x + y = 26
Answer:

Lesson 1.4 Solving Linear Systems

Essential Question
How can you determine the number of solutions of a linear system?
Answer:
A linear system is said to be consistent when it has at least one solution. A linear system is said to be inconsistent when it has no solution.

EXPLORATION 1
Recognizing Graphs of Linear Systems
Work with a partner.
Match each linear system with its corresponding graph. Explain your reasoning. Then classify the system as consistent or inconsistent.
a. 2x – 3y = 3
-4x + 6y = 6
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 83

Answer:
2x – 3y = 3 —- × 2 ⇒ 4x – 6y = 6
-4x + 6y = 6

4x – 6y = 6
-4x + 6y = 6
0 ≠ 6
It has no solution.
So, the linear system is inconsistent.

b. 2x – 3y = 3
x + 2y = 5
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 84

Answer:
Given equations are
2x – 3y = 3
x + 2y = 5 —–eq.2 × 2
2x + 4y = 10

Solve 1 & 2
2x – 3y = 3
(-)2x + 4y = 10
-7y = -7
y = 1
x + 2 = 5
x = 5 – 2
x = 3
So, x = 3 and y = 1
It has two solutions.
So, the system is consistent

c. 2x – 3y = 3
-4x + 6y = 6
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 85

Answer:
Given equations
2x – 3y = 3
-4x + 6y = 6

2x – 3y = 3 —- × 2
-4x + 6y = 6

Solving 1 & 2,
4x – 6y = 6
-4x + 6y = 6
0
So, the linear system is inconsistent.

EXPLORATION 2
Solving Systems of Linear Equations
Work with a partner.
Solve each linear system by substitution or elimination. Then use the graph of the system below to check your solution.
a. 2x + y = 5
x – y = 1
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 86
Answer:
Given equations
2x + y = 5
x – y = 1
Solving Eq. 1 & 2,
2x + y = 5
x – y = 1
3x = 6
x = 6/3
x = 2
Substitute the value of x in the eq. (2)
2 – y = 1
2 – 1 = y
y = 1
The linear system has one solution.

b. x+ 3y = 1
-x + 2y = 4
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 87
Answer:
Given equations are x+ 3y = 1
-x + 2y = 4

Solving 1 & 2,
x+ 3y = 1
-x + 2y = 4
5y = 5
y = 1
x + 3 = 1
x = 1 – 3
x = -2
The linear system has one solution.

c. x + y = 0
3x + 2y = 1
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 88

Answer:
Given equation,
Multiply eq. 1 by 2 to solving the equation
x + y = 0  — × 2 = 2x + 2y = 0
3x + 2y = 1

Solving 1 & 2
2x + 2y = 0
(-)3x + 2y = 1
-x = -1
x = 1
The linear system has one solution.

Communicate Your Answer

Question 3.
How can you determine the number of solutions of a linear system?
Answer:
A linear equation in two variables is an equation of the form ax + by + c = 0 where a, b, c ∈ R, a, and b ≠ 0. A system of linear equations usually has a single solution, but sometimes it can have no solution or an infinite solution. In the system of linear equations, we can find the number of solutions by comparing the coefficients of the variables of the given linear equations.

Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 89
Question 4.
Suppose you were given a system of three linear equations in three variables. Explain how you would approach solving such a system.
Answer:

  1. Solve one equation for one of its variables.
  2. Substitute the expression from point 1 in the other two equations to obtain a linear system in two variables.
  3. Solve the new linear system for both of its variables.
  4. Substitute the values found in point 3 into one of the original equations and solve for the remaining variable.

Question 5.
Apply your strategy in Question 4 to solve the linear system.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 90
Answer:
Given three equations
x + y + z = 1 — eq.1
x – y – z = 3 — eq. 2
-x – y – z = -1 —- eq. 3
Solving 1 & 2
x + y + z = 1
x – y – z = 3
2x = 4
x = 4/2
x = 2
Solving 1 & 3
x – y – z = 1
-x – y – z = -1
-2y -2z = 0
y + z = 0
y = -z

1.4 Lesson

Monitoring Progress

Question 1.
x – 2y + z = -11
3x + 2y – z = 7
-x + 2y + 4z = -9
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-ideas-math-Algebra-2-chapter-10-probability-exercise-10.2-Answer-no-8.jpghttps://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-Exercise-1.4-1.jpg

Question 2.
x + y – z = -1
4x + 4y – 4z = -2
3x + 2y + z = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-Exercise-1.4-2.jpg

Question 3.
x + y + z = 8
x – y + z = 8
2x + y + 2z = 16
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-Exercise-1.4-3.jpg

Question 4.
In Example 3, describe the solutions of the system using an ordered triple in terms of y.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-Exercise-1.4-4.jpg

Question 5.
WHAT IF? On the first day, 10,000 tickets sold, generating $356,000 in revenue. The number of seats sold in Sections A and B are the same. How many lawn seats are still available?
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Monitoring-progress-Exercise-1.4-5.jpg

Solving Linear Systems 1.4 Exercises

Vocabulary and Core Concept Check

Question 1.
VOCABULARY The solution of a system of three linear equations is expressed as a(n)__________.
Answer:
The solution of a system of three linear equations is expressed as an ordered triple.

Question 2.
WRITING Explain how you know when a linear system in three variables has infinitely many solutions.
Answer:
The system has infinitely many solutions when you have an identity such as 0 = 0.

Monitoring Progress and Modeling with Mathematics

In Exercises 3–8, solve the system using the elimination method.

Question 3.
x + y – 2z = 5
-x + 2y + z = 2
2x + 3y – z = 9
Answer:
Given equations are
x + y – 2z = 5 — eq. 1
-x + 2y + z = 2 — eq. 2
2x + 3y – z = 9 — eq. 3
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 3.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 3.2

Question 4.
x + 4y – 6z = -1
2x – y + 2z = -7
-x + 2y – 4z = 5
Answer:
Given equations are
x + 4y – 6z = -1— (eq.1)
2x – y + 2z = -7 —- (eq. 2)
-x + 2y – 4z = 5 —-(eq. 3)
Solving eq.1 & eq.3
x + 4y – 6z = -1
-x + 2y – 4z = 5
6y – 10z = 4
3y – 5z = 2 — (eq. 4)
Solving (1) & (2)
x + 4y – 6z = -1 — × 2 ⇒ 2x + 6y – 12z = -2
2x – y + 2z = -7

2x + 6y – 12z = -2
2x – y + 2z = -7
–    +    –        +
7y – 14z = 5 —- (eq. 5)
Solving (2) & (3)
2x – y + 2z = -7 —- (2)
-x + 2y – 4z = 5 —-(3)—–×2 ⇒ -2x + 4y – 8z = 10

2x – y + 2z = -7
-2x + 4y – 8z = 10
3y – 5z = 3 — (eq. 6)
Solving 5 & 6
7y – 14z = 5 — × 3 ⇒ 21y – 42z = 15 — (eq. 7)
3y – 5z = 3 — × 7 ⇒ 21y – 35z = 21 — (eq. 8)

21y – 42z = 15
21y – 35z = 21
-7z = -6
z = 6/7 or 0.85

7y – 14z = 5
7y – 14(6/7) = 5
7y – 14(6) = 5
7y – 84 = 5
7y = 5 + 84
7y = 89
y = 89/7
y = 12.7

-x + 2y – 4z = 5
-x + 2(12.7) – 4(0.85) = 5
-x + 25.4 – 3.4 = 5
-x + 22 = 5
-x = 5 – 22
-x = -17
x = 17
The solution is x = 17, y = 12.7 and z = 0.85

Question 5.
2x + y – z = 9
-x + 6y + 2z = -17
5x + 7y + z = 4
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 5.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 5.2

Question 6.
3x + 2y – z = 8
-3x + 4y + 5z = -14
x – 3y + 4z = -14

Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-6.jpg

Question 7.
2x + 2y + 5z = -1
2x – y + z = 2
2x + 4y – 3z = 14
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 7.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 7.2

Question 8.
3x + 2y – 3z = -2
7x – 2y + 5z = -14
2x + 4y + z = 6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.3-8-1.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-8a.jpg

ERROR ANALYSIS In Exercises 9 and 10, describe and correct the error in the first step of solving the system of linear equations.

Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 90.1

Question 9.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 91
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 9

Question 10.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 92
Answer:

In Exercises 11–16, solve the system using the elimination method.

Question 11.
3x – y + 2z = 4
6x – 2y + 4z = -8
2x – y + 3z = 10
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 11

Question 12.
5x + y – z = 6
x + y + z = 2
12x + 4y = 10
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-12.jpg

Question 13.
x + 3y – z = 2
x + y – z = 0
3x + 2y – 3z = -1
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 13.1

Question 14.
x + 2y – z = 3
-2x – y + z = -1
6x – 3y – z = -7
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-14.jpg

Question 15.
x + 2y + 3z = 4
-3x + 2y – z = 12
-2x – 2y – 4z = -14
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 15.1

Question 16.
-2x – 3y + z = -6
x + y – z = 5
7x + 8y – 6z = 31
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-16.jpg

Question 17.
MODELING WITH MATHEMATICS Three orders are placed at a pizza shop. Two small pizzas, a liter of soda, and a salad cost $14; one small pizza, a liter of soda, and three salads cost $15; and three small pizzas, a liter of soda, and two salads cost $22. How much does each item cost?
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 93
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 17.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 17.2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 17.3

Question 18.
MODELING WITH MATHEMATICS Sam’s Furniture Store places the following advertisement in the local newspaper. Write a system of equations for the three combinations of furniture. What is the price of each piece of furniture? Explain.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 94
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-18.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-18a.jpg

In Exercises 19–28, solve the system of linear equations using the substitution method.

Question 19.
-2x + y + 6z = 1
3x + 2y + 5z = 16
7x + 3y – 4z = 11
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 19.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 19.2

Question 20.
x – 6y – 2z = -8
-x + 5y + 3z = 2
3x – 2y – 4z = 18
Answer:

Question 21.
x + y + z = 4
5x + 5y + 5z = 12
x – 4y + z = 9
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 21.1

Question 22.
x + 2y = -1
-x + 3y + 2z = -4
-x + y – 4z = 10
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-22.jpg

Question 23.
2x – 3y + z = 10
y + 2z = 13
z = 5
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 23.1

Question 24.
x = 4
x + y = -6
4x – 3y + 2z = 26
Answer:

Question 25.
x + y – z = 4
3x + 2y + 4z = 17
-x + 5y + z = 8
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 25.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 25.2

Question 26.
2x – y – z = 15
4x + 5y + 2z = 10
-x – 4y + 3z = -20
Answer:

Question 27.
4x + y + 5z = 5
8x + 2y + 10z = 10
x – y – 2z = -2
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 27.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 27.2

Question 28.
x + 2y – z = 3
2x + 4y – 2z = 6
-x – 2y + z = -6
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-28.jpg

Question 29.
PROBLEM SOLVING The number of left-handed people in the world is one-tenth the number of right-handed people. The percent of right-handed people is nine times the percent of left-handed people and ambidextrous people combined. What percent of people are ambidextrous?
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 95
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 29.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 29.2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 29.3

Question 30.
MODELING WITH MATHEMATICS Use a system of linear equations to model the data in the following newspaper article. Solve the system to find how many athletes finished in each place.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 96
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-30.jpg

Question 31.
WRITING Explain when it might be more convenient to use the elimination method than the substitution method to solve a linear system. Give an example to support your claim.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 31

Question 32.
REPEATED REASONING Using what you know about solving linear systems in two and three variables, plan a strategy for how you would solve a system that has four linear equations in four variables.
Answer:
First, you have to eliminate one variable by using the substitution method. Then it becomes three linear equations in three variables, then you will know how to solve it.

MATHEMATICAL CONNECTIONS In Exercises 33 and 34, write and use a linear system to answer the question.

Question 33.
The triangle has a perimeter of 65 feet. What are the lengths of sides ℓ, m, and n?
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 97
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 33.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 33.2

Question 34.
What are the measures of angles A, B, and C?
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 97.1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-34.jpg

Question 35.
OPEN-ENDED Consider the system of linear equations below. Choose nonzero values for a, b, and c so the system satisfies the given condition. Explain your reasoning.
x + y + z = 2
ax + by + cz = 10
x – 2y + z = 4
a. The system has no solution.
b. The system has exactly one solution.
c. The system has infinitely many solutions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 35

Question 36.
MAKING AN ARGUMENT A linear system in three variables has no solution. Your friend concludes that it is not possible for two of the three equations to have any points in common. Is your friend correct? Explain your reasoning.
Answer:

Question 37.
PROBLEM SOLVING A contractor is hired to build an apartment complex. Each 840-square-foot unit has a bedroom, kitchen, and bathroom. The bedroom will be the same size as the kitchen. The owner orders 980 square feet of tile to completely cover the floors of two kitchens and two bathrooms. Determine how many square feet of carpet is needed for each bedroom.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 98
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 37.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 37.2

Question 38.
THOUGHT PROVOKING Does the system of linear equations have more than one solution? Justify your answer.
4x + y + z = 0
2x + \(\frac{1}{2}\)y – 3z = 0
-x – \(\frac{1}{4}\)y – z = 0
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-Exercise-1.4-38.jpg

Question 39.
PROBLEM SOLVING A florist must make 5 identical bridesmaid bouquets for a wedding. The budget is $160, and each bouquet must have 12 flowers. Roses cost $2.50 each, lilies cost $4 each, and irises cost $2 each. The florist wants twice as many roses as the other two types of flowers combined.
a. Write a system of equations to represent this situation, assuming the florist plans to use the maximum budget.
b. Solve the system to find how many of each type of flower should be in each bouquet.
c. Suppose there is no limitation on the total cost of the bouquets. Does the problem still have exactly one solution? If so, find the solution. If not, give three possible solutions.
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 39.1
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 39.2
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 39.3

Question 40.
HOW DO YOU SEE IT? Determine whether the system of equations that represents the circles has no solution, one solution, or infinitely many solutions. Explain your reasoning.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 99
Answer:
a. The given three circles in the graph has one common point. So, the system of equations has one solution.
b. Inside the small circle all the points are common. So, it has an infinite number of solutions.

Question 41.
CRITICAL THINKING Find the values of a, b, and c so that the linear system shown has (-1, 2, -3) as its only solution. Explain your reasoning.
x + 2y – 3z = a
– x – y + z = b
2x + 3y – 2z = c
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 41

Question 42.
ANALYZING RELATIONSHIPS Determine which arrangement(s) of the integers -5, 2, and 3 produce a solution of the linear system that consist of only integers. Justify your answer.
x – 3y + 6z = 21
_x + _y + _z = -30
2x – 5y + 2z = -6
Answer:

Question 43.
ABSTRACT REASONING Write a linear system to represent the first three pictures below. Use the system to determine how many tangerines are required to balance the apple in the fourth picture. Note:The first picture shows that one tangerine and one apple balance one grapefruit.
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 100
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 43

Maintaining Mathematical Proficiency

Simplify. (Skills Review Handbook)

Question 44.
(x – 2)2

Answer:

Question 45.
(3m + 1)2

Answer:
(3m + 1)2
It is in the form of (a + b)² = a² + b² + 2ab
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 45

Question 46.
(2z – 5)2

Answer:

Question 47.
(4 – y)2

Answer:

Write a function g described by the given transformation of f(x) =∣x∣− 5.(Section 1.2)

Question 48.
translation 2 units to the left
Answer:

Question 49.
reflection in the x-axis
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 49

Question 50.
translation 4 units up
Answer:

Question 51.
vertical stretch by a factor of 3
Answer:
Big Ideas Math Algebra 2 Answers Chapter 1 Linear Functions 1.4 Question 51

Linear Functions Performance Task: Secret of the Hanging Baskets

1.3–1.4 What Did You Learn?

Core Vocabulary
Big Ideas Math Algebra 2 Answer Key Chapter 1 Linear Functions 101

Core Concepts
Section 1.3
Writing an Equation of a Line, p. 22
Finding a Line of Fit, p. 24
Section 1.4
Solving a Three-Variable System, p. 31
Solving Real-Life Problems, p. 33

Mathematical Practices

Question 1.
Describe how you can write the equation of the line in Exercise 7 on page 26 using only one of the labeled points.

Question 2.
How did you use the information in the newspaper article in Exercise 30 on page 35 to write a system of three linear equations?

Question 3.
Explain the strategy you used to choose the values for a, b, and c in Exercise 35 part (a) on page 35.

Performance Task

Secret of the Hanging Baskets
A carnival game uses two baskets hanging from springs at different heights. Next to the higher basket is a pile of baseballs. Next to the lower basket is a pile of golf balls. The object of the game is to add the same number of balls to each basket so that the baskets have the same height. But there is a catch—you only get one chance. What is the secret to winning the game?
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 102
To explore the answers to this question and more, go to BigIdeasMath.com.
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 103

Linear Functions Chapter Review

Graph the function and its parent function. Then describe the transformation.

Question 1.
f(x) = x + 3
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-1.jpg

Question 2.
g(x) = | x | – 1
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-2.jpg

Question 3.
h(x) = \(\frac{1}{2}\)x2

Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-3.jpg

Question 4.
h(x) = 4
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-4.jpg

Question 5.
f(x) = -| x | – 3
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-5.jpg

Question 6.
g(x) = -3(x + 3)2

Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-6.jpg

Write a function g whose graph represents the indicated transformations of the graph of f. Use a graphing calculator to check your answer.

Question 7.
f(x) = | x |; reflection in the x-axis followed by a translation 4 units to the left
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-7.jpg

Question 8.
f(x) = | x | ; vertical shrink by a factor of \(\frac{1}{2}\) followed by a translation 2 units up
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-8.jpg

Question 9.
f(x) = x; translation 3 units down followed by a reflection in the y-axis

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-9.jpg

Question 10.
The table shows the total number y (in billions) of U.S. movie admissions each year for x years. Use a graphing calculator to find an equation of the line of best fit for the data.
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 103.1

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-10.jpg

Question 11.
You ride your bike and measure how far you travel. After 10 minutes, you travel 3.5 miles. After 30 minutes, you travel 10.5 miles. Write an equation to model your distance. How far can you ride your bike in 45 minutes?
Answer:
Given,
(x1, y1) = (10, 3.5)
(x2, y2) = (30, 10.5)
m = (10.5 – 3.5)/30 – 10
m = 7/20
y – 3.5 = 7/20(x – 10)
y – 3.5 = 7/20 x – 3.5
y = 7/20x
x = 45
y = 7/20 (45)
y = 15.75 miles
Therefore you can 15.75 miles in 45 minutes.

Question 12.
x + y + z = 3
-x + 3y + 2z = -8
x = 4z
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-12.jpg

Question 13.
2x – 5y – z = 17
x + y + 3z = 19
-4x + 6y + z = -20
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-13.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-13a.jpg

Question 14.
x + y + z = 2
2x – 3y + z = 11
-3x + 2y – 2z = -13
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-14.jpg

Question 15.
x + 4y – 2z = 3
x + 3y + 7z = 1
2x + 9y – 13z = 2
Answer:

Question 16.
x – y + 3z = 6
x – 2y = 5
2x – 2y + 5z = 9
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-16.jpg

Question 17.
x + 2y = 4
x + y + z = 6
3x + 3y + 4z = 28
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-17.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-17.jpg

Question 18.
A school band performs a spring concert for a crowd of 600 people. The revenue for the concert is $3150. There are 150 more adults at the concert than students. How many of each type of ticket are sold?
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 104
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-review-18.jpg

Linear Functions Chapter Test

Write an equation of the line and interpret the slope and y-intercept.

Question 1.
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 105
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-1.jpg

Question 2.
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 106
Answer:

Solve the system. Check your solution, if possible.

Question 3.
-2x + y + 4z = 5
x + 3y – z = 2
4x + y – 6z = 11
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-3.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-3a.jpg

Question 4.
y = \(\frac{1}{2}\)z
x + 2y + 5z = 2
3x + 6y – 3z = 9
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-4.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-4a.jpg
Question 5.
x – y + 5z = 3
2x + 3y – z = 2
-4x – y – 9z = -8
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-5.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-5a.jpg

Graph the function and its parent function. Then describe the transformation.

Question 6.
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-6.jpg

Question 8.
f(x) = 4
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-8.jpg

Match the transformation of f(x) = x with its graph. Then write a rule for g.

Question 9.
g(x) = 2f(x) + 3
Big Ideas Math Algebra 2 Solutions Chapter 1 Linear Functions 107
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-9.jpg

Question 10.
g(x) = 3f(x) – 2
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 108

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-.10JPG.jpg

Question 11.
g(x) = -2f(x) – 3
Answer:
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 109

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-11.jpg

Question 12.
A bakery sells doughnuts, muffins, and bagels. The bakery makes three times as many doughnuts as bagels. The bakery earns a total of $150 when all 130 baked items in stock are sold. How many of each item are in stock? Justify your answer.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 110
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-12.jpg

Question 13.
A fountain with a depth of 5 feet is drained and then refilled. The water level (in feet) after t minutes can be modeled by f(t) = \(\frac{1}{4}\)|t – 20 |. A second fountain with the same depth is drained and filled twice as quickly as the first fountain. Describe how to transform the graph of f to model the water level in the second fountain after t minutes. Find the depth of each fountain after 4 minutes. Justify your answers.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-chapter-test-13.jpg

Linear Functions Cumulative Assessment

Question 1.
Describe the transformation of the graph of f(x) = 2x – 4 represented in each graph.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 111
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-1.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-1a.jpg
Question 2.
The table shows the tuition costs for a private school between the years 2010 and 2013.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 112
a. Verify that the data show a linear relationship. Then write an equation of a line of fit.
b. Interpret the slope and y-intercept in this situation.
c. Predict the cost of tuition in 2015.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-2.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functionhttps://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-2a.jpgs-assessments-2a.jpg

Question 3.
Your friend claims the line of best fit for the data shown in the scatter plot has a correlation coefficient close to 1. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 113
Answer:
Use the graphing calculator to find the equation.
From the graph, the correlation coefficient is r = -0.86
So, my friend is not correct, since the correlation coefficient is close to -1.

Question 4.
Order the following linear systems from least to greatest according to the number of solutions.
A. 2x + 4y – z = 7
14x + 28y – 7z = 49
-x + 6y + 12z = 13
B. 3x – 3y + 3z = 5
-x + y – z = 5
-x + y – z = 8
14x – 3y + 12z = 108
C. 4x – y + 2z = 18
-x + 2y + z = 11
3x + 3y – 4z = 44
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-4.jpg
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-4a.jpg

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-4b.jpg

Question 5.
You make a DVD of three types of shows: comedy, drama, and reality-based. An episode of a comedy lasts 30 minutes, while a drama and a reality-based episode each last 60 minutes. The DVDs can hold 360 minutes of programming.
a. You completely fill a DVD with seven episodes and include twice as many episodes of a drama as a comedy. Create a system of equations that models the situation.
b. How many episodes of each type of show are on the DVD in part (a)?
c. You completely fill a second DVD with only six episodes. Do the two DVDs have a different number of comedies? dramas? reality-based episodes? Explain.
Answer:
https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-5.jpg

Question 6.
The graph shows the height of a hang glider over time. Which equation models the situation?
A. y + 450 = 10x
B. 10y = -x+ 450
C. \(\frac{1}{10}\)y = -x + 450
D. 10x + y = 450
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 114
Answer:

Question 7.
Let f(x) = x and g(x) = -3x – 4. Select the possible transformations (in order) of the graph of f represented by the function g.
A. reflection in the x-axis
B. reflection in the y-axis
C. vertical translation 4 units down
D. horizontal translation 4 units right
E. horizontal shrink by a factor of \(\frac{1}{3}\)
F. vertical stretch by a factor of 3
Answer:

Question 8.
Choose the correct equality or inequality symbol which completes the statement below about the linear functions f and g. Explain your reasoning.
Big Ideas Math Answer Key Algebra 2 Chapter 1 Linear Functions 115
Answer:

https://ccssanswers.com/wp-content/uploads/2021/02/Big-idea-math-algerbra-2-chapter-1-linear-functions-assessments-8.jpg

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles

The students of middle school can get the Solution Key for Big Ideas Math Grade 8 Chapter 3 Angles and Triangles on this page. With the help of this Big Ideas Math Book 8th Grade Answer Key Chapter 3 Angles and Triangles you can finish your homework in time and also improve your performance in the exams. Get free step by step solutions for all the questions in Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles.

Big Ideas Math Book 8th Grade Answer Key Chapter 3 Angles and Triangles

Download Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles pdf for free of cost. The solutions for each and every question is prepared in an easy manner. Go through the table of contents shown in the below section to know the topics covered in Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles.

Performance

Lesson: 1 Parallel Lines and Transversals

Lesson: 2 Angles of Triangles

Lesson: 3 Angles of Polygons

Lesson: 4 Using Similar Triangles

Chapter 3 – Angles and Triangles

Angles and Triangles STEAM Video/Performance

STEAM Video

Each cell in a honeycomb is in the shape of a regular hexagon. Why might bees use this shape?
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 1

Watch the STEAM Video “Honeycombs.” Then answer the following questions.

Question 1.
Enid and Tony show regular tilings made out of squares, equilateral triangles, and regular hexagons. What is the sum of the interior angle measures of the tiling made from equilateral triangles, outlined below in yellow?
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 2

Answer:
The sum of interior angles of the equilateral triangle = 180°
x + x + x = 180°
3x° = 180°
x = 180/3
x° = 60°

Question 2.
The cells in a honeycomb use a tiling pattern of the regular hexagon shown. A cell is 10 millimeters deep. About how much honey can one cell hold? Explain.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 3

Performance Task

Turtle Shells

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given angle measures of shapes seen on a turtle shell.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 4
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 5
You will be asked to find angle sums and missing angle measures. What other animals have features that resemble geometric shapes?

Angles and Triangles Getting Ready for Chapter 3

Chapter Exploration

When an object is transverse, it is lying or extending across something. In the drawing, the fallen tree lying across the railroad track is transverse to the track.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 6
1. Work with a partner.
• Discuss what it means for two lines to be parallel. Decide on a strategy for drawing parallel lines. Then draw two parallel lines.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 7
• Draw a third line that intersects the parallel lines. This line is called a transversal
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 8
a. How many angles are formed by the parallel lines and the transversal? Label each angle.

Answer: 8 angles are formed by the parallel lines and the transversal
b. Which of these angles have equal measures? Explain your reasoning.

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what the terms might mean and record your thoughts.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 9

Lesson 3.1 Parallel Lines and Transversals

EXPLORATION 1

Work with a partner. Use geometry software and the lines A and B shown.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 10
a. Are line A and B line parallel? Explain your reasoning.
b. Draw a line that intersects both line A and line B. What do you notice about the measures of the angles that are created?
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 11
c. Rotate line A or line B until the angles created by the intersection of line A and line C are congruent to the angles created by the intersection of line B and line C. What do you notice about line A and line B?
d. Rotate line to create different angle measures. Are the angles that were congruent in part(c) still congruent?
e. Make a conjecture about the measures of the angles created when a line intersects two parallel lines.

3.1 Lesson

Try It

Use the figure to find the measure of the angle. Explain your reasoning

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 12

Question 1.
∠1

Answer: 63°

Explanation:
Big Ideas Math Grade 8 Answers Chapter 3 img_1
∠1 and 63° angle are corresponding angles are formed by transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠1 is 63°

Question 2.
∠2

Answer: 117°

Explanation:
BIM Grade 8 Chapter 3 Angles and triangles answer key img_2
∠1 and ∠2 are supplementary angle.
∠1 + ∠2 = 180°
63° + ∠2 = 180°
∠2 = 180° – 63°
∠2 = 117°
So, the measure of ∠2 = 117°

Try It

Question 3.
Use the figureto find the measures of the numbered angles.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 13

Answer:
∠1 and 59° are the supplementary angles
∠1 + 59° = 180°
∠1 = 180° – 59°
∠1 = 121°
∠2 and 59° are vertical angles. They are congruent.
So, the measure of ∠1 is 121°
∠3 and 59° are supplementary angles.
∠3 + 59° = 180°
∠3 = 180° – 59°
∠3 = 121°
∠4, ∠5, ∠6, ∠7 corresponding angles are congruent because they are formed by a transversal intersecting parallel side.
the measure of  ∠4 is 121°
the measure of ∠5 is 59°
the measure of  ∠6 is 121°
the measure of ∠7 is 59°

Try It

In Example 3, the measure of ∠4 is 84°. Find the measure of the angle. Explain your reasoning.

Question 4.
∠3

Answer:
Bigideas Math Answers Grade 8 Chapter 3 Angles and Triangles img_3
The measure of ∠4 is 84°.
∠3 and ∠4 are supplementary angles.
∠3 + ∠4 = 180°
∠3 + 84° = 180°
∠3 = 180° – 84°
∠3 = 96°

Question 5.
∠5

Answer:
∠4 and ∠5 are alternate interior angles formed by transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠5 is 84°

Question 6.
∠6

Answer:
∠3 and ∠6 are alternate exterior angles formed by transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠6 is 96°

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

FINDING ANGLE MEASURES
Use the figure to find the measures of the numbered angles.

Question 7.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 14

Answer:
∠1 and 120° are the supplementary angles.
∠1 + 120° = 180°
∠1 = 180 – 120
∠1 = 60°
Thus the measure of ∠1 is 60°
∠2 and 120° are the vertical angles. They are congruent.
Thus the measure of ∠2 is 120°
∠3 and 120° are the supplementary angles.
∠3 + 120° = 180°
∠3 = 180 – 120
∠3 = 60°
∠4, ∠5, ∠6, ∠7 are corresponding angles are formed by transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠4 is 60°
Thus the measure of ∠5 is 120°
Thus the measure of ∠6 is 120°
Thus the measure of ∠7 is 60°

Question 8.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 15

Answer:
∠1 and 35° are the supplementary angles.
∠1 + 35° = 180°
∠1 = 180 – 35
∠1 = 145°
Thus the measure of ∠1 is 145°
∠2 and 35° are the supplementary angles.
∠2 + 35° = 180°
∠2 = 180 – 35
∠2 = 145°
Thus the measure of ∠2 is 145°
∠3 and 35° are the vertical angles. They are congruent.
Thus the measure of ∠3 is 35°
∠4, ∠5, ∠6, ∠7 are corresponding angles are formed by transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠4 is 35°
Thus the measure of ∠5 is 145°
Thus the measure of ∠6 is 145°
Thus the measure of ∠7 is 35°

Question 9.
WHICH ONE DOESN’T BELONG?
Which angle measure does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 16

Answer:
∠2, ∠6 are corresponding angles are formed by transversal intersecting parallel lines.
∠6, ∠8 are vertical angles are formed by transversal intersecting parallel lines.
∠5 does not belong to the other three because all the other three measure are equal.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 10.
A cross section of a pier is shown. Find the value of a. Justify your answer.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 17

Answer:
The angle a and the angle of 70 degrees are complementary angles because they belong to a right triangle, where the third angle is the right angle.
∠a + 70 = 90
∠a = 90 – 70
∠a = 20°

Question 11.
The head tube angle of a bike determines how easy the bike is to steer. A bike frame with angle approximations is shown. What is the head tube angle of the bike?
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 18

Answer:
The lines AB and CD are parallel.
ABC and BCD are the corresponding angles formed by transversal intersecting parallel lines.
∠BCD = 55°
∠BAC + ∠ABC + ∠ACB = 180°
The sum of the angles in a triangle is 180°
∠BAC + 55°+ 52° = 180°
∠BAC + 107° = 180°
∠BAC = 180° – 107°
∠BAC = 73°
So, the head tube angle of a bike is 73°

Parallel Lines and Transversals Homework & Practice 3.1

Review & Refresh

Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Question 1.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 19

Answer:
perimeter of red hexagon/perimeter of blue hexagon = \(\frac{3}{5}\)
The values of the ratios of the perimeter is \(\frac{3}{5}\)
Area of red hexagon/Area of blue hexagon = (\(\frac{3}{5}\))² = \(\frac{9}{25}\)
The values of the ratios of the area is \(\frac{9}{25}\)

Question 2.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 20

Answer:
perimeter of red trapezium /perimeter of blue trapezium = \(\frac{7}{6}\)
The values of the ratios of the perimeter is \(\frac{7}{6}\)
Area of red hexagon/Area of blue hexagon = (\(\frac{7}{6}\))² = \(\frac{49}{36}\)
The values of the ratios of the area is \(\frac{49}{36}\)

Evaluate the expression.

Question 3.
4 + 32

Answer:
4 + 9 = 13

Question 4.
5(2)2 – 6

Answer:
5(4) – 6
20 – 6 = 14

Question 5.
11 + (-7)2 – 9

Answer:
11 + 49 – 9
11 + 40 = 50

Concepts, Skills, & Problem Solving
EXPLORING INTERSECTIONS OF LINES
Use a protractor to determine whether lines a and b are parallel. (See Exploration 1, p. 103.)

Question 6.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 21

Answer:
Use a protractor to measure ∠1 and ∠2
∠1 ≈ 60°
∠2 ≈ 60°
∠1 and ∠2, it means the two angles are congruent. The angles are exterior alternate angles.
According to the converse of the exterior alternate angles theorem, the two lines are parallel.
a || b

Question 7.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 22

Answer:
Use a protractor to measure ∠1 and ∠2
∠1 ≈ 50°
∠2 ≈ 60°
∠1 and ∠2, it means the two angles are not congruent. The angles are exterior alternate angles.
According to the converse of the exterior alternate angles theorem, the two lines are not parallel.

FINDING ANGLE MEASURES
Use the figure to find the measures of the numbered angles. Explain your reasoning.

Question 8.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 23

Answer:
∠1 and 107° are corresponding angles. They are congruent.
So, the measure of ∠1 is 107°.
∠1 and ∠2 are supplementary angles.
∠1 + ∠2 = 180°
107° + ∠2 = 180°
∠2 = 180° – 107°
∠2 = 73°
So, the measure of ∠2 is 73°

Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 24

Answer:
∠3 and 95° are corresponding angles. They are congruent.
Thus the measure of ∠3 is 95°
∠3 and ∠4 are supplementary angles.
∠3 + ∠4 = 180°
95° + ∠4 = 180°
∠4 = 180 – 95
∠4 = 85°
So the measure of ∠4 is 85°

Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 25

Answer:
∠5 and 49° are corresponding angles. They are congruent.
So, the measure of ∠5 is 49°
∠5 and ∠6 are supplementary angles.
∠5 + ∠6 = 180°
49° + ∠6 = 180°
∠6 = 180° – 49°
∠6 = 131°
So, the measure of ∠6 is 131°

Question 11.
YOU BE THE TEACHER
Your friend describes a relationship between the angles shown. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 26

Answer:
Since the two lines are not parallel. Hence ∠5 is not congruent to ∠6.
By this, we can say that your friend is not correct.

Question 12.
PROBLEM SOLVING
The painted lines that separate parking spaces are parallel. The measure of ∠1 is 60°. What is the measure of ∠2? Explain.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 27

Answer:
∠1 and ∠2 are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent.
The measure of ∠1 is 60° so the measure of ∠2 is 60°

Question 13.
OPEN-ENDED
Describe two real-life situations that use parallel lines.

Answer:
Example 1: The railroad tracks and the tram tracks are parallel lines.
Example 2:  The shelves of a bookcase.

USING CORRESPONDING ANGLES
Use the figure to find the measures of the numbered angles.

Question 14.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 28

Answer:
∠1 and 60° are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent.
∠1 and ∠2 are supplementary angles.
∠1 + ∠2 = 180°
60° + ∠2 = 180°
∠2 = 180° – 60°
∠2 = 119°
So, the measure of ∠2 is 119°
∠3 and ∠1 are vertical angles. They are congruent.
So, the measure of ∠3 is 61°
∠4 and ∠2 are vertical angles. They are congruent.
∠5, ∠6, ∠7 corresponding angles are congruent because they are formed by a transversal intersecting parallel lines.
So, the measure of ∠5 is 119°
So, the measure of ∠6 is 61°
So, the measure of ∠7 is 119°

Question 15.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 29

Answer:
∠1 and 99° are supplementary angles.
∠1 + 99° = 180°
∠1 = 180° – 99°
∠1 = 81°
Thus the measure of ∠1 is 81°
∠2 and 99° are vertical angles. They are congruent.
The measure of ∠2 is 99°
∠3 and ∠1 are vertical angles. They are congruent.
So, the measure of ∠3 is 81°
∠4, ∠5, ∠6, ∠7 corresponding angles are congruent because they are formed by a transversal intersecting parallel lines.
So, the measure of ∠4 is 99°
So, the measure of ∠5 is 81°
So, the measure of ∠6 is 99°
So, the measure of ∠7 is 81°

Question 16.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 30

Answer:
∠1 and 90° are supplementary angles.
∠1 + 90° = 180°
∠1 = 180° – 90°
∠1 = 90°
Thus the measure of ∠1 is 90°
∠2 and 90° are vertical angles. They are congruent.
Thus the measure of ∠2 is 90°
∠3 and ∠1 are vertical angles. They are congruent.
So, the measure of ∠3 is 90°
∠4, ∠5, ∠6, ∠7 corresponding angles are congruent because they are formed by a transversal intersecting parallel lines.
So, the measure of ∠4 is 90°
So, the measure of ∠5 is 90°
So, the measure of ∠6 is 90°
So, the measure of ∠7 is 90°

USING CORRESPONDING ANGLES
Complete the statement. Explain your reasoning.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 30.1

Question 17.
If the measure of ∠1 = 124°, then the measure of Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 31

Answer:
∠1 and ∠8 are corresponding angles. They are congruent.
The measure of ∠1 = 124°, then the measure of ∠8 is 124°
∠8 and ∠4 are supplementary angles.
∠8 + ∠4 = 180°
124° + ∠4 = 180°
∠4 = 180° – 124°
∠4 = 56°
So, the measure of ∠4 is 56°

Question 18.
If the measure of ∠2 = 48°, then the measure of Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 32

Answer:
∠2 and ∠7 are corresponding angles. They are congruent.
The measure of ∠2 = 48°, then the measure of ∠7 is 48°
∠7 and ∠3 are supplementary angles.
∠7 + ∠3 = 180°
48° + ∠3 = 180°
∠3 = 180° – 48°
∠3 = 132°
Thus the measure of ∠3 = 132°

Question 19.
If the measure of ∠4 = 55°, then the measure of Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 33

Answer:
∠4 and ∠2 are alternate interior angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠2 is 55°

Question 20.
If the measure of ∠6 = 120°, then the measure of Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 34

Answer:
∠6 and ∠8 are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠8 is 120°

Question 21.
If the measure of ∠7 = 50.5°, then the measure of Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 35

Answer:
∠7 and ∠2 are corresponding angles. They are congruent.
The measure of ∠7 is 50.5°, so the measure of ∠2 is 50.5°
∠2 and ∠6 are supplementary angle.
∠2 + ∠6 = 180°
50.5° + ∠6 = 180°
∠6 = 180° – 50.5°
∠6 = 129.5°
So, the measure of ∠6 is 129.5°

Question 22.
If the measure of ∠3 = 118.7°, then the measure of Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 35.1

Answer:
∠3 and ∠6 are corresponding angles. They are congruent.
The measure of ∠3 is 118.7°
So, the measure of ∠6 is 118.7°
∠6 and ∠2 are supplementary angles.
∠6 + ∠2 = 180°
118.7° + ∠2 = 180°
∠2 = 180° – 118.7°
∠2 = 61.3°
So, the measure of ∠2 is 61.3°

Question 23.
MODELING REAL LIFE
A rainbow forms when sunlight reflects of raindrops at different angles. For blue light, the measure ∠2 is 40°. What is the measure of ∠1?
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 31.1

Answer:
∠4 and ∠5 are alternate interior angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠1 is 40°

Question 24.
REASONING
Is there a relationship between exterior angles that lie on the same side of a transversal? interior angles that lie on the same side of a transversal? Explain.

Answer:
∠7 and ∠8 are Exterior angles that lie on the same side of the transversal. Hence ∠7 and ∠8 are supplementary angles.
Big Ideas Math Answers grade 8 Chapter 3 Angles and Triangles img_4
∠3 and ∠4 are interior angles that lie on the same side of the transversal. Hence ∠3 and ∠4 are supplementary angles.

Question 25.
REASONING
When a transversal is perpendicular to two parallel lines, all the angles formed measure 90°. Explain why.

Answer:
When a transversal is perpendicular to two parallel lines, all the angles formed measure 90°.
Big Ideas Math Grade 8 Answers Chapter 3 img
All the angles formed are 90 degrees because perpendicular lines form 90 degree angles.

Question 26.
REASONING
Two horizontal lines are cut by a transversal. What is the least number of angle measures you need to know to find the measure of every angle? Explain your reasoning.

Answer:
BIM Grade 8 Chapter 3 Angles and triangles answers img_6
In the below example two horizontal lines a and b are cut by a transversal line t, we know only one angle 61 the measure of all the angles are find with the help of one known measure, 61°
∠1 and 61° are corresponding angles formed by a transversal intersecting parallel lines. the angles are congruent.
Thus the measure of ∠1 is 61°
∠1 and ∠2 are supplementary angles.
∠1 + ∠2 = 180°
61° + ∠2 = 180°
∠2 = 180° – 61°
∠2 = 119°
Thus the measure of ∠2 is 119°
∠3 and ∠1 are vertical angles. They are congruent.
Thus the measure of ∠3 is 61°
∠4 and ∠2 are vertical angles. They are congruent.
Thus the measure of ∠4 is 119°
∠5, ∠6, ∠7 corresponding angles are formed by transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠5 is 119°
Thus the measure of ∠6 is 61°
Thus the measure of ∠7 is 119°

Question 27.
LOGIC
Describe two ways you can show that ∠1 is congruent to ∠7.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 32.1

Answer:
∠1 and ∠7 are alternate exterior angles formed by a transeversal intersecting parallel lines. So, ∠1 and ∠7 are congruent.
∠1 and ∠5 are corresponding angles formed by a transeversal intersecting parallel lines. So, ∠1 and ∠5 are congruent.
∠5 and ∠7 are vertical angles so they are congruent.
Hence ∠1 and ∠7 are congruent.

FINDING A VALUE
Find the value of x.

Question 28.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 33.1

Answer:
∠1 and 50° are alternate interior angles. They are congruent.
So, the measure of ∠1 is 50°
∠2 and ∠1 are corresponding angles. They are congruent.
So, the measure of ∠2 is 50°
∠2 and x are supplementary angle.
∠2 + x = 180°
50° + x = 180°
x = 180° – 50°
x = 130°
So, the measure of x is 130°

Question 29.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 34.1

Answer:
∠1 and 115° are corresponding angles. They are congruent.
So, the measure of ∠1 is 115°
∠1 and x are alternate exterior angles. They are congruent.
So, the measure of x is 115°

Question 30.
PROJECT
Trace line p and line t on a piece of paper. Label ∠1. Move the paper so that ∠1 aligns with ∠8. Describe the transformations that you used to show that ∠1 is congruent to ∠8.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 35.2

Answer: 180° rotation, translation about line t.

Question 31.
OPEN-ENDED
Refer to the figure.
a. Do the horizontal lines appear to be parallel? Explain.

Answer: The three horizontal lines seem to spread apart, even though in reality they are parallel.

b. Draw your own optical illusion using parallel lines.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 36

Answer:
Big Ideas Math Solution Key Grade 8 Chapter 3 img_7

Question 32.
DIG DEEPER!
The figure shows the angles used to make a shot on an air hockey table.
a. Find the value of x.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 37

Answer:
As the lines AB and CD are parallel and ∠BCD are alternate interior angles transversal BC, they are congruent.
∠ABC ≅ ∠BCD
x = 64
b. How does the angle the puck hits the edge of the table relate to the angle it leaves the edge of the table?

Answer:
m∠MBA + m∠ABC + m∠CBN = 180°
58 ° + 64° + m∠CBN = 180°
122 ° + m∠CBN = 180°
m∠CBN = 180° – 122 °
m∠CBN = 58°

Lesson 3.2 Angles of Triangles

EXPLORATION 1

Exploring Interior and Exterior Angles of Triangles
Work with a partner.
a. Draw several triangles using geometry software. What can you conclude about the sums of the angle measures?
b. You can extend one side of a triangle to form an exterior angle as shown.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 38
Use geometry software to draw a triangle and an exterior angle. Compare the measure of the exterior angle with the measures of the interior angles. Repeat this process for several different triangles. What can you conclude?

Answer:

EXPLORATION 2
Using Parallel Lines and Transversals
Work with a partner. Describe what is shown in the figure below. Then use what you know about parallel lines and transversals to justify your conclusions in Exploration 1.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 39

Answer:

3.2 Lesson

Try It

Find the measures of the interior angles of the triangle.

Question 1.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 40

Answer: 81°, 25°, 74°

Explanation:
Sum of all the angles in a triangle = 180°
x° + 81° + 25° = 180°
x° = 180° – 81° – 25°
x = 74°
Thus the measure of the interior angle is 74°

Question 2.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 41

Answer: 43°, 51°, 86°

Explanation:
Sum of all the angles in a triangle = 180°
x° + (x – 35)° + 43° = 180°
x° + x° – 35° + 43° = 180°
2x° + 8° = 180°
2x° = 180° – 8°
2x° = 172°
x° = 172°/2
x° = 86°
The measure of the interior angle of the triangle
(x – 35)° = 86 – 35
(x – 35)° = 51°
x° = 51° + 35°
x° = 86°

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
VOCABULARY
How many exterior angles does a triangle have at each vertex? Explain.

Answer:
BIM 8th Grade Answers Chapter 3 img_8
At each vertex of a triangle, there are two exterior angles, which are congruent.

FINDING ANGLE MEASURES
Find the value of x.

Question 5.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 42

Answer:
Sum of all the angles in a triangle = 180°
x° + 25° + 40° = 180°
x° + 65° = 180°
x° = 180°
x° = 180° – 65°
x° = 115°
Thus the value of x is 115°

Question 6.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 43

Answer:
x° = 50° + 55°
x° = 105°
Thus the value of x is 105°

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
The Historic Triangle in Virginia connects Jamestown, Williamsburg, and Yorktown. The interior angle at Williamsburg is 120°. The interior angle at Jamestown is twice the measure of the interior angle at Yorktown. Find the measures of the interior angles at Jamestown and Yorktown. Explain your reasoning.

Answer:
The interior angle at Williamsburg is 120°.
Let the interior angles at Jamestown be x°
Let the interior angles at Yorktown be 2x°
Big Ideas math Answers Grade 8 ch3 angles and triangle img_9
Sum of all the angles in a triangle = 180°
x° + 2x° + 120° = 180°
3x° + 120° = 180°
3x° = 180° – 120°
3x° = 60°
x° = 60/3
x° = 20°
The interior angles at Jamestown be x° = 20°
The interior angles at Yorktown be 2x° = 2(20)° = 40°

Question 8.
A helicopter travels from point C to point A to perform a medical supply drop. The helicopter then needs to land at point B. How many degrees should the helicopter turn at point A to travel towards point B? Justify your answer.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 44

Answer:
Given,
A helicopter travels from point C to point A to perform a medical supply drop.
The helicopter then needs to land at point B.
A = 90° + 32°
A = 122°
Thus the helicopter should turn 122° at point A to travel towards point B.

Angles of Triangles Homework & Practice 3.2

Review & Refresh

Use the figure to find the measure of the angle. Explain your reasoning.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 45

Question 1.
∠2

Answer: 82°

∠2 and 82° are alternate exterior angles formed by transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠2 is 82°

Question 2.
∠6

Answer: 82°

∠6 and 82° are vertical angles formed by transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠6 is 82°

Question 3.
∠4

Answer: 82°

∠4 and 82° are corresponding angles formed by transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠4 is 82°

Question 4.
∠1

Answer: 98°

∠4 and 82° are corresponding angles formed by transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠4 is 82°
∠4 and ∠1 are supplementary angles
∠4 + ∠1 = 180°
82° + ∠1 = 180°
∠1 = 180° – 82°
∠1 = 98°

You spin the spinner shown.

Question 5.
What are the favorable outcomes of spinning a number less than 4?

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 46

Answer: 1, 2, 3

Explanation:
The favorable outcome of spinning a number less than 4 is 1, 2, and 3.

Question 6.
In how many ways can spinning an odd number occur?

Answer: two ways
Odd numbers = 1 and 3
So, in two ways spinning an odd number can occur.

Concepts, Skills, & Problem Solving

USING PARALLEL LINES AND TRANSVERSALS
Consider the figure below. (See Exploration 2, p. 111.)

Question 7.
Use a protractor to find the measures of the labeled angles.

Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 47

Answer:
Use a protractor to determine the measures of the angles A, B, C.
m∠A = 30°
m∠B = 105°
m∠C = 45°
m∠D = 150°
m∠E = 75°
m∠F = 105°
m∠G = 30°

Question 8.
Is ∠F an exterior angle of Triangle ABC ? Justify your answer.

Answer:
An exterior angle is the angle between one side of a triangle and the extension of an adjacent side. ∠F is not an exterior angle of triangle ABC because it has a side of triangle ABC, but not the extension of the adjacent side DF.

USING INTERIOR ANGLE MEASURES
Find the measures of the interior angles of the triangle.

Question 9.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 48

Answer:
Sum of all the angles in a triangle = 180°
x° + 90° + 30° = 180°
x° + 120° = 180°
x° = 180° – 120°
x° = 60°

Question 10.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 49

Answer:
Sum of all the angles in a triangle = 180°
x° + 65° + 40° = 180°
x° + 105° = 180°
x° = 180° – 105°
x° = 75°

Question 11.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 50

Answer:
Sum of all the angles in a triangle = 180°
x° + 35° + 45° = 180°
x° + 80° = 180°
x° = 180° – 80°
x° = 100°

Question 12.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 51

Answer:
Sum of all the angles in a triangle = 180°
x° + (x + 65)° + 25° = 180°
x° + x° + 65° + 25° = 180°
2x° + 90° = 180°
2x° = 180° – 90°
2x° = 90°
x° = 90°/2
x° = 45°
(x + 65)° = 45 + 65 = 110
x° = 25°

Question 13.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 52

Answer:
Sum of all the angles in a triangle = 180°
x° + (x – 44)° + 48° = 180°
x° + x° – 44° + 48° = 180°
2x° + 4° = 180°
2x° = 180° – 4°
2x° = 176°
x° = 176°/2
x° = 88°
(x – 44)° = 88 – 44 = 44
x° = 44°

Question 14.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 53

Answer:
Sum of all the angles in a triangle = 180°
x° + (x – 11)° + 73° = 180°
x° + x° – 11° + 73° = 180°
2x° + 62° = 180°
2x° = 180° – 62°
2x° = 118°
x° = 118°/2
x° = 59°
(x – 11)° = 59 – 11 = 48
x° = 48°

FINDING EXTERIOR ANGLE MEASURES
Find the measure of the exterior angle.

Question 15.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 54

Answer:
x° = 38° + 90°
x° = 128°
The measure of exterior angle is 128°

Question 16.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 55

Answer:
k° = 64° + 76°
k° = 140°
The measure of an exterior angle is 140°

Question 17.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 56

Answer:
2a° = (a + 10°) + 44°
2a° = a + 54°
2a° – a° = 54
a° = 54
The measure of the exterior angle = 2a = 2(54°) = 108°

Question 18.
MODELING REAL LIFE
A tornado is located between city hall and a cell phone tower and is heading towards the cell phone tower. By what angle does the tornado’s direction need to change so that it passes over the radar station instead? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 57

Answer:
Sum of all the angles in a triangle = 180°
x° + 75° + 75° = 180°
x° + 150° – 150° = 180° – 150°
x° = 30°
Thus the angle that tornado direction needs to change is 30°.

Question 19.
YOU BE THE TEACHER
Your friend finds the measure of the exterior angle shown. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 58

Answer:
Your friend is not correct because the measure of the exterior angle is equal to the sum of two non-adjacent interior angles.

Question 20.
REASONING
The ratio of the interior angle measures of a triangle is 2 : 3 : 5. What are the angle measures?

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 59

Answer:
Sum of all the angles in a triangle = 180°
2x° + 3x° + 5x° = 180°
10x° = 180°
x = 180/10
x = 18°
2x° = 2(18°) = 36°
3x° = 3(18) = 54°
5x° = 5(18) = 90°

Question 21.
PROBLEM SOLVING
The support for a window air-conditioning unit forms a triangle and an exterior angle. What is the measure of the exterior angle?

Answer:
The measure of the exterior angle DBC is
m∠DBC = m∠ABC + m∠ACB
m∠ABC + m∠ACB = 90°
5x – 6 + 3x = 90
8x – 6 = 90
8x = 90 + 6
8x = 96
x = 96/8
x = 12
m∠DBC = m∠BAC+ m∠ACB
= 90° + 3(12)°
= 126°

Question 22.
REASONING
A triangle has an exterior angle with a measure of 120°. Can you determine the measures of the interior angles? Explain.

Answer:
A triangle has an exterior angle with a measure of 120°
m∠ACB = m∠A + m∠B
m∠A + m∠B = 120°
According to the exterior angles
We have
m∠C + m∠ACD = 180°
m∠C + 120° = 180°
m∠C = 180° – 120°
m∠C = 60°

ANGLES OF TRIANGLES

Determine whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 23.
Given three angle measures, you can construct a triangle.

Answer:
We can construct a triangle if the sum of the measure of the 3 angles is 180°.
As a matter of fact, if the sum of the measures of the 3 angles is 180°
We can build an infinity of triangles that are similar.

Question 24.
The acute interior angles of a right triangle are complementary.

Answer:
Let A, B, C be the angles of a right triangle with m∠A = 90°
m∠A + m∠B + m∠C = 180°
90° + m∠B + m∠C = 180°
m∠B + m∠C = 180° – 90°
m∠B + m∠C = 90°
This means ∠B and ∠C are complementary.

Question 25.
A triangle has more than one vertex with an acute exterior angle.

Answer:
An exterior angle of a triangle and the adjacent triangle’s angle are complementary. If an exterior angle is acute, it means the adjacent triangle’s angle is obtuse.
Since we are given that more than one exterior angle is acute, it means the triangle would have more than one obtuse angle, which is impossible.
The statement is never true.

Question 26.
DIG DEEPER!
Using the figure at the right, show that z = x + y. (Hint: Find two equations involving w.)
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 60

Answer:
The angles z and w are supplementary
z + w = 180°
The sum of a triangle is 180°
x + y + w = 180°
z = 180° – w
x + y = 180° – w
z = x + y

Lesson 3.3 Angles of Polygons

EXPLORATION 1
Work with a partner. In parts (a)-(f), use what you know about the interior angle measures of triangles to find the sum of the interior angle measures of each figure.
Big Ideas Math Answer Key Grade 8 Chapter 3 Angles and Triangles 61
g. REPEATED REASONING
Use your results in parts (a)–(f) to complete the table. Then write an equation that represents the sum of the interior angle measures of a polygon with n sides.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 62

Answer:

3.3 Lesson

Try It

Find the sum of the interior angle measures of the green polygon.

Question 1.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 63

Answer:
S = (n – 2) . 180°
S = (7 – 2) . 180°
S = 5 . 180°
S = 900°
Thus the sum of the interior angle measure is 900°

Question 2.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 64

Answer:
S = (n – 2) . 180°
S = (6 – 2) . 180°
S = 4 . 180°
S = 720°
Thus the sum of the interior angle measure is 720°

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 5.
WRITING
Explain how to find the sum of the interior measures of a polygon.

Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 65

Answer:
Steps to find the sum of the interior measurements of the polygon:
1. Count the number of sides of the polygon.
2. Subtract the number of sides by 2.
3. Multiply the result of the subtraction by 180°

Question 6.
FINDING THE SUM OF INTERIOR ANGLE MEASURES
Find the sum of the interior angle measures of the green polygon.

Answer:
S = (n – 2) . 180°
S = (4 – 2) . 180°
S = 2 . 180°
S = 360°
Thus the sum of the interior angle measure is 360°

FINDING AN INTERIOR ANGLE MEASURE

Find the value of x.

Question 7.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 66

Answer:
S = (n – 2) . 180°
S = (5 – 2) . 180°
S = 3 . 180°
S = 540°
Thus the sum of the interior angle measure is 540°
x° + 160° + 110° + 105° + 95° = 540°
x° + 470° = 540°
x° = 540° – 470°
x° = 70°
Thus the value of x is 70°.

Question 8.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 67

Answer:
S = (n – 2) . 180°
S = (9 – 2) . 180°
S = 7 . 180°
S = 1260°
Thus the sum of the interior angle measure is 1260°
x° + 165° + 155° + 150° + 140° + 135° + 130° + 125° + 110° = 1260°
x° + 1105° = 1260°
x° = 1260° – 1105°
x° = 155°
Thus the value of x is 155°

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 9.
A company installs an octagonal swimming pool.
a. Find the value of a for the pool shown at the left.

Answer:
S = (n – 2) . 180°
S = (8 – 2) . 180°
S = 6 . 180°
S = 1080°
Thus the sum of the interior angle measure is 1080°
a° + 120° + a° + 120° + a° + 120° + a° + 120° = 1080°
4a° + 480° = 1080°
4a° = 1080° – 480°
4a° = 600°
a° = 600/4
a° = 150°
Thus the value of x is 150°.

b. The company installs a different pool that is also in the shape of an octagon. The second pool has twice the length and one-third the width of the first pool. Are the sums of the interior angles of the pools different? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 67.1

Answer:
No for any octagon the sum of the interior angles is 1080 degrees.

Question 10.
DIG DEEPER!
A Bronze Star Medal A is shown.
a. How many interior angles are there?

Answer: 10 interior angles are there

b. What is the sum of the interior angle measures?
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 68

Answer:
S = (n – 2) . 180°
S = (10 – 2) . 180°
S = 8 . 180°
S = 1440°
Thus the sum of the interior angle measure is 1440°

Angles of Polygons Homework & Practice 3.3

Find the value of x.

Question 1.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 69

Answer: 60°

Explanation:
Sum of all the angles = 180°
x° + 60° + 60° = 180°
x° + 120° = 180°
x° = 180° – 120°
x° = 60°

Question 2.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 70

Answer: 45°

Explanation:
Sum of all the angles = 180°
x° + x° + 90° = 180°
2x° + 90° = 180°
2x° = 180° – 90°
2x° = 90°
x° = 45°

Question 3.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 71

Answer: 113°

Explanation:
x° = 65° + 48°
x° = 113°
Thus the measure of an exterior angle is 113°

Solve the proportion.

Question 4.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 72

Answer: 9

Explanation:
\(\frac{x}{12}\) = \(\frac{3}{4}\)
12 . \(\frac{x}{12}\) = \(\frac{3}{4}\) . 12
x = 3 . 3
x = 9

Question 5.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 73

Answer: 2

Explanation:
\(\frac{14}{21}\) = \(\frac{x}{3}\)
3 . \(\frac{14}{21}\) = \(\frac{x}{3}\) . 3
x = 2

Question 6.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 74

Answer: 3

Explanation:
\(\frac{9}{x}\) = \(\frac{6}{2}\)
2. \(\frac{9}{x}\) = 6
18 = 6x
x = 3

Concepts, Skills, & Problem Solving
EXPLORING INTERIOR ANGLES OF POLYGONS
Use triangles to find the sum of the interior angle measures of the polygon. (See Exploration 1, p. 117.)

Question 7.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 75

Answer: 360°

Explanation:
Number of sides = 4
Number of interior triangles in the given figure = 2
The Sum of the measures of the interior angles using triangle = 2 . 180° = 360°

Question 8.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 76

Answer: 1260°

Explanation:
Number of sides = 9
Number of interior triangles in the given figure = 7
The Sum of the measures of the interior angles using triangle = 7 . 180° = 1260°

Question 9.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 77

Answer: 540°

Explanation:
Number of sides = 5
Number of interior triangles in the given figure = 3
The Sum of the measures of the interior angles using triangle = 3 . 180° = 540°

FINDING THE SUM OF INTERIOR ANGLE MEASURES
Find the sum of the interior angle measures of the polygon.

Question 10.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 78

Answer: 360°

Explanation:
S = (n – 2) . 180°
S = (4- 2) . 180°
S = 2 . 180°
S = 360°
Thus the sum of the interior angle measure is 360°

Question 11.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 79

Answer: 1080°

Explanation:
S = (n – 2) . 180°
S = (8- 2) . 180°
S = 6 . 180°
S = 1080°
Thus the sum of the interior angle measure is 1080°

Question 12.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 80

Answer: 1260°

Explanation:
S = (n – 2) . 180°
S = (9- 2) . 180°
S = 7 . 180°
S = 1260°
Thus the sum of the interior angle measure is 1260°

Question 13.
YOU BE THE TEACHER
Your friend finds the sum of the interior angle measures of a 13-gon. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 81

Answer:
To find the sum of the interior angle measures he should subtract 2 from the number of sides of the polygon and then multiply by 180°
S = (n – 2) . 180°
By this, we can say that your friend is not correct.

FINDING AN INTERIOR ANGLE MEASURE
Find the value of x.

Question 14.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 82

Answer:
S = (n – 2) . 180°
S = (4- 2) . 180°
S = 2 . 180°
S = 360°
Thus the sum of the interior angle measure is 360°
x° + 155° + 25° + 137° = 360°
x° + 317° = 360°
x° = 360° – 317°
x° = 43°

Question 15.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 83

Answer:
S = (n – 2) . 180°
S = (6- 2) . 180°
S = 4 . 180°
S = 720°
Thus the sum of the interior angle measure is 720°
x° + x° + x° + x° + 90° + 90° = 720°
4x° + 180° = 720°
4x° = 720° – 180°
4x° = 540°
x° = 540/4
x° = 135°

Question 16.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 84

Answer:
S = (n – 2) . 180°
S = (6- 2) . 180°
S = 4 . 180°
S = 720°
Thus the sum of the interior angle measure is 720°
3x° + 45° + 135° + x° + 135° + 45° = 720°
4x° + 360° = 720°
4x° = 720° – 360°
4x° = 360°
x° = 360/4
x° = 90°

FINDING A MEASURE
Find the measure of each interior angle of the regular polygon.

Question 17.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 85

Answer:
S = (n – 2) . 180°
S = (3- 2) . 180°
S = 1 . 180°
S = 180°
Thus the sum of the interior angle measure is 180°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 3.
180 ÷ 3 = 60°

Question 18.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 86

Answer:

S = (n – 2) . 180°
S = (9 – 2) . 180°
S = 7 . 180°
S = 1260°
Thus the sum of the interior angle measure is 1260°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 9.
1260 ÷ 9 = 140°

Question 19.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 87

Answer:

S = (n – 2) . 180°
S = (12 – 2) . 180°
S = 10 . 180°
S = 1800°
Thus the sum of the interior angle measure is 1800°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 12.
1800 ÷ 12 = 150°

Question 20.
YOU BE THE TEACHER
Your friend finds the measure of each interior angle of a regular 20-gon. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 88

Answer:
No, my friend is not correct because to find the measure of each interior angle of a regular 20-gon, he should divide the sum of the measured interior angles by the number of interior angles, in this case, 20 but your friend divide it by 18 so he is not correct.

Question 21.
MODELING REAL LIFE
A firehydrant bolt is in the shape of a regular pentagon.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 89
a. What is the measure of each interior angle?

Answer:
S = (n – 2) . 180°
S = (5- 2) . 180°
S = 3 . 180°
S = 540°
Thus the sum of the interior angle measure is 540°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 5.
540÷ 5 = 108°

b. RESEARCH
Why are firehydrants made this way?

Question 22.
PROBLEM SOLVING
The interior angles of a regular polygon each measure 165°. How many sides does the polygon have?

Answer:
(n – 2) . 180 = 165 . n
180n – 360 = 165n
180n – 360 + 360 – 165n = 165n + 360 – 165n
15n = 360
n = 360/15
n = 24
Therefore the polygon has 24 sides

Question 23.
STRUCTURE
A molecule can be represented by a polygon with interior angles that each measure 120°. What polygon represents the molecule? Does the polygon have to be regular? Justify your answers.

Answer:
(n – 2) . 180 = 120 . n
180n – 360 = 120n
180n – 120n = 360
60n = 360
n = 6

Question 24.
PROBLEM SOLVING
The border of a Susan B. Anthony dollar is in the shape of a regular polygon.
a. How many sides does the polygon have?

Answer: The polygon has 11 sides.

b. What is the measure of each interior angle of the border? Round your answer to the nearest degree.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 91

Answer:
S = (n – 2) . 180°
S = (11 – 2) . 180°
S = 9 . 180°
S = 1620°
Thus the sum of the interior angle measure is 1620°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 11.
1620 ÷ 11 = 147°

Question 25.
REASONING
The center of the stained glass window is in the shape of a regular polygon. What are the measures of the interior angles of the green triangle?

Answer:
(n-2)180°/n = (8-2)180°/8 = 135°
m∠OAB = m∠OBA = 135/2 = 67.5°
m∠AOB + m∠OAB + m∠OBA = 180°
m∠AOB + 67.5° + 67.5° = 180°
m∠AOB + 135° = 180°
m∠AOB = 180° – 135°
m∠AOB = 45°

Question 26.
GEOMETRY
Draw a pentagon that has two right interior angles, two 45° interior angles, and one 270° interior angle.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 92

Answer:
Big Ideas Math 8th Grade Solution Key Ch 3 Angles and Triangles img_9
Big Ideas Math 8th Grade Solution Key Ch 3 Angles and Triangles img_10

Question 27.
DIG DEEPER!
The floor of a gazebo is in the shape of a heptagon, a seven-sided polygon. Four of the interior angles measure 135°. The other interior angles have equal measures. Find their measures.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 93

Answer:
The given polygon has 7 sides.
S = (n – 2) . 180°
S = (7 – 2) . 180°
S = 5 . 180°
S = 900°
Thus the sum of the interior angle measure is 900°
4 . 135° + 3 . x° = 900°
540° + 3x° = 900°
3x° = 900° – 540°
3x° = 360°
x° = 360/3
x° = 120°

Lesson 3.4 Using Similar Triangles

Using Similar Triangles

EXPLORATION 1

Work with a partner. Use geometry software.
a. Draw a triangle that has a 50° angle and a 30° angle. Then draw a triangle that is either larger or smaller that has the same two angle measures. Are the triangles congruent? similar? Explain your reasoning.
a. Choose any two angle measures whose sum is less than 180°. Repeat part(a) using the angle measures you chose.
c. Compare your results in parts (a) and (b) with other pairs of students. Make a conjecture about two triangles that have two pairs of congruent angles.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 93.1

Answer:

EXPLORATION 2
Using Indirect Measurement
Work with a partner. Use the fact that two rays from the Sun are parallel to make a plan for how to find the height of the flagpole. Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 94

Answer:

3.4 Lesson

Try It

Tell whether the triangles are similar. Explain.

Question 1.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 95

Answer: No

Explanation:
x° + 28° + 80° = 180°
x° + 108° = 180°
x° = 180° – 108°
x° = 72°
BIM Grade 8 Answer Key Chapter 3 Angles and triangles img_11
y° + 28° + 71° = 180°
y° + 99° = 180°
y° = 180° – 99°
y° = 81°
The triangles do not have two pairs of congruent angles.
So, the triangles are not similar.

Question 2.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 96

Answer: Yes

Explanation:
x° + 66° + 90° = 180°
x° + 156° = 180°
x° = 180° – 156°
x° = 24°
y° + 24° + 90° = 180°
y° + 114° = 180°
y° = 180° – 114°
y° = 66°
The triangles have two pairs of congruent angles.
Thus the triangles are similar.

Try It

Question 3.
Can you determine whether △PQR and △TSRand are similar? Explain.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 97

Answer: No

Explanation:
We are not given any information about the lengths of the sides either, therefore with only a pair of congruent angles, we cannot tell whether the triangles are similar.

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 4.
IDENTIFYING SIMILAR TRIANGLES
Tell whether the triangles are similar. Explain.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 98

Answer: Yes

Explanation:
x° + 54° + 63° = 180°
x° + 107° = 180°
x° = 180° – 107°
x° = 63°

Question 5.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 98.1
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 99

Answer: Option B

Explanation:
ΔPQR and ΔTSR are congruent as TS || PQ leads to two pairs of correspondent congruent angles.
ΔPQR is a dilation of ΔTSR because their sides are proportional, the constant of proportionality being greater than 1.
ΔPQR is a scale drawing of ΔTSR because their sides are proportional.
The question that does not fit is
“Are ΔPQR and ΔTSR the same size and shape?”
because the triangles do not have the same size, but they have the same shape.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 6.
DIG DEEPER!
Engineers plan to construct an aqueduct to transport water from the top of a ridge to farmland. A portion of the project is complete. Find the length of the entire aqueduct.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 100
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 101

Answer:
Aqueduct/2.6 = 5/1
Aqueduct = 5 × 2.6
Aqueduct = 13
Thus the length of the Aqueduct is 13 km.

Question 7.
You want to go on a swamp tour. How long does it take a swamp vehicle that travels at 3.2 miles per hour to travel across the swamp, from point Z to point Y? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 102

Answer:
a/10 = 3/6
6 × a = 3 × 10
6a = 30
a = 30/6
a = 5
The length from point Z to point Y is 5 miles.
Time to travel from point Z to point Y = 5/3.5 = 1.56 hour

Using Similar Triangles Homework & Practice 3.4

Review & Refresh

Find the measure of each interior angle of the regular polygon.

Question 1.
octagon

Answer: The measure of each interior angle is 135°

Explanation:
S = (n – 2) . 180°
S = (8- 2) . 180°
S = 6 . 180°
S = 1080°
Thus the sum of the interior angle measure is 1080°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 8.
1080÷ 8= 135°

Question 2.
decagon

Answer: The measure of each exterior angle is 144°

Explanation:
S = (n – 2) . 180°
S = (10 – 2) . 180°
S = 8 . 180°
S = 1440°
Thus the sum of the interior angle measure is 1440°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 10.
1440÷ 10= 144°

Question 3.
18-gon

Answer: The measure of each interior angle is 160°

Explanation:
S = (n – 2) . 180°
S = (18- 2) . 180°
S = 16 . 180°
S = 2880°
Thus the sum of the interior angle measure is 2880°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 18.
2880 ÷ 18= 160°

Solve the equation. Check your solution.

Question 4.
3.5 + y = -1

Answer:
Given the equation
3.5 + y = -1
y = -1 – 3.5
y = -4.5

Question 5.
9x = 54

Answer:
Given the equation
9x = 54
x = 54/9
x = 6

Question 6.
-4 = \(\frac{2}{7}\)p

Answer:
Given the equation
-4 = \(\frac{2}{7}\)p
-4 × 7 = 2p
2p = -28
p = -28/2
p = -14

Concepts, Skills, & Problem Solving
CREATING SIMILAR TRIANGLES
Draw a triangle that is either larger or smaller than the one given and has two of the same angle measures. Explain why the new triangle is similar to the original triangle. (See Exploration 1, p. 123.)

Question 7.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 103

Answer:
BIM Answers for Grade 8 chapter 3 img_12
The above triangle and the given triangle are similar because the two angles in both the triangles are the same so the third angle is also the same.

Question 8.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 104

Answer:
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 104
The above triangle and the given triangle are similar because the two angles in both the triangles are the same so the third angle is also the same.

IDENTIFYING SIMILAR TRIANGLES
Tell whether the triangles are similar. Explain.

Question 9.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 105

Answer: The triangles have two pairs of congruent angles.
So, the third angles are congruent, and the triangles are similar.

Question 10.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 106

Answer:
x° + 36° + 72° = 180°
x° + 108° = 180°
x° = 180° – 108°
x° = 72°
y° + 33° + 72° = 180°
y° + 105° = 180°
y° = 180° – 105°
y° = 72°
The triangles do not have two pairs of congruent angles.
Therefore the triangles are not similar.

Question 11.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 107

Answer:
x° + 64° + 85° = 180°
x° + 149° = 180°
x° = 180° – 149°
x° = 31°
y° + 26° + 85° = 180°
y° + 111° = 180°
y° = 180° – 111°
y° = 69°
The triangles do not have two pairs of congruent angles.
Therefore the triangles are not similar.

Question 12.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 108

Answer:
x° + 48° + 81° = 180°
x° + 129° = 180°
x° = 180° – 129°
x° = 51°
y° + 48° + 51° = 180°
y° + 99° = 180°
y° = 180° – 99°
y° = 81°
The triangles have two pairs of congruent angles.
Therefore the triangles are similar.

Question 13.
GEOMETRY
Which of the rulers are similar in shape? Explain.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 109

Answer:
2x + 90 = 180°
2x = 180 – 90°
2x = 90°
x = 90/2
x = 45°
The ruler on the left and the ruler on the right both have the shape of a right triangle with 45° angles, therefore they are similar in shape, while the middle ruler has 60°, 30° angles.

STRUCTURE
Tell whether the triangles are similar. Explain.

Question 14.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 110

Answer:
m∠APB + m∠B = 90°
m∠APB + 51° = 90°
m∠APB = 90° – 51°
m∠APB = 39°
m∠APB + m∠BPD + m∠DPC = 180°
39° + 102° + m∠DPC = 180°
m∠DPC + 141° = 180
m∠DPC = 180 – 141°
m∠DPC = 39°
m∠A = m∠C
m∠APB = m∠DPC

Question 15.
Big Ideas Math Answers 8th Grade Chapter 3 Angles and Triangles 111

Answer:
∠APB ≅ ∠CPD
m∠APB = m∠CPD
m∠APB = 29°
m∠A + m∠B + m∠APB = 180°
m∠A + 88° + 29° = 180°
m∠A  + 117° = 180°
m∠A = 180° – 117°
m∠A = 63°
m∠PDC + m∠PDE = 180°
m∠PDC + 91° = 180°
m∠PDC = 180° – 91°
m∠PDC = 89°

IDENTIFYING SIMILAR TRIANGLES
Can you determine whether the triangles are similar? Explain.

Question 16.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 112

Answer:
PS || QR
∠PSQ and ∠SQR are interior angles using the transversal QS, thus they are congruent.
∠PSQ ≅ ∠SQR

Question 17.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 113

Answer:
As AB || DE there are two pairs of congruent alternate interior angles, using the transversals AE and BD.
∠A≅ ∠E
∠B≅ ∠D
The two pairs of congruent angles are enough to prove that the triangles are similar.
ΔABC ∼ ΔEDC

Question 18.
PROBLEM SOLVING
A water sample must be taken from water atleast 20 feet deep. Find the depth of the water 50 feet from shore. Is this an appropriate location for a water sample?
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 114

Answer:
ΔAMN ∼ ΔABC
MN/BC = AM/AB
1.5/d = 5/50
d = 1.5 × 10
d = 15 feet
Therefore 15 feet is not an appropriate location.

Question 19.
MODELING REAL LIFE
A map shows the number of steps you must take to get to a treasure. However, the map is old, and the last dimension is unreadable. Explain why the triangles are similar. How many steps do you take from the pyramids to the treasure?
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 115

Answer:
The two triangles are similar because they are right triangles and ∠AXB ≅ ∠PXQ because they are vertical angles.
PQ/300 = 80/240
240PQ = 24000
PQ = 24000/240
PQ = 100 steps

Question 20.
PROBLEM SOLVING
A person who is 6 feet tall casts a 3-foot-long shadow. A nearby pine tree casts a 15-foot-long shadow. What is the height h of the pine tree?
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 116

Answer:
Given,
A person who is 6 feet tall casts a 3-foot-long shadow.
A nearby pine tree casts a 15-foot-long shadow.
ΔXAB ∼ ΔXPQ
AB/PQ = XB/XQ
6/PQ = 3/15
PQ = 30 ft

Question 21.
OPEN-ENDED
You place a mirror on the ground 6 feet from the lamppost. You move back 3 feet and see the top of the lamppost in the mirror. What is the height of the lamppost?

Answer:
Grade 8 BIM Chapter 3 Answer Key img_13
ΔXAB ∼ ΔXPQ
AB/PQ = XB/XQ
h/PQ = 6/3
h = 2PQ

Question 22.
DIG DEEPER!
In each of two right triangles, one angle measure is two times another angle measure. Can you determine that the triangles are similar? Explain your reasoning.

Answer:
We are given the right triangle ABC
m∠A = 2m∠B
Case 1:
m∠A = 90°
90° = 2m∠B
m∠B = 45°
m∠C = 180° – 90° – 45° = 45°
Case 2:
m∠B = 90°
m∠A = 2 × 90° = 180°
Case 3:
m∠C = 90°
m∠A + m∠B = 180 – m∠C = 180° – 90° = 90°
2m∠B + m∠B = 90°
3m∠B = 90°
m∠B = 30°
m∠A = 2 . 30° = 60°

Question 23.
GEOMETRY
In the diagram, \(\overline{B G}\), \(\overline{C F}\) and \(\overline{D E}\) are parallel. The length of \(\overline{B D}\) is 6.32 feet, and the length of \(\overline{D E}\) is 6 feet. Name all pairs of similar triangles in the diagram. Then find the lengths of \(\overline{B G}\) and \(\overline{C F}\)
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 117

Answer:
ΔABG ∼ ΔACF
ΔACF ∼ ΔADE
ΔABG ∼ ΔADE
AB = BC = CD = BD/2 = 6.32/2 = 3.16
AB/CD = BG/DE
3BG = 6
BG = 2 feet
ΔACF ∼ ΔADE
AC/AD = CF/DE
2/3 = CF/6
3CF = 2(6)
CF = 4 feet

Angles and Triangles Connecting Concepts

Using the Problem-Solving Plan

Question 1.
A dog park is divided into sections for large and small dogs. The ratio of the perimeter of the small dog section to the perimeter of the entire dog park is 7 : 12. Find the area of each section.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 118

Understand the problem
You know two dimensions of a dog park and the ratio of the perimeter of the small dog section to the perimeter of the entire park. You are asked to find the area of each section.
Make a plan
Verify that the small triangle and the large triangle are similar. Then use the ratio of the perimeters to find the base or the height of each triangle and calculate the areas.
Solve and check.
Use the plan to solve the problem. Then check your solution.

Answer:

Question 2.
You rotate lines m and t 180° about point P. The image of line m is parallel to the original line. Use the diagram to show that when a transversal intersects parallel lines, each of the following pairs of angles are congruent. Explain your reasoning.
a. alternate interior angles
b. alternate exterior angles
c. corresponding angles
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 119

Answer:

Performance Task

Turtle Shells

At the beginning of this chapter, you watched a STEAM Video called “Honeycombs.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 120

Angles and Triangles Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 121

Graphic Organizers
You can use an Example and Non-Example to list examples and non-examples of a concept. Here is an Example and Non-Example Chart for transversals.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 122

Choose and complete a graphic organizer to help you study the concept.

  1. interior angles formed by parallel lines and a transversal
  2. exterior angles formed by parallel lines and a transversal
  3. interior angles of a triangle
  4. exterior angles of a triangle
  5. polygons
  6. similar triangles

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 123

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 124

3.1 Parallel Lines and Transversals (pp. 103–110)

Use the figure to find the measure of the angle. Explain your reasoning.

Question 1.
∠8

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 125

Answer: 140°

Explanation:
∠8 and 140 degrees angle are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠8 is 140°

Question 2.
∠5

Answer: 140°

Explanation:
∠5 and 140 degrees angle are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠5 is 140°

Question 3.
∠7

Answer: 40°

Explanation:
∠5 and 140 degrees angle are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠5 is 140°
∠5 and ∠7 are supplementary angle.
∠5 + ∠7 = 180°
140° + ∠7 = 180°
∠7 = 180° – 140°
∠7 = 40°
So, the measure of ∠7 is 40°

Question 4.
∠2

Answer: 40°

Explanation:
140 and ∠2 are supplementary angle.
140° + ∠2 = 180°
∠2 = 180° – 140°
∠2 = 40°
So, the measure of ∠2 is 40°

Question 5.
∠6

Answer: 40°

Explanation:
∠5 and 140 degrees angle are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠5 is 140°
∠5 and ∠6 are supplementary angle.
140° + ∠6 = 180°
∠6 = 180° – 140°
∠6 = 40°
So, the measure of ∠6 is 40°

Complete the statement. Explain your reasoning.

Question 6.
If the measure of ∠1 = 123°, then the measure of ∠7 = Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 126

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 127

Answer: 123°

Explanation:
∠1 and ∠7 are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠7 is 123°

Question 7.
If the measure of ∠2 = 58°, then the measure of Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 127.1

Answer: 122°

Explanation:
∠2 and ∠6 are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠6 is 58°
∠5 and ∠6 are supplementary angle.
∠5 + ∠6 = 180°
58° + ∠5 = 180°
∠5 = 180° – 58°
∠5 = 122°
So, the measure of ∠5 is 122°

Question 8.
If the measure of ∠5 = 119°, then the measure of Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 128

Answer: 119°

Explanation:
∠3 and ∠5 are alternate interior angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠3 is 119°

Question 9.
If the measure of ∠4 = 60°, then the measure of Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 129

Answer: 60°

Explanation:
∠4 and ∠6 are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent.
So, the measure of ∠4 is 60°

Question 10.
In Exercises 6–9, describe the relationship between ∠2 and ∠8.

Answer: ∠2 ≅ ∠8

Question 11.
In a park, a bike path and a horse riding path are parallel. In one part of the park, a hiking trail intersects the two paths. Find the measures of ∠1 and ∠2. Explain your reasoning.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 129.1

Answer: ∠1 = 108°, ∠2 = 108°

Explanation:
∠3 and 72° are alternate interior angles. They are congruent.
So, the measure of ∠3 is 72°
∠3 + ∠1 = 180°
72° + ∠1 = 180°
∠1 = 180° – 72°
∠1 = 108°
So, the measure of ∠1 is 108°
∠1 and ∠2 are alternating interior angles. They are congruent.

3.2 Angles of Triangles (pp. 111 – 116)

Find the measures of the interior angles of the triangle.

Question 12.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 129.2

Answer:
Sum of all the angles in a triangle = 180°
x° + 50° + 55° = 180°
x° + 105° = 180°
x° = 180° – 105°
x° = 75°

Question 13.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 130

Answer:
Sum of all the angles in a triangle = 180°
x° + (x + 8)° + 90° = 180°
2x° + 8° + 90° = 180°
2x° + 98° = 180°
2x° = 180° – 98°
2x° = 82
x° = 82/2
x° = 41°
(x + 8)° = (41 + 8)° = 49°

Find the measure of the exterior angle.

Question 14.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 131

Answer:
s° = 50° + 75°
s° = 125°
Thus the measure of the exterior angle is 125°

Question 15.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 132

Answer:
Sum of all the angles in a triangle = 180°
t° + (t + 10)° + (t + 20)° = 180°
3t° + 10° + 20° = 180°
3t° + 30° = 180°
3t° = 180° – 30°
3t° = 150°
t° = 150/3
t° = 50°
Exterior angle:
t° + (t + 10)°
t° + t° + 10°
2t° + 10°
2(50)° + 10°
= 100° + 10°
= 110°
Thus the measure of the exterior angle is 110°.

Question 16.
What is the measure of each interior angle of an equilateral triangle? Explain.

Answer:
S = (n – 2) . 180°
S = (3- 2) . 180°
S = 1 . 180°
S = 180°
Thus the sum of the interior angle measure is 180°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 3.
180 ÷ 3 = 60°

Question 17.
You draw the Leo constellation. You notice that the three stars Denebola, Zosma, and Chertan form a triangle. In your drawing, you find the measure of the interior angle at Denebola is 30° and the measure of the interior angle of the triangle at Zosma is 56°. What is the measure of the interior angle of the triangle at Chertan?
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 133

Answer:
Sum of all the angles in a triangle = 180°
x° + 30° + 56° = 180°
x° + 86° = 180°
x° = 180° – 86°
x° = 94°
Thus the measure of the interior angle of the triangle at Chertan = 94°

3.3 Angles of Polygons (pp. 117–122)

Find the sum of the interior angle measures of the polygon.

Question 18.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 134

Answer:
The polygon has 13 sides
S = (n – 2) . 180°
S = (13- 2) . 180°
S = 11 . 180°
S = 1980°
Thus the sum of the interior angle measure is 1980°

Question 19.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 135

Answer:
The polygon has 9 sides
S = (n – 2) . 180°
S = (9- 2) . 180°
S = 7 . 180°
S = 1260°
Thus the sum of the interior angle measure is 1260°

Find the value of x.

Question 20.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 136

Answer:
S = (n – 2) . 180°
S = (4 – 2) . 180°
S = 2 . 180°
S = 360°
Thus the sum of the interior angle measure is 360°
x° + 60° + 128° + 95° = 360°
x° + 283° = 360°
x° = 360° – 283°
x° = 77°
Thus the value of x is 77°.

Question 21.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 137

Answer:
S = (n – 2) . 180°
S = (7 – 2) . 180°
S = 5 . 180°
S = 900°
Thus the sum of the interior angle measure is 900°
x° + 135° + 125° + 135° + 105° + 150° + 140° = 900°
x° + 790° = 900°
x° = 900° – 790°
x° = 110°

Question 22.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 138

Answer:
S = (n – 2) . 180°
S = (6 – 2) . 180°
S = 4 . 180°
S = 720°
Thus the sum of the interior angle measure is 720°
x° + 120° + 140° + 92° + 125° + 130° = 720°
x° + 607° = 720°
x° = 720° – 607°
x° = 113°
The value of x° is 113°

Question 23.
Find the measure of each interior angle of the regular polygon.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 139

Answer:
The given polygon is an octagon. It has 8 sides.
S = (n – 2) . 180°
S = (8 – 2) . 180°
S = 6 . 180°
S = 1080°
Thus the sum of the interior angle measure is 1080°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 8.
1080 ÷ 3 = 135°

3.4 Using Similar Triangles (pp. 123–128)

Tell whether the triangles are similar. Explain.

Question 24.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 140

Answer:
x° + 68° + 90° = 180°
x° = 180° – 158°
x° = 22°
y° + 22° + 90° = 180°
y° + 112° = 180°
y° = 180° – 112°
y° = 68°
The triangles have two pairs of congruent angles.
So, the triangles are similar.

Question 25.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 141

Answer:
x° + 100° + 30° = 180°
x° + 130° = 180°
x° = 180° – 130°
x° = 50°
y° + 100° + 50° = 180°
y° + 150° = 180°
y° = 180° – 150°
y° = 30°
The triangles have two pairs of congruent angles.
So, the triangles are similar.

Question 26.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 142

Answer:
x° + 50° + 85° = 180°
x° + 135° = 180°
x° = 180° – 135°
x° = 45°
y° + 85° + 35° = 180°
y° + 120° = 180°
y° = 180° – 120°
y° = 60°
The triangles do not have two pairs of congruent angles.
So, the triangles are not similar.

Question 27.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 143

Answer:
∠B ≅ ∠D
∠A ≅ ∠C
∠AXB ≅ ∠CXD
∠AXB and ∠CXD are vertical angles.
ΔAXB ∼ ΔCXD

Question 28.
A person who is 5 feet tall casts a shadow that is 4 feet long. A nearby building casts a shadow that is 24 feet long. What is the height of the building?

Answer:
Given,
A person who is 5 feet tall casts a shadow that is 4 feet long.
A nearby building casts a shadow that is 24 feet long.
Let the height of the building = x ft
x/24 = 5/4
24 . x/24 = 5/4 . 24
x = 30
Thus the height of the building is 30 ft.

Angles and Triangles Practice Test

Practice Test

Use the figure to find the measure of the angle. Explain your reasoning.

Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 144

Question 1.
∠7

Answer: 47°

Explanation:
∠7 and 47° angles are alternate exterior angles formed by a transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠7 is 47°

Question 2.
∠6

Answer: 47°

Explanation:
∠6 and 47° angles are corresponding angles formed by a transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠6 is 47°

Question 3.
∠4

Answer: 133°

Explanation:
∠4 and 47° are supplementary angles.
47° + ∠4 = 180°
∠4 = 180° – 47°
∠4 = 133°
Thus the measure of ∠4 = 133°

Question 4.
∠5

Answer: 133°

Explanation:
∠6 and 47° angles are corresponding angles formed by transversal intersecting parallel lines. The angles are congruent.
Thus the measure of ∠6 is 47°
∠6 + ∠5 = 180°
47° + ∠5 = 180°
∠5 = 180° – 47°
∠5 = 133°
Thus the measure of ∠5 = 133°

Question 5.
Find the value of x.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 144.1

Answer: 28°

Explanation:
Sum of all the angles in a triangle = 180°
x° + 129° + 23° = 180°
x° + 152° = 180°
x° = 180° – 152°
x° = 28°
Thus the value of x° is 28°

Question 6.
Find the measures of the interior angles.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 145

Answer: 68°

Explanation:
Sum of all the angles in a triangle = 180°
x° + (x – 24)° + 68° = 180°
x° + x° – 24° + 68° = 180°
2x° + 44° = 180°
2x° = 180° – 44°
2x° = 136°
x° = 68°
(x – 24)° = (68 – 24)° = 44°

Question 7.
Find the measure of the exterior angle.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 146

Answer:
j° = 40° + 90°
j° = 130°
The measure of an exterior angle is 130°.

Question 8.
Find the measure of the interior angle measures of the border of the coin.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 147

Answer:
The coin has 7 sides.
S = (n – 2) . 180°
S = (7 – 2) . 180°
S = 5 . 180°
S = 900°
Thus the sum of the interior angle measure is 900°

Question 9.
Find the value of x.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 148

Answer:
S = (n – 2) . 180°
S = (5 – 2) . 180°
S = 3 . 180°
S = 540°
Thus the sum of the interior angle measure is 540°
2x° + 125° + 90° + 2x° + 125° = 540°
4x° + 340° = 540°
4x° = 540° – 340°
4x° = 200°
x° = 200/4
x° = 50°
The value of x° is 50°

Question 10.
Find the measure of each interior angle of the regular polygon.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 149

Answer:
S = (n – 2) . 180°
S = (6 – 2) . 180°
S = 4 . 180°
S = 720°
Thus the sum of the interior angle measure is 720°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 6.
720 ÷ 6 = 120°

Tell whether the triangles are similar. Explain.

Question 11.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 150

Answer:
To find x°:
x° + 61° + 70° = 180°
x° + 131° = 180°
x° = 180° – 131°
x° = 49°
To find y°:
x° + 39° + 70° = 180°
x° + 109° = 180°
x° = 180° – 109°
x° = 71°
The triangles do not have two pairs of congruent angles.
So, the triangles are not similar.

Question 12.
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 151

Answer:
∠A ≅ ∠QPB
∠C ≅ ∠PQB
ΔBPQ ∼ ΔBAC

Question 13.
Describe two ways you can find the measure of ∠5.
Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles 152

Answer:
One way: ∠3 and 65° are supplementary angles. ∠5 and ∠3 are alternate interior angles.
Another way: ∠8 and 65° are alternate exterior angles. ∠5 and ∠8 are supplementary angles.

Question 14.
You swim 3.6 kilometers per hour. How long (in minutes) will it take you to swim the distance d across the pond?
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 153

Answer:
Given,
You swim 3.6 kilometers per hour.
d/105 = 80/140
105 . d/105 = 80/140 . 105
d = 60
The length of the pond is 60 m.
Speed = 3.6 km per hour = 1 m sec
Distance = d = 60m
Time it will take to swim across the pond
= distance/speed
= 60/1
= 60 sec = 1 min

Angles and Triangles Cumulative Practice

Question 1.
The border of a Canadian one-dollar coin is shaped like an 11-sided regular polygon. The shape was chosen to help visually impaired people identify the coin. How many degrees are in each interior angle along the border? Round your answer to the nearest degree.
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 154
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 154.1

Answer:
S = (n – 2) . 180°
S = (11 – 2) . 180°
S = 9 . 180°
S = 1620°
Thus the sum of the interior angle measure is 1620°
In a regular polygon, each interior angle is congruent. So, divide the sum of the interior angle measures by the number of interior angles, 11.
1620 ÷ 11 = 147°

Question 2.
A public utility charges its residential customers for natural gas based on the number of therms used each month. The formula shows how the monthly cost tin dollars is related to the number of terms used.
C = 11 + 1.6t
Solve this formula for t.
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 155

Answer:
C = 11 + 1.6t
C – 11 = 1.6t
1.6t = C – 11
t = (C – 11)/1.6
Thus the correct answer is option B.

Question 3.
What is the value of x?
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 156

Answer:
5(x – 4) = 3x
5x – 20 = 3x
5x – 3x = 20
2x = 20
x = 20/2
x = 10
Thus the correct answer is option I.

Question 4.
In the figures, △PQR is similar to △STU. What is the value of x?
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 157
A. 9.6
B. 10\(\frac{2}{3}\)
C. 13.5
D. 15

Answer:
△PQR is similar to △STU
PQ = 12
ST = 16
SU = 20
TU = 18
PQ/ST = QR/TU
12/16 = X/18
16X = 12 × 18
X = 216/16
X = 13.5 cm
Thus the correct answer is option C.

Question 5.
What is the value of x?
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 158

Answer:
∠y and 125° are supplementary angles.
125° + ∠y = 180°
∠y = 180° – 125°
∠y = 55°
So, the measure of ∠y = 55°
∠x and ∠y are alternate interior angles. They are congruent.
So, the measure of ∠x = 55°

Question 6.
Your friend was solving an equation in the box shown.
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 159
What should your friend do to correct the error that she made?
F. Multiply both sides by –\(\frac{5}{2}\) instead of –\(\frac{2}{5}\).
G. Multiply both sides by \(\frac{2}{5}\) instead of –\(\frac{2}{5}\).
H. Distribute –\(\frac{2}{5}\) to get 4x – 6.
I. Add 15 to -30.

Answer:
My friend made the error by multiplying both sides by –\(\frac{2}{5}\).
To correct the error she should multiply both sides by –\(\frac{5}{2}\) instead of –\(\frac{2}{5}\)
Thus the correct answer is option F.

Question 7.
In the coordinate plane below, ∆XYZ is plotted and its vertices are labeled.
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 160
Which of the following shows ∆X’Y’Z’, the image of ∆XYZ after it is reflected in the y-axis?
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 161

Answer:
Given,
X(-6,-1)
Y(-3,-5)
X(-2,-3)
Reflecting a point (x,y) in the y-axis.
(x, y) = (-x, y)
X(-6,-1) = X'(6, -1)
Y(-3,-5) = Y'(3, -5)
X(-2,-3) = Z'(2, -3)
Thus the correct answer is option B.

Question 8.
The sum S of the interior angle measures of a polygon with n sides can be found by using a formula.
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 162
Part A
Write the formula.

Answer:
S = (n – 2) . 180°
Part B
A quadrilateral has angles measuring 100°, 90°, and 90°. Find the measure of its fourth angle. Show your work and explain your reasoning.

Answer:
The quadrilateral has 4 sides
S = (n – 2) . 180°
S = (4 – 2) . 180°
S = 2 . 180°
S = 360 °
Thus the sum of the interior angles is 360 °
x° + 100° + 90° + 90° = 360°
x° + 280° = 360°
x° = 360° – 280°
x° = 80°
Thus the value of x° is 80°

Part C
The sum of the measures of the angles of the pentagon shown is 540°. Divide the pentagon into triangles to show why this must be true. Show your work and explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 3 Angles and Triangles 163

Answer:
Number of sides = 3
The number of interior triangles in the given figure = 3
Sum of the interior angles measure using triangle = 3 × 180° = 540

Conclusion:

I wish the details prevailed in the above article is beneficial for all the 8th grade students. Hope our Big Ideas Math Answers Grade 8 Chapter 3 Angles and Triangles helped you a lot to overcome the difficulties in this chapter. Feel free to post your comments in the comment box. Stay tuned to our ccssanswers.com to get step by step explanation for all the Grade 8 chapters.

Parametric Equations of a Parabola Formula, Examples | How to find Parametric Form of Parabola?

Parametric Equations of a Parabola

Parabolas describe many natural phenomena like the motion of objects affected by gravity, increase or decrease in the population, amount of reagents in a chemical equation, etc. At times, you need to evaluate how variables change with respect to time. To track how variables change over time, you can put equations into Parametric Form.

Different Parametric Equations can be used to represent a Parabola. We have listed the simple and easiest way on How to find the Parametric Equations of a Parabola in the below modules. Refer to the Solved Examples on Parametric Equations of Parabola for a better understanding of the concept.

Standard Forms of Parabola and their Parametric Equations

Let us discuss in detail the Parametric Coordinates of a Point on Standard Forms of Parabola and their Parametric Equations

Standard Equation of Parabola y2 = 4ax

  • Parametric Coordinates of the Parabola y2 = 4ax are (at2, 2at)
  • Parametric Equations of Parabola y2 = 4ax are x = at2 and y = 2at

Standard Equation of Parabola y2 = -4ax

  • Parametric Coordinates of the Parabola y2 = -4ax are (-at2, 2at)
  • Parametric Equations of Parabola y2 = -4ax are x = -at2 and y = 2at

Standard Equation of Parabola x2 = 4ay

  • Parametric Coordinates of the Parabola x2 = 4ay are (2at, at2)
  • Parametric Equations of Parabola x2 = 4ay are x = 2at, y = at2

Standard Equation of Parabola x2 = -4ay

  • Parametric Coordinates of the Parabola x2 = 4ay are (2at, -at2)
  • Parametric Equations of Parabola x2 = 4ay are x = 2at, y = -at2

Standard Equation of Parabola (y-k)2 = 4a(x-h)

Parametric Equations of Parabola (y-k)2 = 4a(x-h) are x=h+at2, and y = k+2at

Solved Examples on finding the Parametric Equations of a Parabola

1. Write the Parametric Equations of the Parabola y2 = 16x?

Solution:

Given Equation is in the form of y2 = 4ax

On Comparing the terms we have the 4a = 16

a = 4

The formula for Parametric Equations of the given parabola is x = at2 and y = 2at

Substitute the value of a to get the parametric equations i.e. x = 4t2 and y = 2*4*t = 8t

Therefore, Parametric Equations of Parabola y2 = 16x are x= 4t2 and y = 8t

2. Write the Parametric Equations of Parabola x2 = 12y?

Solution:

Given Equation is in the form of x2 = 4ay

On Comparing the terms we have the 4a = 12

a = 3

The formula for Parametric Equations of the given parabola is x = 2at, and y =  at2

Substitute the value of a to get the parametric equations i.e. x = 2*3*t and y = 3t2

Therefore, Parametric Equations of Parabola x2 = 12y are x = 6t and y = 3t2

3. Write the Parametric Equations of the Parabola (y-3)2 =8(x-2)?

Solution:

Given Equation is in the form of (y-k)2 = 4a(x-2)

Comparing the two equations we have k = 3, h = 2 and 4a = 8 i.e. a =2

The Formula for Parametric Equations of Parabola (y-k)2 = 4a(x-h) are x=h+at2, and y = k+2at

substitute the values of k, a in the formula and obtain the parametric equation

x = 2+2t2 and y = 3+2*2t

x = 2+2t2 and y = 3+4t

Therefore, Parametric Equations of the Parabola (y-3)2 =8(x-2) are x = 2+2t2 and y = 3+4t

Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations

Check out the detailed solutions to Big Ideas Math Grade 8 Chapter 5 Systems of Linear equations here. By solving the questions from Big Ideas Math 8th Grade Chapter 5 Systems of Linear Equations Answer Key, you can understand solving systems of linear equations by graphing, solving systems of linear equations by substitution, solving systems of linear equations by elimination, and solving special systems of linear equations. We have included the answers to all the problems in an easy manner. Learn the strategies and simple ways to solve the systems of linear equations with examples on from Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations.

Big Ideas Math Book 8th Grade Answer Key Chapter 5 Systems of Linear Equations

Quick and easy learning is possible with our best Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations. Get Big Ideas Math Book 8th Grade Solution Key Chapter 5 Systems of Linear Equations and understand the concept. This answer key is prepared and well explained by our math experts. Just download Big Ideas Math Book Grade 8 5th Chapter Systems of Linear Equations Answer Key and prepare well, score highest marks in the exam.

Performance

Lesson: 1 Solving Systems of Linear Equations by Graphing

Lesson: 2 Solving Systems of Linear Equations by Substitution

Lesson: 3 Solving Systems of Linear Equations by Elimination

Lesson: 4 Solving Special Systems of Linear Equations

Chapter: 5 – Systems of Linear Equations

Systems of Linear Equations STEAM Video/Performance

STEAM Video

Gold Alloys

An alloy is a mixture of different metals melted together at high temperatures. A dental filling is created using a gold alloy. What are other uses of alloys?
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 1.1

Watch the STEAM Video “Gold Alloys.” Then answer the following questions.

Question 1.
Enid says that the proportion of gold in an alloy can be measured in karats. For example, 24 karats represent 100% gold and 18 karats represent 75% gold.
a. A dental filling is 9 karats. What percent of the filling is gold?
b. A watch is 60% gold. How many karats is the watch?

Answer:
a. 37.5% of the filling is gold.
b. 14.4 karats are the watch

Explanation:
a. A dental filling is 9 karats.
24 karats represent 100% gold
So, 9 karats = (100 * 9)/24
= 900/24
= 37.5%
37.5% of the filling is gold.
b. A watch is 60% gold
24 karats represent 100% gold
So, watch = (60 * 24)/100
= 1440/100
= 14.4
14.4 karats are the watch

Question 2.
What percent gold is each described alloy?
a. A mixture of 2 grams 10-karat gold and 2 grams 14-karat gold
b. A mixture of 6 grams 24-karat gold and 4 grams 9-karat gold

Answer:
a. 200%
b. 750%

Explanation:
a. 24 karats represents 100% gold
10 karat gold is 125/3 %
14 -karat gold gold is 175/3 %
2 grams 10-karat gold = 2(125/3) = 250/3
2 grams 14-karat gold = 2(175/3) = 350/3
The alloy mixture = (250 + 350)/3
= 600/3 = 200
b. 6 grams 24-karat gold
6 grams = 600%
9-karat gold = 75/2
4 grams 9-karat gold = 4(75)/2 = 150%
The alloy mixture = 600% + 150% = 750%

Performance Task

Mixing Alloys

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given a list of gold alloys available at a jewelry store.
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 1
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 2
You will use a system of equations to determine the amounts of the given alloys that a jeweler needs to create a new alloy. Why might a jeweler need to create a mixture with a specific proportion of gold?

Systems of Linear Equations Getting Ready for Chapter 5

Getting Ready for Chapter 5

Chapter Exploration

Question 1.
Work with a partner. Your family starts a bed-and-breakfast. You spend $500 fixing up a bedroom to rent. The cost for food and utilities is $10 per night. Your family charges $60 per night to rent the bedroom.
a. Write an equation that represents the costs.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 3
b. Write an equation that represents the revenue (income).
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 4
c. A set of two (or more) linear equations is called a system of linear equations. Write the system of linear equations for this problem.

Answer:
a. C = 10x + 500
b. R = 60x
c. The system of linear equations are
C = 10x + 500
R = 60x

Explanation:
a. Cost, C = $10 per night . Number of nights, x + $500
C = 10x + 500
b.
Revenue, R = $60 per night . Number of nights x
R = 60x
c. The system of linear equations are
C = 10x + 500
R = 60x

Question 2.
Work with a partner. Use a graphing calculator to solve the system.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 5
a. Enter the equations into your calculator. Then graph the equations. What is an appropriate window?
b. On your graph, how can you determine which line is the graph of which equation? Label the equations on the graph shown.
c. Visually estimate the point of intersection of the graphs.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 6
d. To find the solution, use the intersect feature to find the point of intersection. The solution is Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 7

Answer:
The solution is (10, 600)

Explanation:
 Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations
The solution is (10, 600)

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 7.1

Lesson 5.1 Solving Systems of Linear Equations by Graphing

EXPLORATION 1
Work with a partner. You charge your headphones and your phone. The equations below represent the battery powers p% of the devices after x minutes of charging.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 8
a. You check the battery power of each device every 10 minutes. Copy and complete the table. How do the devices’ battery powers compare?
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 8.1
b. After how much time do the devices have the same battery power? What is the battery power at that time? Justify your answer.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 9
c. The solutions of a linear equation are all the points on its graph. How many solutions can two linear equations share? Explain your reasoning.
d. Graph the battery power equations in the same coordinate plane. What do you notice?
e. Use a graphing calculator to check your answers in part(b). Explain your method.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 10

Answer:
a. Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 30
d. The solution is (75/2, 125/2)

Explanation:
a. Headphones equation is p = 5/3 x
p1 = 5/3 (10) = 50/3, p2 = 5/3 (20) = 100/3, p3 = 5/3 (30) = 50, p4 = 5/3 (40) = 200/3, p5 = 5/3 (50) = 250/3
p6 = 5/3 (60) = 100
Phone equation is p = x + 25
p1 = 10 + 25 = 35, p2 = 20 + 25 = 45, p3 = 30 + 25 = 55, p4 = 40 + 25 = 65
p5 = 50 + 25 = 75, p6 = 60 + 25 = 85
b. Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 31
The solution is (75/2, 125/2)
p = 5/3 x, p = x + 25
Put x = 75/2
p = 5/3 (75/2)
= 125/2
p = 75/2 + 25
= 125/2

5.1 Lesson

Try It

Solve the system by graphing.

Question 1.
y = x – 1
y  = -x + 3

Answer:
The solution is (1, 2).

Explanation:
The given systems of linear equations are y = x – 1, y  = -x + 3
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 1
The graphs appear to intersect at (1, 2)
So, the solution is (1, 2)

Question 2.
y = -5x + 14
y = x – 10

Answer:
The solution is (4, -6)

Explanation:
The given systems of linear equations are y = -5x + 14, y = x – 10
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 2
The lines intersect at (4, -6)
So, the solution is (4, -6)

Question 3.
y = x
y = 2x + 1

Answer:
The solution is (-1, -1)

Explanation:
The given systems of linear equations are y = x, y = 2x + 1
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 3
The lines intersect at (-1, -1)
So, the solution is (-1, -1)

Question 4.
y = -4x – 7
x + y = 2

Answer:
The solution is (-3, 5)

Explanation:
The given systems of linear equations are y = -4x – 7, x + y = 2
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 4
The lines intersect at (-3, 5)
So, the solution is (-3, 5)

Question 5.
x – y = 5
-3x +  y = -1

Answer:
The solution is (-2, -7)

Explanation:
The given systems of linear equations are x – y = 5, -3x +  y = -1
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 5
The lines intersect at (-2, -7)
So, the solution is (-2, -7)

Question 6.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 16

Answer:
The solution is (4, -8)

Explanation:
The given systems of linear equations are 1/2 x + y = -6, 6x + 2y = 8
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 7
The lines intersect at (4, -8)
So, the solution is (4, -8)

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

SOLVING A SYSTEM OF LINEAR EQUATIONS
Solve the system by graphing.

Question 7.
y = x + 1
y = 4x +1

Answer:
The solution is (0, 1).

Explanation:
The given systems of linear equations are y = x + 1, y = 4x +1
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 8
The lines intersect at (0, 1)
So, the solution is (0, 1)

Question 8.
3x – y = -1
y = -x + 5

Answer:
The solution is (1, 4).

Explanation:
The given systems of linear equations are 3x – y = -1, y = -x + 5
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 9
The lines intersect at (1, 4)
So, the solution is (1, 4).

Question 9.
x + 2y = 3
-x + 3y = 7

Answer:
The solution is (-1, 2).

Explanation:
The given systems of linear equations are x + 2y = 3, -x + 3y = 7
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 10
The lines intersect at (-1, 2)
So, the solution is (-1, 2).

Question 10.
WRITING
Explain why the solution of a system of linear equations is the point of intersection of their graphs.

Answer:
The solution of a system of linear equations in two variables is an ordered pair that is a solution of each equation in the system. The ordered pair is obtained by drawing a graph for two equations and the point of intersection.

Question 11.
DIFFERENT WORDS, SAME QUESTION
Which is different? Find “both” answers.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 20

Answer:
The solution of the system is (1, 6)
The graphs of the equations intersect at (1, 6)
Ordered pair (1, 6) makes both equations true

Explanation:
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 32
The point of intersection is the solution.
The solution is (1, 6).

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 12.
Your family attends a comic convention. Each autograph costs $20 and each photograph costs $50. Your family buys a total of 5 autographs and photographs for $160. How many photographs does your family buy?
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 21

Answer:
The number of photographs the family buys will be 2.

Explanation:
Each autograph costs $20 and photograph costs $50
According to the question, the family buys a total of 5 items including both autographs and photographs for $160.
Let’s assume the number of autographs bought be x, while the number of photographs is 5 – x
Thus, cost to buy autographs will be 20* x = 20x and photographs will be 50(5 – x) = 250 – 50x
now, total cost becomes 20x + 250 – 50x, which will be equal to 160. lets find number of each items x
20x + 250 – 50x = 160
-30x = 160 – 250
-30x = -90
x = 3
Hence, the cost of buying autographs will be $20 * 3 = $60 and the cost to buy photographs will be (250 – 50) * 3 = 250 – 150 = $100
The number of photographs the family buys will be 5 – 3 = 2

Question 13.
DIG DEEPER!
Two apps on your phone take away points for using your phone at school. You have 140 points on the first app and 80 points on the second app when a school day begins. Each time you check your phone, you lose 10 points on your first app and p points on your second app. After you check your phone ten times, you have the same number of points on each app. Find the value of p.

Answer:
p = 4

Explanation:
App one has 140 points and 80 points on the second app
As per the question, app one loses 10 points for each check while app second loses p points. We check the phone 10 times
So, total points lost by app 1 will be 10 * 10 = 100, and points lost by app 2 will be p * 10 = 10p
After 10 times both apps are left with the same points.
So let’s find the value of p
Thus, points left in the first app will be 140 – 100 = 40 while in the second app will be 80 – 10p
40 = 80 – 10p
10p = 80 – 40
10p = 40
p = 40/10
p = 4

Solving Systems of Linear Equations by Graphing Homework & Practice 5.1

Review & Refresh

Write an equation in point-slope form of the line that passes through the given point and has the given slope.

Question 1.
(3, -4); m = 1

Answer:
y = x – 7

Explanation:
Given that,
(3, -4); m = 1
x₁ = 3, y₁ = -4
Slope intercept form of a line is (y – y₁) = m(x – x₁)
Therefore, (y – (-4)) = 1(x – 3)
(y + 4) = (x – 3)
y = x – 3 – 4
y = x – 7

Question 2.
(5, 6); m = \(\frac{3}{5}\)

Answer:
5y = 3x + 15

Explanation:
Given that,
(5, 6); m = \(\frac{3}{5}\)
x₁ = 5, y₁ = 6
Slope intercept form of a line is (y – y₁) = m(x – x₁)
Therefore, (y – 6) = 3/5(x – 5)
5(y – 6) = 3(x – 5)
5y – 30 = 3x – 15
5y = 3x – 15 + 30
5y = 3x + 15

Question 3.
(1, 10); m = –\(\frac{1}{4}\)

Answer:
4y = 41 – x

Explanation:
Given that,
(1, 10); m = –\(\frac{1}{4}\)
x₁ = 1, y₁ = 10
Slope intercept form of a line is (y – y₁) = m(x – x₁)
Therefore, (y – 10) = -1/4(x – 1)
4(y – 10) = -1(x – 1)
4y – 40 = -x + 1
4y = -x + 1 + 40
4y = 41 – x

Solve the equation. Check your solution

Question 4.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 21.1

Answer:
c = 8

Explanation:
Given equation is 3/4 c – 1/4 c + 3 = 7
2/4 c + 3 = 7
1/2 c = 7 – 3
1/2 c = 4
c = 4 * 2
c = 8
Substituting c = 8 in 3/4 c – 1/4 c + 3 = 7
3/4 (8) – 1/4 (8) + 3 = 6 – 2 + 3
= 9 – 2 = 7

Question 5.
5(2 – y) + y = -6

Answer:
y = 4

Explanation:
Given equation is 5(2 – y) + y = -6
10 – 5y + y = -6
10 – 4y = -6
10 + 6 = 4y
4y = 16
y = 16/4
y = 4
Substituting y = 4 in 5(2 – y) + y = -6
5(2 – 4) + 4 = 5(-2) + 4
= -10 + 4 = -6

Question 6.
6x – 3(x + 8) = 9

Answer:
x = 11

Explanation:
Given equation is 6x – 3(x + 8) = 9
6x – 3x – 24 = 9
3x – 24 = 9
3x = 9 + 24
3x = 33
x = 33/3
x = 11
Substituting x = 11 in 6x – 3(x + 8) = 9
6(11) – 3(11 + 8) = 66 – 3(19)
= 66 – 57 = 9

Concepts, Skills, &Problem Solving
USING A GRAPH TO SOLVE A PROBLEM
The equations below represent the numbers y of tickets sold after x weeks for two different local music festivals. (See Exploration 1, p. 199.)
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 22

Question 7.
You check the ticket sales for both festivals each week for 10 weeks. Create a table for the ticket sales each week. How do the festivals’ ticket sales compare?

Answer:
We can say that tickets for the country Music Festival sold more by mid the 4th week. After the mid of 4th-week tickets for the Pop Music Festival sold more than the country music festival.

Explanation:
Ticket sales for a country music festival and pop music festival are calculated in the tables below for 10 weeks
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 33
From the above tables, we can say that tickets for the country Music Festival sold more by mid of the 4th week. After the mid of 4th-week tickets for the Pop Music Festival sold more than the country music festival.

Question 8.
After how much time have the same number of tickets been sold for both festivals? What is the number of tickets sold at that time?

Answer:
The number of tickets sold for both festivals each is 185.

Explanation:
Country music festival y = 10x + 150
Pop musuic festival y = 20x + 115
As per the question, we will equate both the equation to get x value
10x + 150 = 20x + 115
10x – 20x = 115 – 150
-10x = -35
x = 35/10
x = 3.5
Thus, we get 3.5 weeks after which both festivals would sold equal number of tickets.
Let us find the number of tickets at that time
Put x = 3.5 in y = 10x + 150
y = 10(3.5) + 150
= 35 + 150 = 185
Hence, the number of tickets sold for both festivals each are 185.

SOLVING A SYSTEM OF LINEAR EQUATIONS
Solve the system by graphing.

Question 9.
y = 2x + 9
y = 6 – x

Answer:
The solution is (-1, 7)

Explanation:
The given systems of linear equations are y = 2x + 9, y = 6 – x
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 11
The lines intersect at (-1, 7)
So, the solution is (-1, 7)

Question 10.
y = -x – 4
y = \(\frac{3}{5}\)x + 4

Answer:
The solution is (-5, 1).

Explanation:
The given systems of linear equations are y = -x – 4, y = \(\frac{3}{5}\)x + 4
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 12
The lines intersect at
So, the solution is (-5, 1)

Question 11.
y = 2x + 5
y = \(\frac{1}{2}\)x – 1

Answer:
The solution is (-4, -3).

Explanation:
The given systems of linear equations are y = 2x + 5, y = \(\frac{1}{2}\)x – 1
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 13
The lines intersect at (-4, -3)
So, the solution is (-4, -3).

Question 12.
x + y = 27
y = x + 3

Answer:
The solution is (12, 15).

Explanation:
The given systems of linear equations are x + y = 27, y = x + 3
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 14
The lines intersect at (12, 15)
So, the solution is (12, 15).

Question 13.
y – x = 17
y = 4x + 2

Answer:
The solution is (5, 22)

Explanation:
The given systems of linear equations are y – x = 17, y = 4x + 2
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 15
The lines intersect at (5, 22)
So, the solution is (5, 22)

Question 14.
x – y = 7
0.5x + y =5

Answer:
The solution is (8, 1).

Explanation:
The given systems of linear equations are x – y = 7, 0.5x + y =5
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 16
The lines intersect at (8, 1)
So, the solution is (8, 1)

USING A GRAPHING CALCULATOR
Use a graphing calculator to solve the system.

Question 15.
2.2x + y = 12.5
1.4x – 4y =1

Answer:
The solution is (5, 1.5).

Explanation:
The given systems of linear equations are 2.2x + y = 12.5, 1.4x – 4y = 1
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 17
The lines intersect at (5, 1.5)
So, the solution is (5, 1.5)

Question 16.
2.1x + 4.2y = 14.7
-5.7x – 1.9y = -11.4

Answer:
The solution is (1, 3)

Explanation:
The given systems of linear equations are 2.1x + 4.2y = 14.7, -5.7x – 1.9y = -11.4
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 18
The lines intersect at (1, 3)
So, the solution is (1, 3)

Question 17.
-1.1x – 5.5y = -4.4
0.8x – 3.2y = -11.2

Answer:
The solution is (-6, 2)

Explanation:
The given systems of linear equations are -1.1x – 5.5y = -4.4, 0.8x – 3.2y = -11.2
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 19
The lines intersect at (-6, 2)
So, the solution is (-6, 2)

Question 18.
YOU BE THE TEACHER
Your friend solves the system of linear equations below. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 23
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 24

Answer:
Correct.

Explanation:
The given system of equations are y = 0.5x + 1, y = – + 7
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 34
The point of intersection is (4, 3)
So, the solution is (4, 3)

Question 19.
MODELING REAL LIFE
You have a total of 42 math and science problems for homework. You have 10 more math problems than science problems. How many problems do you have in each subject? Use a system of linear equations to justify your answer.

Answer:
26 math and 16 science problems.

Explanation:
Write the system of equations where m and s are the number of math and science problems
m + s = 42 —- (1)
m = 10 + s —- (2)
Substitute equation (2) in equation (1)
10 + s + s = 42
10 + 2s = 42
2s = 42 – 10
2s = 32
s = 32/2
s = 16
Put s = 16 in equation (2)
m = 10 + 16
m = 26
So, 26 math and 16 science problems.

Question 20.
PROBLEM SOLVING
A generator contains 60 gallons of fuel and uses 2.5 gallons per hour. A more efficient power generator contains 40 gallons of fuel and uses 1.5 gallons per hour. After how many hours do the generators have the same amount of fuel? Which generator runs longer? Justify your answers.

Answer:
After 20 hours both generators will be having an equal amount of fuel. The generator has 40 gallons of fuel that will run for a longer time.

Explanation:
A generator contains 60 gallons of fuel and uses 2.5 gallons per hour. A more efficient power generator contains 40 gallons of fuel and uses 1.5 gallons per hour
Let’s assume that after x hours, both generate will have an equal amount of fuel
Thus, fuel consumed by the first generator will be 2.5x gallons and for the more efficient generator will be 1.5x gallons
Fuel left in the less efficient generator will be (60 – 2.5x) gallons. Fuel left in a more efficient generator will be (40 – 1.5x) gallons.
60 – 2.5x = 40 – 1.5x
-2.5x + 1.5x = 40 – 60
-x = -20
x = 20
Thus, fuel left in each generator will be 60 – 2.5 * 20 = 10 gallons
Now lets find which generator will run longer.
The number of hours less efficient generator would run is equal to 60/2.5 = 24 hours
The number of hours more efficient generator would run is equal to 40/1.5 = 26.66 hours
Hence, After 20 hours both generators will be having an equal amount of fuel.
The generator has 40 gallons of fuel that will run for a longer time.

Question 21.
PROBLEM SOLVING
You and your friend are in a canoe race. Your friend is a half-mile in front of you and paddling 3 miles per hour. You are paddling 3.4 miles per hour.
a. You are 8.5 miles from the finish line. How long will it take you to catch up with your friend? your friend
b. You both maintain your paddling rates for the remainder of the race. How far ahead of your friend will you be when you cross the finish line?
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 25

Answer:
a. It takes 1 hour 25 minutes for you to catch up to your friend.
b. You 0.5 miles ahead of your friend when you finish the race.

Explanation:
a. Let d be the distance traveled by your friend, h=then you have to travel d + 0.5 to catch up since you are currently 0.5 miles behind your friends.
distance = rate * time
you is d + 0.5 = 3.4t
a friend is d = 3t
Substitute the equation for your friend into the equation for y to solve d
3t + 0.5 = 3.4t
0.5 = 3.4t – 3t
0.4t = 0.5
t = 0.5/0.4
t = 1.25
b. Time it takes you to complete the race
8.5 = 3.4t
t = 8.5/3.4
t = 2.5
Distance traveled by a friend in that time 3(2.5) = 7.5
Your friend started 8 miles from the finishing line so you will be 8 – 7.5 = 0.5 miles ahead of your friend when you finish the race.

OPEN-ENDED
Write a system of linear equations that fits the description. Use a graph to justify your answer.

Question 22.
The solution of the system is a point on the line y = -9x + 1.

Answer:
The equations have infinitely many solutions.

Explanation:
The given equation is y = -9x + 1
Multiply both sides by 2
2y = 2(-9x + 1)
2y = -18x + 2
Draw a graph for the system of equations y = -9x + 1, 2y = -18x + 2
The equations have infinitely many solutions.

Question 23.
The solution of the system is (3, -1).

Answer:
m = -1/3

Explanation:
The slope intercept form is y = mx + c
-1 = 3m
m = -1/3

Question 24.
DIG DEEPER!
A graph of a system of two linear equations is shown. Write the system of linear equations represented by the graph. What is the solution to the system?
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 25.1

Answer:
The systems of linear equations are y = -x + 2, y = 2x – 1
The solution is (1, 1)

Explanation:
for line 1,
x₁ = 0, y₁ = 2, x₂ = 2, y₂ = 0
Slope m = (y₂ – y₁)/(x₂ – x₁)
= (0 – 2)/(2 – 0)
= -2/2 = -1
Slope intercept form of a line is y = mx + b
The line passes through (0, 2)
2 = 0(x) + b
b = 2
y = -1(x) + 2
y = -x + 2
For line 2
x₁ = 2, y₁ = 3, x₂ = 0, y₂ = -1
Slope m = (-1 – 3)/(0 – 2)
= -4/-2
= 2
y = 2x + b
The line passes through (0, -1)
-1 = 2(0) + b
n = -1
The slope intercept form of a line is y = 2x – 1
So, the systems of linear equations are y = -x + 2, y = 2x – 1
The point of intersection in the graph is (1, 1)

Question 25.
CRITICAL THINKING
Your friend is trying to grow her hair as long as her cousin’s hair. The table shows their hair lengths (in inches) in different months.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 26
a. Write a system of linear equations that represents this situation. Let x = 1 represent January.
b. Will your friend’s hair ever be as long as her cousin’s hair? If so, in what month?

Answer:
a. The equation of the line for a friend’s hair is y = 1/2 x + 5/2, the equation of the line for a friend’s cousin hair is y =2/5x + 29/5
b. After 33 months length of hair of both girls would be the same.

Explanation:
a. The length of hair is represented by y coordinate and month is represented as x coordinate
The slope formula when we know two points is m = (y₂ – y₁)/(x₂ – x₁)
x₁ = 3, y₁ = 4, x₂ = 8, y₂ = 6.5
m = (6.5 – 4)/(8 -3)
= 2.5/5
m = 1/2
The slope form of a line is (y – y₁) = m(x – x₁)
y – 4 = 1/2 (x – 3)
2(y – 4) = (x – 3)
2y – 8 = x – 3
2y = x – 3 + 8
2y = x + 5
y = 1/2 x + 5/2
Hence, the equation of line for friend’s hair is y = 1/2 x + 5/2
Equations of line for friend’s cousin hair
x₁ = 3, y₁ = 7, x₂ = 8, y₂ = 9
m = (y₂ – y₁)/(x₂ – x₁)
= (9 – 7)/(8 – 3)
= 2/5
Now to find the equation of line for friends cousin hair
(y – y₁) = m(x – x₁)
(y – 7) = 2/5 (x – 3)
5(y – 7) = 2(x – 3)
5y – 35 = 2x – 6
5y = 2x – 6 + 35
5y = 2x + 29
y = 2/5x + 29/5
Hence, the equation of line for friend’s cousin hair is y =2/5x + 29/5
b.
2/5x x + 29/5 = 1/2 x + 5/2
2/5 x – 1/2x =5/1 – 29/5
(4x – 5x)/10 = (25 – 58)/10
-x/10 = -33/10
x = 33
Hence, after 33 months length of hair of both girls would be same.

Question 26.
REASONING
Is it possible for a system of two linear equations to have multiple solutions? Explain your reasoning.

Answer:
No, it is not possible for a system of two linear equations to have multiple solutions. Because the system of linear equations is the straight lines and those lines intersect at only one point.

Question 27.
GEOMETRY
The length of a rectangle is 8 feet more than its width. The perimeter of the rectangle is 72 feet. Find the width of the rectangle.

Answer:
The width of the rectangle is 14 ft.

Explanation:
Let us say rectangle length is l, its width is l – 8
Rectangle perimeter = 2(l + b)
72 = 2(l + l – 8)
72/2 = 2l – 8
36 = 2l – 8
2l = 36 + 8
2l = 44
l = 44/2
l = 22
rectangle width is 22 – 8 = 14 ft

Lesson 5.2 Solving Systems of Linear Equations by Substitution

EXPLORATION 1
Work with a partner.
a. Find the value of each symbol in the systems below. Compare your solution methods with other pairs of students.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 26.1
b. Use a method similar to your method in part(a) to solve the system below. Then explain how to solve a system of linear equations in two variables algebraically.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 27
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 28

EXPLORATION 2
Writing and Solving Systems of Equations
Work with a partner. Roll two number cubes that are different colors. Then write the ordered pair shown by the number cubes.
a. Write a system of linear equations that has your ordered pair as its solution. Explain how you found your system.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 29
b. Exchange systems with another pair of students. Use a method from Exploration 1 to solve the system.

5.2 Lesson

Try It

Solve the system by substitution. Check your solution.

Question 1.
y = 2x + 3
y = 5x

Answer:
The solution is (1, 1)

Explanation:
The given systems of linear equations are
y = 2x + 3 —- (i)
y = 5x —– (ii)
Substitute equation (ii) in equation (i)
5x = 2x + 3
5x – 2x = 3
3x = 3
x = 3/3
x = 1
Substitute x = 1 in equation (ii)
y = 5(1)
y = 1
So, the solution is (1, 1)

Question 2.
4x + 2y = 0
y = \(\frac{1}{2}\)x – 5

Answer:
The solution is (0, -5)

Explanation:
The given systems of linear equations are
4x + 2y = 0 —– (i)
y = \(\frac{1}{2}\)x – 5 —– (ii)
Substitute equation (ii) in equation (i)
4x + 2(1/2 x) = 0
4x + x = 0
5x = 0
x = 0
putting x = 0 in equation (ii)
y = 1/2 (0) – 5
y = -5
So, the solution is (0, -5)

Question 3.
x = 5y + 3
2x + 4y = -1

Answer:
The solution is (1/2, -1/2)

Explanation:
The given systems of linear equations are
x = 5y + 3 —- (i)
2x + 4y = -1 —- (ii)
Substitute equation (i) in equation (ii)
2(5y + 3) + 4y = -1
10y + 6 + 4y = -1
14y = -1 – 6
14y = -7
y = -7/14
y = -1/2
Put y = -1/2 in equation (i)
x = 5(-1/2) + 3
= -5/2 + 3
= 1/2
So, the solution is (1/2, -1/2)

Try It

Solve the system. Explain your choice of method.

Question 4.
y = -3x + 2
y = 2

Answer:
The solution set is (0, 2)

Explanation:
The given systems of linear equations are
y = -3x + 2 —- (1)
y = 2 —- (2)
substitute equation (2) in (1)
2 = -3x + 2
2 – 2 = -3x
-3x = 0
x = 0
So, the solution set is (0, 2)

Question 5.
4y = x
x + 4y = -8

Answer:
T4y = x
x + 4y = -8

Explanation:
The given systems of linear equations are
4y = x —– (i)
x + 4y = -8 —– (ii)
Substitute x = 4y in equation (ii)
4y + 4y = 8
8y = 8
y = 1
Put y = 1 in equation (i)
4(1) = x
x = 4
So, the solution set is (4, 1)

Question 6.
2x + 2y = 1
-x + 2y = -3

Answer:
The solution set is (4/3, -5/6)

Explanation:
The given systems of linear equations are
2x + 2y = 1 —- (i)
-x + 2y = -3
2y + 3 = x —- (ii)
Substitute equation (ii) in equation (i)
2(2y + 3) + 2y = 1
4y + 6 + 2y = 1
6y + 6 = 1
6y = 1 – 6
6y = -5
y = -5/6
Put y = -5/6 in equation (ii)
2(-5/6) + 3 = x
x = -5/3 +3
x = 4/3
So, the solution set is (4/3, -5/6)

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 7.
REASONING
Does solving a system of linear equations by graphing give the same solution as solving by substitution? Explain.

Answer:
Yes.

SOLVING A SYSTEM OF LINEAR EQUATIONS
Solve the system by substitution. Check your solution.

Question 8.
y = x – 8
y = 2x – 14

Answer:
The solution set is (6, -2)

Explanation:
The given systems of linear equations are
y = x – 8 —- (i)
y = 2x – 14 —- (ii)
Substitute equation (i) in (ii)
x – 8 = 2x – 14
-8 + 14 = 2x – x
x = 6
Putting x = 6 in equation (i)
y = 6 – 8
y = -2
Substitute x = 6, y = -2 in equation (i)
-2 = 6 – 8
So, the solution set is (6, -2)

Question 9.
x = 2y + 2
2x – 5y = 1

Answer:
The solution set is (8, 3)

Explanation:
The given systems of linear equations are
x = 2y + 2 —- (i)
2x – 5y = 1 —- (ii)
Substituting equation (i) in (ii)
2(2y + 2) – 5y = 1
4y + 4 – 5y = 1
4 – 1 = y
y = 3
Put y = 3 in equation (i)
x = 2(3) + 2
x = 8
Substitute x = 8, y = 3 in 2x – 5y = 1
2(8) – 5(3) = 16 – 15 = 1
So, the solution set is (8, 3)

Question 10.
x – 5y = 1
-2x + 9y = -1

Answer:
The solution is (-4, -1)

Explanation:
The given systems of linear equations are
x – 5y = 1
x = 1 + 5y —– (i)
-2x + 9y = -1 —- (ii)
Substitute equation (i) in (ii)
-2(1 + 5y) + 9y = -1
-2 – 10y + 9y = -1
-2 – y = -1
y = -2 + 1
y = -1
Put y = -1 in (i)
x = 1 + 5(-1)
x = 1 – 5
x = -4
Put x = -4, y = -1 in (ii)
-2(-4) + 9(-1) = 8 – 9 = -1
So, the solution is (-4, -1)

CHOOSING A SOLUTION METHOD

Solve the system. Explain your choice of method.

Question 11.
y = -x + 3
y = 2x

Answer:
The solution set is (1, 2).

Explanation:
The given systems of linear equations are
y = -x + 3
y = 2x
Equating both equations
-x + 3 = 2x
3 = 2x + x
3x = 3
x = 1
Substitute x = 1 in y = 2x
y = 2(1)
y = 2
So, the solution set is (1, 2).

Question 12.
0.5x + y = 2
0.5x = 1 + y

Answer:
The solution set is (3, 1/2).

Explanation:
The given systems of linear equations are
0.5x + y = 2 —— (i)
0.5x = 1 + y
x = 1/0.5 + y/0.5
x = 2 + 2y —- (ii)
Substitute equation (ii) in (i)
0.5(2 + 2y) + y = 2
1 + y + y = 2
2y + 1 = 2
2y = 2 – 1
2y = 1
y = 1/2
Put y = 1/2 in equation (i)
0.5x + 1/2 = 2
0.5x = 2 – 0.5
0.5x = 1.5
x = 1.5/0.5
x = 3
So the solution set is (3, 1/2).

Question 13.
x = 5y
y = 22 – 2x

Answer:
The solution set is (10, 2)

Explanation:
The given systems of linear equations are
x = 5y —- (i)
y = 22 – 2x —– (ii)
Substitute x = 5y in equation (ii)
y = 22 – 2(5y)
y = 22 – 10y
y + 10y = 22
11y = 22
y = 2
Put y = 2 in x = 5y
x = 5(2)
x = 10
So, the solution set is (10, 2)

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 14.
To stock your school store, you buy a total of 25 sweatshirts and hats for $172.50. You pay $8.00 per sweatshirt and $2.50 per hat. How many of each item do you buy?

Answer:
The number of sweatshirts you buy would be 20 and the hat you buy would be 5.

Explanation:
Let’s assume the number of sweatshirts bought be x, while the number of hats will be 25 – x
Thus, cost to buy sweatshirts is 8 * x = 8x, and for hats will be 2.5 * (25 – x) = 62.50 – 2.50x
Therefore, total cost is 8x + 62.50 – 2.5x which is equal to 172.50
8x + 62.50 – 2.5x = 172.50
5.5x = 172.50 – 62.50
5.5x = 110
x = 110/5.5
x = 20
Hence, the cost of buying sweatshirts will be $20 * 8 = $160 and the cost to buy hats will be (25 – 20)2.50 = 12.5
The number of sweatshirts you buy would be 20 and the hat you buy would be (25 – 20) = 5

Question 15.
DIG DEEPER!
The length of a volleyball court is twice its width. The perimeter of the court is 180 feet. Find the area of the volleyball court. Justify your answer.

Answer:
The area of the rectangle will be 1800 sq ft.

Explanation:
The perimeter of the volleyball court is 180 feet. The length of the court is twice its width
Let us take the length and width of the court be 2x, x
Perimeter = 2(l + b)
As per the question
2(l + b) = 180
2x + x = 90
3x = 90
x = 90/3
x = 30
Thus, the length of court is 2 . 30 = 60 and width is 30 ft
Area = length * width
= 60 * 30
= 1800
Hence, the area of the rectangle will be 1800 sq ft.

Solving Systems of Linear Equations by Substitution Homework & Practice 5.2

Review & Refresh

Solve the system by graphing.

Question 1.
y = 2x – 3
y = -x + 9

Answer:
The solution is (4, 5)

Explanation:
The given systems of linear equations are y = 2x – 3, y = -x + 9
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 20
The lines intersect at (4, 5)
So, the solution is (4, 5)

Question 2.
6x + y = -2
y = -3x + 1

Answer:
The solution is (-1, 4)

Explanation:
The given systems of linear equations are 6x + y = -2, y = -3x + 1
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 21
The lines intersect at (-1, 4)
So, the solution is (-1, 4)

Question 3.
4x + 2y = 2
3x = 4 – y

Answer:
The solution is (3, -5).

Explanation:
The given systems of linear equations are 4x + 2y = 2, 3x = 4 – y
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 22
The lines intersect at (3, -5)
So, the solution is (3, -5)

Question 4.
Use the figure to find the measure of ∠2
A. 17°
B. 73°
C. 83°
D. 107°

Answer:
B. 73°

Explanation:
Given that,
∠1 = 107 degrees
∠1 + ∠2 = 180
107 + ∠2 = 180
∠2 = 180 – 107
= 73 degrees

Concepts, Skills, &Problem Solving
SOLVING A SYSTEM ALGEBRAICALLY
Find the value of each symbol in the system. (See Exploration 1, p. 205.)

Question 5.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 30

Question 6.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 31

SOLVING A SYSTEM OF LINEAR EQUATIONS
Solve the system by substitution. Check your solution.

Question 7.
y = x – 4
y = 4x – 10

Answer:
The solution set is (2, -2).

Explanation:
The given systems of linear equations are
y = x – 4 —- (i)
y = 4x – 10 —– (ii)
Substitute (ii) in (i)
4x – 10 = x – 4
4x – x = -4 + 10
3x = 6
x = 2
Put x = 2 in (i)
y = 2 – 4
y = -2
Substitute x = 2, y = -2 in equation (ii)
-2 = 4(2) – 10
= 8 – 10
So, the solution set is (2, -2).

Question 8.
y = 2x + 5
y = 3x – 1

Answer:
The solution set is (6, 17).

Explanation:
The given systems of linear equations are
y = 2x + 5 —– (i)
y = 3x – 1 —— (ii)
Substitute (ii) in (i)
3x – 1 = 2x + 5
3x – 2x = 5 + 1
x = 6
Substitute x = 6 in (i)
y = 2(6) + 5
y = 12 + 5
y = 17
Put x = 6, y = 17 in (ii)
17 = 3(6) – 1
= 18 – 1
So, the solution set is (6, 17)

Question 9.
x = 2y + 7
3x – 2y = 3

Answer:
The solution set is (-2, -9/2).

Explanation:
The given systems of linear equations are
x = 2y + 7 —– (i)
3x – 2y = 3  —– (ii)
Substitute (i) in (ii)
3(2y + 7) – 2y = 3
6y + 21 – 2y = 3
4y = 3 – 21
4y = -18
y = -9/2
Substitute y = -9/2 in (i)
x = 2(-9/2) + 7
x = -9 + 7
x = -2
Put x = -2, y = -9/2 in (ii)
3(-2) – 2(-9/2) = -6 + 9 = 3
So, the solution set is (-2, -9/2)

Question 10.
4x – 2y =14
y = \(\frac{1}{2}\)x – 1

Answer:
The solution set is (4, 1)

Explanation:
The given systems of linear equations are
4x – 2y =14 —– (i)
y = \(\frac{1}{2}\)x – 1 —– (ii)
Substitute equation (i) in (ii)
4x – 2(0.5x – 1) = 14
4x – x + 2 = 14
3x = 14 – 2
3x = 12
x = 4
Substitute x = 4 in (i)
4(4) – 2y = 14
16 – 14 = 2y
2 = 2y
y = 1
Put x = 4, y = 1 in (ii)
1 = 1/2 (4) – 1
= 2 – 1
So, the solution set is (4, 1)

Question 11.
2x = y – 10
2x + 7 = 2y

Answer:
The solution set is (-13/2, -3)

Explanation:
The given systems of linear equations are
2x = y – 10
y = 2x + 10 —- (i)
2x + 7 = 2y —— (ii)
Substitute (i) in (ii)
2x + 7 = 2(2x + 10)
2x + 7 = 4x + 20
4x – 2x = 7 – 20
2x = -13
x = -13/2
Substitute x = -13/2 in (i)
y = 2(-13/2) + 10
= -13 + 10
= -3
Put x = -13/2, y = -3 in (ii)
2(-13/2) + 7 = 2(-3)
-13 + 7 = -6
So, the solution set is (-13/2, -3)

Question 12.
8x – \(\frac{1}{3}\)y = 0
12x + 3 =y

Answer:
The solution set is (1/4, 6).

Explanation:
The given systems of linear equations are
8x – \(\frac{1}{3}\)y = 0 —– (i)
12x + 3 =y —— (ii)
Substitute (ii) in (i)
8x – 1/3(12x + 3) = 0
8x – 4x – 1 = 0
4x – 1 = 0
4x = 1
x = 1/4
Substitute x = 1/4 in (ii)
12(1/4) + 3 = y
3 + 3 = y
y = 6
Put x = 1/4, y = 6 in (ii)
12(1/4) + 3 = 3 + 3 = 6
So, the solution set is (1/4, 6).

Question 16.
MODELING REAL LIFE
There are a total of 64 students in a film making club and a yearbook club. The filmmaking club has 14 more students than the yearbook club.
a. Write a system of linear equations that represents this situation.
b. How many students are in the film making club? the yearbook club?
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 32

Answer:
a. x + y = 64, x = y + 14
b. The number of students in the filmmaking club and yearbook club are 39 and 25 respectively.

Explanation:
a. Let us take the number of students in film making club to be x and the number of students in the yearbook club be y
x + y = 64 —- (i)
It is given that the number of students in the filmmaking club is greater than students in the yearbook club by 14
So, x = y + 14 —- (ii)
b. Put equation (ii) in (i)
y + 14 + y = 64
2y + 14 = 64
2y = 64 – 14
2y = 50
y = 25
Substitute y = 25 in (i)
x + 25 = 64
x = 64 – 25
x = 39
Hence, the number of students in filmmaking club and yearbook club are 39 and 25 respectively.

Question 17.
MODELING REAL LIFE
A drama club earns $1040 from production by selling 64 adult tickets and 132 student tickets. An adult ticket costs twice as much as a student ticket.
a. Write a system of linear equations that represents this situation.
b. What is the cost of each ticket?

Answer:
a. 64a + 132s = 1040, a = 2s
b. The price of a student ticket is $4, adult ticket is $8.

Explanation:
Write the system of equations that models the problem where a is the price of an adult ticket and s is the price of a student ticket
64a + 132s = 1040
a = 2s
b. Substitute a = 2s in 64a + 132s = 1040
64(2s) + 132s = 1040
128s + 132s = 1040
260s = 1040
s = 1040/260
s = $4 per student ticket
a = 2(4) = $8 per adult ticket.

Question 18.
OPEN-ENDED
Write a system of linear equations that has the ordered pair (1, 6) as its solution.

Answer:
The system of linear equations that pass through (1, 6) is y – 6 = m(x – 1)

Explanation:
We know that the equation of a line that passes through a point is (y – y₁) = m(x – x₁)
The line pass through (1,6 )
So, y – 6 = m(x – 1)
We can get system of linear equations by inserting different values of m in above equation
If we put m = 1, then equation is y – 6 = 1(x – 1)
y – 6 = x – 1
y = x – 1 + 6
y = x + 5
Hence, the system of linear equation is y – 6 = m(x – 1) that pass through (1, 6).

CHOOSING A SOLUTION METHOD
Solve the system. Explain your choice of method.

Question 19.
y – x = 4
x + y = 6

Answer:
The solution set is (1, 5)

Explanation:
The given system of linear equations are
y – x = 4
y = 4 + x —– (i)
x + y = 6 —– (ii)
Substitute (i) in (ii)
x + 4 + x = 6
2x + 4 = 6
2x = 6 – 4
2x = 2
x = 1
Substitute x = 1 in (i)
y = 4 + 1
y = 5
So, the solution set is (1, 5)

Question 20.
0.5x + y = 4
0.5x – y =-1

Answer:
The solution set is (3, 2.5)

Explanation:
The given system of linear equations are
0.5x + y = 4 —- (i)
0.5x – y =-1
0.5x + 1 = y —- (ii)
Substitute (ii) in (i)
0.5x + 0.5x + 1 = 4
x + 1 = 4
x = 4 – 1
x = 3
Substitute x = 3 in (ii)
y = 0.5(3) + 1
= 2.5
So, the solution set is (3, 2.5)

Question 21.
y = 2x + 5
y = -3x

Answer:
The solution set is (-1, 3)

Explanation:
The given system of linear equations are
y = 2x + 5 —- (i)
y = -3x —– (ii)
Substitute (i) in (ii)
-3x = 2x + 5
-3x – 2x = 5
-5x = 5
x = -1
Substitute x = -1 in (i)
y = 2(-1) + 5
y = -2 + 5
y = 3
So, the solution set is (-1, 3)

Question 22.
CRITICAL THINKING
A system consists of two different proportional relationships. What is the solution to the system? Justify your answer.

Answer:
The solution is (0, 0).

Explanation:
The proportional relationships in the system of linear equations mean that there is no constant available in the equation.
Therefore, the lines will always pass through the origin (0, 0)
Hence, the solution is (0, 0).

Question 23.
GEOMETRY
The measure of the obtuse angle in the isosceles triangle is two and a half times the measure of one of the acute angles. Write and solve a system of linear equations to find the measure of each angle.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 33

Answer:
The systems of linear equations are x + 2y = 180°, x = 2.5y
x = 100°, y = 40°

Explanation:
In an isosceles triangle, there are two angles that are the same and one that is different. We know there is one angle that is more than 90° because that is the definition of an obtuse angle. We also know that the sum of all the angles of a triangle equals 180°.
Let’s put all this together and find the measures of the angles.
x + 2y = 180°
x = 2.5y
Put x = 2.5 y in x + 2y = 180°
2.5y + 2y = 180°
4.5y = 180
y = 180/4.5
y = 40
Put y = 40 in x = 2.5y
x = 2.5(40)
x = 100

Question 24.
NUMBER SENSE
The sum of the digits of a two-digit number is 8. When the digits are reversed, the number increases by 36. Find the original number.

Answer:
The original number is 26.

Explanation:
Let the two-digit number be xy
The sum of the digits of a two-digit number is 8
x + y = 8
y = 8 – x —- (i)
So, the number is 10x + y
When the digits are reversed, the number increases by 36
10y + x = 36 + 10x + y
10y – y = 36 + 10x – x
9y = 36 + 9x
Put y = 8 – x
9(8 – x) = 36 + 9x
72 – 9x = 36 + 9x
72 – 36 = 18x
18x = 36
x = 36/18
x = 2
Put x = 2 in equation (i)
y = 8 – 2
y = 6
So, the original number is 26.

Question 25.
DIG DEEPER!
A hospital employs a total of 77 nurses and doctors. The ratio of nurses to doctors is 9 : 2. How many nurses are employed at the hospital? How many doctors are employed at the hospital?

Answer:
There are 63 nurses, 14 doctors are employed at the hospital.

Explanation:
Let x be the number of nurses and y be the number of doctors
A hospital employs a total of 77 nurses and doctors
x + y = 77 —- (i)
The ratio of nurses to doctors is 9:2.
x/y = 9/2
x = 9/2y
Put x = 9/2y in (i)
9/2 y + y = 77
11/2 y = 77
11y = 77 * 2
y = 14
x = 9/2 (14)
y = 63
So, there are 63 nurses, 14 doctors are employed at the hospital.

Question 26.
REPEATED REASONING
A DJ has a total of 1075 dance, rock, and country songs on her system. The dance selection is three times the rock selection. The country selection has 105 more songs than the rock selection. How many songs on the system is dance? rock? country?
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 34

Answer:
582 dance, 299 countries, and 194 ock songs on the system.

Explanation:
Let d, r, c be the number of dance, rock and country songs
A DJ has a total of 1075 dance, rock, and country songs on her system
d + r + c = 1075 —- (i)
The dance selection is three times the size of the rock selection
d = 3r —- (ii)
The country selection has 105 more songs than the rock selection
c = 105 + r —– (iii)
Put (ii), (iii) in (i)
3r + r + 105 + r = 1075
5r = 1075 – 105
5r = 970
r = 194
d = 3(194) = 582
c = 105 + 194 = 299
So, 582 dance, 299 country, and 194 ock songs on the system.

Lesson 5.3 Solving Systems of Linear Equations by Elimination

EXPLORATION 1
Work with a partner. A student found the value of in the system using substitution as shown.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 35
a. Find another way to obtain the equation 4x = -4 from the original system. Does your method produce an equation in one variable for any system? Explain.
b. Can you use your method in part(a) to solve each system below? If so, solve the system. If not, replace one of the equations with an equivalent equation that allows you to use your method in part(a). Then solve the system.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 36
c. Compare your solution methods in part(b) with other pairs of students.

5.3 Lesson

Try It

Solve the system by elimination. Check your solution.

Question 1.
2x – y = 9
4x + y =21

Answer:
The solution set is (5, 1).

Explanation:
The given system of linear equations are
2x – y = 9 —- (i)
4x + y =21 —– (ii)
Add both equations
2x – y + 4x + y = 9 + 21
6x = 30
x = 30/6
x = 5
Put x = 5 in (i)
2(5) – y = 9
10 – y = 9
10 – 9 = y
y = 1
Substitute x = 5, y = 1 in (ii)
4(5) + 1 = 20 + 1 = 21
So, the solution set is (5, 1).

Question 2.
-5x + 2y = 13
5x + y = -1

Answer:
The solution set is (-1, 4).

Explanation:
The given system of linear equations are
-5x + 2y = 13 —– (i)
5x + y = -1 —— (ii)
Add both equations
-5x + 2y + 5x + y = 13 – 1
3y = 12
y = 4
Substitute y = 4 in (ii)
5x + 4 = -1
5x = -1 – 4
5x = -5
x = -1
Substitute x = -1, y = 4 in (i)
-5(-1) + 2(4) = 5 + 8 = 13
So, the solution set is (-1, 4).

Question 3.
3x + 4y = -6
7x + 4y = -14

Answer:
The solution set is (-2, 0).

Explanation:
The given system of linear equations are
3x + 4y = -6 —- (i)
7x + 4y = -14 —– (ii)
Subtract (ii) from (i)
3x + 4y – (7x + 4y) = -6 + 14
3x + 4y – 7x – 4y = 8
-4x = 8
x = -8/4
x = -2
Substitute x = -2 in (i)
3(-2) + 4y = -6
-6 + 4y = -6
4y = -6 + 6
y = 0
Substitute x = -2, y = 0 in (ii)
7(-2) + 4(0) = -14
So, the solution set is (-2, 0).

Try It

Solve the system by elimination. Check your solution.

Question 4.
3x + y = 11
6x + 3y = 24

Answer:
The solution set is (3, 2)

Explanation:
The given system of linear equations are
3x + y = 11 —— (i)
6x + 3y = 24 ——- (ii)
Divide equation (ii) by 1/3
1/3(6x + 3y = 24)
2x + y = 8 —- (iii)
Subtract (iii) from (i)
3x + y – (2x + y) = 11 – 8
3x + y – 2x – y = 3
x = 3
Put x = 3 in (i)
3(3) + y = 11
9 + y = 11
y = 11 – 9
y = 2
Substitute x = 3, y = 2 in (iii)
2(3) + 2 = 6 + 2 = 8
So, the solution set is (3, 2)

Question 5.
4x – 5y = -19
-x – 2y = 8

Answer:
The solution set is (-6, -1)

Explanation:
The given system of linear equations are
4x – 5y = -19 —– (i)
-x – 2y = 8 —— (ii)
Multiply both sides of equation (ii) by 4
4(-x – 2y = 8)
-4x – 8y = 32 —- (iii)
Add (i) & (iii)
4x – 5y – 4x – 8y = -19 + 32
-13y = 13
y = -1
Substitute y = -1 in (ii)
-x – 2(-1) = 8
-x + 2 = 8
-x = 8 – 2
x = -6
Substitute x = -6, y = -1 in (ii)
-(-6) – 2(-1) = 6 + 2 = 8
So, the solution set is (-6, -1)

Question 6.
5y = 15 – 5x
y = -2x + 3

Answer:
The solution set is (0, 3)

Explanation:
The given system of linear equations are
5y = 15 – 5x —– (i)
y = -2x + 3 —– (ii)
Divide equation (i) by 1/5
1/5(5y = 15 – 5x)
y = 3 – x —– (iii)
Subtract (iii) from (ii)
y – y = -2x + 3 – (3 – x)
0 = -2x + 3 – 3 + x
-x = 0
x = 0
Substitute x = 0 in (iii)
y = 3 – 0
y = 3
Substitute x = 0, y = 3 in (ii)
3 = -2(0) + 3
So, the solution set is (0, 3)

Try It

Question 7.
Change one word in Choice B so that it represents an efficient approach to solving the system.

Answer:
Multiply equation (i) by -1 and subtract the equations.

Explanation:
The given system of linear equations are
x – 2y = 6 —- (i)
-x + 4y = 6 —- (ii)
multiply equation (i) by -1 and subtract the equations.
-1(x – 2y = 6)
-x + 2y = -6
-x + 2y – (-x + 4y) = -6 – 6
-x + 2y + x – 4y = -12
-2y = -12
y = 6
Put y = 6 in (i)
x – 2(6) = 6
x – 12 = 6
x = 6 + 12
x = 18

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

SOLVING A SYSTEM OF LINEAR EQUATIONS
Solve the system by elimination. Check your solution.

Question 8.
2x + y = 4
-2x + 2y = 5

Answer:
The solution set is (1/2, 3)

Explanation:
The given system of linear equations are
2x + y = 4 —— (i)
-2x + 2y = 5 —– (ii)
Add equations
2x + y – 2x + 2y = 4 + 5
3y = 9
y = 3
Substitute y = 3 in (i)
2x + 3 = 4
2x = 4 – 3
2x = 1
x = 1/2
Substitute x = 1/2, y = 3 in (ii)
-2(1/2) + 2(3) = -1 + 6 = 5
So, the solution set is (1/2, 3)

Question 9.
-x + y = 1
-3x + y =7

Answer:
The solution set is (-3, -2).

Explanation:
The given system of linear equations are
-x + y = 1 —- (i)
-3x + y =7 —– (ii)
Subtract equations
-x + y – (-3x + y) = 1 – 7
-x + y + 3x – y = -6
2x = -6
x = -3
Substitute x = -3 in (ii)
-3(-3) + y = 7
9 + y = 7
y = 7 – 9
y = -2
Substitute x = -3, y = -2 in (i)
-(-3) – 2 = 3 – 2 = 1
So, the solution set is (-3, -2).

Question 10.
y = -2x + 3
4x – 5y = 13

Answer:
The solution set is (2, -1).

Explanation:
The given system of linear equations are
y = -2x + 3
2x + y = 3 —– (i)
4x – 5y = 13 —– (ii)
Multiply equation (i) by 2 and subtract
2(2x + y = 3)
4x + 2y = 6
4x + 2y – (4x – 5y) = 6 – 13
4x + 2y – 4x + 5y = -7
7y = -7
y = -1
Substitute y = -1 in (i)
2x – 1 = 3
2x = 4
x = 2
Substitute x = 2, y = -1 in (ii)
4(2) – 5(-1) = 8 + 5 = 13
So, the solution set is (2, -1).

CHOOSING A SOLUTION METHOD
Solve the system. Explain your choice of method.

Question 11.
y = 6x – 1
y = 3x – 4

Answer:
The solution set is (-1, -7)

Explanation:
The given system of linear equations are
y = 6x – 1 —- (i)
y = 3x – 4 —– (ii)
Equating both the equations
6x – 1 = 3x – 4
6x – 3x = -4 + 1
3x = -3
x = -3/3
x = -1
Substitute x = -1 in equation (i)
y = 6(-1) – 1
y = -7
So, the solution set is (-1, -7)

Question 12.
3x = y + 2
3x + 2y = 5

Answer:
The solution set is (1, 1).

Explanation:
The given system of linear equations are
3x = y + 2 —- (i)
3x + 2y = 5 —– (ii)
Substitute equation (i) in (ii)
y + 2 + 2y = 5
3y = 5 – 2
3y = 3
y = 1
Substitute y = 1 in (i)
3x = 1 + 2
3x = 3
x = 1
So, the solution set is (1, 1).

Question 13.
2x – y = 7
x + y = 5

Answer:
The solution set is (4, 1).

Explanation:
The given system of linear equations are
2x – y = 7 —- (i)
x + y = 5 —- (ii)
Add equations
2x – y + x + y = 7 + 5
3x = 12
x = 4
Substitute x = 4 in (ii)
4 + y = 5
y = 5 – 4
y = 1
So, the solution set is (4, 1).

Question 14.
WHICH ONE DOESN’T BELONG?
Which system does not belong with the other three? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 37

Answer:
2x + 3y = 11, 3x – 2y = 10 is does not belong to other three.

Explanation:
1. 3x + 3y = 3 —- (i)
2x – 3y = 7 —- (ii)
Add equations
3x + 3y + 2x – 3y = 3 + 7
5x = 10
x = 2
Substitute x = 2 in (ii)
2(2) – 3y = 7
4 – 3y = 7
-3y = 7 – 4
-3y = 3
y = -1
The solution set is (2, -1).
2. -2x + y = 6 — (i)
2x – 3y = -10 — (ii)
Add equations
-2x + y + 2x – 3y = 6 – 10
-2y = -4
y = 2
Substitute y = 2 in (i)
-2x + 2 = 6
-2x = 4
x = -2
The solution set is (-2, 2)
3. 2x + 3y = 11 —- (i)
3x – 2y = 10
3x = 10 + 2y
x = (10 + 2y)/3
Substitute x = (10 + 2y)/3 in (i)
2(10 + 2y)/3 + 3y = 11
20 + 4y + 9y = 33
20 + 13y = 33
13y = 33 – 20
y = 1
Put y = 1 in (i)
2x + 3 = 11
2x = 11 – 3
2x = 8
x = 4

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 15.
A fitness instructor purchases exercise bikes and treadmills for two gyms. For the first gym, 2 exercise bikes and 3 treadmills cost $2200. For the second gym, 3 exercise bikes and 4 treadmills cost $3000. How much does a treadmill cost?

Answer:
The cost of a treadmill is $600.

Explanation:
Let x and y be the price of exercise bikes and treadmills, respectively
Using the fact that
Total cost = cost of exercise bike . Number of bikes in a gym + Cost of treadmill . Number of treadmills in a gym
2x + 3y = 2200 —- (i)
3x + 4y = 3000 —– (ii)
Multiply (i) by 3 and (ii) by 2
3(2x + 3y = 2200)
6x + 9y = 6600 —- (iii)
2(3x + 4y = 3000)
6x + 8y = 6000 —– (iv)
Subtract equations
6x + 9y – 6x – 8y = 6600 – 6000
y = 600
The cost of treadmill is $600.

Question 16.
DIG DEEPER!
At your school, cooking club members raise $5 per member for a charity, and woodshop club members raise $10 per member for a different charity. The cooking club has three times as many members as the woodshop club. The difference in the number of members in the two clubs is 12 members. How much does each club raise?
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 38

Answer:
Money raised for charity by the cooking club is $90, Money raised for charity by the woodshop club $60.

Explanation:
Let x and y be the number of members of the cooking club and woodshop club
The cooking club has three times as many members as the woodshop club.
x = 3y
The difference in the number of members in the two clubs is 12 members.
x – y = 12
3y – y = 12
2y = 12
y = 6
So, x = 3(6) = 18
Money raised for charity by cooking club = Number of members in a cooking club. Charity per member of cooking club
= x . 5
= 18 . 5
= $90
Money raised for charity by woodshop club = Number of members in woodshop club. Charity per member of woodshop club
= y . 10
= 6 . 10
= $60

Solving Systems of Linear Equations by Elimination Homework & Practice 5.3

Review & Refresh

Solve the system by substitution. Check your solution.

Question 1.
x = 5 – y
x – y = 3

Answer:
The solution set is (4, 1).

Explanation:
The given system of linear equations are
x = 5 – y —- (i)
x – y = 3 —- (ii)
Substitute equation (i) in (ii)
5 – y – y = 3
5 – 2y = 3
5 – 3 = 2y
2 = 2y
y = 1
Substitute y = 1 in (i)
x = 5 – 1
x = 4
Substitute x = 4, y = 1 in (ii)
4 – 1 = 3
So, the solution set is (4, 1).

Question 2.
x – 5y = 1
-x + y = 7

Answer:
The solution set is (-9, -2)

Explanation:
The given system of linear equations are
x – 5y = 1
x = 1 + 5y —- (i)
-x + y = 7 —- (ii)
Substitute (i) in (ii)
-1 – 5y + y = 7
-1 – 4y = 7
-4y = 7 + 1
-4y = 8
y = -2
Substitute y = -2 in (i)
x = 1 + 5(-2)
x = 1 – 10
x = -9
Substitute x = -9, y = -2
-(-9) – 2 = 9 – 2 = 7
So, the solution set is (-9, -2)

Question 3.
x + 6y = -2
-x = 3y – 10

Answer:
The solution set is (22, -4).

Explanation:
The given system of linear equations are
x + 6y = -2 —- (i)
-x = 3y – 10 —- (ii)
Substitute (ii) in (i)
-3y + 10 + 6y = -2
3y + 10 = -2
3y = -2 – 10
3y = -12
y = -4
Substitute y = -4 in (ii)
-x = 3(-4) – 10
-x = -12 – 10
-x = -22
x = 22
Substitute x = 22, y = -4 in (i)
22 + 6(-4) = 22 – 24 = -2
So, the solution set is (22, -4).

The vertices of a triangle are given. Draw the triangle and its image after a dilation with the given scale factor. Identify the type of dilation.

Question 4.
A(1, 1), B(1, 3), C(3, 1); k = 2

Answer:
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 35
The new triangle is larger than the original triangle so it’s an increase.

Explanation:
The vertices of a triangle are A(1, 1), B(1, 3), C(3, 1)
Multiply the coordinates by 2 and then graph the original and new coordinates.
New coordinates are D(2, 2), E(2, 6), F(6, 2)
The new triangle is larger than the original triangle so it’s an increase.

Question 5.
D(-8, -4), E(-4, 8), F(0, 0); k = 0.5

Answer:
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 36
The new triangle is smaller than the original triangle so it’s a reduction.

Explanation:
The vertices of a triangle are D(-8, -4), E(-4, 8), F(0, 0)
Multiply the coordinates by 0.5 and then graph the original and new coordinates
The new coordinates are A(-4, -2), B(-2, 4), C(0, 0)
The new triangle is smaller than the original triangle so it’s a reduction.

Concepts, Skills, &Problem Solving
SOLVING A SYSTEM ALGEBRAICALLY
Explain how to obtain the equation 3x = 6 from the given system. (See Exploration 1, p. 211.)

Question 6.
2x + y = 5
x – y = 1

Answer:
Add both equations

Explanation:
The given system of linear equations are
2x + y = 5 —– (i)
x – y = 1 —– (ii)
Add both equations
2x + y + x – y = 5 + 1
3x = 6

Question 7.
5x + 2y = 2
x + y = -2

Answer:
Multiply the equation 2 by 2 and Subtract equation obtained equation from (i)

Explanation:
The given system of linear equations are
5x + 2y = 2 —– (i)
x + y = -2 —— (ii)
Multiply the equation 2 by 2
2(x + y) = 2(-2)
2x + 2y = -4
Subtract equation obtained equation from (i)
5x + 2y – 2x – 2y = 2 + 4
3x = 6

Question 8.
-x + y = -3
6x – 3y =15

Answer:
Multiply the equation (i) by 3, Add equations (i) & (iii)

Explanation:
The given system of linear equations are
-x + y = -3 —- (i)
6x – 3y =15 —- (ii)
Multiply the equation (i) by 3
3(-x + y = -3)
-3x + 3y = -9 —- (ii)
Add equation (i) & (iii)
6x – 3y – 3x + 3y = 15 – 9
3x = 6

SOLVING A SYSTEM OF LINEAR EQUATIONS
Solve the system by elimination. Check your solution.

Question 9.
x + 3y = 5
-x – y = -3

Answer:
The solution set is (2, 1).

Explanation:
The given system of linear equations are
x + 3y = 5 —- (i)
-x – y = -3 —– (ii)
Add equations (i) & (ii)
x + 3y – x – y = 5 – 3
2y = 2
y = 1
Substitute y = 1 in (ii)
-x – 1 = -3
-x = -3 + 1
-x = -2
x = 2
Substitute x = 2, y = 1 in (i)
2 + 3(1) = 2 + 3 = 5
So, the solution set is (2, 1).

Question 10.
x – 2y = -7
3x + 2y = 3

Answer:
The solution set is (-1, 3).

Explanation:
The given system of linear equations are
x – 2y = -7 —- (i)
3x + 2y = 3 —- (ii)
Add equations (i), (ii)
x – 2y + 3x + 2y = -7 + 3
4x = -4
x = -1
Substitute x = -1 in (i)
-1 – 2y = -7
-2y = -7 + 1
-2y = -6
y = 3
Substitute x = -1, y = 3 in (i)
-1 – 2(3) = -1 – 6 = -7
So, the solution set is (-1, 3).

Question 11.
4x + 3y = -5
-x + 3y = -10

Answer:
The solution set is(1, -3).

Explanation:
The given system of linear equations are
4x + 3y = -5 —- (i)
-x + 3y = -10 —- (ii)
Subtract equations (i), (ii)
4x + 3y – (-x + 3y) = -5 – (-10)
4x + 3y + x – 3y = -5 + 10
5x = 5
x = 1
Substitute x = 1 in (ii)
-1 + 3y = -10
3y = -10 + 1
3y = -9
y = -3
Substitute x = 1, y = -3 in (i)
4(1) + 3(-3) = 4 – 9 = -5
So, the solution set is(1, -3).

Question 12.
2x + 7y = 1
2x – 4y = 12

Answer:
The solution set is (4, -1).

Explanation:
The given system of linear equations are
2x + 7y = 1 —- (i)
2x – 4y = 12 —– (ii)
Subtract equations (i), (ii)
2x – 4y – (2x + 7y) = 12 – 1
2x – 4y – 2x – 7y = 11
-11y = 11
y = -1
Substitute y = -1 in (ii)
2x – 4(-1) = 12
2x + 4 = 12
2x = 12 – 4
2x = 8
x = 4
Substitute x = 4, y = -1 in (i)
2(4) + 7(-1) = 8 – 7 = 1
So, the solution set is (4, -1).

Question 13.
2x + 5y = 16
3x – 5y = -1

Answer:
The solution set is (3, 2).

Explanation:
The given system of linear equations are
2x + 5y = 16 —- (i)
3x – 5y = -1 —– (ii)
Add both equations
2x + 5y + 3x – 5y = 16 – 1
5x = 15
x = 3
Substitute x = 3 in (i)
2(3) + 5y = 16
5y = 16 – 6
5y = 10
y = 2
Substitute x = 3, y = 2 in (ii)
3(3) – 5(2) = 9 – 10 = -1
So, the solution set is (3, 2).

Question 14.
3x – 2y = 4
6x – 2y = -2

Answer:
The solution set is (-2, -5).

Explanation:
The given system of linear equations are
3x – 2y = 4 —– (i)
6x – 2y = -2 —– (ii)
Subtract equations
3x – 2y – (6x – 2y) = 4 – (-2)
3x – 2y – 6x + 2y = 4 + 2
-3x = 6
x = -2
Substitute x = -2 in (i)
3(-2) – 2y = 4
-6 – 2y = 4
-2y = 4 + 6
-2y = 10
y = -5
Substitute x = -2, y = -5 in (ii)
6(-2) – 2(-5) = -12 + 10 = -2
So, the solution set is (-2, -5).

Question 15.
YOU BE THE TEACHER
Your friend solves the system. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 39

Answer:
Wrong.

Explanation:
The given equations are 5x + 2y = 9 —- (i)
3x – 2y = -1 —- (ii)
Add both equations
5x + 2y + 3x – 2y = 9 – 1
8x = 8
x = 1
Put x = 1 in (ii)
3(1) – 2y = -1
3 – 2y = -1
-2y = -1 – 3
-2y = -4
y = 2
So, the solution set is (1, 2).

Question 16.
MODELING REAL LIFE
You and your friend are selling raffle tickets for a new laptop. You sell 14 more tickets than your friend sells. Together, you and your friend sell 58 tickets.
a. Write a system of linear equations that represents this situation.
b. How many tickets do each of you sell?

Answer:
a. x = 14 + y, x + y = 58
b. The number of tickers you sell is 36, the number of tickets your friend sells is 22.

Explanation:
a. Let x be the number of tickets you sell and let y be the number of tickets your friend sells.
You sell 14 more tickets than your friend sells
x = 14 + y —- (i)
Together, you and your friend sell 58 tickets.
x + y = 58 —- (ii)
b. Substitute equation (i) in (ii)
14 + y + y = 58
14 + 2y = 58
2y = 58 – 14
2y = 44
y = 44/2
y = 22
Substitute y = 22 in (i)
x = 14 + 22
x = 36
So, The number of tickers you sell is 36, the number of tickets your friend sells is 22.

Question 17.
MODELING REAL LIFE
You can jog around your block twice and the park once in 10 minutes. You can jog around your block twice and the park 3 times in 22 minutes. Write a system of linear equations that represents this situation. How long does it take you to jog around the park?
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 40

Answer:
The system of linear equations are 2x + y = 10, 2x + 3y = 22
It takes 6 minutes to jog around the park.

Explanation:
Let x be the number of minutes it takes to jog around the block and y be the number of minutes it takes to jog around the park
You can jog around your block twice and the park once in 10 minutes
2x + y = 10 —- (i)
You can jog around your block twice and the park 3 times in 22 minutes
2x + 3y = 22 —- (ii)
Subtract equations
2x + y – 2x – 3y = 10 – 22
-2y = -12
y = 6
Substitute y = 6 in (i)
2x + 6 = 10
2x = 10 – 6
2x = 4
x = 2
So, it takes 6 minutes to jog around the park.

SOLVING A SYSTEM OF LINEAR EQUATIONS
Solve the system by elimination. Check your solution.

Question 18.
2x – y = 0
3x – 2y = -3

Answer:
The solution set is (3, 6).

Explanation:
The systems of linear equations are
2x – y = 0 —- (i)
3x – 2y = -3 —– (ii)
Multiply equation (i) by 2
2(2x – y = 0)
4x – 2y = 0 —- (iii)
Subtract equation (ii) from (iii)
4x – 2y – (3x – 2y) = 0 – (-3)
4x – 2y – 3x + 2y = 3
x = 3
Substitute x = 3 in (i)
2(3) – y = 0
6 – y = 0
y = 6
Substitute x = 3, y = 6 in (i)
2(3) – 6 = 6 – 6 = 0
So, the solution set is (3, 6).

Question 19.
x + 4y = 1
3x + 5y = 10

Answer:
The solution set is (5, -1).

Explanation:
The systems of linear equations are
x + 4y = 1 —- (i)
3x + 5y = 10 —- (ii)
Multiply equation (i) by 3
3(x + 4y = 1)
3x + 12y = 3 —- (iii)
Subtract equation (iii) from (ii)
3x + 5y – (3x + 12y) = 10 – 3
3x + 5y – 3x – 12y = 7
-7y = 7
y = -1
Substitute y = -1 in (i)
x + 4(-1) = 1
x – 4 = 1
x = 1 + 4
x = 5
Substitute x = 5, y = -1 in (i)
5 + 4(-1) = 5 – 4 = 1
So, the solution set is (5, -1).

Question 20.
-2x + 3y = 7
5x + 8y = -2

Answer:
The solution set is (-2, 1).

Explanation:
The systems of linear equations are
-2x + 3y = 7 —- (i)
5x + 8y = -2 —– (ii)
Multiply equation (i) by 5 and equation (ii) by 2
5(-2x + 3y = 7)
-10x + 15y = 35 —- (iii)
2(5x + 8y = -2)
10x + 16y = -4 —- (iv)
Add equations (iii) & (iv)
-10x + 15y + 10x + 16y = 35 – 4
31y = 31
y = 1
Substitute y = 1 in (i)
-2x + 3(1) = 7
-2x = 7 – 3
-2x = 4
x = -2
Substitute x = -2, y = 1 in (ii)
5(-2) + 8(1) = -10 + 8 = -2
So, the solution set is (-2, 1).

Question 21.
3x + 3 = 3y
2x – 6y = 2

Answer:
The solution set is (-2, -1).

Explanation:
The systems of linear equations are
3x + 3 = 3y —- (i)
2x – 6y = 2 —- (ii)
Multiply equation (i) by 2
2(3x + 3 = 3y)
6x + 6 = 6y
6x – 6y = -6 —- (iii)
Subtract equations (ii) & (iii)
2x – 6y – 6x + 6y = 2 + 6
-4x = 8
x = -2
Substitute x = -2 in (i)
3(-2) + 3 = 3y
-6 + 3 = 3y
-3 = 3y
y = -1
Substitute x = -2, y = -1 in (i)
3(-2) + 3 = 3(-1)
-6 + 3 = -3
So, the solution set is (-2, -1).

Question 22.
2x – 6 = 4y
7y =-3x + 9

Answer:
The solution set is (3, 0).

Explanation:
The systems of linear equations are
2x – 6 = 4y
2x – 4y = 6 —- (i)
7y = -3x + 9
3x + 7y = 9 —- (ii)
Multiply (i) by 7 and (ii) by 4
7(2x – 4y = 6)
14x – 28y = 42 —- (iii)
4(3x + 7y = 9)
12x + 28y = 36 —- (iv)
Add equations (iii) & (iv)
14x – 28y + 12x + 28y = 42 + 36
26x = 78
x = 78/26
x = 3
Substitute x = 3 in (i)
2(3) – 4y = 6
6 – 4y = 6
6 – 6 = 4y
y = 0
Substitute x = 3, y = 0 in (i)
2(3) – 4(0) = 6 – 0 = 6
So, the solution set is (3, 0).

Question 23.
5x = 4y + 8
3y = 3x – 3

Answer:
The solution set is (4, 3).

Explanation:
The systems of linear equations are
5x = 4y + 8
5x – 4y = 8 — (i)
3y = 3x – 3
3x – 3y = 3 —- (ii)
Multiply (i) by 3 and (ii) by 4
3(5x – 4y = 8)
15x – 12y = 24 — (iii)
4(3x – 3y = 3)
12x – 12y = 12 —- (iv)
Subtract obtained equations
15x – 12y – 12x + 12y = 24 – 12
3x = 12
x = 4
Substitute x = 4 in (ii)
3(4) – 3y = 3
12 – 3y = 3
-3y = 3 – 12
-3y = -9
y = 3
Substitute x = 4, y = 3 in (ii)
3(4) – 3(3) = 12 – 9 = 3
So, the solution set is (4, 3).

Question 24.
YOU BE THE TEACHER
Your friend solves the system. Is your friend correct? Explain your reasoning.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 41

Answer:
Wrong.

Explanation:
Given equations are
x + y = 1 — (i)
5x + 3y = -3 (ii)
Multiply equation (i) by -5
-5(x + y = 1)
-5x – 5y = -5 — (iii)
Add equations (iii) & (ii)
-5x – 5y + 5x + 3y = -3 – 5
-2y = -8
y = 4
Put y = 4 in (i)
x + 4 = 1
x = 1 – 4 = -3
So, the solution set is (-3, 4).

CHOOSING A SOLUTION METHOD
Solve the system. Explain your choice of method.

Question 25.
x + y = 4
x – y = 4

Answer:
The solution set is (4, 0).

Explanation:
The systems of linear equations are
x + y = 4 — (i)
x – y = 4 — (ii)
Add equations
x + y + x – y = 4 + 4
2x = 8
x = 4
Substitute x = 4 in (i)
4 + y = 4
y = 4 – 4
y = 0
So, the solution set is (4, 0).

Question 26.
y = x – 3
y = -2x + 3

Answer:
The solution set is (2, -1).

Explanation:
The systems of linear equations are
y = x – 3 — (i)
y = -2x + 3 — (ii)
Equate both equations
x – 3 = -2x + 3
x + 2x = 3 + 3
3x = 6
x = 2
Substitute x = 2 in (i)
y = 2 – 3
y = -1
So, the solution set is (2, -1).

Question 27.
x + 2y = 0
2x – y = 4

Answer:
The solution set is (8/5, -4/5)

Explanation:
The systems of linear equations are
x + 2y = 0 —- (i)
2x – y = 4 —- (ii)
y = 2x – 4
Substitute y = 2x – 4 in (i)
x + 2(2x – 4) = 0
x + 4x – 8 = 0
5x = 8
x = 8/5
Substitute x = 8/5 in (i)
8/5 + 2y = 0
2y = -8/5
y = -4/5
So, the solution set is (8/5, -4/5)

Question 28.
y + 5x = 1
5y – x = 5

Answer:
The solution set is (0, 1).

Explanation:
The systems of linear equations are
y + 5x = 1 —- (i)
5y – x = 5 —- (ii)
5y – 5 = x
Substitute x = 5y – 5 in (i)
y + 5(5y – 5) = 1
y + 25y – 25= 1
26y = 1 + 25
26y = 26
y = 1
Substitute y = 1 in (i)
1 + 5x = 1
5x = 0
x = 0
So, the solution set is (0, 1).

Question 29.
2 = x – 3y
-2x + y = 4

Answer:
The solution set is (-14/5, -8/5)

Explanation:
The systems of linear equations are
2 = x – 3y
x = 2 + 3y — (i)
-2x + y = 4 —- (ii)
Substitute (i) in (ii)
-2(2 + 3y) + y = 4
-4 – 6y + y = 4
-5y = 4 + 4
y = -8/5
Substitute y = -8/5 in (i)
x = 2 + 3(-8/5)
x = 2 – 24/5
x = -14/5
So, the solution set is (-14/5, -8/5)

Question 30.
8x + 5y = 6
8x = 3 – 2y

Answer:
The solution set is (1/8, 1).

Explanation:
The systems of linear equations are
8x + 5y = 6 —- (i)
8x = 3 – 2y —- (ii)
Substitute (ii) in (i)
3 – 2y + 5y = 6
3y = 6 – 3
3y = 3
y = 1
Substitute y = 1 in (ii)
8x = 3 – 2(1)
8x = 1
x = 1/8
So, the solution set is (1/8, 1).

NUMBER SENSE
For what value of a might you choose to solve the system by elimination? Explain.

Question 31.
4x – y = 3
ax + 10y = 6

Answer:
If a is 4 or -4 it is easy to solve the systems of linear equations using the elimination method as by just subtracting or adding the equations, we can eliminate x.

Explanation:
The systems of linear equations are
4x – y = 3
ax + 10y = 6
If a is 4 or -4 it is easy to solve the systems of linear equations using the elimination method as by just subtracting or adding the equations, we can eliminate x.

Question 32.
x – 7y = 6
-6x + ay = 9

Answer:
If a is 7 or -7 it is easy to solve the systems of linear equations using the elimination method as by just adding or subtracting the equations, we can eliminate x.

Explanation:
The systems of linear equations are
x – 7y = 6
-6x + ay = 9
If a is 7 or -7 it is easy to solve the systems of linear equations using the elimination method as by just adding or subtracting the equations, we can eliminate x.

CRITICAL THINKING
Determine whether the line through the first pair of points intersects the line through the second pair of points. Explain.

Question 33.
Line 1: (-2, 1), (2, 7)
Line 2: (-4, -1), (0, 5)

Answer:
No

Explanation:
Equation of a line passing through points (x₁, y₁) and (x₂, y₂) is y – y₁ = (y₂ – y₁)/(x₂ – x₁) (x – x₁)
So, equation of line 1 is
y – 1 = (7 – 1)/(2 + 2)(x + 2)
y – 1 = 6/4 (x + 2)
y – 1 = 3/2 (x + 2)
2(y – 1) = 3(x + 2)
2y – 2 = 3x + 6
3x – 2y = -2 – 6
3x – 2y = -8
Equation of line 2 is
y + 1 = (5 + 1)/(0 + 4) (x + 4)
y + 1 = 6/4 (x + 4)
y + 1 = 3/2 (x + 4)
2(y + 1) = 3(x + 4)
2y + 2 = 3x + 12
3x – 2y = 2 – 12
3x – 2y = -10
As the slope of the lines are same, but the consant is different, the lines are parallek to each other and there is no point of intersection.

Question 34.
Line 1: (-2, 8), (0, 2)
Line 2: (3, -2), (6, 4)

Answer:
Yes

Explanation:
Equation of a line passing through points (x₁, y₁) and (x₂, y₂) is y – y₁ = (y₂ – y₁)/(x₂ – x₁) (x – x₁)
So, line 1 is
y – 8 = (2 – 8)/(0 + 2) (x + 2)
y – 8 = -6/2 (x + 2)
y – 8 = -3(x + 2)
y – 8 = -3x – 6
3x + y = -6 + 8
3x + y = 2
Line 2 is
y + 2 = (4 + 2)/(6 – 3) (x – 3)
y + 2 = 6/3 (x – 3)
y + 2 = 2(x – 3)
y + 2 = 2x – 6
2x – y = 8
As the slopes are different, so the lines intesect each other.

Question 35.
REASONING
Two airplanes are flying to the same airport. Their positions are shown in the graph. Write a system of linear equations that represents this situation. Solve the system by elimination to justify your answer.
Big Ideas Math Answer Key Grade 8 Chapter 5 Systems of Linear Equations 42

Answer:
x = 6, y = 12.

Explanation:
To write the system we need the slope of each line and at least one point on the line. The two lines to consider will be the lines connecting the location of each plane to the airport they are flying to. It is also worth noting that the coordinates of the airport represent the point of intersection of the two lines and thus the solution to the system.
Airport (6, 12), Airplane 1 (2, 4), Airplane 2 (15, 9)
the slope of the line connecting airplane 1 and the airport = (4 – 12)/(2- 6)
= -8/-4 = 2
The line equation is y – 4 = 2(x -4)
2x – y = 0
slope of the line connecting airplane 2 and the airport = (4 – 9)/(2 – 15)
= -5/-13 = 5/13
The line equation is y – 9 = 5/13 (x – 15)
x + 3y = 42
So, the system of linear equations are
2x – y = 0 —- (i)
x + 3y = 42 —– (ii)
Multiply (ii) by 2
2(x + 3y = 42)
2x + 6y = 84 — (iii)
Subtract (iii) & (i)
2x – y – 2x – 6y = 0 – 84
-7y = -84
y = 12
Put y = 12 in (i)
2x – 12 = 0
x = 6
We have proven that the location of the airport is in fact the solution to our system.

Question 36.
MODELING REAL LIFE
A laboratory uses liquid nitrogen tanks of two different sizes. The combined volume of 3 large tanks and 2 small tanks is 24 liters. The combined volume of 2 large tanks and 3 small tanks is 21 liters. What is the volume of each size of tank? Justify your answer.

Answer:
The volume of the large tank is 6 lit, the volume of the small tank is 3 lit.

Explanation:
Let x and y be the volume of the large and small tank
The combined volume of 3 large tanks and 2 small tanks is 24 liters. The combined volume of 2 large tanks and 3 small tanks is 21 liters
Combined volume = No of large tanks x volume of large tank + No of small tanks x volume of small tank
3x + 2y = 24 — (i)
2x + 3y = 21 —- (ii)
Multiply (i) by -2 and (ii) by 3
-2(3x + 2y = 24)
-6x – 4y = -48
3(2x + 3y = 21)
6x + 9y = 63
Add obtained equations
-6x – 4y + 6x + 9y = -48 + 63
5y = 15
y = 3
Put y = 3 in (i)
3x + 6 = 24
3x = 24 – 6
3x = 18
x = 6
So, the volume of the large tank is 6 lit, volume of the small tank is 3 lit.

Question 37.
PROBLEM SOLVING
The table shows the numbers of correct answers on a practice standardized test. You score 86 points on the test and your friend scores 76 points. How many points is each type of question worth?
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 43

Answer:
The marks for correct multiple choice questions, short response questions are 2 and 4.

Explanation:
Let x, y be the marks of multiple choice & short response questions
Total marks = no of correct MCQs x marks for correct MCQs + no of short response questions x marks for short response questions
23x + 10y = 86 — (i)
28x + 5y = 76 —- (ii)
Multiply (ii) by 2
2(28x + 5y = 76)
56x + 10y = 152 —- (iii)
Subtract (i) & (iii)
23x + 10y – 56x – 10y = 86 – 152
-33x = -66
x = 2
Put x = 2 in (i)
23(2) + 10y = 86
10y = 86 – 46
10y = 40
y = 4
So, the marks for correct multiple choice questions, short response questions are 2 and 4.

Question 38.
LOGIC
You solve a system of equations in which x represents the number of adult memberships sold and y represents the number of student memberships sold. Can (-6, 24) be the solution to the system? Explain your reasoning.

Answer:
If x represents the number of adult tickets sold, then x must be a non-negative number since there can’t be a negative number of tickets sold.
Therefore (-6, 24) can’t be a solution since that would give x = 6.

Explanation:
If x represents the number of adult tickets sold, then x must be a non-negative number since there can’t be a negative number of tickets sold.
Therefore (-6, 24) can’t be a solution since that would give x = 6.

Question 39.
PROBLEM SOLVING
The table shows the activities of two tourists at a vacation resort. You want to go parasailing for 1 hour and horseback riding for 2 hours. How much do you expect to pay?
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 44
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 44.1

Answer:
The cost of parasailing for 1 hour and horseback riding for hours is $90.

Explanation:
Let x and y represent per hour price for parasailing and horseback riding
Total cost = no of hours of parasailing x per hour price for parasailing + no of hours of horseback riding x per hour price for horseback riding
2x + 5y = 205 —- (i)
3x + 3y = 240 — (ii)
Multiply (i) by 3 and (ii) by 2
3(2x + 5y = 205)
6x + 15y = 615 — (iii)
2(3x + 3y = 240)
6x + 6y = 480 —- (iv)
Subtract (iii) & (iv)
6x + 15y – 6x – 6y = 615 – 480
9y = 135
y = 15
Put y = 15 in (i)
2x + 15(5) = 205
2x = 205 – 75
2x = 130
x = 65
So, the Cost of parasailing for 1 hour and harseback riding for hours is x + 2y
= 65 + 2(15)
= $95.

Question 40.
REASONING
Write a system of linear equations containing 2x + y = 0 and that has the solution (2, -4).

Answer:
The system of linear equations are 2x + y = 0, x – y = 6.

Explanation:
The equation of a line that passes through (2, -4) is y = mx + b
-4 = 2m + b
Let us take m slope as 1
-4 = 2 + b
b = -4 – 2
b = -6
So, line is y = x – 6
So, the system of linear equations are 2x + y = 0, x – y = 6.

Question 41.
REASONING
A metal alloy is a mixture of two or more metals. A jeweler wants to make 8 grams of 18-karat gold, which is 75% gold. The jeweler has an alloy that is 90% gold and an alloy that is 50% gold. old. How much of each alloy should the jeweler use?
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 45

Answer:
The jeweler should use 5, 3 grams of first and second alloy.

Explanation:
Let x, y be the amount of first & second alloy
Amount of gold = percentage of gold in first alloy x amount of first alloy + percentage of gold in second alloy x amount of the second alloy
x + y = 8
0.9x + 0.5y = 6
y = 8 – x
0.9x + 0.5(8 – x) = 6
0.4x + 4 = 6
0.4x = 2
x = 5
y = 8 – 5
y = 3
The jeweler should use 5, 3 grams of first and second alloy.

Question 42.
PROBLEM SOLVING
It takes a powerboat traveling with the current 30 minutes to go 10 miles. The return trip takes 50 minutes traveling against the current. What is the speed of the current?

Answer:
The current speed is 4 miles per hour.

Explanation:
Let r be the speed of the boat and w be the speed of water current
downstream 10 = (r + w)30
1 = 3(r + w)
3r + 3w = 1 — (i)
Upstream 10 = (r – w)50
1 = 5(r – w)
5r – 5w = 1 —- (ii)
Multiply (i) by 5 and (ii) by 3 and add them
15r + 15w + 15r – 15w = 5 + 3
30r = 8
r = 4/15
substitute r = 4/15 in (ii)
5(4/15) – 5w = 1
4/3 – 5w = 1
-5w = -1/3
w = 1/15 miles per inute
= 1/15 (60) = 4 miles per hour
The current speed is 4 miles per hour.

Question 43.
DIG DEEPER!
Solve the system of equations by elimination.
2x – y + 3z = -1
x + 2y – 4z = -1
y – 2z = 0

Answer:
The solution set is (-1, 2, 1).

Explanation:
The given systems of linear equations are
2x – y + 3z = -1 —– (i)
x + 2y – 4z = -1 —- (ii)
y – 2z = 0 —– (iii)
Substitute y = 2z in (i) & (ii)
2x – 2z + 3z = -1
2x+ z = -1 —- (iv)
x + 4z – 4z = -1
x = -1
Substitute x = -1 in (iv)
2(-1) + z = -1
z = -1 + 2
z = 1
Substitute z = 1 in (iii)
y – 2 = 0
y = 2
So, the solution set is (-1, 2, 1).

Lesson 5.4 Solving Special Systems of Linear Equations

EXPLORATION 1
Exploring Solutions of Systems
Work with a partner. You spend $50 on a sewing machine to makedog backpacks. Each backpack costs you $15 for materials.
a. Represent the cost (in dollars) to make backpacks in the coordinate plane.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 46
b. You charge $25 per backpack. How many backpacks do you have to break even sell to ? Use a graph to justify your answer.
c. Can you break even when you sell each backpack for $20? $15? Use graphs to justify your answers.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 47
d. Explain whether it is possible for a system of linear equations to have the numbers of solutions below.

  • no solution
  • exactly one solution
  • exactly two solutions
  • infinitely many solutions

Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 48

5.4 Lesson

Try It

Solve the system. Explain your choice of method.

Question 1.
y = -x + 3
y = -x + 5

Answer:
There is no solution.

Explanation:
The given systems of linear equations are
y = -x + 3
y = -x + 5
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 37
The two lines are parallel to each other.
So, there is no solution.

Question 2.
y = -5x – 2
5x + y = 0

Answer:
The system has no solution.

Explanation:
The given systems of linear equations are y = -5x – 2, 5x + y = 0
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 38
The two lines are parallel to each other.
So, the system has no solution.

Question 3.
x = 2y + 10
2x + 3y = -1

Answer:
The solution set is (4, -3)

Explanation:
The given systems of linear equations are x = 2y + 10, 2x + 3y = -1
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 39
The point of intersection of two lines is (4, -3)
So, the solution set is (4, -3)

Try It

Solve the system. Explain your choice of method.

Question 4
x + y = 3
x = y – 3

Answer:
The solution set is (0, 3).

Explanation:
The given systems of linear equations are
x + y = 3 — (i)
x = y – 3 —- (ii)
Substitute (ii) in (i)
y – 3 + y = 3
2y = 6
y = 3
Substitute y = 3 in (ii)
x = 3 – 3
x = 0
So, the solution set is (0, 3).

Question 5.
2x + y = 5
4x + 2y = 0

Answer:
The system has no solution.

Explanation:
The given systems of linear equations are
2x + y = 5 —- (i)
4x + 2y = 0 —- (ii)
Solve by graphing
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 40
The lines are parallel so the system has no solution.

Question 6.
2x – 4y = 10
-12x + 24y = -60

Answer:
The system has infinitely many solutions.

Explanation:
The given systems of linear equations are
2x – 4y = 10 —- (i)
-12x + 24y = -60 — (ii)
Multiply (i) by 6
6(2x – 4y = 10)
12x – 24y = 60 — (iii)
Add (ii) & (iii)
-12x + 24y + 12x – 24y = -60 + 60
0 = 0
So, the system has infinitely many solutions.

Self-Assessment for Concepts & Skills
Solve each exercise. Then rate your understanding of the success criteria in your journal.

STRUCTURE
Without graphing or solving, determine the number of solutions of the system. Explain your reasoning.

Question 7.
y = 5x – 9
y = 5x + 9

Answer:
The system has infinitely many solutions.

Explanation:
The given systems of linear equations are
y = 5x – 9
y = 5x + 9
the slope of the two lines are the same i.e 5
So, the system has infinitely many solutions.

Question 8.
y = 6x + 2
y = 3x + 1

Answer:
The system has one solution.

Explanation:
The given systems of linear equations are
y = 6x + 2
slope1 = 6
y = 3x + 1
slope2 = 3
Slopes are different
So, the system has one solution.

Question 9.
y = 8x – 2
y – 8x = -2

Answer:
The system has infinitely many solutions.

Explanation:
The given systems of linear equations are
y = 8x – 2
slope 1 = 8
y – 8x = -2
y = -2 + 8x
slope 2 = 8
the slope of the two lines are the same i.e 8
So, the system has infinitely many solutions.

CHOOSING A METHOD
Solve the system. Explain your choice of method.

Question 10.
2x + y = 6
x – y = 3

Answer:
The solution set is (3, 0).

Explanation:
The given systems of linear equations are
2x + y = 6 —- (i)
x – y = 3 — (ii)
Add equations
2x + y + x – y = 3 + 6
3x = 9
x = 3
Substitute x = 3 in (ii)
3 – y = 3
y = 0
So, the solution set is (3, 0).

Question 11.
4y – 4x = 8
y = x + 2

Answer:
The system has infinitely many solutions.

Explanation:
The given systems of linear equations are
4y – 4x = 8 — (i)
y = x + 2 —- (ii)
Substitute 2 in 1
4(x + 2) – 4x = 8
4x + 8 – 4x = 8
8 = 8
So, the system has infinitely many solutions.

Question 12.
5x – 4y = 12
7.5x = 6(y – 1)

Answer:
The system has no solution.

Explanation:
The given systems of linear equations are
5x – 4y = 12
7.5x = 6(y – 1)
By graphing
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 41
The lines are parrel
So, the system has no solution.

Question 13.
-6x = 9
6x – y = 3

Answer:
The solution set is (-3/2, -12)

Explanation:
The given systems of linear equations are
-6x = 9
6x – y = 3
-6x = 9
x = -9/6
x = -3/2
Substitute x = -3/2 in 6x – y = 3
6(-3/2) – y = 3
-18/2 – 3 = y
y = -24/2
y = -12
So, the solution set is (-3/2, -12)

Question 14.
0.5x + 4y = -11
-1.5x – 12y = 33

Answer:
The system has infinitely many solutions.

Explanation:
The given systems of linear equations are
0.5x + 4y = -11
-1.5x – 12y = 33
By graphing
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 42
Two equations are on the same line.
So, the system has infinitely many solutions.

Question 15.
x = y + 2
3x = 6(y + 2)

Answer:
The solution set is (0, -2).

Explanation:
The given systems of linear equations are
x = y + 2
3x = 6(y + 2)
By graphing
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 43
The point of intersection of two equations is (0, -2)
so, the solution set is (0, -2).

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 16.
Your friend wants to sell painted rocks. He spends $10.00 on startup costs, and each painted rock costs him $0.75 to make. A store offers to pay your friend’s startup costs and buy his painted rocks for $0.75 each. How many painted rocks does your friend need to sell to make a profit?

Answer:
No profit.

Explanation:
The cost on startup is $10 and each painted rock costs around $0.75 to make
A store pays startup cost and buys his painted rocks for $0.75 each
There will be no profit because the cost price and selling price is the same. He buys for $0.75 and sells the same for $0.75.

Question 17.
DIG DEEPER!
The difference in age of two orangutans is 6 years. In 4 years, is it possible for the older orangutan to be twice as old as the younger orangutan? three times as old? Justify your answers.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 49

Answer:
Both are possible

Explanation:
Let us take the age of two orangutans as x and y
The difference in age of two orangutans is 6 years
x – y = 6
So, x is the older one
x = 6 + y
If the older orangutan to be twice as old as the younger orangutan
x = 2y
2y – y = 6
y = 6
If the older orangutan to be three times as old as the younger orangutan
x = 3y
3y – y = 6
2y = 6
y = 3
Both are possible, the older orangutan to be twice as old as the younger orangutan when younger age is 6 years and the older orangutan to be three times as old as the younger orangutan when the younger age is 3 years.

Solving Special Systems of Linear Equations Homework & Practice 5.4

Review & Refresh

Solve the system by elimination. Check your solution.

Question 1.
x + 2y = 4
-x – y = 2

Answer:
The solution set is (-8, 6)

Explanation:
The given systems of linear equations are
x + 2y = 4 — 1
-x – y = 2 — 2
Add both equations
x + 2y – x – y = 4 + 2
y = 6
Substitute y = 6 in 2
-x – 6 = 2
-x = 2 + 6
x = -8
Substitute x = -8, y = 6 in 2
8 – 6 = 2
So, the solution set is (-8, 6)

Question 2.
2x – y = 1
x + 3y – 4 = 0

Answer:
The solution set is (1, 1).

Explanation:
The given systems of linear equations are
2x – y = 1 —- (i)
x + 3y – 4 = 0 —- (ii)
Multiply (i) by 3
3(2x – y = 1)
6x – 3y = 3 —- (iii)
Add (iii) & (ii)
6x – 3y + x + 3y – 4 = 3
7x = 3 + 4
7x = 7
x = 1
Substitute x = 1 in (i)
2(1) – y = 1
2 – y = 1
2 – 1 = y
y = 1
Substitute x = 1, y = 1 in (i)
2(1) – 1 = 2 – 1 = 1
So, the solution set is (1, 1).

Question 3.
3x = -4y + 10
4x + 3y = 11

Answer:
The solution set is (2, 1).

Explanation:
The given systems of linear equations are
3x = -4y + 10
3x + 4y = 10 —- (i)
4x + 3y = 11—- (ii)
Multiply (i) by 4 and (ii) by 3
4(3x + 4y = 10)
12x + 16y = 40 —- (iii)
3(4x + 3y = 11)
12x + 9y = 33 —- (iv)
Subtract (iii) from (iv)
12x + 16y – 12x – 9y = 40 – 33
7y = 7
y = 1
Substitute y = 1 in (i)
3x + 4 = 10
3x = 6
x = 2
Substitute x = 2, y = 1 in (i)
3(2) + 4 = 6 + 4 = 10
So, the solution set is (2, 1).

Write an equation of the line that passes through the given points.

Question 4.
(0, 0), (2, 6)

Answer:
y = 3x

Explanation:
The equation of a line when two points given is
(y – y₁) = [(y₂ – y₁)/(x₂ – x₁)](x – x₁)
x₁ = 0, y₁ = 0, x₂ = 2, y₂ = 6
So, (y – 0) = [(6 – 0)/(2 – 0)] (x – 0)
y = 3x

Question 5.
(0, -3), (3, 3)

Answer:
y = 2x – 3

Explanation:
The equation of a line when two points given is
(y – y₁) = [(y₂ – y₁)/(x₂ – x₁)](x – x₁)
x₁ = 0, y₁ = -3, x₂ = 3, y₂ = 3
So, (y + 3) = [(3 + 3)/(3 – 0)](x – 0)
(y + 3)  = 2x
y = 2x – 3

Question 6.
(-6, 5), (0, 2)

Answer:
x + 2y = 6

Explanation:
The equation of a line when two points are given is
(y – y₁) = [(y₂ – y₁)/(x₂ – x₁)](x – x₁)
x₁ = -6, y₁ = 5, x₂ = 0, y₂ = 2
So, (y – 5) = [(2 – 5)/(0 + 6)](x + 6)
y – 6 = -3/6 (x + 6)
y – 6 = -1/2 (x + 6)
2(y – 6) = -1(x + 6)
2y – 12 = -x – 6
2y = -x – 6 + 12
x + 2y = 6

Concepts, Skills, &Problem Solving
EXPLORING SOLUTIONS OF SYSTEMS
Use a graph to determine the number of solutions of the system. (See Exploration 1, p. 219.)

Question 7.
y = 2x + 1
y = 2x + 5

Answer:
The system has no solution.

Explanation:
The given systems of linear equations are y = 2x + 1, y = 2x + 5
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 44
The lines are parallel
So, the system has no solution.

Question 8.
y + 8 = 0
y = 8

Answer:
The system has no solution.

Explanation:
The given systems of linear equations are y + 8 = 0, y = 8
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 45
The lines are parallel
So, the system has no solution.

Question 9.
x + y = 2
5x + y = 9

Answer:
The solution is (7/4, 1/4)

Explanation:
The given systems of linear equations are x + y = 2, 5x + y = 9
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 46
The lines intersect at (7/4, 1/4)
So, the solution is (7/4, 1/4)

SOLVING A SYSTEM
Solve the system. Explain your choice of method.

Question 10.
y = 2x – 2
y = 2x + 9

Answer:
The system has no solution.

Explanation:
The given systems of linear equations are
y = 2x – 2 — (i)
y = 2x + 9 — (ii)
Equate equations
2x – 2 = 2x + 9
So, the system has no solution.

Question 11.
y = 3x + 1
-x + 2y = -3

Answer:
The solution is (-1, -2).

Explanation:
The given systems of linear equations are y = 3x + 1, -x + 2y = -3
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 47
The lines intersect at (-1, -2)
So, the solution is (-1, -2).

Question 12.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 50

Answer:
The system has no solution.

Explanation:
The given systems of linear equations are
y = π/3 x + π —- (i)
-πx + 3y = -6π —- (ii)
Substitute (i) in (ii)
-πx + 3(π/3 x + π) = -6π
-πx + πx + 3π = -6π
So, the system has no solution.

Question 13.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 51

Answer:
The system has infinitely many solutions.

Explanation:
The given systems of linear equations are
y = -1/6 x + 5 —-(i)
x + 6y = 30 —- (ii)
Substitute (i) in (ii)
x + 6(-1/6 x + 5) = 30
x – x + 30 = 30
So, the system has infinitely many solutions.

Question 14.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 52

Answer:
The system has infinitely many solutions.

Explanation:
The given systems of linear equations are
1/3 x + y = 1 —- (i)
2x + 6y = 6 —- (ii)
Divide (ii) by 1/6
1/6(2x + 6y = 6)
1/3 x + y = 1 —- (iii)
equation (i) & (iii) are same
So, the system has infinitely many solutions.

Question 15.
-2x + y = 1.3
2(0.5x – y) = 4.6

Answer:
The solution is (-2.4, -3.5)

Explanation:
The given systems of linear equations are -2x + y = 1.3, 2(0.5x – y) = 4.6
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 48
The lines intersect at (-2.4, -3.5)
So, the solution is (-2.4, -3.5)

Question 16.
2(x + y) = 9
1 = -4(x + y)

Answer:
The system has no solution.

Explanation:
The given systems of linear equations are 2(x + y) = 9, 1 = -4(x + y)
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 49
The lines are parallel
So, the system has no solution.

Question 17.
y = 9x
x + y = 1

Answer:
The solution is (1/10, 9/10)

Explanation:
The given systems of linear equations are y = 9x, x + y = 1
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 50
The lines intersect at (1/10, 9/10)
So, the solution is (1/10, 9/10)

Question 18.
0.2y = 4.6x + 1.2
-2.3x = -0.1y + 0.6

Answer:
The system has infinitely many solutions

Explanation:
The given systems of linear equations are 0.2y = 4.6x + 1.2, -2.3x = -0.1y + 0.6
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 51
Two equations lies on the same line
So, the system has infinitely many solutions

Question 19.
YOU BE THE TEACHER
Your friend finds the number of solutions of the system. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 53

Answer:
Correct.

Explanation:
The given systems of linear equations are
y = -2x + 4
y = -2x + 6
Two equations have the same slope
So, the system has infinitely many solutions.

Question 20.
REASONING
In a pig race, your pig has a head start of 3 feet and runs at a rate of 2 feet per second. Your friend’s pig also runs at a rate of 2 feet per second. A system of linear equations that represents this situation is y = 2x + 3 and y = 2x. Does your friend’s pig catch up to your pig? Explain.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 54

Answer:
No.

Explanation:
y = 2x + 3 and y = 2x
Substitute y = 2x in y = 2x + 3
2x = 2x + 3
2x – 2x = 3
0 = 3
The system has no solution
Your friend’s pig never catch up to your pig

Question 21.
REASONING
One equation in a system of linear equations has a slope of 3. The other equation has a slope of 4. How many solutions does the system have? Explain.

Answer:
The system has one solution.

Explanation:
One equation in a system of linear equations has a slope of 3. The other equation has a slope of 4
As the slopes are different. The system has one solution.

Question 22.
LOGIC
How can you use the slopes and the y-intercepts of equations in a system one solution, infinitely many solutions, or no solution?

Answer:
The slope-intercept form of a line is y = mx + c
If the slope is the same, different y-intercept for the two equations, then the system has no solution.
If the slope is different for two equations, then the system has one solution.
If the slope is the same, the same y-intercept for two equations, then the system has infinitely many solutions.

Question 23.
PROBLEM SOLVING
You and a friend both work two different jobs. The system of linear equations represents the total earnings (in dollars) for x hours worked at the first job and y hours worked at the second job. Your friend earns twice as much as you.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 55
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 56
a. One week, both of you work 4 hours at the first job. How many hours do you and your friend work at the second job?
b. Both of you work the same number of hours at the second job. Compare the number of hours you and your friend work at the first job.

Answer:
a. 16 hours you and your friend work at the second job
b. If both work the same y hours at a second job, then both will same x hours at the first job.

Explanation:
Let x represent the first job and y represents the second job
The system of linear equations for the total number of hours for the first job & second job for you & your friend is
4x + 8y = 64
8x + 16y = 128
a. If x = 4
8(4) + 16y = 128
32 + 16y = 128
16y = 128 – 32
16y = 96
y = 16 hours
b. If both work the same y hours at the second job, then both will same x hours at the first job.

Question 24.
MODELING REAL LIFE
You download a digital album for $10.00. Then you and your friend each download the same number of individual songs for $0.99 each. Write a system of linear equations that represents this situation. Will you and your friend spend the same amount of money? Explain.

Answer:
No

Explanation:
Write the equations for you and your friends’ total cost y where x is the number of songs
you y = 10 + 0.99x
your friend y = 0.99x
The two equations have the same slope but different y-intercepts so the system has no solution.

Question 25.
MODELING REAL LIFE
The table shows the research activities of two students at an observatory. How much does a student pay to use the telescope for one hour?
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 57

Answer:
A student must pay $7.5 for using a telephone per hour and $11 for using a supercomputer per hour.

Explanation:
Let the cost of telephonic use be x, supercomputer use be y. The system of equations are
5x + 3y = 70.50 — (i)
6x + 2y = 67 —- (ii)
Multiply (i) by 2 and (ii) by 3 and subtract them
10x + 6y = 141
18x + 6y = 201
10x + 6y – 18x – 6y = 141 – 201
-8x = -60
x = 7.5
Substitute x = 7.5 in (ii)
6(7.5) + 2y = 67
45 + 2y = 67
2y = 67 – 45
2y = 22
y = 11
So, a student must pay $7.5 for using telephone per hour and $11 for using supercomputer per hour.

Question 26.
REASONING
Does the system shown always, sometimes, or never have a solution when a = b? a ≥ b? a < b? Explain your reasoning.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 58

Answer:
If a = b, the system will always have no solution
If a ≥ b, the system will sometimes have no solution.
If a < b, the system will never have no solution.

Explanation:
If a = b, then two linear equations will have the same slope but different y-intercepts. So, they will always have no solution
If a ≥ b, then the slopes may be different or the same so they may intersect once or not at all. Therefore, they will sometimes have no solution.
If a < b, then the equations have a different slope so they will intersect once. so, they will never have no solution.

Question 27.
LOGIC
The table shows the numbers of lift tickets and ski rentals sold to different groups. Is it possible to determine how much each lift ticket costs using the information for Groups 1 and 2? Groups 1 and 3? Justify your answers.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 89

Answer:
a. It is not possible to find out the cost of lift tickets using the given information.
b. The cost of a lift ticket is $14, cost of ski rental is $10.

Explanation:
Let the cost for lift tickets and cost for sky rentals be y. The system of equations are
36x + 18y = 684 — (i)
24x + 12y = 456 —- (ii)
18x + 18y = 432 —-(iii)
3(12x + 6y) = 684
12x + 6y = 228
2(12x + 6y) = 456
12x + 6y = 228
Both the equations are same. So, they have infinately many solutions.
Hence, it is not possible to find out the cost of lift tickets using the given information.
b. Subtract 1 from 2
36x + 18y – 18x – 18y = 684 – 432
18x = 252
x = 14
Put x = 14
36(14) + 18y = 684
504 + 18y = 684
18y = 180
y = 10
The cost of lift ticket is $14, cost of ski rentaks is $10.

Question 28.
DIG DEEPER!
Find the values of a and b so the system is shown has the solution (2, 3). Does the system have any other solutions for these values of a and b? Explain.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 90

Answer:
(a, b) = (2, 2)

Explanation:
Given that,
12x – 2by = 12
3ax – by = 6
The value of a, b if (x, y) = (2, 3)
12(2) – 2b(3) = 12
24 – 6b = 12
6b = 24 – 12
6b = 12
b = 2
3a(2) – 3b = 6
6a – 3(2) = 6
6a – 6 = 6
6a = 12
a = 2
The system of equations are
12x – 4y = 12
2(6x – 2y) = 12
6x – 2y = 6 — (i)
6x – 2y = 6 — (ii)
As both the equations are same, they have infinately many solutions
(a, b) = (2, 2)

Systems of Linear Equations Connecting Concepts

Connecting Concepts

Using the Problem-Solving Plan

Question 1.
An animal shelter has a total of 65 cats and dogs. The ratio of cats to dogs is 6:7. Find the number of cats and the number of dogs in the shelter.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 91
Understand the problem.
You know the total number of cats and dogs in an animal shelter, and the ratio of cats to dogs. You are asked to find the number of cats and the number of dogs in the shelter.
Make a plan.
Write a system of equations. Use the total number of cats and dogs to write an equation relating the number of cats and the number of dogs. Use the ratio of cats to dogs to write a second equation. Then solve the system.
Solve and check.
Use the plan to solve the problem. Then check your solution.

Answer:
The number of cats in the shelter is 30, the number of dogs are 35.

Explanation:
Let us take the number of dogs as x, cats as y
An animal shelter has a total of 65 cats and dogs.
x + y = 65 — (i)
The ratio of cats to dogs is 6:7
x : y = 6 : 7
6x = 7y
x = 7y/6
Substitute x = 7y/6 in (i)
7y/6 + y = 65
7y + 6y = 65(6)
13y = 390
y = 30
Substitute y = 30 in (i)
x + 30 = 65
x = 65 – 30
x = 35
So, the number of cats in the shelter are 30, number of dogs are 35.

Question 2.
The measure of ∠1 is 15 degrees less than two times the measure of ∠2. Find the measure of each of the four angles formed by the intersecting lines. Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 92

Answer:
The angles are ∠1 = 125, ∠2 = 55, ∠3 = 125, ∠4 = 55.

Explanation:
The measure of ∠1 is 15 degrees less than two times the measure of ∠2.
∠1 – 15 = 2(∠2) — (i)
∠1 + ∠2 = 180 degrees — (ii)
∠1 = 180 – ∠2
180 – ∠2 – 15 = 2(∠2)
165 = 2∠2 + ∠2
165 = 3∠2
∠2 = 55 degrees
Substitute ∠2 = 55 in (i)
∠1 – 15 = 2(55)
∠1 = 110 + 15
∠1 = 125 degrees
So, the angles are ∠1 = 125, ∠2 = 55, ∠3 = 125, ∠4 = 55.

Question 3.
A landscaper plants grass seed over the entire area of two parks that are similar in shape. The ratio of the perimeter of Park A to the perimeter of Park B is 2 : 1. The parks have a combined area of 9000 square feet. How many square feet does the landscaper cover with grass seed at ParkA? Park B? Justify your answer.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 93

Answer:
The landscaper cover with grass seed at ParkA is 60√2 at park B is 30√2.

Explanation:
let the side of park A is x, park b is y.
The ratio of the perimeter of Park A to the perimeter of Park B is 2 : 1
x : y = 2 : 1
x = 2y — (i)
The parks have a combined area of 9000 square feet
x² + y² = 9000 —- (ii)
Substitute x = 2y in (ii)
(2y)² + y² = 9000
4y² + y² = 9000
5y² = 9000
y² = 1800
y = 30√2
Substitute y = 30√2 in (i)
x = 2(30√2)
= 60√2
The landscaper cover with grass seed at ParkA is 60√2 at park B is 30√2.

Performance Task

Mixing Alloys

At the beginning of this chapter, you watched a STEAM Video called “Gold Alloys.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 94

Systems of Linear Equations Chapter Review

Review Vocabulary

Write the definition and give an example of each vocabulary term.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 95

Graphic Organizers

You can use a Four Square to organize information about a concept. Each of the four squares can be a category, such as definition, vocabulary, example, non-example, words, algebra, table, numbers, visual, graph, or equation. Here is an example of a Four Square for solving systems of linear equations by graphing.
Big Ideas Math Answers 8th Grade Chapter 5 Systems of Linear Equations 96

Choose and complete a graphic organizer to help you study the concept.

  1. solving systems of linear equations by substitution
  2. solving systems of linear equations by elimination
  3. systems of linear equations with no solution
  4. systems of linear equations with infinitely many solutions

Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 97

Chapter Self-Assessment

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 98

5.1 Solving Systems of Linear Equations by Graphing (pp. 199-204)

Solve the system by graphing.

Question 1.
y = 2x – 3
y = x + 2

Answer:
The solution is (5, 7)

Explanation:
The given systems of linear equations are y = 2x – 3, y = x + 2
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 24
The lines intersect at (5, 7)
So, the solution is (5, 7)

Question 2.
y = -x + 4
x + 2y = 0

Answer:
The solution is (8, -4)

Explanation:
The given systems of linear equations are y = -x + 4, x + 2y = 0
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 25
The lines intersect at (8, -4)
So, the solution is (8, -4)

Question 3.
x – y = -2
2x – 3y = -2

Answer:
The solution is (-4, -2).

Explanation:
The given systems of linear equations are x – y = -2, 2x – 3y = -2
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 26
The lines intersect at (-4, -2)
So, the solution is (-4, -2)

Use a graphing calculator to solve the system.

Question 4.
y = -0.5x
y = 0.75x + 1.25

Answer:
The solution is (-1, 0.5)

Explanation:
The given systems of linear equations are y = -0.5x, y = 0.75x + 1.25
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 27
The lines intersect at (-1, 0.5)
So, the solution is (-1, 0.5)

Question 5.
y = 0.2x – 3
10x + 3y = 5

Answer:
The solution is (1.3, -2.7)

Explanation:
The given systems of linear equations are y = 0.2x – 3, 10x + 3y = 5
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 28
The lines intersect at (1.3, -2.7)
So, the solution is (1.3, -2.7)

Question 6.
2.6x + 1.3y = 7.8
1.2x – 3.6y = 12

Answer:
The solution is (4, -2)

Explanation:
The given systems of linear equations are 2.6x + 1.3y = 7.8, 1.2x – 3.6y = 12
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 29
The lines intersect at (4, -2)
So, the solution is (4, -2)

Question 7.
The sum of the two numbers is 38. Find each number when one number is 8 more than the other number. Use a system of linear equations to justify your answer.

Answer:
The numbers are 15, 28.

Explanation:
Let the two numbers be x, y
The sum of two numbers is 38
x + y = 38 — (i)
One number is 8 more than the other number
x + 8 = y — (ii)
Substitute y = x + 8 in (i)
x + x + 8  = 38
2x + 8 = 38
2x = 38 – 8
2x = 30
x = 15
Substitute x = 15 in (ii)
y = 15 + 8
y = 23
So, the numbers are 15, 28.

Question 8.
You observe the heights of two plants for an experiment. Plant A has a height of 8 centimeters and grows 1 centimeter each week. Plant B has a height of 4 centimeters and grows 2 centimeters each week.
a. Write a system of linear equations that represents this situation.
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 99
b. Will the plants ever have the same height? If so, what is the height?

Answer:
a. y = x + 8, y = 2x + 4
b. The same height will be 12 cm after4 weeks.

Explanation:
Let the total height be y, growth of the plant each week be x.
Plant A has a height of 8 centimeters and grows 1 centimeter each week. Plant B has a height of 4 centimeters and grows 2 centimeters each week.
a. y = x + 8 — (i)
y = 2x + 4 —- (ii)
b. If plants ever have the same height
x + 8 = 2x + 4
2x – x = 8 – 4
x = 4
y = 4 + 8
y = 12
The same height will be 12 cm after4 weeks.

Question 9.
Write a system of linear equations containing the equation y = -3x + 2 and that has a solution of (-1, 5). Use a graph to justify your answer.

Answer:
The equation is y = x + 6

Explanation:
Equation is y = -3x + 2
The point is (-1, 5)
The slope intercept form is y = mx + c
5 = -m + c
If m = 1
then c = 5 = -1 + c
c = 6
Then the equation is y = x + 6
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 52

5.2 Solving Systems of Linear Equations by Substitution (pp. 205–210)

Solve the system by substitution. Check your solution.

Question 10.
y = -3x – 7
y = x + 9

Answer:
The solution set is (-4, 5).

Explanation:
The given system of linear equations are
y = -3x – 7 —- (i)
y = x + 9 —- (ii)
Substitute (i) in (ii)
-3x – 7 = x + 9
-3x – x = 9 + 7
-4x = 16
x = -4
Substitute x = -4 in (i)
y = -3(-4) – 7
y = 12 – 7
y = 5
Substitute x = -4, y = 5 in (i)
5 = -3(-4) – 7 = 12 – 7
So, the solution set is (-4, 5).

Question 11.
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 100

Answer:
The solution set is (-8, 0)

Explanation:
The given system of linear equations are
1/2 x + y = -4 —- (i)
y = 2x + 16 —-(ii)
Substitute (ii) in (i)
1/2 x + 2x + 16 = -4
2.5x = -4 – 16
2.5x = -20
x = -8
Substitute x = -8 in (ii)
y = 2(-8) + 16
y = -16 + 16
y = 0
Substitute x = -8, y = 0 in (ii)
0 = 2(-8) + 16 = -16 + 16
So, the solution set is (-8, 0)

Question 12.
-x + 5y = 28
x + 3y = 20

Answer:
The solution set is (2, 6)

Explanation:
The given system of linear equations are
-x + 5y = 28
x + 3y = 20 —- (ii)
x = 5y – 28 —- (i)
Substitute (i) in (ii)
5y – 28 + 3y = 20
8y – 28 = 20
8y = 48
y = 6
Substitute y = 6 in (i)
x = 5(6) – 28
= 30 – 28
x = 2
Substitute y = 6, x = 2 in (ii)
2 + 3(6) = 2 + 18 = 20
So, the solution set is (2, 6)

Question 13.
Zoo admission costs $6 for children and $9 for adults. On Monday, 2200 people visit the zoo and the zoo collects $14,850 in admissions.
a. Write a system of linear equations that represents this situation.
b. How many zoo visitors are children? adults?
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 101

Answer:
a. 6x + 9y = 14850, x + y = 2200
b. There are 1650 child visitors and 550 adult visitors.

Explanation:
The cost for the zoo is $6 for children & $9 for adults. On Monday, 2200 people visit the zoo and the zoo collects $14,850 in admissions.
a. The system of equations are
6x + 9y = 14850 — (i)
x + y = 2200 — (ii)
b. Multiply (ii) by 6
6x + 6y = 13200
6x + 6y – 6x – 9y = 13200 – 14850
-3y = -1650
y = 550
Substitute y = 550 in (ii)
x + 550 = 2200
x = 2200 – 550
x = 1650
There are 1650 child visitors and 550 adult visitors.

Solve the system. Explain your choice of method.

Question 14.
y = x – 2
y = -2x + 1

Answer:
The solution set is (1, -1).

Explanation:
The given system of linear equations are
y = x – 2 — (i)
y = -2x + 1 — (ii)
x – 2 = -2x + 1
x + 2x = 1 + 2
3x = 3
x = 1
Substitute x = 1 in (i)
y = 1 – 2
y = -1
So, the solution set is (1, -1).

Question 15.
3y + 9 = 3x
y = –\(\frac{1}{3}\)x + 1

Answer:
The solution set is (3, 0).

Explanation:
The given system of linear equations are
3y + 9 = 3x — (i)
y = –\(\frac{1}{3}\)x + 1 —- (ii)
Substitute (ii) in (i)
3(-1/3 x + 1) + 9 = 3x
-x + 3 + 9 = 3x
4x = 12
x = 3
Substitute x = 3 in (i)
9 = 9 + 3y
y = 0
So, the solution set is (3, 0).

Question 16.
-x + 2y = -4
4y = x

Answer:
The solution set is (8, 2).

Explanation:
The given system of linear equations are
-x + 2y = -4 —- (i)
4y = x —- (ii)
Substitute (ii) in (i)
-4y + 2y = -4
-2y = -4
y = 2
Substitute y = 2 in (ii)
x = 4(2)
x = 8
So, the solution set is (8, 2).

Question 17.
The measure of an acute angle in a right triangle is one-fourth the measure of the other acute angle. Write a system of linear equations that represents this situation and use it to find the measures of the acute angles of the triangle.

Answer:
x = 72 degrees, y = 18 degrees

Explanation:
Let two acute angles be x, y
The measure of an acute angle in a right triangle is one-fourth the measure of the other acute angle
x + y = 90 — (i)
y = 1/4 x — (ii)
Substitute (ii) in (i)
x + 1/4 = 90
5/4 x = 90
x = 90 . (4/5)
x = 72 degrees
So, y = 1/4 (72)
y = 18 degrees

5.3 Solving Systems of Linear Equations by Elimination (pp. 211–218)

Solve the system by elimination. Check your solution.

Question 18.
2x + 5y = 60
2x – 5y = -20

Answer:
The solution set is (10, 8).

Explanation:
The given system of linear equations are
2x + 5y = 60 — (i)
2x – 5y = -20 — (ii)
Add both equations
2x + 5y + 2x – 5y = 60 – 20
4x = 40
x = 10
Substitute x = 10 in (i)
2(10) + 5y = 60
20 + 5y = 60
5y = 60 – 20
5y = 40
y = 8
Substitute x = 10, y = 8 in (i)
2(10) + 5(8) = 20 + 40 = 60
So, the solution set is (10, 8).

Question 19.
4x – 3y = 15
2x + y = -5

Answer:
The solution set is (3/2, -3).

Explanation:
The given system of linear equations are
4x – 3y = 15 —- (i)
2x + y = -5 —- (ii)
Multiply (ii) by 3
6x + 3y = -15 —- (iii)
Add (ii) & (i)
4x – 3y + 6 + 3y = 15 – 15
4x – 6 = 0
4x = 6
x = 3/2
Substitute x = 3/2 in (i)
4(3/2) – 3y = 15
6 – 3y = 15
-3y = 15 – 6
-3y = 9
y = -3
Substitute x = 3/2, y = -3 in (i)
4(3/2) – 3(-3) = 6 + 9 = 15
So, the solution set is (3/2, -3).

Question 20.
A gift basket that contains jars of jam and packages of bread mix costs $45. There are 8 items in the basket. Jars of jam cost $6 each, and packages of bread mix cost $5 each. Write and solve a system of linear equations to find the number of each item in the gift basket.

Answer:
5 jars of jam and 3 packages of bread mix.

Explanation:
Let j be the no of jars of jam, b bethe no of packages of bread mix
j + b = 8
6j + 5b = 45
5j + 5b = 40
6j + 5b – 5j – 5b = 45 – 40
j = 5
5 + b = 8
b = 3
So, 5 jars of jam and 3 packages of bread mix.

Question 21.
When might it be easier to solve a system by elimination instead of graphing?

Answer:
Substitution is easier to solve a system by elimination instead of graphing.

Question 22.
You have a total of 10 coins consisting of nickels and dimes in your pocket. The value of the coins is $0.70. Write and solve a system of linear equations to find the numbers of nickels and dimes in your pocket.

Answer:
There are 6 nickels and 4 dimes.

Explanation:
The no of coins of nickles and dimes 10.
x + y = 10 — (i)
0.05x + 0.1y = 0.70 — (ii)
Multiply (i) by 0.05
0.05x + 0.05y = 0.5
0.05x + 0.1y – 0.05x – 0.05y = 0.70 – 0.50
0.05y = 0.2
y = 4
x + 4 = 10
x = 6
There are 6 nickles and 4 dimes.

5.4 Solving Special Systems of Linear Equations (pp. 219–224)

Solve the system. Explain your choice of method.

Question 23.
x + 2y = -5
x – 2y = -5

Answer:
The solution set is (-5, 0)

Explanation:
The given system of linear equations are
x + 2y = -5
x – 2y = -5
add both equations
x + 2y + x – 2y = -5 – 5
2x = -10
x = -5
Substitute x = -5 in x – 2y = -5
-5 – 2y = -5
-2y = 0
y = 0
So, the solution set is (-5, 0)

Question 24.
3x – 2y = 1
9x – 6y = 3

Answer:
The system has infinitely many solutions.

Explanation:
The given system of linear equations are
3x – 2y = 1 —- (i)
9x – 6y = 3 —- (ii)
Multiply (ii) by 1/3
3x – 2y = 1 —- (iii)
Both (iii) & (ii) are samem so the system has infinitely many solutions.

Question 25.
8x – 2y = 16
-4x + y = 8

Answer:
The system has no solution.

Explanation:
The given system of linear equations are
8x – 2y = 16 — (i)
-4x + y = 8 — (ii)
Multiply (i) by 1/2
4x – y = 8 — (iii)
Add (iii) & (ii)
-4x + y – 4x + y = 8 + 8
So, the system has no solution.

Question 26.
4y = x – 8
–\(\frac{1}{4}\)x + y = -1

Answer:
The system has no solution.

Explanation:
The given system of linear equations are
4y = x – 8
–\(\frac{1}{4}\)x + y = -1
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 53
Lines are parallel
So, the system has no solution.

Question 27.
-2x + y = -2
3x + y = 3

Answer:
The solution set is (1, 0).

Explanation:
The given system of linear equations are
-2x + y = -2 —(i)
3x + y = 3 — (ii)
Subtract equations
-2x + y – 3x – y = -2 – 3
-5x = -5
x = 1
Substitute x = 1 in (i)
-2(1) + y = -2
-2 + y = -2
y = 0
Substitute x = 1, y = 0 in (i)
-2(1) + 0 = -2
So, the solution set is (1, 0).

Question 28.
3x = \(\frac{1}{3}\)y + 2
9x – y =-6

Answer:
The system has no solution.

Explanation:
The given system of linear equations are
3x = \(\frac{1}{3}\)y + 2
9x – y =-6
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 54
The lines are parallel
So, the system has no solution.

Question 29.
You have $50 in your savings account and plan to deposit $10 each week. Your friend has $25 in her savings account and plans to also deposit $10 each week.
a. Write a system of linear equations that represents this situation.
b. Will your friend’s account ever have the same amount of money as your account? Explain.
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 102

Answer:
a. y = 10x + 50, y = 10x + 25
b. No, the amount will never be equal.

Explanation:
Let the total amount be y and the number of weeks is x
a. You have $50 in your savings account and plan to deposit $10 each week.
y = 10x + 50
Your friend has $25 in her savings account and plans to also deposit $10 each week.
y = 10x + 25
b. No, the account will never have the same amounts. By inspection, the lines are parallel and have the same slopes but different y-intercepts. So, the amount will always be $25 greater than the friend account but will not be the same.

Write a system of linear equations that fits the description. Use a graph to justify your answer.

Question 30.
The system has no solution.

Answer:
The possible systems of linear equations can be y = -5x – 2, 5x + y = 0.

Explanation:
The conditions where two equations have no solution are The lines should be parallel having the same slopes but different y-intercepts.
So, let us consider the slope as -5 and different y-intercepts.
Then, the possible systems of linear equations can be y = -5x – 2, 5x + y = 0.
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 38

Question 31.
The system has infinitely many solutions.

Answer:
The system of linear equations can be 0.5x + 4y = -11, -1.5x – 12y = 33.

Explanation:
The condition is lines are the same and have the same slope, y-intercepts.
So, the system of linear equations can be 0.5x + 4y = -11, -1.5x – 12y = 33
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 42

Question 32.
The system has one solution.

Answer:
The system of linear equations can be -2x + y = 1.3, 2(0.5x – y) = 4.6

Explanation:
The system has only one solution means the lines intersect and have different slopes.
So, the system of linear equations can be -2x + y = 1.3, 2(0.5x – y) = 4.6
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 48

Question 33.
Solve the system by graphing, by substitution, and by elimination. Which method do you prefer? Explain your reasoning.
5x + y = 8
2y = -10x + 8

Answer:
The system has no solution.
Out of all methods, I feel graphing is easier.

Explanation:
The given system of linear equations are
5x + y = 8 —- (i)
2y = -10x + 8 —- (ii)
Graph the above equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 55
The lines are parallel. So, the system has no solution.
Using substitution method
Multiply equation (ii) by 1/2
2y = -10x + 8
y = -5x + 4
Substitute y = -5x + 4 in (i)
5x + -5x + 4 = 8
4 = 8
So, the system has no solution
Using elimination method
y = -5x + 4
5x + y = 4 — (iii)
Subtract (i) & (iii)
5x + y – -5x – y = 8 – 4
0 = 4
So, the system has no solution.

Question 34.
Your friend chooses to solve the system of equations by graphing. Would you choose the same method? Why or why not?
5x + 2y = 12
y = x – 8

Answer:
The solution set is (4, -4).

Explanation:
The given system of linear equations are
5x + 2y = 12, y = x – 8
Yes, I will also choose graphing to solve the equations.
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 56
The solution set is (4, -4).

Systems of Linear Equations Practice Test

5 Practice Test

Question 1.
Solve the system by graphing.
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 103

Answer:
The solution is (4, 12).

Explanation:
The given systems of linear equations are y = 1/2 x + 10, y = 4x – 4
Graph the equations
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 23
The lines intersect at (4, 12)
So, the solution is (4, 12)

Question 2.
Solve the system by substitution.
-3x + y = 2
-x + y – 4 = 0

Answer:
The solution set is (1, 5)

Explanation:
The given system of linear equations are
-3x + y = 2
y = 2 + 3x —- (i)
-x + y – 4 = 0 —- (ii)
Substitute (i) in (ii)
-x + 2 + 3x – 4 = 0
2x – 2 = 0
2x = 2
x = 1
Substitute x = 1 in (i)
y = 2 + 3(1)
y = 2 + 3
y = 5
So, the solution set is (1, 5)

Question 3.
Solve the system by elimination. Solve the system. Check your solution.
x + y = 12
3x = 2y + 6

Answer:
The solution set is (-18, 30)

Explanation:
The given system of linear equations are
x + y = 12 —- (i)
3x = 2y + 6
3x – 2y = 6 —- (ii)
Multiply equation (i) by 2
2(x + y = 12)
2x + 2y = 24 —- (iii)
Subtract (ii) from (iii)
3x – 2y – 2x – 2y = 6 – 24
x = -18
Substitute x = -18 in (i)
-18 + y = 12
y = 12 + 18
y = 30
So, the solution set is (-18, 30)

Question 4.
Solve the system. Explain your choice of method.
-2x + y + 3 = 0
3x + 4y = -1

Answer:
The solution set is (1, -1).

Explanation:
The given system of linear equations are
-2x + y + 3 = 0
3x + 4y = -1
By graphing
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 57
The solution set is (1, -1).

Without graphing or solving, determine whether the system of linear equations has one solution, infinitely many solutions, or no solution. Explain your reasoning.

Question 5.
y = 4x + 8
y = 5x + 1

Answer:
The system has one solution.

Explanation:
The given system of linear equations are
y = 4x + 8 — (i)
y = 5x + 1 —- (ii)
Equate both equations
4x + 8 = 5x + 1
5x – 4x = 8 – 1
x = 7
Substitute x = 7 in (i)
y = 4(7) + 8
y = 28 + 8
y = 36
So, the system has one solution.

Question 6.
2y = 16x – 2
y = 8x – 1

Answer:
The system has infinitely many solutions

Explanation:
The given system of linear equations are
2y = 16x – 2 — (i)
y = 8x – 1 — (ii)
Substitute (ii) in (i)
2(8x – 1) = 16x – 2
16x – 2 = 16x – 2
So, the system has infinitely many solutions.

Question 7.
y = -3x + 2
6x + 2y = 10

Answer:
The system has  no solution.

Explanation:
The given system of linear equations are
y = -3x + 2 — (i)
6x + 2y = 10 — (ii)
Substitute (i) in (ii)
6x + 2(-3x + 2) = 10
6x – 6x + 4 = 10
4 = 10
So, the system has  no solution.

Question 8.
In the diagram, the measure of ∠1 is three times the measure of ∠2. Find the measure of each angle.
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 104

Answer:
∠1 = 3(∠2) = ∠6
∠3 = 180 – ∠2 = ∠5 = ∠7
∠4 = ∠2

Explanation:
The measure of ∠1 is three times the measure of ∠2
∠1 = 3(∠2)
∠5 + ∠2 = 180
∠5 = 180 – ∠2
∠6 + ∠1 = 180
∠6 + 3(∠2) = 180
∠6 = 180 – 3(∠2)
∠2 + ∠3 = 180
∠3 = 180 – ∠2 = ∠5
So, ∠4 = ∠2

Question 9.
The price of 2 pears and 6 apples is $14. The price of 3 pears and 9 apples is $21. Can you determine the unit prices for pears and apples? Explain.

Answer:
They have infinitely many solutions.

Explanation:
Let the price of 1 pears be x, price of 1 apple be y
The price of 2 pears and 6 apples is $14
2x + 6y = 14 — (i)
x + 3y = 7
The price of 3 pears and 9 apples is $21
3x + 9y = 21 —- (ii)
x + 3y = 7
Both the equations are the same. So they have infinitely many solutions.

Question 10.
A bouquet of lilies and tulips has 12 flowers. Lilies cost $3 each, and tulips cost $2 each. The bouquet costs $32. Write and solve a system of linear equations to find the numbers of lilies and tulips in the bouquet.
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 105

Answer:
8 lilies and 4 tulips.

Explanation:
Let l be the no of lilies, t be the no of tulips
A bouquet of lilies and tulips has 12 flowers
l + t = 12 —- (i)
Lilies cost $3 each, and tulips cost $2 each. The bouquet costs $32.
3l + 2t = 32 — (ii)
Multiply (i) by 3
3(l + t = 12)
3l + 3t = 36 — (iii)
Subtract (iii) from (ii)
3l + 3t – (3l + 2t) = 36 – 32
3l + 3t – 3l – 2t = 4
t = 4
Substitute t = 4 in (i)
l + 4 = 12
l = 12 – 4
l = 8
8 lilies and 4 tulips.

Question 11.
How much does it cost for 2 specials and 2 glasses of milk?
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 106

Answer:
The cost of 2 specials & 2 glasses of milk is $16.1.

Explanation:
Let specials be x and glasses of milk be y
4x + 2y = 28 — (i)
3x + 4y = 26.25 — (ii)
2(4x + 2y = 28)
8x + 4y = 56
subtract equations
8x + 4y – (3x + 4y) = 56 – 26.25
8x + 4y – 3x – 4y = 29.75
5x = 29.75
x = 5.95
Substitute x = 5.95 in (i)
4(5.95) + 2y = 28
23.8 + 2y = 28
2y = 28 – 23.8
2y = 4.2
y = 2.1
The cost of 2 specials & 2 glasses of milk is 2x + 2y
= 2(5.95) + 2(2.1)
= 11.9 + 4.2
= 16.1
The cost of 2 specials & 2 glasses of milk is $16.1.

Systems of Linear Equations Cumulative Practice

Question 1.
What is the solution of the system of equations?
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 107
B. (0, -1)
C. no solution
D. infinitely many solutions

Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 108

Answer:
B. (0, -1)

Explanation:
The given system of linear equations are
y = 2/3 x – 1
4x + 6y = -6
By graphing
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 58
so, the solution set is (-1, 0)

Question 2.
What is the value of x?
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 109

Answer:
x = 40 degrees

Explanation:
x + 140 = 180
x = 180 – 140
x = 40 degrees

Question 3.
Which of the following shows Rectangle E’F’G’H’, the image of Rectangle EFGH after it is translated 4 units down?
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 110
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 111

Answer:
F.

Explanation:
By observing all the images we can say that F is the answer.

Question 4.
Which point is a solution of the system of equations?
x + 3y = 10
x = 2y – 5
A. (1, 3)
B. (3, 1)
C. (55, -15)
D. (-35, -15)

Answer:
The solution set is (1, 3).

Explanation:
The given system of linear equations are
x + 3y = 10
x = 2y – 5
Substitute x = 2y – 5 in x + 3y = 10
2y – 5 + 3y = 10
5y = 10 + 5
5y = 15
y = 3
Substitute y = 3 in x = 2y – 5
x = 2(3) – 5
x = 6 – 5
x = 1
So, the solution set is (1, 3).

Question 5.
The graph of a system of two linear equations is shown. Which point is the solution of the system?
F. (-1, 2)
G. (0, 4)
H. (2, -1)
I. (0, 0)
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 112

Answer:
F. (-1, 2)

Explanation:
From the graph, the point of intersection of two equations is (-1, 2)
So, the solution set is (-1, 2).

Question 6.
A scenic train ride has one price for adults and one price for children. One family of two adults and two children pays $62 for the train ride. Another family of one adult and four children pays $70. Which system of linear equations can you use to find the price x for an adult and the price y for a child?
A. 2x + 2y = 70
x + 4y = 62
B. x + y = 62
x + y = 70
C. 2x + 2y = 62
4x + y = 70
D. 2x + 2y = 62
x + 4y = 70

Answer:
D. 2x + 2y = 62
x + 4y = 70

Explanation:
If x is the cost of an adult ticket, y is the cost of a child ticket, then
One family of two adults and two children pays $62 for the train ride.
2x + 2y = 62
Another family of one adult and four children pays $70.
x + 4y = 70.

Question 7.
Which of the following is true about the graph of the linear equation y =-7x + 5?
F. The slope is 5, and the y-intercept is -7.
G. The slope is -5, and the y-intercept is -7.
H. The slope is -7, and the y-intercept is -5.
I. The slope is -7, and the y-intercept is 5.

Answer:
H. The slope is -7, and the y-intercept is -5.

Explanation:
The given equation is y =-7x + 5
The equation in the form of y = mx + c
So, slope m = -7, y-intercept c = 5

Question 8.
What is the measure (in degrees) of the exterior angle of the triangle?
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 113

Answer:
The exterior angle of the triangle is 127 degrees

Explanation:
The Sum of angles of a triangle is 180
x + 64 + y = 180
x + y = 180 – 64
x + y = 116
y = 116 – x — (i)
Sum of unknown angle and exterior angle is 180
y + 2x + 1 = 180
y + 2x = 180 – 1
2x + y = 179 —(ii)
Substitute (i) in (ii)
2x + 116 – x = 179
x + 116 = 179
x = 179 – 116
x = 63
So, the exterior angle is (2x + 1) = 2(63) + 1
= 126 + 1 = 127 degrees

Question 9.
The graph of which equation is parallel to the line that passes through the points (-1, 5) and (4, 7)?
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 114

Answer:
c.

Explanation:
The equation of a line pass through two points is (y – y₁) = [ =(y₂ – y₁)/(x₂ – x₁)](x – x₁)
x₁ = -1, y₁ = 5, x₂ = 4, y₂ = 7
So, (y – 5) = (7 – 5)/(4 + 1)(x + 1)
y – 5 = 2/5(x + 1)
5(y – 5) = 2(x + 1)
5y – 25 = 2x + 2
5y = 2x + 2 + 25
5y = 2x + 27
y = 2/5 x + 27/5

Question 10.
You buy 3 T-shirts and 2 pairs of shorts for $42.50. Your friend buys 5 T-shirts and 3 pairs of shorts for $67.50. Use a system of linear equations to find the cost of each T-shirt. Show your work and explain your reasoning.
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 115

Answer:
The cost of each T-shirt is $7.5.

Explanation:
Let t be the price of a t-shirt, s be the price of a pair of shorts
You buy 3 T-shirts and 2 pairs of shorts for $42.50.
3t + 2s = 42.50
Your friend buys 5 T-shirts and 3 pairs of shorts for $67.50
5t + 3s = 67.50
Multiply the first equation by 3 and second by 2 and then subtract two equations
3(3t + 2s = 42.50) ➝ 9t + 6s = 127.50
2(5t + 3s = 67.50) ➝ 10t + 6s = 135
10t + 6s – 9t – 6s = 135 – 127.50
t = 7.5
The cost of each T-shirt is $7.5.

Question 11.
The red figure is congruent to the blue figure. Which of the following is a sequence of rigid motions between the figures?
F.Translate the red triangle 6 units left and then 4 units down.
G. Reflect the red triangle in the x-axis, and then translate 4 units down.
H. Reflect the red triangle in the y-axis, and then translate 4 units down.
I. Rotate the red triangle 180° clockwise about the origin.
Big Ideas Math Solutions Grade 8 Chapter 5 Systems of Linear Equations 116

Answer:
I. Rotate the red triangle 180° clockwise about the origin.

Explanation:
The vertices of the red triangle is (1, 1), (4, 1), (3, 4)
The vertices of the blue triangle are (-1, -3), (-4, -3), (-3, 0)

Question 12.
Which of the following is true about the graph of the linear equation y = 2?
A. The graph is a vertical line that passes through (2, 0).
B. The graph is a vertical line that passes through (0, 2).
C. The graph is a horizontal line that passes through (2, 0).
D. The graph is a horizontal line that passes through (0, 2).

Answer:
C. The graph is a horizontal line that passes through (2, 0).

Explanation:
The graph for the linear equation y = 2 is
Big Ideas Math Answers Grade 8 Chapter 5 Systems of Linear Equations 59

Question 13.
The sum of one-third of a number and 10 is equal to 13. What is the number?
F. \(\frac{8}{3}\)
G. 9
H. 29
I. 69

Answer:
G. 9

Explanation:
Let the number be n
The sum of one-third of a number and 10 is equal to 13
1/3 n + 10 = 13
1/3 n = 13 – 10
1/3 n = 3
n = 3 x 3
n = 9

Question 14.
Solve the equation 4x + 7y = 16 for x.
A. x = 4 + \(\frac{7}{4}\)y
B. x = 4 – \(\frac{7}{4}\)y
C. x = 4 + \(\frac{4}{7}\)y
D. x = 16 – 7y

Answer:
B. x = 4 – \(\frac{7}{4}\)y

Explanation:
The given equation is 4x + 7y = 16
4x = 16 – 7y
x = (16 – 7y)/4
x = 4 – 7y/4

Conclusion:

The solutions given in this Big Ideas Math Grade 8 Chapter 5 are prepared by the subject experts. So, don’t worry about the answers just go through the answers and try to solve the problems. Test your knowledge by practicing the questions in the practice test section. After solving them cross check the answers. Follow our ccssanswers.com to get the latest updates regarding all Grade 8 Chapters.

Big Ideas Math Geometry Answers Chapter 10 Circles

Big Ideas Math Book Geometry Chapter 10 Circles Answers are provided here. Students who have been looking for the BIM Geometry Chapter 10 Circles Answers can read the following sections. The high school students can find a direct link to download Big Ideas Math Geometry Answers Chapter 10 Circles pdf for free of cost. With the help of this answer key, you can prepare well for the exam.

Big Ideas Math Book Geometry Answer Key Chapter 10 Circles

The different chapters included in Big Ideas Math Geometry Solutions are Lines and Segments That Intersect Circles, Finding Arc Measures, Inscribed Angles and Polygons, Angle Relationships in Circles, Segment Relationships in Circles, Circles in the Coordinate Plane, and Using Chords. Students have to practise all the questions from Big Ideas Math Textbook Geometry Chapter 10 Circles.

This Big Ideas Math Book Geometry Answer Key Chapter 10 Circles helps the students while doing the assignments. Get the solutions for all the questions through the quick links provided in the following sections. Test your skills through performance task, chapter review, and maintaining mathematical proficiency.

Circles Maintaining Mathematical Proficiency

Find the Product.

Question 1.
(x + 7) (x + 4)

Solution:
x² + 4x + 28

Explanation:
Given,
(x + 7) (x + 4)
= x(x + 4) + 7(x + 4)
= x² + 4x + 7x + 28
= x² + 4x + 28

Question 2.
(a + 1) (a – 5)

Solution:
a² – 4a – 5

Explanation:
Given,
(a + 1) (a – 5)
= a(a – 5) + 1(a – 5)
= a² – 5a + a – 5
= a² – 4a – 5

Question 3.
(q – 9) (3q – 4)

Solution:
3q² – 31q + 32

Explanation:
Given,
(q – 9) (3q – 4)
= q(3q – 4) – 9(3q – 4)
= 3q² – 4q – 27q + 32
= 3q² – 31q + 32

Question 4.
(2v – 7) (5v + 1)

Solution:
10v² – 33v – 7

Explanation:
Given,
(2v – 7) (5v + 1)
= 2v(5v + 1) – 7(5v + 1)
= 10v² + 2v – 35v – 7
= 10v² – 33v – 7

Question 5.
(4h + 3) (2 + h)

Solution:
4h² + 11h + 6

Explanation:
Given,
(4h + 3) (2 + h)
= 4h(2 + h) + 3(2 + h)
= 8h + 4h² + 6 + 3h
= 4h² + 11h + 6

Question 6.
(8 – 6b) (5 – 3b)

Solution:
18b² – 54b + 40

Explanation:
Given,
(8 – 6b) (5 – 3b)
= 8(5 – 3b) – 6b(5 – 3b)
= 40 – 24b – 30b + 18b²
= 18b² – 54b + 40

Solve the equation by completing the square. Round your answer to the nearest hundredth, if necessary.

Question 7.
x2 – 2x = 5

Solution:
x = √6 + 1, x = 1 – √6

Explanation:
Given,
x² – 2x = 5
x² – 2x + 1² = 5 + 1²
(x – 1)² = 6
x – 1 = ±√6
x = ±√6 + 1
So, x = √6 + 1, x = -√6 + 1

Question 8.
r2 + 10r = -7

Solution:
r = √18 – 5, r = 5 – √18

Explanation:
Given,
r2 + 10r = -7
r² + 10r + 5² = -7 + 5²
(r + 5)² = -7 + 25 = 18
r + 5 = ±√18
r = ±√18 – 5
So, r = √18 – 5, r = 5 – √18

Question 9.
w2 – 8w = 9

Solution:
w = 9, w = -1

Explanation:
Given,
w2 – 8w = 9
w2 – 8w + 4² = 9 + 4²
(w – 4)² = 9 + 16 = 25
w – 4 = ±5
w = 5 + 4, w = -5 + 4
So, w = 9, w = -1

Question 10.
p2 + 10p – 4 = 0

Solution:
p = √29 – 5, p = 5 – √29

Explanation:
Given,
p2 + 10p – 4 = 0
Transformation of -4 from L.H.S to R.H.S
p2 + 10p = 4
p² + 10p + 5² = 4 + 5²
(p + 5)² = 4 + 25
(p + 5)² = 29
p + 5 = ±√29
p = ±√29 – 5
So, p = √29 – 5, p = 5 – √29

Question 11.
k2 – 4k – 7 = 0

Solution:
k = √11 + 2, k = 2 – √11

Explanation:
Given,
k2 – 4k – 7 = 0
Transformation of -7 from L.H.S to R.H.S
k² – 4k = 7
k² – 4k + 2² = 7 + 4
(k – 2)² = 11
k – 2 = ±√11
k = √11 + 2, k = 2 – √11
So, k = √11 + 2, k = 2 – √11

Question 12.
– z2 + 2z = 1

Solution:
z = 1

Explanation:
Given,
-z² + 2z = 1
z² – 2z = -1
z² – 2z + 1 = -1 + 1
(z – 1)² = 0
z = 1

Question 13.
ABSTRACT REASONING
write an expression that represents the product of two consecutive positive odd integers. Explain your reasoning.

Solution:
x(x + 2) = x² + 2x

Explanation:
Given to write an expression that represents the product of two consecutive positive odd integers.
Let us take two consecutive odd integers are x and (x + 2)
The product of two consecutive odd integers is x • (x + 2)
So, the product of consecutive integers is x(x + 2) = x² + 2x

Circles Mathematical Practices

Monitoring progress

Let ⊙A, ⊙B, and ⊙C consist of points that are 3 units from the centers.

Circles Maintaining Mathematical Proficiency Find the Product. Question 1. (x + 7) (x + 4) Answer: Question 2. (a + 1) (a - 5) Answer: Question 3. (q - 9) (3q - 4) Answer: Question 4. (2v - 7) (5v + 1) Answer: Question 5. (4h + 3) (2 + h) Answer: Question 6. (8 - 6b) (5 - 3b) Answer: Solve the equation by completing the square. Round your answer to the nearest hundredth, if necessary. Question 7. x<sup>2</sup> - 2x = 5 Answer: Question 8. r<sup>2</sup> + 10r = -7 Answer: Question 9. w<sup>2</sup> - 8w = 9 Answer: Question 10. p<sup>2</sup> + 10p - 4 = 0 Answer: Question 11. k<sup>2</sup> - 4k - 7 = 0 Answer: Question 12. - z<sup>2</sup> + 2z = 1 Answer: Question 13. ABSTRACT REASONING write an expression that represents the product of two consecutive positive odd integers. Explain your reasoning. Answer: Circles Mathematical Practices Monitoring progress Let ⊙A, ⊙B, and ⊙C consist of points that are 3 units from the centers. im - 1 Question 1. Draw ⊙C so that it passes through points A and B in the figure at the right. Explain your reasoning. Answer: Question 2. Draw ⊙A, ⊙B, and OC so that each is tangent to the other two. Draw a larger circle, ⊙D, that is tangent to each of the other three circles. Is the distance from point D to a point on ⊙D levss than, greater than, or equal to 6? Explain. Answer:

Question 1.
Draw ⊙C so that it passes through points A and B in the figure at the right. Explain your reasoning.

Solution:

Explanation:
As we know,
A circle is the set of all points in a plane that are equidistant from a given point is known as the center of the circle.
Circle with center A is called “circle A” and can be written as ⊙A.
Circles ⊙A, ⊙B, and ⊙C consist of points that are 3 units from the centers.
⊙C is passes through points A and B as shown in the above figure.

Question 2.
Draw ⊙A, ⊙B, and ⊙C so that each is tangent to the other two. Draw a larger circle, ⊙D, that is tangent to each of the other three circles. Is the distance from point D to a point on ⊙D less than, greater than, or equal to 6? Explain.

Solution:
The distance from point D to a point on ⊙D  greater then 6 units.

Explanation:
As we know,
A circle is the set of all points in a plane that are equidistant from a given point is called the center of the circle.
A circle with center D is called “circle D” and can be written as ⊙D.
Coplanar circles that intersect in one point are called tangent circles.

10.1 Lines and Segments that Intersect Circles

Exploration 1

Lines and Line Segments That Intersect Circles

Big Ideas Math Geometry Answers Chapter 10 Circles 2

Work with a partner: The drawing at the right shows five lines or segments that intersect a circle. Use the relationships shown to write a definition for each type of line or segment. Then use the Internet or some other resource to verify your definitions.
Chord: _________________
Secant: _________________
Tangent: _________________
Radius: _________________
Diameter: _________________

Solution:
Chord:
A chord of a circle is a straight line segment whose endpoints both lie on a circular arc.
Secant:
A straight line that intersects a circle in two points is called a secant line.
Tangent:
Tangent line is a line that intersects a curved line at exactly one point.
Radius:
It is the distance from the center of the circle to any point on the circle.
Diameter:
It the straight that joins two points on the circle and passes through the center of the circle.

Exploration 2

Using String to Draw a Circle

Work with a partner: Use two pencils, a piece of string, and a piece of paper.

a. Tie the two ends of the piece of string loosely around the two pencils.
Answer:

b. Anchor one pencil of the paper at the center of the circle. Use the other pencil to draw a circle around the anchor point while using slight pressure to keep the string taut. Do not let the string wind around either pencil.
Big Ideas Math Geometry Answers Chapter 10 Circles 3
Answer:

c. Explain how the distance between the two pencil points as you draw the circle is related to two of the lines or line segments you defined in Exploration 1.
REASONING ABSTRACTLY
To be proficient in math, you need to know and flexibly use different properties of operations and objects.
Answer:

Communicate Your Answer

Question 3.
What are the definitions of the lines and segments that intersect a circle?
Answer:

Question 4.
Of the five types of lines and segments in Exploration 1, which one is a subset of another? Explain.
Answer:

Question 5.
Explain how to draw a circle with a diameter of 8 inches.
Answer:

Lesson 10.1 Lines and Segments that Intersect Circles

Monitoring progress

Question 1.
In Example 1, What word best describes \(\overline{AG}\)? \(\overline{CB}\)?

Solution:
\(\overline{A G}\) is secant because it is a line that intersects the circle at two points.
\(\overline{C B}\) is the radius as it is the distance from the center to the point of a circle.

Explanation:

secant is a line that intersects a circle in two points.
\(\overline{AG}\) is a line that intersects the circle at two points.

A segment whose endpoints are the center and any point on a circle is a radius.
\(\overline{CB}\) is the radius as it is the distance from the center to the point of a circle.

Question 2.
In Example 1, name a tangent and a tangent segment.

Solution:
\(\overline{D E}\) is the tangent of the circle
\(\overline{D E}\) is the tangent segment of the circle.

Explanation:
A tangent is a line in the plane of a circle that intersects the circle in exactly one point, the
point of tangency.
The tangent ray \(\overline{AB}\) and the tangent segment \(\overline{AB}\) are also called tangents.

Tell how many common tangents the circles have and draw them. State whether the tangents are external tangents or internal tangents.

Question 3.
Big Ideas Math Geometry Answers Chapter 10 Circles 4

Solution:
4 tangents.

Explanation:
Big Ideas Math Geometry Answers Chapter 10 Circles 2
A tangent is a line segment that intersects the circle at exactly one point.
Internal tangents are the lines that intersect the segments joining the centers of two circles.
External tangents are the lines that do not cross the segment joining the centers of the circles.
Blue lines represent the external tangents and red lines represent the internal tangents.

Question 4.
Big Ideas Math Geometry Answers Chapter 10 Circles 5

Solution:
One tangent.

Explanation:
External tangents are the lines that do not cross the segment joining the centers of the circles.
Big Ideas Math Geometry Answers Chapter 10 Circles 3
One external tangent.

Question 5.
Big Ideas Math Geometry Answers Chapter 10 Circles 6

Solution:
No tangent.

Explanation:
As circles are of different size with common center, there is no overlapping circles.
It is not possible to draw a common tangent for this type of circles.

Question 6.
Is \(\overline{DE}\) tangent to ⊙C?
Big Ideas Math Geometry Answers Chapter 10 Circles 7

Solution:
Yes,

Explanation:
By using the converse of Pythagorean theorem,
CE² = CD² + DE²
(r + 2)² = 3² + 4²
(3 + 2)² = 3² + 4²
25 = 9 + 16
So, the Tangent Line to Circle Theorem that CD⊥ DE, so △CDE is a right triangle.
By the tangent line to the circle theorem, \(\overline{DE}\) is a tangent to ⊙C

Question 7.
\(\overline{S T}\) is tangent to ⊙Q.
Find the radius of ⊙Q.
Big Ideas Math Geometry Answers Chapter 10 Circles 8

Solution:
The radius of ⊙Q is 7 units.

Explanation:
Given,
The Tangent Line to Circle Theorem that QS⊥ ST, so △QST is a right triangle.
By using the tangent line to the circle theorem,
\(\overline{ST}\) is a tangent to ⊙Q.
By using the Pythagorean theorem,
(18 + r)² = r² + 24²
324 + 36r + r² = r² + 576
36r = 576 – 324
36r = 252
r = 7 units.

Question 8.
Points M and N are points of tangency. Find the value(s) of x.
Big Ideas Math Geometry Answers Chapter 10 Circles 9

Solution:
The values of x are 3 or -3.

Explanation:
Given that, Points M and N are points of tangency.
x² = 9
x = ±3

Exercise 10.1 Lines and Segments that Intersect Circles

Vocabulary and Core Concept Check

Question 1.
WRITING
How are chords and secants alike? How are they different?

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 1

Question 2.
WRITING
Explain how you can determine from the context whether the words radius and diameter are referring to segments or lengths.

Solution:
Radius and diameter are the lengths of the line segments that pass through the center of a circle.
So, radius is half of the diameter.

Question 3.
COMPLETE THE SENTENCE
Coplanar circles that have a common center are called ____________ .

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 3

Question 4.
WHICH ONE DOESNT BELONG?
Which segment does not belong with the other three? Explain your reasoning.
chord radius tangent diameter

Solution:
A chord, a radius and a diameter are segments and they intersect a circle in two points.
Where as tangent is a line that intersects a circle at one point.

Monitoring Progress and Modeling with Mathematics

In Exercises 5 – 10, use the diagram.

Big Ideas Math Geometry Answers Chapter 10 Circles 10

Question 5.
Name the circle.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 5

Question 6.
Name two radii.

Solution:
The name of the two radii is CD and AC.

Explanation:
Radius:
It is the distance from the center of the circle to any point on the circle.
In the above given figure the two radii are CD and AC.

Question 7.
Name two chords.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 7
Explanation:
A chord is a segment whose endpoints are on a circle.
There are two chords in the given diagram,
\(\overline{BH}\) and \(\overline{AD}\)

Question 8.
Name a diameter.

Solution:
The name of diameter is AD.

Explanation:
Diameter:
It the straight that joins two points on the circle and passes through the center of the circle.
In the given figure the name of diameter is AD.
\(\overline{AD}\)

Question 9.
Name a secant.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 9

Explanation:
Secant:
A straight line that intersects a circle in two points is called a secant line.
In the above given figure \(\overline{KG}\) is a secant.

Question 10.
Name a tangent and a point of tangency

Solution:
GE is the tangent and F is the point of tangency.

Explanation:
Tangent:
Tangent line is a line that intersects a curved line at exactly one point.
In the given figure, \(\overline{GE}\) is the tangent and F is the point of tangency.

In Exercises 11 – 14, copy the diagram. Tell how many common tangents the circles have and draw them.

Question 11.
Big Ideas Math Geometry Answers Chapter 10 Circles 11

Solution:

Explanation:
Based on the above figure,
Tangent: Tangent line is a line that intersects a curved line at exactly one point.
\(\overline{AB}\) is the tangent and A and B are the points of tangency of circles X and Y.
\(\overline{CD}\) is the tangent and C and D are the points of tangency of circles X and Y.
\(\overline{GH}\) is the tangent and A and B are the points of tangency of circles X and Y.
\(\overline{EF}\) is the tangent and A and B are the points of tangency of circles X and Y.

Question 12.
Big Ideas Math Geometry Answers Chapter 10 Circles 12

Solution:
No common tangent because two circles do not intersect at one point.

Explanation:
With reference to the above figure, circles do not overlap each other.
Circle to be overlap at common point for common tangent of circles.

Question 13.
Big Ideas Math Geometry Answers Chapter 10 Circles 13

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 13

Explanation:
With reference to the above figure,
There are two common tangents, as circles are of same size and shape due to same radius.

Question 14.
Big Ideas Math Geometry Answers Chapter 10 Circles 14

Solution:
One common tangent.
Big Ideas Math Geometry Answers Chapter 10 Circles 3
Explanation:
The above two circles given in the figure are of two different size and intersected at one point.
So, there is one tangent drawn as shown in the above diagram.

In Exercises 15 – 18, tell whether the common tangent is internal or external.

Question 15.
Big Ideas Math Geometry Answers Chapter 10 Circles 15

Solution:
External tangent.

Explanation:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 15

Question 16.
Big Ideas Math Geometry Answers Chapter 10 Circles 16

Solution:
Internal tangent.

Explanation:
The common tangent is the internal tangent because it intersects the segment that joins the centers of two circles as shown in the given figure.

Question 17.
Big Ideas Math Geometry Answers Chapter 10 Circles 17

Solution:
Internal tangent.

Explanation:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 17

Question 18.
Big Ideas Math Geometry Answers Chapter 10 Circles 18

Solution:
Internal tangent.

Explanation:
The common tangent is the internal tangent because it intersects the segment that joins the centers of two circles as shown in the above figure.

In Exercises 19 – 22, tell whether \(\overline{A B}\) is tangent to ⊙C. Explain your reasoning.

Question 19.
Big Ideas Math Geometry Answers Chapter 10 Circles 19

Solution:
\(\overline{AB}\) is the tangent to ⊙C at point A.

Explanation:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 19

Question 20.
Big Ideas Math Geometry Answers Chapter 10 Circles 20

Solution:
△ ACB is not a right angled triangle.
So, \(\overline{A B}\) is not tangent to ⊙C at B.

Explanation:
Use the converse of the Pythagorean theorem,
18² _____________ 15² + 9²
324 _____________ 225 + 81
324 ≠ 304
△ ACB is not a right angled triangle.
So, \(\overline{A B}\) is not tangent to ⊙C at B.

Question 21.
Big Ideas Math Geometry Answers Chapter 10 Circles 21

Solution:
△ ABD is not a right angled triangle.
So, \(\overline{AB}\) is not tangent to ⊙C at A.

Explanation:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 21

Question 22.
Big Ideas Math Geometry Answers Chapter 10 Circles 22

Solution:
\(\overline{AB}\) is tangent to ⊙C at A.

Explanation:
Use the converse of the Pythagorean theorem,
(8+12)² _____________ 12² + 16²
20 x 20 _____________ 144 + 256
400 = 400
△ ACB is a right angled triangle.
So, \(\overline{AB}\) is tangent to ⊙C at A.

In Exercises 23 – 26, point B is a point of tangency. Find the radius r of ⊙C.

Question 23.
Big Ideas Math Geometry Answers Chapter 10 Circles 23

Solution:
r = 10 units.

Explanation:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 23

Question 24.
Big Ideas Math Geometry Answers Chapter 10 Circles 24

Solution:
r = \(\frac {15}{4} \) units.

Explanation:
(r + 6)² = r² + 9²
r² + 12r + 36 = r² + 81
12r = 81 – 36
12r = 45
r = \(\frac { 15 }{ 4 } \)
Therefore, the radius of ⊙C is \(\frac { 15 }{ 4 } \)

Question 25.
Big Ideas Math Geometry Answers Chapter 10 Circles 25

Solution:
r = 10.5 units.

Explanation:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 25

Question 26.
Big Ideas Math Geometry Answers Chapter 10 Circles 26

Solution:
r = 16 units.

Explanation:
(r + 18)² = r² + 30²
r² + 36r + 324 = r² + 900
36r = 900 – 324
36r = 576
r = 16
Therefore, the radius of ⊙C is 16

CONSTRUCTION
In Exercises 27 and 28. construct ⊙C with the given radius and point A outside of ⊙C. Then construct a line tangent to ⊙C that passes through A.

Question 27.
r = 2 in.

Solution:

Explanation:
⊙C circle, the given radius 2 in and point A outside of ⊙C.
The construct  line tangent to ⊙C that passes through A.

Question 28.
r = 4.5 cm

Solution:

Explanation:
⊙C circle, the given radius 4.5 cm and point A outside of ⊙C.
The construct  line tangent to ⊙C that passes through A.

In Exercises 29 – 32, points B and D are points of tangency. Find the value(s) of x.

Question 29.
Big Ideas Math Geometry Answers Chapter 10 Circles 27
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 29

Question 30.
Big Ideas Math Geometry Answers Chapter 10 Circles 28

Solution:
x = 4

Explanation:
With reference to above figure,
3x + 10 = 7x – 6
7x – 3x = 10 + 6
4x = 16
x = 4

Question 31.
Big Ideas Math Geometry Answers Chapter 10 Circles 29

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 31

Question 32.
Big Ideas Math Geometry Answers Chapter 10 Circles 30

Solution:
x = ±2

Explanation:
2x + 5 = 3x² + 2x – 7
3x² = 5 + 7
3x² = 12
x² = 4
x = ±2

Question 33.
ERROR ANALYSIS
Describe and correct the error in determining whether \(\overline{X Y}\) is tangent to ⊙Z.
Big Ideas Math Geometry Answers Chapter 10 Circles 31
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 33

Question 34.
ERROR ANALYSIS
Describe and correct the error in finding the radius of ⊙T.
Big Ideas Math Geometry Answers Chapter 10 Circles 32

Solution:
The radius is \(\frac { 15 }{ 2 } \).

Explanation:
39² = 36² + 15²
So, 15 is the diameter.
The radius is \(\frac { 15 }{ 2 } \).

Question 35.
ABSTRACT REASONING
For a point outside of a circle, how many lines exist tangent to the circle that pass through the point? How many such lines exist for a point on the circle? inside the circle? Explain your reasoning.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 35

Question 36.
CRITICAL THINKING
When will two lines tangent to the same circle not intersect? Justify your answer.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 6

Explanation:
When two tangents are drawn at the two ends of the same diameter of the same circle they will not intersect.
Since at this point of contact both the angles will be right angles.
So, these tangents will be parallel lines as shown in the above figure.

Using tangent line to circle theorem, it follow that the angle between tangent and radius is a right angle. Let’s draw these tangents at the two ends of the same diameter. We can observe a diameter AD like a transversal of these tangents.

Question 37.
USING STRUCTURE
Each side of quadrilateral TVWX is tangent to ⊙Y. Find the perimeter of the quadrilateral.
Big Ideas Math Geometry Answers Chapter 10 Circles 33
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 37

Question 38.
LOGIC
In ⊙C, radii \(\overline{C A}\) and \(\overline{C B}\) are perpendicular. Big Ideas Math Geometry Answers Chapter 10 Circles 34 are tangent to ⊙C.

a. Sketch ⊙C, \(\overline{C A}\), \(\overline{C B}\), Big Ideas Math Geometry Answers Chapter 10 Circles 34.
Answer:

b. What type of quadrilateral is CADB? Explain your reasoning.
Answer:

Question 39.
MAKING AN ARGUMENT
Two hike paths are tangent to an approximately circular pond. Your class is building a nature trail that begins at the intersection B of the bike paths and runs between the bike paths and over a bridge through the center P of the pond. Your classmate uses the Converse of the Angle Bisector Theorem (Theorem 6.4) to conclude that the trail must bisect the angle formed by the bike paths. Is your classmate correct? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 10 Circles 35

Solution:

Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 39

Question 40.
MODELING WITH MATHEMATICS
A bicycle chain is pulled tightly so that \(\overline{M N}\) is a common tangent of the gears. Find the distance between the centers of the gears.
Big Ideas Math Geometry Answers Chapter 10 Circles 36

Solution:
The distance between the center of the gear is 17.8 in.

Explanation:
Given,
height h = 4.3 – 1.8
h = 2.5 in
x² = MN² + h²
x² = 17.6² + 2.5²
x² = 316.01
x = 17.8
Therefore, the distance between the center of the gear is 17.8 in.

Question 41.
WRITING
Explain why the diameter of a circle is the longest chord of the circle.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 41

Question 42.
HOW DO YOU SEE IT?
In the figure, \(\vec{P}\)A is tangent to the dime. \(\vec{P}\)C is tangent to the quarter, and \(\vec{P}\)B is a common internal tangent. How do you know that \(\overline{P A} \cong \overline{P B} \cong \overline{P C}\)
Big Ideas Math Geometry Answers Chapter 10 Circles 37
Answer:

Question 43.
PROOF
In the diagram, \(\overline{R S}\) is a common internal tangent to ⊙A and ⊙B. Prove that \(\frac{\Lambda C}{B C}=\frac{R C}{S C}\)
Big Ideas Math Geometry Answers Chapter 10 Circles 38

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 43.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 43.2

Question 44.
THOUGHT PROVOKING
A polygon is circumscribed about a circle when every side of the polygon is tangent to the circle. In the diagram. quadrilateral ABCD is circumscribed about ⊙Q. Is it always true that AB + CD = AD + BC? Justify your answer.
Big Ideas Math Geometry Answers Chapter 10 Circles 39
Answer:

Question 45.
MATHEMATICAL CONNECTIONS
Find the values of x and y. Justify your answer.
Big Ideas Math Geometry Answers Chapter 10 Circles 40
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 45

Question 46.
PROVING A THEOREM
Prove the External Tangent Congruence Theorem (Theorem 10.2).
Big Ideas Math Geometry Answers Chapter 10 Circles 41
Given \(\overline{S R}\) and \(\overline{S T}\) are tangent to ⊙P.
Prove \(\overline{S R} \cong \overline{S T}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 7

Explanation:
Given that,
\(\overline{S R}\) and \(\overline{S T}\) are tangent to ⊙P.
According to the external tangent theorem,
if from one external point, two tangents are drawn to a circle then they have equal tangent segments. Tangent segment means line joining to the external point and the point of tangency.
We observe that,
∠PRS and ∠PTS are the right angles.
So, the legs of circles are congruent.
Therefore, \(\overline{S R} \cong \overline{S T}\)

Question 47.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Tangent Line to Circle Theorem (Theorem 10.1 ).
Big Ideas Math Geometry Answers Chapter 10 Circles 42
a. Prove indirectly that if a line is tangent to a circle, then it is perpendicular to a radius. (Hint: If you assume line m is not perpendicular to \(\overline{Q P}\), then the perpendicular segment from point Q to line m must intersect line m at some other point R.)
Ghen Line m is tangent to ⊙Q at point P.
Prove m ⊥ \(\overline{Q P}\)
b. Prove indirectly that if a line is perpendicular to a radius at its endpoint, then the line is tangent to the circle.
Gien m ⊥ \(\overline{Q P}\)
Prove Line m is tangent to ⊙Q.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 47.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 47.2

Question 48.
REASONING
In the diagram, AB = AC = 12, BC = 8, and all three segments are Langent to ⊙P. What is the radius of ⊙P? Justify your answer.
Big Ideas Math Geometry Answers Chapter 10 Circles 43

Answer:

Maintaining Mathematical Proficiency

Find the indicated measure.

Question 49.
m∠JKM
Big Ideas Math Geometry Answers Chapter 10 Circles 44

Solution:

Big Ideas Math Geometry Answers Chapter 10 Circles 10.1 Ans 49

Question 50.
AB
Big Ideas Math Geometry Answers Chapter 10 Circles 45

Solution:
AB = 3

Explanation:
Given,
AC = 10
We know that,
AC = AB + BC
10 = AB + 7
AB = 10 – 7
AB = 3

10.2 Finding Arc Measures

Exploration 1

Measuring Circular Arcs

Work with a partner: Use dynamic geometry software to find the measure of \(\widehat{B C}\). Verify your answers using trigonometry.

a.
Big Ideas Math Answers Geometry Chapter 10 Circles 46
Points
A(0, 0)
B(5, 0)
C(4, 3)

Solution:
30 degrees.

Explanation:
We know that the length of an arc in a circle is to plot two lines from the arc’s endpoints to the center of the circle as shown in the given figure.
Then measure the angle where the two lines meet the center.
measure of angle in degrees/360° = L/circumference.

b.
Big Ideas Math Answers Geometry Chapter 10 Circles 47
Points
A(0, 0)
B(5, 0)
C(3, 4)

Solution:
60 degrees.

Explanation:
We know that the length of an arc in a circle is to plot two lines from the arc’s endpoints to the center of the circle as shown in the given figure.
Then measure the angle where the two lines meet the center.
measure of angle in degrees/360° = L/circumference.

c.
Big Ideas Math Answers Geometry Chapter 10 Circles 48
Points
A(0, 0)
B(4, 3)
C(3, 4)

Solution:
15 degrees.

Explanation:
We know that the length of an arc in a circle is to plot two lines from the arc’s endpoints to the center of the circle as shown in the given figure.
Then measure the angle where the two lines meet the center.
measure of angle in degrees/360° = L/circumference.

d.
Big Ideas Math Answers Geometry Chapter 10 Circles 49
Points
A(0, 0)
B(4, 3)
C(- 4, 3)

Solution:
90 degrees.
Explanation:
We know that the length of an arc in a circle is to plot two lines from the arc’s endpoints to the center of the circle as shown in the given figure.
Then measure the angle where the two lines meet the center.
measure of angle in degrees/360° = L/circumference.

Communicate Your Answer

Question 2.
How are circular arcs measured?

Solution:
If an arc in a circle is to plot two lines from the arc’s endpoints to the center of the circle,
Then measure the angle where the two lines meet the center.
measure of angle in degrees/360° = L/circumference.

Question 3.
Use dynamic geometry software to draw a circular arc with the given measure.
USING TOOLS STRATEGICALLY
To be proficient in math, you need to use technological tools to explore and deepen your understanding of concepts.
a. 30°

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 8

b. 45°

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 9

c. 60°
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10

d. 90°

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 11

Lesson 10.2 Finding Arc Measures

Monitoring Progress

Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Big Ideas Math Answers Geometry Chapter 10 Circles 50

Question 1.
\(\widehat{T Q}\)

Solution:
\(\widehat{T Q}\) = 120°

Explanation:
With reference to the above figure,
\(\widehat{T Q}\) is a minor arc.
\(\widehat{T Q}\) = 120°

Question 2.
\(\widehat{Q R T}\)

Solution:
\(\widehat{Q R T}\) = 240°

Explanation:
With reference to the above figure,
Given,
\(\widehat{Q R T}\)
RS = QR + RS + ST
RS = 360° – (60 + 120 + 80)
So, 360 – 260 = 100°
So, \(\widehat{Q R T}\) = 60° + 100° + 80°
\(\widehat{Q R T}\) = 240°

Question 3.
\(\widehat{T Q R}\) is a major arc.

Solution:
\(\widehat{Q R T}\) = 240°

Explanation:
With reference to the above figure,
Given,
\(\widehat{Q R T}\) is a major arc.
RS = QR + RS + ST
RS = 360° – (60 + 120 + 80)
So, 360 – 260 = 100°
So, \(\widehat{Q R T}\) = 60° + 100° + 80°
\(\widehat{Q R T}\) = 240°

Question 4.
\(\widehat{Q S}\)

Solution:
\(\widehat{Q S}\) = 160°

Explanation:
With reference to the above figure,
Given,
\(\widehat{Q S}\)
QS = QR + RS
So, 60 + 100 = 160°
Therefore, \(\widehat{Q S}\) = 160° and it is a minor arc.

Question 5.
\(\widehat{T S}\)

Solution:
\(\widehat{T S}\) = 80° and it is a minor arc.

Explanation:
With reference to the above figure,
\(\widehat{T S}\) is a minor arc.
\(\widehat{T S}\) = 80°

Question 6.
\(\widehat{R S T}\)

Solution:
\(\widehat{R S T}\) = 180°

Explanation:
With reference to the above figure,
Given,
\(\widehat{R S T}\)
RS + TS = 100 + 80 = 180
Therefore, \(\widehat{R S T}\) = 180° and it is a minor arc.

Tell whether the red arcs are congruent. Explain why or why not.

Question 7.
Big Ideas Math Answers Geometry Chapter 10 Circles 51

Solution:
\(\widehat{A B}\), \(\widehat{C D}\) are congruent as they measure same radius and same arc length.

Explanation:
We know that,
Congruent angles are two or more angles that are identical to each other.
Therefore, the measure of these angles is equal to each other as in the above figure.

Question 8.
Big Ideas Math Answers Geometry Chapter 10 Circles 52

Solution:
\(\widehat{M N}\), \(\widehat{P Q}\) are not congruent as they measure different radius.

Explanation:
We know that,
Congruent radius are two or more radius that are identical to each other.
Therefore, the measure of these radii also equal to each other,
as in the above figure the measure of radii is different.
Hence the given circles are not congruent.

Exercise 10.2 Finding Arc Measures

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
Copy and complele:
If ∠ACB and ∠DCE are congruent central angles of ⊙C, then \(\widehat{A B}\) and \(\widehat{D E}\) arc.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 1

Question 2.
WHICH ONE DOESNT BELONG?
Which circle does not belong with the other three? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 53

Solution:
We know that 1 ft = 12 in
So, the fourth circle does not belong to the other three as its diameter is different.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, name the red minor arc and find its measure. Then name the blue major arc and find its measure.

Question 3.
Big Ideas Math Answers Geometry Chapter 10 Circles 54
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 3

Question 4.
Big Ideas Math Answers Geometry Chapter 10 Circles 55

Solution:
The minor arc \(\widehat{E F}\) = 68°
The major arc \(\widehat{F G E}\) = 360° – 68° = 292°.

Explanation:
An arc whose measure is less than 180 degrees is called a minor arc. An arc whose measure is greater than 180 degrees is called a major arc.

Question 5.
Big Ideas Math Answers Geometry Chapter 10 Circles 56
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 5

Question 6.
Big Ideas Math Answers Geometry Chapter 10 Circles 57

Solution:
The minor arc is \(\widehat{M N}\) = 170°,
major arc \(\widehat{N P M}\) = 360° – 170° = 190°.

Explanation:
An arc whose measure is less than 180 degrees is called a minor arc. An arc whose measure is greater than 180 degrees is called a major arc.

In Exercises 7 – 14. identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Big Ideas Math Answers Geometry Chapter 10 Circles 58

Question 7.
\(\widehat{B C}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 7

Question 8.
\(\widehat{D C}\)

Solution:
\(\widehat{D C}\) is a minor arc and it has a measure of 65°.

Explanation:
With reference to the above figure,
An arc whose measure is less than 180 degrees is called a minor arc.

Question 9.
\(\widehat{E D}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 9

Question 10.
\(\widehat{A E}\)

Solution:
\(\widehat{A E}\) is a minor arc and it has a measure of 70°.

Explanation:
With reference to the above figure,
An arc whose measure is less than 180 degrees is called a minor arc.

Question 11.
\(\widehat{E A B}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 11

Question 12.
\(\widehat{A B C}\)

Solution:
\(\widehat{A B C}\) is a semicircle and it has a measure of 180°.

Explanation:
With reference to the above figure,
An arc whose measure is less than 180 degrees is called a minor arc.

Question 13.
\(\widehat{B A C}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 13

Question 14.
\(\widehat{E B D}\)

Solution:
\(\widehat{E B D}\) is a major arc and it has a measure of 315°.

Explanation:
With reference to the above figure,
An arc whose measure is less than 180 degrees is called a minor arc.

In Exercises 15 and 16, find the measure of each arc.

Question 15.
Big Ideas Math Answers Geometry Chapter 10 Circles 59

a. \(\widehat{J L}\)

b. \(\widehat{K M}\)

c. \(\widehat{J L M}\)

d. \(\widehat{J M}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 15

Question 16.
Big Ideas Math Answers Geometry Chapter 10 Circles 60

a. \(\widehat{R S}\)

Solution:
\(\widehat{R S}\) = 138°

Explanation:
Given,
\(\widehat{R S}\)
\(\widehat{R S}\) = \(\widehat{Q R S}\) – \(\widehat{Q R}\)
= 180 – 42
= 138°
So, \(\widehat{R S}\) = 138°

b. \(\widehat{Q R S}\)

Solution:
\(\widehat{Q R S}\) = 180°

Explanation:
With reference to the above figure,

c. \(\widehat{Q S T}\)

Solution:
\(\widehat{Q S T}\) = 222°

Explanation:
\(\widehat{Q S T}\) = \(\widehat{Q R S}\) + \(\widehat{S T}\)
= 180 + 42 = 222
So, \(\widehat{Q S T}\) = 222°

d. \(\widehat{Q T}\)

Solution:
\(\widehat{Q T}\) = 138°

Explanation:
\(\widehat{Q T}\) = 360 – (42 + 138 + 42)
= 360 – (222)
= 138°
\(\widehat{Q T}\) = 138°

Question 17.
MODELING WITH MATHEMATICS
A recent survey asked high school students their favorite type of music. The results are shown in the circle graph. Find each indicated arc measure.
Big Ideas Math Answers Geometry Chapter 10 Circles 61
a. m\(\widehat{A E}\)
b. m\(\widehat{A C E}\)
c. m\(\widehat{G D C}\)
d. m\(\widehat{B H C}\)
e. m\(\widehat{F D}\)
f. m\(\widehat{F B D}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 17

Question 18.
ABSTRACT REASONING
The circle graph shows the percentages of students enrolled in fall Sports at a high school. Is it possible to find the measure of each minor arc? If so, find the measure of the arc for each category shown. If not, explain why it is not possible.
Big Ideas Math Answers Geometry Chapter 10 Circles 62

Solution:
Soccer angle = 30% of 360
(30/100) x 360
3 x 36= 108°
Volleyball angle = 15% of 360
(15/100) x 360= 54°
Cross-country angle = 20% of 360
(20/100) x 360 = 72°
None angle = 15% of 360
(15/100) x 360 = 54°
Football angle = 20% of 360
(20/100) x 360= 72°

In Exercises 19 – 22, tell whether the red arcs are congruent. Explain why or why not.

Question 19.
Big Ideas Math Answers Geometry Chapter 10 Circles 63

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 19

Question 20.
Big Ideas Math Answers Geometry Chapter 10 Circles 64

Solution:
\(\widehat{L P}\) and \(\widehat{M N}\) are not congruet because they are not in the same circle.

Explanation:
We know that,
Congruent angles are two or more angles that are identical to each other.
Therefore, the measure of these angles also equal to each other,
as in the above figure the measure of angles are different.
Hence the given circles are not congruent.

Question 21.
Big Ideas Math Answers Geometry Chapter 10 Circles 65

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 21

Question 22.
Big Ideas Math Answers Geometry Chapter 10 Circles 66

Solution:
\(\widehat{R S Q}\), \(\widehat{F G H}\) are not congruent because those two circles have different radii.

Explanation:
We know that,
Congruent radius are two or more radius that are identical to each other.
Therefore, the measure of these radii also equal to each other,
as in the above figure the measure of radii is different.
Hence the given circles are not congruent.

MATHEMATICAL CONNECTIONS
In Exercises 23 and 24. find the value of x. Then find the measure of the red arc.

Question 23.
Big Ideas Math Answers Geometry Chapter 10 Circles 67

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 23

Question 24.
Big Ideas Math Answers Geometry Chapter 10 Circles 68

Solution:
m\(\widehat{R S T}\) = 195°

Explanation:
Given,
4x + 6x + 7x + 7x = 360
24x = 360°
x = 15°
m\(\widehat{R S T}\) = 6(15) + 7(15)
= 90 + 105 = 195°
So, m\(\widehat{R S T}\) = 195°

Question 25.
MAKING AN ARGUMENT
Your friend claims that any two arcs with the same measure are similar. Your cousin claims that an two arcs with the same measure are congruent. Who is correct? Explain.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 25

Question 26.
MAKING AN ARGUMENT
Your friend claims that there is not enough information given to find the value of x. Is your friend correct? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 69

Solution:
x = 36°

Explanation:
Given,
4x + x + x + 4x = 360°
10x = 360°
x = 36°

Question 27.
ERROR ANALYSIS
Describe and correct the error in naming the red arc.
Big Ideas Math Answers Geometry Chapter 10 Circles 70

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 27

Question 28.
ERROR ANALYSIS
Describe and correct the error in naming congruent arc.
Big Ideas Math Answers Geometry Chapter 10 Circles 71

Solution:
\(\widehat{J K}\), \(\widehat{N P}\) are not congruent because those two arcs are form different circles.

Explanation:
We know that,
Congruent arcs are two or more arcs that are identical to each other.
Therefore, the measure of these arcs also equal to each other,
as in the above figure the measure of arc is different.
Hence the given circles are not congruent.

Question 29.
ATTENDING TO PRECISION
Two diameters of ⊙P are \(\widehat{A B}\) and \(\widehat{C D}\). Find m\(\widehat{A C D}\) and m\(\widehat{A C}\) when m\(\widehat{A D}\) = 20°.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 29

Question 30.
REASONING
In ⊙R, m\(\widehat{A B}\) = 60°, m\(\widehat{B C}\) = 25°. m\(\widehat{C D}\) = 70°, and m\(\widehat{D E}\) = 20°. Find two possible measures of \(\widehat{A E}\).

Solution:
The two possibilities of \(\widehat{A E}\) are 185°, 175°

Explanation:
Given,
\(\widehat{A E}\) = 360 – (\(\widehat{A B}\) + \(\widehat{B C}\) + \(\widehat{C D}\) + \(\widehat{D E}\))
= 360 – (60 + 25 + 70 + 20)
= 360 – (175)
= 185°
\(\widehat{A E}\) = \(\widehat{A B}\) + \(\widehat{B C}\) + \(\widehat{C D}\) + \(\widehat{D E}\)
= 60 + 25 + 70 + 20 = 175°
So, the two possibilities of \(\widehat{A E}\) are 185°, 175°

Question 31.
MODELING WITH MATHEMATICS
On a regulation dartboard, the outermost circle is divided into twenty congruent sections. What is the measure of each arc in this circle?
Big Ideas Math Answers Geometry Chapter 10 Circles 72
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 31

Question 32.
MODELING WITH MATHEMATICS
You can use the time zone wheel to find the time in different locations across the world. For example, to find the time in Tokyo when it is 4 P.M. in San Francisco, rotate the small wheel until 4 P.M. and San Francisco line up, as shown. Then look at Tokyo to see that it is 9 A.M. there.
Big Ideas Math Answers Geometry Chapter 10 Circles 73
a. What is the arc measure between each time zone 0n the wheel?

Solution:
As the circle is divided into 24 sectors,
each time zone angle = \(\frac { 360 }{ 24 } \) = 15°

b. What is the measure of the minor arc from the Tokyo zone to the Anchorage zone?

Solution:
The measure of the minor arc from the Tokyo zone to the Anchorage zone,
15 + 15 + 15 + 15 + 15 + 15 = 90°

c. If two locations differ by 180° on the wheel, then it is 3 P.M. at one location when it is _________ at the other location.

Solution:
Kuwaiti city.

Question 33.
PROVING A THEOREM
Write a coordinate proof of the Similar Circles Theorem (Theorem 10.5).
Given ⊙O with center O(0, 0) and radius r.
⊙A with center A(a, 0) and radius s
Prove ⊙O ~ ⊙A
Big Ideas Math Answers Geometry Chapter 10 Circles 74
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 33

Question 34.
ABSTRACT REASONING
Is there enough information to tell whether ⊙C ≅ ⊙D? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 75

Solution:
Both circles ⊙C and ⊙D have the same radius so those circles are congruent.

Explanation:
We know that the term congruent means exactly equal shape and size.

Question 35.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Circles Theorem (Theorem 10.3).
Big Ideas Math Geometry Solutions Chapter 10 Circles 297
a. Given \(\overline{A C} \cong \overline{B D}\)
Prove ⊙A ≅ ⊙B
b. Given ⊙A ≅ ⊙B
prove \(\overline{A C} \cong \overline{B D}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 35

Question 36.
HOW DO YOU SEE IT?
Are the circles on the target similar or congruent? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 76
Answer:

Question 37.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Central Angles Theorem (Theorem 10.4).
Big Ideas Math Answers Geometry Chapter 10 Circles 77
a. Given ∠ABC ≅ ∠DAE
Prove \(\widehat{B C}\) ≅ \(\widehat{D E}\)
b. Given \(\widehat{B C}\) ≅ \(\widehat{D E}\)
Prove ∠ABC ≅ ∠DAE

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 37

Question 38.
THOUGHT PROVOKING
Write a formula for the length of a circular arc. Justify your answer.

Solution:
The formula to find the length of a circular arc is radius x angle.

Explanation:
Arc length formula is used to calculate the measure of the distance along the curved line making up the arc.

Maintaining Mathematical Proficiency

Find the value of x. Tell whether the side lengths form a Pythagorean triple.

Question 39.
Big Ideas Math Answers Geometry Chapter 10 Circles 78

Solution:

Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 39

Question 40.
Big Ideas Math Answers Geometry Chapter 10 Circles 81

Solution:
x = 13√2

Explanation:
According to Pythagoras theorem,
AC² = AB² + BC²
x² = 13² + 13²
x² = 169 + 169
x² = 338
x = 13√2

Question 41
Big Ideas Math Answers Geometry Chapter 10 Circles 79

Solution:

Big Ideas Math Geometry Answers Chapter 10 Circles 10.2 Ans 41

Question 42.
Big Ideas Math Answers Geometry Chapter 10 Circles 80

Solution:
x = 4√6

Explanation:
According to Pythagoras theorem,
AC² = AB² + BC²
14² = x² + 10²
196 = x² + 100
x² = 196 – 100
x² = 96
x = 4√6

10.3 Using Chords

Exploration 1

Drawing Diameters

Work with a partner: Use dynamic geometry software to construct a circle of radius 5 with center at the origin. Draw a diameter that has the given point as an endpoint. Explain how you know that the chord you drew is a diameter.
a. (4, 3)
b. (0, 5)
c. (-3, 4)
d. (-5, 0)

Answer:
Exploration 2

Writing a Conjecture about Chords

Work with a partner. Use dynamic geometry software to construct a chord \(\overline{B C}\) of a circle A. Construct a chord on the perpendicular bisector of \(\overline{B C}\). What do you notice? Change the original chord and the circle several times. Are your results always the same? Use your results to write a conjecture.
LOOKING FOR STRUCTURE
To be proficient in math, you need to look closely to discern a pattern or structure.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 82
Answer:

Exploration 3

A Chord Perpendicular to a Diameter

Work with a partner. Use dynamic geometry software to construct a diameter \(\overline{B C}\) of a circle A. Then construct a chord \(\overline{D E}\) perpendicular to \(\overline{B C}\) at point F. Find the lengths DF and EF. What do you notice? Change the chord perpendicular to \(\overline{B C}\) and the circle several times. Do you always get the same results? Write a conjecture about a chord that is perpendicular to a diameter of a circle.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 83
Answer:

Communicate Your Answer

Question 4.
What are two ways to determine when a chord is a diameter of a circle?

Solution:
If a chord passes through the center of the circle, then it is known as the diameter of a circle.
The longest chord of the circle is the diameter of a circle.

Lesson 10.3 Using Chords

Monitoring Progress

In Exercises 1 and 2, use the diagram of ⊙D.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 84

Question 1.
If m\(\widehat{A B}\) = 110°. find m\(\widehat{B C}\).

Solution:
m\(\widehat{B C}\) = 110°

Explanation:
With reference to the above figure,
AB and BC are congruent chords in congruent circles.
So, the corresponding minor arcs \(\widehat{A B}\), \(\widehat{B C}\) are congruent according to the congruent corresponding chords theorem.
So, \(\widehat{A B}\) = \(\widehat{B C}\)
\(\widehat{B C}\) = 110°

Question 2.
If m\(\widehat{A C}\) = 150° find m\(\widehat{A B}\).

Solution:
m\(\widehat{A B}\) = 105°

Explanation:
With reference to the above figure,
\(\widehat{A C}\) = 360 – (\(\widehat{A B}\) + \(\widehat{B C}\))
Given,
m\(\widehat{A C}\) = 150°
150 = 360 – 2(\(\widehat{A B}\))
2(\(\widehat{A B}\)) = 360 – 150 = 210
\(\widehat{A B}\) = 210/2
\(\widehat{A B}\) = 105°

In Exercises 3 and 4. find the indicated length or arc measure.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 85

Question 3.
CE

Solution:
10 units.

Explanation:
With reference to the above given figure,
CD and DE = 5
CE = CD + DE
CE = 5 + 5
CE = 10 units

Question 4.
m\(\widehat{C E}\)

Solution:
m\(\widehat{C E}\) = 180 – 8x

Explanation:
With reference to the above given figure,
m\(\widehat{C E}\) = 9x + 180 – x
m\(\widehat{C E}\) = 180 – 8x

Question 5.
In the diagram, JK = LM = 24, NP = 3x, and NQ = 7x – 12. Find the radius of ⊙N
Big Ideas Math Geometry Answer Key Chapter 10 Circles 86

Answer:

Exercise 10.3 Using Chords

Vocabulary and Core Concept Check

Question 1.
WRITING
Describe what it means to bisect a chord.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 1

Question 2.
WRITING
Two chords of a circle are perpendicular and congruent. Does one of them have to be a diameter? Explain your reasoning.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 14
Imagine a line segment of length 3 units, AB.
Big Ideas Math Geometry Answers Chapter 10 Circles 15
A second congruent segment of length 3 that is perpendicular to AB called CD.
Big Ideas Math Geometry Answers Chapter 10 Circles 16
Explanation:
Circumscribe both these line segments and note that AB and CD are new chords.
While both chords are perpendicular and congruent, none of the chord is a diameter.
So, it is possible to have two chords of this type with one diameter of the circle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the measure of the red arc or chord in ⊙C.

Question 3.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 87

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 3

Question 4.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 88

Solution:
170°

Explanation:
Given,
radius = 5 units, angle = 34°
Arc length = radius x angle
= 5 x 34 = 170°

Question 5.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 89

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 5

Question 6.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 90

Solution:
Chord length is 11 units.

Explanation:
With reference to the above data given in the figure,
we observe that the  two circles radius is the same and the angle is also the same.
So, the chord length is 11 units.

In Exercise 7-10, find the value of x.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 91
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 7

Question 8.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 92

Solution:
x = 40°

Explanation:
According to the perpendicular bisector theorem,
if a point is on the perpendicular bisector of a segment,
then it is equidistant from the segment’s endpoints.
RS = ST
x = 40°

Question 9.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 93

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 9

Question 10.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 94

Solution:
x = 7

Explanation:
According to the perpendicular bisector theorem,
if a point is on the perpendicular bisector of a segment,
then it is equidistant from the segment’s endpoints.
5x + 2 = 7x – 12
7x – 5x = 2 + 12
2x = 14
x = 7

Question 11.
ERROR ANALYSIS
Describe and correct the error in reasoning.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 95

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 11

Question 12.
PROBLEM SOLVING
In the cross section of the submarine shown, the control panels are parallel and the same length. Describe a method you can use to find the center of the cross section. Justify your method.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 96

Answer:

In Exercises 13 and 14, determine whether \(\overline{A B}\) is a diameter of the circle. Explain your reasoning.

Question 13.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 97

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 13

Question 14.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 98

Solution:
AB is not diameter of the circle.

Explanation:
With reference to the above given figure,
AD² = AC² + ED²
Let ED be x
5² = 3² + x²
25 = 9 + x²
x² = 25 – 9
x = 4
So, AB is not diameter of the circle.

In Exercises 15 and 16, find the radius of ⊙Q.

Question 15.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 99

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 15

Question 16.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 100

Solution:
radius = 13

Explanation:
With reference to the above given figure,
we know that:
AD = BC
Given,
4x + 4 = 6x – 6
6x – 4x = 4 + 6
2x = 10
x = 10/2
x = 5
BC = 6(5) – 6
BC = 30 – 6
BC = 24
QC² = 5² + 12²
QC² = 25 + 144
QC² = 169
QC = 13
Therefore, the radius is 13.

Question 17.
PROBLEM SOLVING
An archaeologist finds part of a circular plate. What was the diameter of the plate to the nearest tenth of an inch? Justify your answer.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 101

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 17

Question 18.
HOW DO YOU SEE IT?
What can you conclude from each diagram? Name a theorem that justifies your answer.
a.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 102

Solution:
Perpendicular chord bisector converse theorem.

Explanation:
With reference to the above figure:
If a point is on the perpendicular bisector of a segment,
then it is equidistant from the segment’s endpoints.
So, the given figure shows perpendicular chord bisector converse theorem.ws

b.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 103

Solution:
Congruent Corresponding Chords theorem.

Explanation:
With reference to the given figure,
if two minor arcs are congruent then their corresponding chords are also congruent.
So, it the given figure proves Congruent Corresponding Chords theorem.

c.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 104

Solution:
Perpendicular chord bisector theorem.

Explanation:
With reference to the given figure,
if the line passes through the center of the circle and is perpendicular to the chord also bisects that chord is known as Perpendicular chord bisector theorem.

d.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 105

Solution:
Equidistant chords theorem.

Explanation:
With reference to the given figure,
If a point is on the perpendicular bisector of a segment,
then it is equidistant from the endpoints of that segment as shown in the above given figure.
So, the above given figure proves Equidistant chords theorem.

Question 19.
PROVING A THEOREM
Use the diagram to prove each part of the biconditional in the Congruent Corresponding Chords Theorem (Theorem 10.6).
Big Ideas Math Geometry Answer Key Chapter 10 Circles 106
a. Given \(\overline{A B}\) and \(\overline{C D}\) are congruent chords.
Prove \(\widehat{A B} \cong \widehat{C D}\)
b. Given \(\widehat{A B} \cong \widehat{C D}\)
Prove \(\overline{A B}\) ≅ \(\overline{C D}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 19

Question 20.
MATHEMATICAL CONNECTIONS
In ⊙P, all the arcs shown have integer measures. Show that x must be even.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 109

Answer:

Question 21.
REASONING
In ⊙P. the lengths of the parallel chords are 20, 16, and 12. Find m\(\widehat{A B}\). Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 109

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 21

Question 22.
PROVING A THEOREM
Use congruent triangles to prove the Perpendicular Chord Bisector Theorem (Theorem 10.7).
Big Ideas Math Geometry Answer Key Chapter 10 Circles 109
Given \(\overline{E G}\) is a diameter of ⊙L.
\(\overline{E G}\) ⊥ \(\overline{D F}\)
Prove \(\overline{D C}\) ≅ \(\overline{F C}\), \(\widehat{D G} \cong \widehat{F G}\)

Solution:
According to the Perpendicular Chord Bisector Theorem,
if any point on the perpendicular bisector is equidistant from both the endpoints of the line segment.
Let L be the center of the circle.
Draw a chord from any point as DF on the circle.
As DF passes through LG.
The length of DC is the same as FC as shown in the given figure.
Therefore it prove the Perpendicular Chord Bisector Theorem.

Question 23.
PROVING A THEOREM
Write a proof of the Perpendicular Chord Bisector Converse (Theorem 10.8).
Big Ideas Math Geometry Answer Key Chapter 10 Circles 110
Given \(\overline{Q S}\) is a perpendicular bisector of \(\overline{R T}\).
Prove \(\overline{Q S}\) is a diameter of the circle L.
(Hint: Plot the center L and draw △LPT and △LPR.)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 23

Question 24.
THOUGHT PROVOKING
Consider two chords that intersect at point P. Do you think that \(\frac{A P}{B P}=\frac{C P}{D P}\)? Justify your answer.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 111

Answer:

Question 25.
PROVING A THEOREM
Use the diagram with the Equidistant Chords Theorem (Theorem 10.9) to prove both parts of the biconditional of this theorem.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 112
\(\overline{A B}\) ≅ \(\overline{C D}\) if and only if EF = EG

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 25

Question 26.
MAKING AN ARGUMENT
A car is designed so that the rear wheel is only partially visible below the body of the car. The bottom edge of the panel is parallel [o the ground. Your friend claims that the point where the tire touches the ground bisects \(\widehat{A B}\). Is your friend correct? Explain your reasoning.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 113

Answer:

Maintaining Mathematical Proficiency

Find the missing interior angle measure.

Question 27.
Quadrilateral JKLW has angle measures m∠J = 32°, m∠K = 25°, and m∠L = 44°. Find m∠M.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.3 Ans 27

Question 28.
Pentagon PQRST has angle measures m∠P = 85°, m∠Q = 134°, m∠R = 97°, and m∠S =102°.
Find m∠T.

Solution:
m∠T = 122°.

Explanation:
Given, Pentagon PQRST
angle measures m∠P = 85°, m∠Q = 134°, m∠R = 97°, and m∠S =102°.
We know that the sum of interior angles of a pentagon = 540°
m∠T = 540 – (85 + 134 + 97 + 102)
= 540 – 418 = 122
So, m∠T = 122°.

10.1 – 10.3 Quiz

In Exercises 1 – 6, use the diagram. (Section 10.1)

Big Ideas Math Geometry Answer Key Chapter 10 Circles 114

Question 1.
Name the circle.

Solution:
The circle has a chord, diameter, radius, secant and tangent.

Explanation:
With reference to the above given figure,
P is the center, NP is the radius, KN is the diameter, JL is the chord, SN is the secant and QR is the tangent.

Question 2.
Name a radius.

Solution:
NP is the radius of the circle.

Explanation:
The common distance between the points from the center is called a radius.

Question 3.
Name a diameter.

Solution:
KN is the diameter of the circle.

Explanation:
A line that crosses from the center of the circle starting from one point to the other is called a diameter.

Question 4.
Name a chord.

Solution:
JL is the chord.

Explanation:
A chord is the line segment that joins two distinct points of the circle.

Question 5.
Name a secant.

Solution:
SN is the secant.

Explanation:
A straight line that intersects a circle in two points is called a secant line.

Question 6.
Name a tangent.

Solution:
QR is the tangent.

Explanation:
A straight line that touches the curve of the circle at only one point and does not enter the circle’s interior.

Find the value of x.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 115

Solution:
x = 8

Explanation:
With reference to the given figure,
(9 + x)² = x² + 15²
(9 + x)² is in the form of (a + b)²
81 + 18x + x² = x² + 225
18x = 225 – 81
18x = 144
x = 144/18
x = 8

Question 8.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 116
Solution:
x = 7

Explanation:
With reference to the given figure,
6x – 3 = 3x + 18
6x – 3x = 18 + 3
3x = 21
x = 21/3
x = 7

Identify the given arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 117

Question 9.
\(\widehat{A E}\)

Solution:
Minor Arc.

Explanation:
With reference to the above given figure,
\(\widehat{A E}\) = 180 – 36
\(\widehat{A E}\) = 144°
So, an arc whose measure is less than 180 degrees is called a minor arc.

Question 10.
\(\widehat{B C}\)

Solution:
Minor arc.

Explanation:
With reference to the above given figure,
\(\widehat{B C}\) = 180 – (67 + 70)
\(\widehat{B C}\) = 180 – 137
\(\widehat{B C}\) = 43°
So, an arc whose measure is less than 180 degrees is called a minor arc.

Question 11.
\(\widehat{A C}\)

Solution:
Minor Arc.

Explanation:
With reference to the above given figure,
\(\widehat{A C}\) = 43 + 67
\(\widehat{A C}\) = 110°
So, an arc whose measure is less than 180 degrees is called a minor arc.

Question 12.
\(\widehat{A C D}\)

Solution:
Semin circle.

Explanation:
With reference to the above given figure,
\(\widehat{A C D}\) = 180°
So, an arc whose measure is equal to 180 degrees is called a semi circle.

Question 13.
\(\widehat{A C E}\)

Solution:
Major Arc.

Explanation:
With reference to the above given figure,
\(\widehat{A C E}\) = 180 + 36 = 216°
So, an arc whose measure is greater than 180 degrees is called a Major Arc.

Question 14.
\(\widehat{B E C}\)

Solution:
Explanation:
With reference to the above given figure,
\(\widehat{B E C}\) = 70 + 36 + 43 = 149°
So, an arc whose measure is less than 180 degrees is called a minor arc.

Tell whether the red arcs are congruent. Explain why or why not.

Question 15.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 118

Solution:
Those two red arcs are congruent.

Explanation:
A chord is the line segment that joins two distinct points of the circle.

Question 16.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 119
Solution:
Red arcs are not congruent.

Explanation:
When we observe both  radii of the circle the radius of the two circles is different.
So, the given red arcs ae not congruent.

Question 17.
Find the measure of the red arc in ⊙Q.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 120

Answer:

Question 18.
In the diagram. AC = FD = 30, PG = x + 5, and PJ = 3x – 1. Find the radius of ⊙P.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 121

Answer:

Question 19.
A circular clock can be divided into 12 congruent sections.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 122

a. Find the measure of each arc in this circle.

Solution:
The measure of each arc = \(\frac { 360 }{ 12 } \) = 30°

Explanation:
An arc whose measure equals 180 degrees is called a semicircle, since it divides the circle in two. Every pair of endpoints on a circle either defines one minor arc and one major arc, or two semicircles.

b. Find the measure of the minor arc formed by the hour and minute hands when the times is 7:00.

Solution:
150°

Explanation:
Given the time is 7:00 the minute hand is at 12 and hour hand is at 7,
so the minor arc is subtended by 12 – 7 = 5 of these sections and so the angle subtended is 30 x 5 = 150°

c. Find a time at which the hour and minute hands form an arc that is congruent to the arc in part (b).

Solution:
5:00

Explanation:
We know from the above that,
A minor arc is equal to 150° which can be formed by multiplication placements of the hour and the minute hand.
One of them can be the time 5:00 when the minute hand is at 12 and the hour hand is at 5.

10.4 Inscribed Angles and Polygons

Exploration 1

Inscribed Angles and Central Angles

work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Geometry Solutions Chapter 10 Circles 123

a. Construct an inscribed angle in a circle. Then construct the corresponding central angle.
Answer:

b. Measure both angles. How is the inscribed angle related to its intercepted arc?
Answer:

c. Repeat parts (a) and (b) several times. Record your results in a table. Write a conjecture about how an inscribed angle is related to its intercepted arc.

ATTENDING TO PRECISION

To be proficient in math, you need to communicate precisely with others.
Answer:

Exploration 2

A Quadrilateral with Inscribed Angles

work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Geometry Solutions Chapter 10 Circles 124

a. Construct a quadrilateral with each vertex on a circle.
Answer:

b. Measure all four angles. What relationships do you notice?
Answer:

c. Repeat parts (a) and (b) several times. Record your results in a table. Then write a conjecture that summarizes the data.
Answer:

Communicate Your Answer

Question 3.
How are inscribed angles related to their intercepted arcs? How are the angles of an inscribed quadrilateral related to each other?
Answer:

Question 4.
Quadrilateral EFGH is inscribed in ⊙C. and m ∠ E = 80°. What is m ∠ G? Explain.

Solution:
m ∠ G = 100°

Explanation:
Given,
m ∠ E = 80°
Quadrilateral EFGH is inscribed in ⊙C.
We know that the sum of the angles in quadrilateral is 360°
m ∠ E + m ∠ H = 80 + 80 = 160°
m ∠ E + m ∠ H + m ∠ G + m ∠ F = 360°
160° + m ∠ G + m ∠ F = 360°
m ∠ G + m ∠ F = 360 – 160 = 200
m ∠ G = 200/2
m ∠ G = 100°

Lesson 10.4 Inscribed Angles and Polygons

Monitoring Progress

Find the measure of the red arc or angle.

Question 1.
Big Ideas Math Geometry Solutions Chapter 10 Circles 125

Solution:
45°

Explanation:
With reference to the above figure,
The measure of an arc angle is found by dividing the arc length by the circle’s circumference.
m∠G = \(\frac { 90 }{ 2 } \) = 45°

Question 2.
Big Ideas Math Geometry Solutions Chapter 10 Circles 126

Solution:
76°

Explanation:
With reference to the above figure,
The measure of an arc angle is found by dividing the arc length by the circle’s circumference,
\(\widehat{T V}\) = 2 • 38 = 76°

Question 3.
Big Ideas Math Geometry Solutions Chapter 10 Circles 127

Solution:
m∠W = 72°

Explanation:
With reference to the above given figure,
YZ and XW are two arcs of the same circle and form ∠YOZ and ∠XOW at the center.
So, the measure of these two angles are also the same,
then ∠YZ and ∠XW are said to be congruent arcs.
Therefore, m∠W = 72°

Find the value of each variable.

Question 4.
Big Ideas Math Geometry Solutions Chapter 10 Circles 128

Solution:
y° = 50°

Explanation:
With reference to the above given figure,
x° = 90°
We know that the sum of the angles = 180°
y° = 180 – (40 + 90)
y° = 180 – 130
y° = 50°

Question 5.
Big Ideas Math Geometry Solutions Chapter 10 Circles 129

Solution:
x° = 98°
y° = 112°

Explanation:
With reference to the above given figure,
∠B + ∠D = 180
Given,
∠D = 98°, ∠B = x°, ∠C  = 68°
∠B + 82 = 180
∠B = 180 – 82
x° = 98°
∠C + ∠A = 180
68 + y° = 180
y° = 112°

Question 6.
Big Ideas Math Geometry Solutions Chapter 10 Circles 130

Solution:
x° = 10°
c = 62°

Explanation:
With reference to the above given figure
∠S + ∠U = 180°
Given,
∠S = c°, ∠U = 2c – 6
c + 2c – 6 = 180
3c = 186
c = 186/3
c = 62°
∠T + ∠V = 180°
Given,
∠T = 10x, ∠V = 8x
10x + 8x = 180
18x = 180
x = 180/18
x = 10°

Question 7.
In Example 5, explain how to find locations where the left side of the statue is all that appears in your camera’s field of vision.
Answer:

Exercise 10.4 Inscribed Angles and Polygons

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
If a circle is circumscribed about a polygon, then the polygon is an ___________ .

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 1

Question 2.
DIFFERENT WORDS, SAME QUESTION
Which is different?
Find “both” answers.
Big Ideas Math Geometry Solutions Chapter 10 Circles 131

Find m∠ABC.

Solution:
m∠ABC = 60°

Explanation:
With reference to the above given figure it is equilateral triangle.
We know that an equilateral triangle is a triangle in which all three sides have the same length and same angles i.e., 60°.

Find m∠AGC.

Solution:
m∠AGC = 130°

Explanation:
With reference to the above given figure,
m∠AGC = 180 – (∠AF + ∠FC)
m∠AGC = 180 – (25 + 25)
m∠AGC = 180 – 50
m∠AGC  = 130°

Find m∠AEC.

Solution:
m∠AEC = 80°

Explanation:
With reference to the above given figure,
m∠AEC = 180 – (∠AE + ∠CE)
m∠AEC = 180 – (50 + 50)
m∠AEC = 180 – 100
m∠AEC = 80°

Find m∠ADC.

Solution:
m∠AEC = 105°

Explanation:
With reference to the above given figure,
m∠ADC = 180 – (∠AD + ∠CD)
m∠ADC = 180 – (25 + 50)
m∠ADC = 180 – 75
m∠ADC = 105°

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, find the indicated measure.

Question 3.
m∠A
Big Ideas Math Geometry Solutions Chapter 10 Circles 132
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 3

Question 4.
m∠G
Big Ideas Math Geometry Solutions Chapter 10 Circles 133

Solution:
m∠G = 170°

Explanation:
We know that the sum of the angles in circle is 360°
Given,
m∠F = 70°, m∠D = 120°
m∠G = 360 – (70 + 120)
m∠G = 360 – 190
m∠G = 170°

Question 5.
m ∠ N
Big Ideas Math Geometry Solutions Chapter 10 Circles 134

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 5

Question 6.
m\(\widehat{R S}\)
Big Ideas Math Geometry Solutions Chapter 10 Circles 135

Solution:
m\(\widehat{R S}\) = 134°

Explanation:
With reference to the above given figure,
m\(\widehat{R S}\) = 2 x m\(\widehat{R QS}\)
Giveen,
m\(\widehat{R QS}\) = 67°
m\(\widehat{R S}\) = 2 • 67
m\(\widehat{R S}\) = 134°

Question 7.
m\(\widehat{V U}\)
Big Ideas Math Geometry Solutions Chapter 10 Circles 136

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 7

Question 8.
m\(\widehat{W X}\)
Big Ideas Math Geometry Solutions Chapter 10 Circles 137

Solution:
m\(\widehat{W X}\) = 37.5°

Explanation:
With reference to the data given in the above figure,
m\(\widehat{W X}\) = \(\frac { 75 }{ 2 } \)
m\(\widehat{W X}\) = 37.5

In Exercises 9 and 10, name two pairs of congruent angles.

Question 9.
Big Ideas Math Geometry Solutions Chapter 10 Circles 138

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 9

Question 10.
Big Ideas Math Geometry Solutions Chapter 10 Circles 139

Solution:
m∠W = m∠Z, m∠X = m∠Y

Explanation:
With reference to the above given figure,
YZ and XW are two arcs of the same circle and form ∠YOZ and ∠XOW at the center.
So, the measure of these two angles are also the same,
then ∠YZ and ∠XW are said to be congruent arcs.
Therefore, m∠W = m∠Z, m∠X = m∠Y

In Exercises 11 and 12, find the measure of the red arc or angle.

Question 11.
Big Ideas Math Geometry Solutions Chapter 10 Circles 140

Solution:
m∠EHF = 51°

Explanation:
We know that,
if an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 11

Question 12.
Big Ideas Math Geometry Solutions Chapter 10 Circles 141

Solution:
\(\widehat{P S}\) = 80°

Explanation:
With reference to the given figure,
\(\widehat{P S}\) = 2 • 40
\(\widehat{P S}\) = 80°

In Exercises 13 – 16, find the value of each variable.

Question 13.
Big Ideas Math Geometry Solutions Chapter 10 Circles 142

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 13

Question 14.
Big Ideas Math Geometry Solutions Chapter 10 Circles 143
Solution:
k = 60°

Explanation:
With reference to the given figure,
m∠E + m∠G = 180°
m + 60 = 180°
m = 120°
m∠D + m∠F = 180°
60 + 2k = 180°
k = 60°

Question 15.
Big Ideas Math Geometry Solutions Chapter 10 Circles 144

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 15

Question 16.
Big Ideas Math Geometry Solutions Chapter 10 Circles 145

Solution:
y° = 28°

Explanation:
With reference to the given figure,
3x° = 90°
x° = 30°
2y° + 90° + 34° = 180°
2y° + 124° = 180°
2y° = 56°
y° = 28°

Question 17.
ERROR ANALYSIS
Describe and correct the error in finding m\(\widehat{B C}\).
Big Ideas Math Geometry Solutions Chapter 10 Circles 146
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 17

Question 18.
MODELING WITH MATHEMATICS
A carpenter’s square is an L-shaped tool used to draw right angles. You need to cut a circular piece of wood into two semicircles. How can you use the carpenter’s square to draw a diameter on the circular piece of wood?
Big Ideas Math Geometry Solutions Chapter 10 Circles 147

Solution:
Recall that when a right triangle is inscribed in a circle,
then the hypotenuse is the diameter of the circle.
Simply use the carpenter’s square to inscribe it into the circle.
The hypotenuse formed by both legs of the square should provide a diameter for the circle.

MATHEMATICAL CONNECTIONS
In Exercises 19 – 21, find the values of x and y. Then find the measures of the interior angles of the polygon.

Question 19.
Big Ideas Math Geometry Solutions Chapter 10 Circles 148
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 19

Question 20.
Big Ideas Math Geometry Solutions Chapter 10 Circles 149
Solution:
∠A = 130.9°, ∠B = 40°, ∠C = 140°, ∠D = 49°

Explanation:
With reference to the given figure,
∠B + ∠C = 180°
14x + 4x = 180°
18x = 180°
x = 10°
∠A + ∠D = 180°
9y + 24y = 180°
33y = 180°
y = 5.45°
∠A = 130.9°, ∠B = 40°, ∠C = 140°, ∠D = 49°

Question 21.
Big Ideas Math Geometry Solutions Chapter 10 Circles 150

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 21

Question 22.
MAKING AN ARGUMENT
Your friend claims that ∠PTQ ≅ ∠PSQ ≅ ∠PRQ. Is our friend correct? Explain your reasoning.
Big Ideas Math Geometry Solutions Chapter 10 Circles 151

Solution:
Yes, my friend is correct.

Explanation:
∠PTQ ≅ ∠PSQ ≅ ∠PRQ is correct according to the inscribed angles of a circle theorem.
According to the inscribed angle theorem,
if an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle. Therefore, the angle does not change as its vertex is moved to different positions on the circle.

Question 23.
CONSTRUCTION
Construct an equilateral triangle inscribed in a circle.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 23

Question 24.
CONSTRUTION
The side length of an inscribed regular hexagon is equal to the radius of the circumscribed circle. Use this fact to construct a regular hexagon inscribed in a circle.

Solution:
As the side length is equal to the radius.
Draw a line representing the radius and draw a chord different chords in the form of hexagons of the radius of the circle.

REASONING
In Exercises 25 – 30, determine whether a quadrilateral of the given type can always be inscribed inside a circle. Explain your reasoning.

Question 25.
Square

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 25

Question 26.
rectangle

Solution:
yes, angles are right angles.

Explanation:
The opposite angles are always right angles in a rectangle.

Question 27.
parallelogram

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 27

Question 28.
kite

Solution:
No.

Explanation:
The opposite angles in the kite is not supplementary,
because it is a irregular polygon.

Question 29.
rhombus

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 29

Question 30.
isosceles trapezoid

Solution:
Yes,

Explanation:
The opposite angles are always supplementary in isosceles trapezoid.

Question 31.
MODELING WITH MATHEMATICS
Three moons, A, B, and C, are in the same circular orbit 1,00,000 kilometers above the surface of a planet. The planet is 20,000 kilometers in diameter and m∠ABC = 90°. Draw a diagram of the situation. How far is moon A from moon C?

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 31

Question 32.
MODELING WITH MATHEMATICS
At the movie theater. you want to choose a seat that has the best viewing angle, so that you can be close to the screen and still see the whole screen without moving your eyes. You previously decided that seat F7 has the best viewing angle, but this time someone else is already sitting there. Where else can you sit so that your seat has the same viewing angle as seat F7? Explain.
Big Ideas Math Geometry Solutions Chapter 10 Circles 152
Answer:

Question 33.
WRITING
A right triangle is inscribed in a circle, and the radius of the circle is given. Explain how to find the length of the hypotenuse.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 33

Question 34.
HOW DO YOU SEE IT?
Let point Y represent your location on the soccer field below. What type of angle is ∠AYB if you stand anywhere on the circle except at point A or point B?
Big Ideas Math Geometry Solutions Chapter 10 Circles 153
Answer:

Question 35.
WRITING
Explain why the diagonals of a rectangle inscribed in a circle are diameters of the circle.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 35

Question 36.
THOUGHT PROVOKING
The figure shows a circle that is circumscribed about ∆ABC. Is it possible to circumscribe a circle about any triangle? Justify your answer.
Big Ideas Math Geometry Solutions Chapter 10 Circles 154

Solution:
Yes.

Explanation:
With reference to the above given triangle in the figure,
we observe that the circumscribed circle is the circle that passes through all three vertices of the triangle. The center of the circumscribed circle is the circumcenter of the triangle.

Question 37.
PROVING A THEOREM
If an angle is inscribed in ⊙Q. the center Q can be on a side of the inscribed angle, inside the inscribed angle, or outside the inscribed angle. Prove each case of the Measure of an Inscribed Angle Theorem (Theorem 10. 10).

a. Case 1
Given ∠ABC is inscribed in ⊙Q
Let m∠B = x°
Center Q lies on \(\overline{B C}\).
Prove m∠ABC = \(\frac{1}{2}\)m\(\widehat{A C}\)
(Hint: Show that ∆AQB is isosceles. Then write m\(\widehat{A C}\) in terms of x.)
Big Ideas Math Geometry Solutions Chapter 10 Circles 155
b. Case 2
Use the diagram and auxiliary line to write Given and Prove statements for Case 2. Then write a proof
Big Ideas Math Geometry Solutions Chapter 10 Circles 156
c. Case 3
Use the diagram and auxiliary line to write Given and Prove statements for Case 3. Then write a proof.
Big Ideas Math Geometry Solutions Chapter 10 Circles 157

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 37.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 37.2
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 37.3

Question 38.
PROVING A THEOREM
Write a paragraph proof of the Inscribed Angles of a Circle Theorem (Theorem 10.11). First, draw a diagram and write Given and Prove statements.
Answer:

Question 39.
PROVING A THEOREM
The Inscribed Right Triangle Theorem (Theorem 10.12) is written as a conditional statement and its converse. Write a plan for proof for each statement.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 39

Question 40.
PROVING A THEOREM
Copy and complete the paragraph proof for one part of the Inscribed Quadrilateral Theorem (Theorem 10. 13).
Given ⊙C with inscribed quadrilateral DEFG
Prove m ∠ D + m ∠ F = 180°,
m ∠ E + m ∠ G = 180°
Big Ideas Math Geometry Solutions Chapter 10 Circles 158
By the Arc Addition Postulate (Postulate 10. 1),
m\(\widehat{E F G}\) + ________ = 360° and m\(\widehat{F G D}\) + m\(\widehat{D E F}\) = 360°.
Using the ___________ Theorem. m\(\widehat{E D G}\) = 2m ∠ F, m\(\widehat{E F G}\) = 2m∠D, m\(\widehat{D E F}\) = 2m∠G, and m\(\widehat{F G D}\) = 2m ∠ E. By the Substitution Property of Equality, 2m∠D + ________ = 360°, So _________ . Similarly, __________ .

Solution:
m\(\widehat{E F G}\) + m\(\widehat{E D F}\) = 360°
m\(\widehat{F G D}\) + m\(\widehat{D E F}\) = 360°.
According to the the measure of an inscribed angle Theorem,
m\(\widehat{E D G}\) = 2m ∠ F,
m\(\widehat{E F G}\) = 2m∠D,
m\(\widehat{D E F}\) = 2m∠G and
m\(\widehat{F G D}\) = 2m ∠ E.
By the Substitution Property of Equality, 2m∠D + 2m∠G = 360°.

Question 41.
CRITICAL THINKING
In the diagram, ∠C is a right angle. If you draw the smallest possible circle through C tangent to \(\overline{A B}\), the circle will intersect \(\overline{A C}\) at J and \(\overline{B C}\) at K. Find the exact length of \(\overline{J K}\).
Big Ideas Math Geometry Solutions Chapter 10 Circles 159

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 41

Question 42.
CRITICAL THINKING
You are making a circular cutting board. To begin, you glue eight 1-inch boards together, as shown. Then you draw and cut a circle with an 8-inch diameter from the boards.
Big Ideas Math Geometry Solutions Chapter 10 Circles 160
a. \(\overline{F H}\) is a diameter of the circular cutting board. Write a proportion relating GJ and JH. State a theorem in to justify your answer.

Solution:
Each board is 1 inch and FJ spans 6 boards.
\(\overline{F H}\) = 6 inches

b. Find FJ, JH, and GJ. What is the length of the cutting board seam labeled \(\overline{G K}\)?

Solution:
FJ = 6, JH = 2, JG = 2√3, GK = 4√3

Explanation:
We know that,
Each board is 1 inch and JH spans 2 boards.
JH = 2 inches
Equation is \(\frac { JH }{ GJ } \) = \(\frac { GJ }{ FJ } \)
\(\frac { 2 }{ GJ } \) = \(\frac { GJ }{ 6 } \)
12 = GJ²
GJ = 2√3
GK = 2(GJ)
GK = 4√3
So, FJ = 6, JH = 2, JG = 2√3, GK = 4√3

Maintaining Mathematical Proficiency

Solve the equation. Check your solution.

Question 43.
3x = 145

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 43

Question 44.
\(\frac{1}{2}\)x = 63

Solution:
x = 126

Explanation:
Given,
\(\frac{1}{2}\)x = 63
x = 63 • 2
x = 126

Question 45.
240 = 2x

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.4 Ans 45

Question 46.
75 = \(\frac{1}{2}\)(x – 30)

Solution:
x = 180

Explanation:
Given,
75 = \(\frac{1}{2}\)(x – 30)
75 • 2 = x – 30
150 + 30 = x
x = 180

10.5 Angle Relationships in Circles

Exploration 1

Angles Formed by a Chord and Tangent Line

Work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Answer Key Geometry Chapter 10 Circles 161

a. Construct a chord in a circle. At one of the endpoints of the chord. construct a tangent line to the circle.
Answer:

b. Find the measures of the two angles formed by the chord and the tangent line.
Answer:

c. Find the measures of the two circular arcs determined by the chord.
Answer:

d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data.
Answer:

Exploration 2

Angles Formed by Intersecting Chords

Work with a partner: Use dynamic geometry software.

sample
Big Ideas Math Answer Key Geometry Chapter 10 Circles 162

a. Construct two chords that intersect inside a circle.
Answer:

b. Find the measure of one of the angles formed by the intersecting chords.
Answer:

c. Find the measures of the arcs intercepted h the angle in part (b) and its vertical angle. What do you observe?
Answer:

d. Repeat parts (a) – (c) several times. Record your results in a table. Then write a conjecture that summarizes the data.

CONSTRUCTING VIABLE ARGUMENTS

To be proficient in math, you need to understand and use stated assumptions, definitions, and previously established results.
Answer:

Communicate Your Answer

Question 3.
When a chord intersects a tangent line or another chord, what relationships exist among the angles and arcs formed?
Answer:

Question 4.
Line m is tangent to the circle in the figure at the left. Find the measure of ∠1.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 163

Solution:
m∠1 = 74°

Explanation:
Given,
angle 148°
So, m∠1 = \(\frac { 1 }{ 2 } \) • 148
m∠1 = 74°

Question 5.
Two chords intersect inside a circle to form a pair of vertical angles with measures of 55°. Find the sum of the measures of the arcs intercepted by the two angles.

Solution:
27.5°

Explanation:
Given angle = 55°
The sum of the measures of the arcs intercepted by the two angles = \(\frac { 1 }{ 2 } \) • 55
s = 27.5°

Lesson 10.5 Angle Relationships in Circles

Monitoring Progress

Line m is tangent to the circle. Find the indicated measure.

Question 1.
m ∠ 1
Big Ideas Math Answer Key Geometry Chapter 10 Circles 164

Solution:
m ∠ 1 = 105°

Explanation:
Given angle = 210°
The sum of the measures of the arcs intercepted by the two angles
m ∠ 1 = \(\frac { 1 }{ 2 } \) • 210
m ∠ 1 = 105°

Question 2.
m\(\widehat{R S T}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 165

Solution:
m\(\widehat{R S T}\)= 196°

Explanation:
With reference to the data given in the above figure,
m\(\widehat{R S T}\) = 2 • 98
m\(\widehat{R S T}\) = 196°

Question 3.
m\(\widehat{X Y}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 166

Solution:
m\(\widehat{X Y}\) = 40°

Explanation:
With reference to the data given in the above figure,
m\(\widehat{X Y}\) = \(\frac { 1 }{ 2 } \) • 80
m\(\widehat{X Y}\) = 40°

Find the value of the variable.

Question 4.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 167

Solution:
y° = 98.5°

Explanation:
With reference to the data given in the above figure,
y° = \(\frac { 1 }{ 2 } \) • (102 + 95)
y° = \(\frac { 1 }{ 2 } \) • 197
y° = 98.5°

Question 5.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 168

Solution:
a° = 104°

Explanation:
With reference to the data given in the above figure,
a° = 2 • 30° + 44°
a° = 60° + 44°
a° = 104°.

Find the value of x.

Question 6.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 169
Solution:
x° = 60°

Explanation:
With reference to the data given in the above figure,
We know that sum of the angles = 180°
x° = 180° – 120°
x° = 60°

Question 7.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 170

Solution:
x° = 130°

Explanation:
With reference to the data given in the above figure,
We know that sum of the angles = 180°
50° = 180° – x°
x° = 180° – 50°
x° = 130°

Question 8.
You are on top of Mount Rainier on a clear day. You are about 2.73 miles above sea level at point B. Find m\(\widehat{C D}\), which represents the part of Earth that you can see.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 171
Solution:
The part of earth you can see is 31°

Explanation:
With reference to the data given in the above figure,
CB and BD are tangents,
CB is perpendicular to AB and CD is perpendicular to AD by the tangent line to circle theorem.
△ABC is similar to △ABD by the hypotenuese leg congruence theorem.
∠CBA is similar to ∠ABD. So, m∠CBA = 74.5°, m∠CBD = 2 • 74.5° = 149°
m∠CBD = 180° – m∠CAD
m∠CBD = 180° – CD
149° = 180° – CD
CD = 31°

Exercise 10.5 Angle Relationships in Circles

Vocabulary and Core Concept Check

Question 1.
COMPLETE THE SENTENCE
Points A, B, C, and D are on a circle, and Big Ideas Math Answer Key Geometry Chapter 10 Circles 172 intersects Big Ideas Math Answer Key Geometry Chapter 10 Circles 173 at point P.
If m∠APC = \(\frac{1}{2}\)(m\(\widehat{B D}\) – m\(\widehat{A C}\)). then point P is _________ the circle.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 1

Question 2.
WRITING
Explain how to find the measure of a circumscribed angle.

Solution:
A circumscribed angle is the angle formed by two intersecting tangent lines to a circle.
Draw lines from the circle centre to the point of tangency.
The angle between the radius and tangent line is 90°.
The sum of angles of a quadrilateral is 360°.
Angles between radii and tangent lines is 180°.
The angle at two tangent lines meet is circumscribed angle.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, line t is tangent to the circle. Find the indicated measure.

Question 3.
m\(\widehat{A B}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 174

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 3

Question 4.
m\(\widehat{D E F}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 175

Solution:
m\(\widehat{D E F}\) = 234°

Explanation:
With reference to the data given in the above figure,
m\(\widehat{D E F}\) = 2(117°)
m\(\widehat{D E F}\) = 234°

Question 5.
m < 1
Big Ideas Math Answer Key Geometry Chapter 10 Circles 176

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 5

Question 6.
m ∠ 3
Big Ideas Math Answer Key Geometry Chapter 10 Circles 177

Solution:
m ∠ 3 = 70°

Explanation:
With reference to the data given in the above figure,
m ∠ 3 = ½ • 140
m ∠ 3 = 70°

In Exercises 7 – 14, find the value of x.

Question 7.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 178

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 7

Question 8.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 179

Solution:

Question 9.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 180

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 9

Question 10.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 181

Solution:
x° = 38°

Explanation:
With reference to the data given in the above figure,
34° = ½ (3x – 2 – (x + 6))
34° = ½ (3x – 2 – x – 6)
34° = ½ (2x – 8)
34° = x – 4
x° = 34 + 4
x° = 38°

Question 11.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 182
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 11

Question 12.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 183

Solution:
x° = 43.5°

Explanation:
With reference to the data given in the above figure,
6x – 11 = 2  • 125
6x = 250 + 11
6x = 261
x° = 216/6
x° = 43.5°

Question 13.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 184

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 13

Question 14.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 185

Solution:
x° = 4.41°

Explanation:
With reference to the data given in the above figure,
17x° = 75°
x° = 75/17
x° = 4.41°

ERROR ANALYSIS
In Exercises 15 and 16, describe and correct the error in finding the angle measure.

Question 15.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 186
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 15

Question 16.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 187
Solution:
m∠1 = 26°

Explanation:
Given,
m∠1 = ½ (122 – 70)
m∠1 = ½ (52)
m∠1 = 26

In Exercises 17 – 22, find the indicated angle measure. justify your answer.

Big Ideas Math Answer Key Geometry Chapter 10 Circles 188

Question 17.
m ∠ 1

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 17

Question 18.
m ∠ 2

Solution:
m ∠ 2 = 60°

Explanation:
With reference to the above given figure,
m ∠ 3 =30°,
So, m ∠ 2 = 180° – (90° + 30°)
m ∠ 2 = 180° – 120°
m ∠ 2 = 60°

Question 19.
m ∠ 3

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 19

Question 20.
m ∠ 4

Solution:
m ∠ 4 = 90°

Explanation:
With reference to the above given figure,
∠ 4 is a right angle.
So, m∠ 4 = 90°

Question 21.
m ∠ 5

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 21

Question 22.
m ∠ 6

Solution:
m ∠ 6 = 60°

Explanation:
With reference to the above given figure,
m ∠ 6 = 180° – (60° + 30° + 30°)
m ∠ 6 = 180° – 120°
m ∠ 6 = 60°

Question 23.
PROBLEM SOLVING
You are flying in a hot air balloon about 1.2 miles above the ground. Find the measure of the arc that represents the part of Earth you can see. The radius of Earth is about 4000 miles.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 189

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 23

Question 24.
PROBLEM SOLVING
You are watching fireworks over San Diego Bay S as you sail away in a boat. The highest point the fireworks reach F is about 0.2 mile above the bay. Your eyes E are about 0.01 mile above the water. At point B you can no longer see the fireworks because of the curvature of Earth. The radius of Earth is about 4000 miles, and \(\overline{F E}\) is tangent to Earth at point T. Find m\(\widehat{s B}\). Round your answer to the nearest tenth.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 190
Answer:

Question 25.
MATHEMATICAL CONNECTIONS
In the diagram, \(\vec{B}\)A is tangent to ⊙E. Write an algebraic expression for m\(\widehat{C D}\) in terms of x. Then find m\(\widehat{C D}\).
Big Ideas Math Answer Key Geometry Chapter 10 Circles 191

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 25

Question 26.
MATHEMATICAL CONNECTIONS
The circles in the diagram are concentric. Write an algebraic expression for c in terms of a and b.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 192

Solution:
a° = ½(c° – b°)

Explanation:
Concentric circles are circles with a common center.
The region between two concentric circles of different radii is called an annulus.
An algebraic expression for c in terms of a and b,
a° = ½(c° – b°)

Question 27.
ABSTRACT REASONING
In the diagram. \(\vec{P}\)L is tangent to the circle, and \(\overline{K J}\) is a diameter. What is the range of possible angle measures of ∠LPJ? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 193

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 27

Question 28.
ABSTRACT REASONING
In the diagram, \(\overline{A B}\) is an chord that is not a diameter of the circle. Line in is tangent to the circle at point A. What is the range of possible values of x? Explain your reasoning. (The diagram is not drawn to scale.)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 194

Solution:
The possible values of x are less than 180°

Explanation:
With reference to the above given figure,
\(\overline{A B}\) is an chord that is not a diameter of the circle.
Line in is tangent to the circle at point A.
We know that, if the angle is less than 180 degrees it is known as minor arc.

Question 29.
PROOF
In the diagram Big Ideas Math Answer Key Geometry Chapter 10 Circles 195 and Big Ideas Math Answer Key Geometry Chapter 10 Circles 196 are secant lines that intersect at point L. Prove that m∠JPN > m∠JLN.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 197

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 29

Question 30.
MAKING AN ARGUMENT
Your friend claims that it is possible for a circumscribed angle to have the same measure as its intercepted arc. Is your friend correct? Explain your reasoning.

Solution:
Yes,

Explanation:
when the circumscribed angle measures 90°,
the central angle measures are also 90°.
So, the intercepted arc also measures 90°.

Question 31.
REASONING
Points A and B are on a circle, and t is a tangent line containing A and another point C.
a. Draw two diagrams that illustrate this situation.
b. Write an equation for m\(\widehat{A B}\) in terms of m∠BAC for each diagram.
c. For what measure of ∠BAC can you use either equation to find m\(\widehat{A B}\)? Explain.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 31

Question 32.
REASONING
∆XYZ is an equilateral triangle inscribed in ⊙P. AB is tangent to ⊙P at point X, \(\overline{B C}\) is tangent to ⊙P at point Y. and \(\overline{A C}\) is tangent to ⊙P at point Z. Draw a diagram that illustrates this situation. Then classify ∆ABC by its angles and sides. Justify your answer.
Answer:

Question 33.
PROVING A THEOREM
To prove the Tangent and Intersected Chord Theorem (Theorem 10. 14), you must prove three cases.
a. The diagram shows the case where \(\overline{A B}\) contains the center of the circle. Use the Tangent Line to Circle Theorem (Theorem 10.1) to write a paragraph proof for this case.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 198
b. Draw a diagram and write a proof for the case where the center of the circle is in the interior of ∠CAB.
c. Draw a diagram and write a proof for the case where the center of the circle is in the exterior of ∠CAB.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 33.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 33.2

Question 34.
HOW DO YOU SEE IT?
In the diagram, television cameras are Positioned at A and B to record what happens on stage. The stage is an arc of ⊙A. You would like the camera at B to have a 30° view of the stage. Should you move the camera closer or farther away? Explain your reasoning.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 199

Solution:
Move the camera closer.

Explanation:
With reference to the above given data,
25° = ½(80° – 30°)
25° = ½(50°)
25° = 25°
So, you should move the camera closer.

Question 35.
PROVING A THEOREM
Write a proof of the Angles Inside the Circle Theorem (Theorem 10.15).
Big Ideas Math Answer Key Geometry Chapter 10 Circles 200
Given Chords \(\overline{A C}\) and \(\overline{B D}\) intersect inside a circle.
Prove m ∠ 1 = \(\frac{1}{2}\)(m\(\widehat{D C}\) + m\(\widehat{A B}\))

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 35

Question 36.
THOUGHT PROVOKING
In the figure, Big Ideas Math Answer Key Geometry Chapter 10 Circles 201 and Big Ideas Math Answer Key Geometry Chapter 10 Circles 202 are tangent to the circle. Point A is any point on the major are formed by the endpoints of the chord \(\overline{B C}\). Label all congruent angles in the figure. Justify your reasoning.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 203

Solution:
m∠CPB = ½(CAB – CB)

Explanation:
With reference to the data given in the above figure,
Big Ideas Math Answer Key Geometry Chapter 10 Circles 201 and Big Ideas Math Answer Key Geometry Chapter 10 Circles 202 are tangent to the circle.
Point A is any point on the major are formed by the endpoints of the chord \(\overline{B C}\).
So, m∠CPB = ½(CAB – CB)

Question 37.
PROVING A THEOREM
Use the diagram below to prove the Angles Outside the Circle Theorem (Theorem 10.16) for the case of a tangent and a secant. Then copy the diagrams for the other two cases on page 563 and draw appropriate auxiliary segments. Use your diagrams to prove each case.
Big Ideas Math Answer Key Geometry Chapter 10 Circles 204

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 37.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 37.2
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 37.3

Question 38.
PROVING A THEOREM
Prove that the Circumscribed Angle Theorem (Theorem 10.17) follows from the Angles Outside the Circle Theorem (Theorem 10.16).
Answer:

In Exercises 39 and 40, find the indicated measure(s). justify your answer

Question 39.
Find m ∠ P when m\(\widehat{W Z Y}\) = 200°
Big Ideas Math Answer Key Geometry Chapter 10 Circles 205

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 39

Question 40.
Find m\(\widehat{A B}\) and m\(\widehat{E D}\)
Big Ideas Math Answer Key Geometry Chapter 10 Circles 206

Solution:
m\(\widehat{A B}\) = 30°
m\(\widehat{E D}\) = 57.5°

Explanation:
With reference to the data given in the above figure,
m\(\widehat{E D}\) = ½ (115°) = 57.5°
m\(\widehat{E D}\) = 57.5°
m\(\widehat{A B}\) = ½ (60°)
m\(\widehat{A B}\) = 30°
∠GJA = 120° – 60°
∠GJA = 30°

Maintaining Mathematical Proficiency

Solve the equation.

Question 41.
x2 + x = 12

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 41

Question 42.
x2 = 12x + 35

Solution:
x = \(\frac { 12 ± √284  }{ 2 } \)

Explanation:
x² = 12x + 35
x = \(\frac { 12 ± √(144 + 140)  }{ 2 } \)
x = \(\frac { 12 ± √284  }{ 2 } \)
x = \(\frac { 12 + √284  }{ 2 } \), \(\frac { 12 – √284  }{ 2 } \)

Question 43.
– 3 = x2 + 4x

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.5 Ans 43

10.6 Segment Relationships in Circles

Exploration 1

Segments Formed by Two Intersecting Chords

Work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Geometry Answers Chapter 10 Circles 207

a.
Construct two chords \(\overline{B C}\) and \(\overline{D E}\) that intersect in the interior of a circle at a point F.
Answer:

b.
Find the segment lengths BE, CF, DF, and EF and complete the table. What do you observe?
Big Ideas Math Geometry Solutions Chapter 10 Circles 298
Answer:

c. Repeat parts (a) and (b) several times. Write a conjecture about your results.
REASONING ABSTRACTLY
To be proficient in math, you need to make sense of quantities and their relationships in problem situations.
Answer:

Exploration 2

Secants Intersecting Outside a Circle

Work with a partner: Use dynamic geometry software.

Sample
Big Ideas Math Geometry Answers Chapter 10 Circles 208

a. Construct two secant Big Ideas Math Geometry Answers Chapter 10 Circles 209 and Big Ideas Math Geometry Answers Chapter 10 Circles 210 that intersect at a point B outside a circle, as shown.
Answer:

b. Find the segment lengths BE, BC, BF, and BD. and complete the table. What do you observe?
Big Ideas Math Geometry Solutions Chapter 10 Circles 299
Answer:

c. Repeat parts (a) and (b) several times. Write a conjecture about your results.
Answer:

Communicate Your Answer

Question 3.
What relationships exist among the segments formed by two intersecting chords or among segments of two secants that intersect outside a circle?

Question 4.
Find the segment length AF in the figure at the left.
Big Ideas Math Geometry Answers Chapter 10 Circles 211

Solution:
AF = 4

Explanation:
With reference to the data given in the above figure,
EA • AF = AD • AC
18 • AF = 9 • 8
18 • AF = 72
AF = 72/18
AF = 4

Lesson 10.6 Segment Relationships in Circles

Monitoring Progress

Find the value of x.

Question 1.
Big Ideas Math Geometry Answers Chapter 10 Circles 212

Solution:
x = 8

Explanation:
According to the data given in the above figure,
4 • 6 = 3 • x
3x = 24
x = 24/3
x = 8

Question 2.
Big Ideas Math Geometry Answers Chapter 10 Circles 213

Solution:
x = 5

Explanation:
According to the data given in the above figure,
2 • x + 1 = 4 • 3
2x + 1 = 12
x + 1 = 12/2
x + 1 = 6
x = 6 – 1
x = 5

Question 3.
Big Ideas Math Geometry Answers Chapter 10 Circles 214

Solution:
x = \(\frac { 54 }{ 5 } \)

Explanation:
According to the data given in the above figure,
6 • 9 = 5 • x
54 = 5x
x = \(\frac { 54 }{ 5 } \)

Question 4.
Big Ideas Math Geometry Answers Chapter 10 Circles 215

Solution:
x = \(\frac { 3 ± √37 }{ 2 } \)

Explanation:
According to the data given in the above figure,
3 • x + 2 = x + 1 • x – 1
3x + 6 = x² – 1
x² – 3x – 7 = 0
x = \(\frac { 3 ± √(9 + 28) }{ 2 } \)
x = \(\frac { 3 ± √37 }{ 2 } \)

Question 5.
Big Ideas Math Geometry Answers Chapter 10 Circles 216

Solution:
x = ±√3

Explanation:
According to the data given in the above figure,
x² = 3 • 1
x² = 3
x = ±√3

Question 6.
Big Ideas Math Geometry Answers Chapter 10 Circles 217

Solution:
x = \(\frac { 49 }{ 5 } \)

Explanation:
According to the data given in the above figure,
7² = 5 • x
49 = 5x
x = \(\frac { 49 }{ 5 } \)

Question 7.
Big Ideas Math Geometry Answers Chapter 10 Circles 218

Solution:
x = 14.4

Explanation:
According to the data given in the above figure,
12² = 10x
144 = 10x
x = 144/10
x = 14.4

Question 8.
WHAT IF?
In Example 4, CB = 35 feet and CE = 14 feet. Find the radius of the tank.

Solution:
The radius of the tank is 36.75.

Explanation:
According to the data given in the above figure,
CB² = CE CD
Given,
CB = 35 feet and CE = 14 feet.
35² = 14 (2r + 14)
1225 = 28r + 196
28r = 1225 – 196
28r = 1029
r = 1029/28
r = 36.75

Exercise 10.6 Segment Relationships in Circles

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
The part of the secant segment that is outside the circle is called a(n) ______________ .

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 1

Question 2.
WRITING
Explain the difference between a tangent segment and a secant segment.

Solution:
A tangent segment intersects the circle at only one point.
It actually doesn’t go through the circle.
For example;
If a ball is rolling on a table top, then it would be the tangent.
Where as a secant segment intersects the circle in two points.
It goes through the circle. In a tangent, no part is in the interior of the circle.
In a secant, there is a part in the interior called a chord.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 6, find the value of x.

Question 3.
Big Ideas Math Geometry Answers Chapter 10 Circles 219

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 3

Question 4.
Big Ideas Math Geometry Answers Chapter 10 Circles 220

Solution:
x = 23

Explanation:
With reference to the data given in above figure,
10 • 18 = 9 • (x – 3)
180 = x – 3
x – 3 = 180/9
x – 3 = 20
x = 20 + 3
x = 23

Question 5.
Big Ideas Math Geometry Answers Chapter 10 Circles 221

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 5

Question 6.
Big Ideas Math Geometry Answers Chapter 10 Circles 222

Solution:
x = 5

Explanation:
With reference to the data given in above figure,
2x • 12 = 15 • (x + 3)
24x = 15x + 45
24 x – 15x = 45
9x = 45
x = 45/9
x = 5

In Exercises 7 – 10, find the value of x.

Question 7.
Big Ideas Math Geometry Answers Chapter 10 Circles 223

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 7

Question 8.
Big Ideas Math Geometry Answers Chapter 10 Circles 224

Solution:
x = \(\frac { 35 }{ 4 } \)

Explanation:
With reference to the data given in above figure,
5 • 7 = 4 • x
4x = 35
x = \(\frac { 35 }{ 4 } \)

Question 9.
Big Ideas Math Geometry Answers Chapter 10 Circles 225

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 9

Question 10.
Big Ideas Math Geometry Answers Chapter 10 Circles 226

Solution:
x = 30

Explanation:
With reference to the data given in above figure,
45 • x = 50 • 27
45x = 1350
x = 1350/45
x = 30

In Exercises 11 – 14. find the value of x.

Question 11.
Big Ideas Math Geometry Answers Chapter 10 Circles 227

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 11

Question 12.
Big Ideas Math Geometry Answers Chapter 10 Circles 228

Solution:
x = 48

Explanation:
With reference to the data given in above figure,
24² = 12x
576 = 12x
x = 576/12
x = 48

Question 13.
Big Ideas Math Geometry Answers Chapter 10 Circles 229

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 13

Question 14.
Big Ideas Math Geometry Answers Chapter 10 Circles 230

Solution:
x = 1.5

Explanation:
With reference to the data given in above figure,
3 = 2x
x = 3/2
x = 1.5

Question 15.
ERROR ANALYSIS
Describe and correct the error in finding CD.
Big Ideas Math Geometry Answers Chapter 10 Circles 231

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 15

Question 16.
MODELING WITH MATHEMATICS
The Cassini spacecraft is on a mission in orbit around Saturn until September 2017. Three of Saturn’s moons. Tethys. Calypso, and Telesto. have nearly circular orbits of radius 2,95,000 kilometers. The diagram shows the positions of the moons and the spacecraft on one of Cassini’s missions. Find the distance DB from Cassini to Tethys when \(\overline{A D}\) is tangent to the circular orbit.
Big Ideas Math Geometry Answers Chapter 10 Circles 232

Solution:
BD = 579493 km.

Explanation:
Given,
radius 2,95,000 kilometers.
With reference to the data given in above figure,
AD = the distance between Cassini and Telesto i.e., 203,000km
CD = the distance between Calypso and Cassini i.e., 83000
BD = x
AD² = CD x BD
(203,000)² = 83000x
x = 496493
BC = 496493
BD = 496493 + 83000
BD = 579493km.

Question 17.
MODELING WITH MATHEMATICS
The circular stone mound in Ireland called Newgrange has a diameter of 250 feet. A passage 62 feet long leads toward the center of the mound. Find the perpendicular distance x from the end of the passage to either side of the mound.
Big Ideas Math Geometry Answers Chapter 10 Circles 233
Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 17

Question 18.
MODELING WITH MATHEMATICS
You are designing an animated logo for our website. Sparkles leave point C and move to the Outer circle along the segments shown so that all of the sparkles reach the outer circle at the same time. Sparkles travel from point C to point D at 2 centimeters per second. How fast should sparkles move from point C to point N? Explain.
Big Ideas Math Geometry Answers Chapter 10 Circles 234

Solution:
5.33 should sparkles move from point C to point N.

Explanation:
With reference to the data given in the above figure,
4 • 8 = 6 • x
32 = 6x
x = 32/6
x = 5.33

Question 19.
PROVING A THEOREM
Write a two-column proof of the Segments of Chords Theorem (Theorem 10.18).

Plan for Proof:
Use the diagram to draw \(\overline{A C}\) and \(\overline{D B}\). Show that ∆EAC and ∆EDB are similar. Use the fact that corresponding side lengths in similar triangles are proportional.
Big Ideas Math Geometry Answers Ch

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 19.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 19.2

Question 20.
PROVING A THEOREM
Prove the Segments of Secants Theorem (Theorem 10.19). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)
Answer:

Question 21.
PROVING A THEOREM
Use the Tangent Line to Circle Theorem (Theorem 10. 1) to prove the Segments of Secants and Tangents Theorem (Theorem 10.20) for the special case when the secant segment Contains the center of the circle.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 21

Question 22.
PROVING A THEOREM
Prove the Segments of Secants and Tangents Theorem (Theorem 10.20). (Hint: Draw a diagram and add auxiliary line segments to form similar triangles.)
Answer:

Question 23.
WRITING EQUATIONS
In the diagram of the water well, AB, AD, and DE are known. Write an equation for BC using these three measurements.
Big Ideas Math Geometry Answers Chapter 10 Circles 236

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 23

Question 24.
HOW DO YOU SEE IT?
Which two theorems would you need to use to tind PQ? Explain your reasoning.
Big Ideas Math Geometry Answers Chapter 10 Circles 237

Answer:

Question 25.
CRITICAL THINKING
In the figure, AB = 12, BC = 8, DE = 6, PD = 4, and a is a point of tangency. Find the radius of ⊙P.
Big Ideas Math Geometry Answers Chapter 10 Circles 238

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 25

Question 26.
THOUGHT PROVOKING
Circumscribe a triangle about a circle. Then, using the points of tangency, inscribe a triangle in the circle. Must it be true that the two triangles are similar? Explain your reasoning.

Answer:

Maintaining Mathematical Proficiency

Solve the equation by completing the square.

Question 27.
x2 + 4x = 45

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 27

Question 28.
x2 – 2x – 1 = 8

Solution:
x = 1 + √10, x = 1 – √10

Explanation:
According to the information given in the above figure,
x² – 2x – 1 = 8
x² – 2x – 9 = 0
x = \(\frac { 2 ± √(4 + 36) }{ 2 } \)
x = \(\frac { 2 ± √40 }{ 2 } \)
x = 1 + √10, x = 1 – √10

Question 29.
2x2 + 12x + 20 = 34

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.6 Ans 29

Question 30.
– 4x2 + 8x + 44 = 16

Solution:
x = 1 + √8, x = 1 – √8

Explanation:
According to the information given in the above figure,
– 4x² + 8x + 44 = 16
4x² – 8x – 28 =0
x² – 2x – 7 = 0
x = \(\frac { 2 ± √(4 + 28) }{ 2 } \)
x = \(\frac { 2 ± √32 }{ 2 } \)
x = 1 ± √8

10.7 Circles in the Coordinate Plane

Exploration 1

The Equation of a Circle with Center at the Origin

Work with a partner: Use dynamic geometry software to Construct and determine the equations of circles centered at (0, 0) in the coordinate plane, as described below.
Big Ideas Math Geometry Solutions Chapter 10 Circles 300
a. Complete the first two rows of the table for circles with the given radii. Complete the other rows for circles with radii of your choice.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 18

b. Write an equation of a circle with center (0, 0) and radius r.

Solution:
x²  + y²  = r²

Explanation:
According to the information given in the above figure,
(x – 0)² + (y – 0)²  = r²
x²  + y²  = r²

Exploration 2

The Equation of a Circle with Center (h, k)

Work with a partner: Use dynamic geometry software to construct and determine the equations of circles of radius 2 in the coordinate plane, as described below.
Big Ideas Math Geometry Solutions Chapter 10 Circles 301
a. Complete the first two rows of the table for circles with the given centers. Complete the other rows for circles with centers of your choice.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 17

b. Write an equation of a circle with center (h, k) and radius 2.

Solution:
According to the information given in the above figure,
(x – h)² + (y – k)² = 4

c. Write an equation of a circle with center (h, k) and radius r.

Solution:
According to the information given in the above figure,
(x – h)² + (y – k)² = r²

Exploration 3

Deriving the Standard Equation of a Circle

Work with a partner. Consider a circle with radius r and center (h, k).

Write the Distance Formula to represent the distance d between a point (x, y) on the circle and the center (h, k) of the circle. Then square each side of the Distance Formula equation.

How does your result compare with the equation you wrote in part (c) of Exploration 2?

MAKING SENSE OF PROBLEMS
To be proficient in math, you need to explain correspondences between equations and graphs.
Big Ideas Math Answers Geometry Chapter 10 Circles 239

Solution:
According to the information given in the above figure,
(x – h)² + (y – k)² = r²

Communicate Your Answer

Question 4.
What is the equation of a circle with center (h, k) and radius r in the coordinate plane?

Solution:
According to the information given in the above figure,
(x – h)² + (y – k)² = r²

Question 5.
Write an equation of the circle with center (4, – 1) and radius 3.

Solution:
x² + y² – 8x + 2y + 8 = 0

Explanation:
According to the information given in the above figure,
(x – 4)² + (y + 1)² = 9
The above equation is in the form of (a – b)² + (a + b)²
x² – 8x + 16 + y² + 2y + 1 = 9
x² + y² – 8x + 2y + 17 – 9 = 0
x² + y² – 8x + 2y + 8 = 0

Lesson 10.7 Circles in the Coordinate Plane

Monitoring Progress

Write the standard equation of the circle with the given center and radius.

Question 1.
center: (0, 0), radius: 2.5

Solution:
x² + y² = 6.25

Explanation:
Given,
center: (0, 0), radius: 2.5
(x – 0)² + (y – 0)² = 2.5²
x² + y² = 6.25

Question 2.
center: (- 2, 5), radius: 7

Solution:
(x + 2)² + (y – 5)² = 49

Explanation:
Given,
center: (- 2, 5), radius: 7
(x + 2)² + (y – 5)² = 7²
(x + 2)² + (y – 5)² = 49

Question 3.
The point (3, 4) is on a circle with center (1, 4). Write the standard equation of the circle.

Solution:
(x – 1)² + (y – 4)² = 4

Explanation:
Given,
The point (3, 4) is on a circle with center (1, 4).
r = √(3 – 1)² + (4 – 4)²
r = √(2)²
r = 2
(x – 1)² + (y – 4)² = 2²
(x – 1)² + (y – 4)² = 4

Question 4.
The equation of a circle is x2 + y2 – 8x + 6y + 9 = 0. Find the center and the radius of the circle. Then graph the circle.

Solution:
The center of the circle (4, -3) and radius is 4.

Explanation:
Given,
x² + y² – 8x + 6y + 9 = 0
x² – 8x + 16 + y² + 6y + 9 = 16
(x – 4)² + (y + 3)² = 4²
The center of the circle (4, -3) and radius is 4.

Question 5.
Prove or disprove that the point (1, √5) lies on the circle centered at the origin and containing the point (0, 1).

Solution:
Disproved.

Explanation:
Given that the point (1, √5) lies on the circle centerd at the origin and containing the point (0, 1).
Therefore, the radius of the circle r = 1,
let O (0, 0) and B (1, √5).
So the distance between two points is,
OB = √(1 – 0) + (√5 – 0)² = √(1 + 5) = √6
As the radius of the given circle is 1 and distance of the point B from its center is √6,
the point does lie on the given circle.
Hence , it is disproved.

Question 6.
why are three seismographs needed to locate an earthquake’s epicentre?

Answer:

Exercise 10.7 Circles in the Coordinate Plane

Vocabulary and Core Concept Check

Question 1.
VOCABULARY
What is the standard equation of a circle?

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 1

Question 2.
WRITING
Explain why knowing the location of the center and one point on a circle is enough to graph the circle.

Solution:
If we know the location of the center and one point on the circle,
we can graph a circle because the distance from the center to the point is called the radius.

Monitoring Progress and Modeling with Mathematics

In Exercises 3 – 8, write the standard equation of the circle.

Question 3.
Big Ideas Math Answers Geometry Chapter 10 Circles 240

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 3

Question 4.
Big Ideas Math Answers Geometry Chapter 10 Circles 241

Solution:
x² + y² = 36

Explanation:
With reference to the data given in the above figure,
The center is (0, 0) and the radius is 6.
We know that,
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 6²
x² + y² = 36

Question 5.
a circle with center (0, 0) and radius 7

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 5

Question 6.
a circle with center (4, 1) and radius 5

Solution:
(x – 4)² + (y – 1)² = 25

Explanation:
Given,
a circle with center (4, 1) and radius 5.
We know that,
(x – h)² + (y – k)² = r²
(x – 4)² + (y – 1)² = 5²
(x – 4)² + (y – 1)² = 25

Question 7.
a circle with center (- 3, 4) and radius 1

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 7

Question 8.
a circle with center (3, – 5) and radius 7

Solution:
(x – 3)² + (y + 5)² = 49

Explanation:
Given,
a circle with center (3, – 5) and radius 7.
We know that,
(x – h)² + (y – k)² = r²
(x – 3)² + (y + 5)² = 7²
(x – 3)² + (y + 5)² = 49

In Exercises 9 – 11, use the given information to write the standard equation of the circle.

Question 9.
The center is (0, 0), and a point on the circle is (0, 6).

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 9

Question 10.
The center is (1, 2), and a point on the circle is (4, 2).

Solution:
x² + y² = 9

Explanation:
Given,
The center is (1, 2), and a point on the circle is (4, 2).
We know that,
r = √(x – h)² + (y – k)²
r = √(4 – 1)² + (2 – 2)²
r = √3²
r = 3
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 3²
x² + y² = 9

Question 11.
The center is (0, 0). and a point on the circle is (3, – 7).

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 11

Question 12.
ERROR ANALYSIS
Describe and correct the error in writing the standard equation of a circle.
Big Ideas Math Answers Geometry Chapter 10 Circles 242

Solution:
(x + 3)² + (y + 5)² = 9

Explanation:
Given,
the center (-3, -5) and radius is 3.
We know that,
(x – h)² + (y – k)² = r²
(x + 3)² + (y + 5)² = 3²
(x + 3)² + (y + 5)² = 9

In Exercises 13 – 18, find the center and radius of the circle. Then graph the circle.

Question 13.
x2 + y2 = 49

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 13

Question 14.
(x + 5)2 + (y – 3)2 = 9

Solution:
Center is (-5, 3) and radius is 3.

Explanation:
Given,
(x + 5)2 + (y – 3)2 = 9
x2 + 25 + y2 – 9 = 9
center is (-5, 3) and radius is 3.

Question 15.
x2 + y2 – 6x = 7

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 15

Question 16.
x2 + y2 + 4y = 32

Solution:
The center is (0, -2) and radius is 6

Explanation:
Given,
x2 + y2 + 4y = 32
x² + y² + 4y + 4 = 32 + 4
x² + (y + 2)² = 36
(x – 0)² + (y – (-2))² = 6²
The center is (0, -2) and radius is 6
Big Ideas Math Geometry Answers Chapter 10 Circles 21

Question 17.
x2 + y2 – 8x – 2y = – 16

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 17

Question 18.
x2 + y2 + 4x + 12y = – 15

Solution:
The center is (-2, -6) and radius is 5.

Explanation:
Given,
x2 + y2 + 4x + 12y = – 15
x² + 4x + 4 + y² + 12y + 36 = -15 + 36 + 4
(x + 2)² + (y + 6)² = 5²
The center is (-2, -6) and radius is 5

In Exercises 19 – 22, prove or disprove the statement.

Question 19.
The point (2, 3) lies on the circle centered at the origin with radius 8.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 19

Question 20.
The point (4, √5) lies on the circle centered at the origin with radius 3.

Solution:
The point (4, √5) does not lie on the circle.

Explanation:
Given,
The point (4, √5) lies on the circle centered at the origin with radius 3.
We know that,
r² = (x – h)² + (y – k)²
3² ______________ (4 – 0)² + (√5 – 0)²
9 ______________ 16 + 5
9 ≠ 21
Because the radius is 3 and the distance between center and the point is more than the radius.
So, the point does not lie on the circle.

Question 21.
The point (√6, 2) lies on the circle centered at the origin and containing the point (3, – 1).

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 21

Question 22.
The point (√7, 5) lies on the circle centered at the origin and containing the point (5, 2).

Solution:
The point (√7, 5) does not lie on the circle.

Explanation:
Given,
The point (√7, 5) lies on the circle centered at the origin and containing the point (5, 2).

We know that,
r² = (x – h)² + (y – k)²
= (√7 – 0)² + (5 – 0)² = 7 + 25 = 32
r = 5.65
(5.65)² ______________ (5 – 0)² + (2 – 0)²
32 ______________ 25 + 4
32 ≠ 29
Because the radius is 3 and the distance between center and the point is more than the radius.
So, the point does not lie on the circle.

Question 23.
MODELING WITH MATHEMATICS
A City’s commuter system has three zones. Zone I serves people living within 3 miles of the city’s center. Zone 2 serves those between 3 and 7 miles from the center. Zone 3 serves those over 7 miles from the center.
Big Ideas Math Answers Geometry Chapter 10 Circles 243
a. Graph this Situation on a coordinate plane where each unit corresponds to 1 mile. Locate the city’s center at the origin.
b. Determine which zone serves people whose homes are represented by the points (3, 4), (6, 5), (1, 2), (0.3). and (1, 6).

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 23.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 23.2

Question 24.
MODELING WITH MATHEMATICS
Telecommunication towers can be used to transmit cellular phone calls. A graph with units measured in kilometers shows towers at points (0, 0), (0, 5), and (6, 3). These towers have a range of about 3 kilometers.
a. Sketch a graph and locate the towers. Are there any locations that may receive calls from more than one tower? Explain your reasoning.
Answer:

b. The center of City A is located at (- 2, 2.5), and the center of City B is located at (5, 4). Each city has a radius of 1.5 kilometers. Which city seems to have better cell phone coverage? Explain your reasoning.

Solution:
There are three towers at points (0, 0), (0, 5) and (6, 3) with range of about 3 km.

a. Let’s sketch the graph to locate towers. Draw the points (0, 0), (0, 5) and (6, 3). Then draw three circles with centers at these points and radii 3.
There are locations that can receive calls from more that one tower because circles with centers (0, 0) and (0, 5) overlap. Locations in their intersection can receive calls from two towers.
The city A has a center located at (-2, 2.5) and city B has a center located at (5, 4).
Both cities have radius 1.5 km
Let’s draw the city A as a circle with center (-2, 2.5) and radius 1.5 and city B with center (5, 4) and radius 1.5.
From the graph we can conclude that the city B has better cell phone coverage because parts of city A do not have coverage.

Question 25.
REASONING
Sketch the graph of the circle whose equation is x2 + y2 = 16. Then sketch the graph of the circle after the translation (x, y) → (x – 2, y – 4). What is the equation of the image? Make a conjecture about the equation of the image of a circle centered at the origin after a translation m units to the left and n units down.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 25

Question 26.
HOW DO YOU SEE IT?
Match each graph with its equation.

a. Big Ideas Math Answers Geometry Chapter 10 Circles 244 A. x2 + (y + 3) 2 = 4
b. Big Ideas Math Answers Geometry Chapter 10 Circles 245 B. (x – 3) 2 + y2 = 4
c. Big Ideas Math Answers Geometry Chapter 10 Circles 246 C. (x + 3) 2 + y2 = 4
d. Big Ideas Math Answers Geometry Chapter 10 Circles 247 D. x2 + (y – 3) 2 = 4

Solution:
a ➝ C,
b ➝ A,
c ➝ D,
d ➝ B

Question 27.
USING STRUCTURE
The vertices of ∆XYZ are X(4, 5), Y(4, 13), and Z(8, 9). Find the equation of the circle circumscribed about ∆XYZ. Justify your answer.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 27.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 27.2

Question 28.
THOUGHT PROVOKING
A circle has center (h, k) and contains point (a, b). Write the equation of the line tangent to the circle at point (a, b).

Solution:
y – b = \(\frac { h – a }{ b – k } \)(x – a)

Explanation:
It is given that a circle with center C(h, k. A circle with point A(a, b).
We have to write the equation of a tangent that intersects the circle at point A.
According to the tangent line to circle theorem, a tangent is perpendicular to the radius.
Two lines are perpendicular if their slopes are negative reciprocals.
So, find the equation of the line AC to know its slope.
The equation of the line which has two points (a, b), (c, d) is y – b = \(\frac { b – c }{ a – d } \)(x – a)
The equation of the line which has two points A(a, b) and C(h, k) is,
y – b = \(\frac { b – k }{ a – h } \)(x – a)
Therefore, the slope of the line through A and C is \(\frac { b – k }{ a – h } \)
Hence, the slope of the tangent is –\(\frac { a – h }{ b – k } \) = \(\frac { h – a }{ b – k } \)
Use the equation of the line y = kx + n through the point (a, b)
y – b = k(x – a) to find the equation of the tangent.
The equation of the tangent with slope \(\frac { h – a }{ b – k } \) and through the point A(a, b) is y – b = \(\frac { h – a }{ b – k } \)(x – a)

MATHEMATICAL CONNECTIONS
In Exercises 29 – 32, use the equations to determine whether the line is a tangent, a secant a secant that contains the diameter, or name of these. Explain your reasoning.

Question 29.
Circle: (x – 4)2 + (y – 3)2 = 9
Line: y = 6

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 29.1

Question 30.
Circle: (x + 2)2 + (y – 2)2 = 16
Line: y = 2x – 4

Solution:
The line is a secant line.

Explanation:
Given,
Circle: (x + 2)2 + (y – 2)2 = 16,
Line: y = 2x – 4
(x + 2)2 + (2x – 4 – 2)2 = 16
x² + 4x + 4 + (2x – 6)² = 16
x² + 4x + 4 + 4x² – 24x + 36 = 16
5x² – 20x + 40 – 16 = 0
5x² – 20x + 24 = 0
x = \(\frac { 20 ±√-80 }{ 10 } \)
x = 2,
y = 2 • 2 – 4
y = 0, (2, 0)
The system has two solutions and point does not lie on the line.
So, the line is a secant line.

Question 31.
Circle: (x – 5)2 + (y + 1)2 = 4
Line: y = \(\frac{1}{5}\)x – 3\

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 31.1
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 31.2
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 31.3

Question 32.
Circle: (x + 3)2 + (y – 6)2 = 25
Line: y = –\(\frac{4}{3}\)x + 2

Solution:
The line is a secant line.

Explanation:
Given,
Circle: (x + 3)2 + (y – 6)2 = 25,
Line: y = –\(\frac{4}{3}\)x + 2
(x + 3)2 + (-\(\frac{4}{3}\)x + 2 – 6)2 = 25
x² + 6x + 9 + \(\frac { 16x² }{ 9 } \) + \(\frac { 32x }{ 3 } \) + 16 = 25
\(\frac { 25x² }{ 9 } \) + \(\frac { 50x }{ 3 } \) = 0
x(25x + 150) = 0
x = 0 or x = -6
y = 2, y = 10
(0, 2) and (-6, 10)
d = √(0 + 6)² + (2 – 10)²
= √(36 + 64)
= 10 ≠ 5
The system has two solutions and point does not lie on the line.
So, the line is a secant line.

MAKING AN ARGUMENT
Question 33.
Your friend claims that the equation of a circle passing through the points (- 1, 0) and (1, 0) is x2 – 2yk + y2 = 1 with center (0, k). Is your friend correct? Explain your reasoning.

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 33

Question 34.
REASONING
Four tangent circles are centered on the x-axis. The radius of ⊙A is twice the radius of ⊙O, The radius of ⊙B is three times the radius of ⊙O, The radius of ⊙C is four times the radius of ⊙O, All circles have integer radii, and the point (63, 16) is On ⊙C. What is the equation of ⊙A? Explain your reasoning.
Big Ideas Math Answers Geometry Chapter 10 Circles 248
Answer:

Maintaining Mathematical Proficiency

Identify the arc as a major arc, minor arc, or semicircle. Then find the measure of the arc.

Big Ideas Math Answers Geometry Chapter 10 Circles 249

Question 35.
\(\widehat{R S}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 35

Explanation:
The arc which is less than 180 degrees is known as minor arc.

Question 36.
\(\widehat{P R}\)

Solution:
\(\widehat{P R}\) = 25 + 65 = 90°

Explanation:
With reference to the information given in the above data,
\(\widehat{P R}\) is a right angle.
\(\widehat{P R}\) = 25 + 65 = 90°

Question 37.
\(\widehat{P R T}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 37

Question 38.
\(\widehat{S T}\)

Solution:
\(\widehat{S T}\) = 127°

Explanation:
With reference to the information given in the above data,
\(\widehat{S T}\) is a major arc.
We know that major arc is greater than 180 degrees.
\(\widehat{S T}\) = 360 – (90 + 65 +25 + 53)
\(\widehat{S T}\) = 127°

Question 39.
\(\widehat{R S T}\)

Solution:
Big Ideas Math Geometry Answers Chapter 10 Circles 10.7 Ans 39

Explanation:
When the arc is equal to 180 degrees is known as semicircle.

Question 40.
\(\widehat{Q S}\)

Solution:
\(\widehat{Q S}\) = 78°

Explanation:
With reference to the information given in the above data,
\(\widehat{Q S}\) is a minor arc.
We know that minor arc is less than 180 degrees.
\(\widehat{Q S}\) = 25 + 53 = 78°
\(\widehat{Q S}\) = 78°

Circles Review

10.1 Lines and Segments That Intersect Circles

Tell whether the line, ray, or segment is best described as a radius, chord, diameter, secant, or tangent of ⊙P.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 250

Question 1.
\(\overline{P K}\)

Solution:
\(\overline{P K}\) is radius.

Explanation:
With reference to the information given in the above figure,
\(\overline{P K}\) is radius.
We know that the radius is half the length of the diameter and is joined to the center at one end.

Question 2.
\(\overline{N M}\)

Solution:
\(\overline{N M}\) is chord.

Explanation:
With reference to the information given in the above figure,
\(\overline{N M}\) is chord.
A chord is a straight line inside the circle that touches the circumference at 2 points,
but does not touch the center of a circle.

Question 3.
\(\vec{J}\)L

Solution:
\(\vec{J}\)L is tangent.

Explanation:
With reference to the information given in the above figure,
\(\vec{J}\)L is tangent.
A line that intersects a circle exactly at one point is called a tangent and the point where the intersection occurs is called the point of tangency.
The tangent is always perpendicular to the radius drawn to the point of tangency.

Question 4.
\(\overline{K N}\)

Solution:
\(\overline{K N}\) is diameter.

Explanation:
With reference to the information given in the above figure,
\(\overline{K N}\) is diameter.
We know that,
The diameter is the length of the line through the center that touches two points on the edge of the circle as shown in the above given figure.

Question 5.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 251

Solution:
NL is secant.

Explanation:
We know that,
A secant is a line that intersects a circle in exactly two points.
So, NL is secant.

Question 6.
\(\overline{P N}\)

Solution:
\(\overline{P N}\) is radius.

Explanation:
With reference to the information given in the above figure,
\(\overline{P N}\) is radius.
We know that the radius is half the length of the diameter and is joined to the center at one end.

Tell whether the common tangent is internal or external.

Question 7.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 252

Solution:
Internal common tangent.

Explanation:
We know that the lines are said to be internal tangents,
if they intersect the segment joining the centers of two circles.
So, the above given figure has a common internal tangent.

Question 8.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 253

Solution:
External common tangent.

Explanation:
The lines that do not cross the segment joining the centers of the circles is said to be External tangents.
So, the above given figure has a common external tangent.

Points Y and Z are points of tangency. Find the value of the variable.

Question 9.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 254

Solution:
a = \(\frac { 3 ± 33 }{ 18 } \)

Explanation:
Given,
Points Y and Z are points of tangency.
With reference to the information given in the above figure,
3a = 9a² – 30
9a² – 3a – 30 = 0
a = \(\frac { 3 ± 33 }{ 18 } \)

Question 10.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 255

Solution:
c = 2

Explanation:
With reference to the information given in the above figure,
2c² + 9c + 6 = 9c + 14
2c² – 8 = 0
c² – 4 = 0
c = 2

Question 11.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 256

Solution:
r = 12

Explanation:
With reference to the information given in the above figure,
(3 + r)² = r² + 9²
9 + 6r + r² = r² + 81
6r = 72
r = 72/6
r = 12

Question 12.
Tell whether \(\overline{B D}\) is tangent to ⊙C. Explain.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 257

Solution:
\(\overline{B D}\) is not tangent to ⊙C

Explanation:
With reference to the information given in the above figure,
52² = 10² + 48²
2704 = 100 + 2304
2704 = 2404
So, \(\overline{B D}\) is not tangent to ⊙C

10.2 Finding Arc Measures

Use the diagram above to find the measure of the indicated arc.

Question 13.
\(\widehat{K L}\)

Solution:
\(\widehat{K L}\) = 100°

Explanation:
According ot the information given in the above figure,
\(\widehat{K L}\) = ∠KPL = 100°
So, \(\widehat{K L}\) = 100°

Question 14.
\(\widehat{L M}\)

Solution:
\(\widehat{L M}\) = 60°

Explanation:
According ot the information given in the above figure,
\(\widehat{L M}\) = 180° – 120°
\(\widehat{L M}\) = 60°

Question 15.
\(\widehat{K M}\)

Solution:
\(\widehat{K M}\) = 160°

Explanation:
According ot the information given in the above figure,
\(\widehat{K M}\) = 100° + 60°
\(\widehat{K M}\) = 160°

Question 16.
\(\widehat{K N}\)

Solution:
\(\widehat{K N}\) = 80°

Explanation:
According ot the information given in the above figure,
\(\widehat{K N}\) = 360 – (120 + 100 + 60)
\(\widehat{K N}\) = 360 – 280
\(\widehat{K N}\) = 80°

Tell whether the red arcs are congruent. Explain why or why not.

Question 17.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 258

Solution:
\(\widehat{S T}\), \(\widehat{Y Z}\) are not congruent.

Explanation:
We know that,
congruent means exactly equal shape and size.
This shape and size should remain equal, even when we flip, turn, or rotate the shapes.
So, \(\widehat{S T}\), \(\widehat{Y Z}\) are not congruent.
Because both arcs are from different circles and having different radii.

Question 18.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 259

Solution:
\(\widehat{A B}\), \(\widehat{E F}\) are congruent.

Explanation:
We know that,
congruent means exactly equal shape and size.
This shape and size should remain equal, even when we flip, turn, or rotate the shapes.
So,\(\widehat{A B}\), \(\widehat{E F}\) are congruent.
Because those circles have same radii.

10.3 Using Chords

Find the measure of \(\widehat{A B}\).

Question 19.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 260

Solution:
\(\widehat{A B}\) = 61°

Explanation:
With reference to the above given figure,
The inscribed triangles are congruent to each other.
\(\widehat{A B}\) = 61°
If ED = AB,
then \(\widehat{A B}\) = \(\widehat{E D}\)

Question 20.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 261

Solution:
\(\widehat{A B}\) = 65°

Explanation:
With reference to the above given figure,
\(\widehat{A B}\) = \(\widehat{A D}\)
So, \(\widehat{A B}\) = 65°

Question 21.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 262

Solution:
\(\widehat{A B}\) = 91°

Explanation:
With reference to the above given figure,
\(\widehat{A B}\) = \(\widehat{E D}\)
So, \(\widehat{A B}\) = 91°

Question 22.
In the diagram. QN = QP = 10, JK = 4x, and LM = 6x – 24. Find the radius of ⊙Q.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 263

Solution:
The radius of ⊙Q is 26.

Explanation:
With reference to the above given figure,
6x – 24 = 4x
6x – 4x = 24
2x = 24
x = 24/2
x = 12
ML = 6(12) – 24 = 48
JN = \(\frac { 48 }{ 2 } \) = 24
JQ² = JN² + NQ²
JQ² = 24² + 10²
JQ² = 576 + 100
JQ² = 676
JQ = 26
Therefore, the radius of ⊙Q is 26.

10.4 Inscribed Angles and Polygons

Find the value(s) of the variable(s).

Question 23.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 264

Solution:
x° = 80°

Explanation:
With reference to the data given in the above figure,
x° = 2θ
x° = 2 • 40°
x° = 80°

Question 24.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 265

Solution:
q° = 100°, r° = 20°

Explanation:
With reference to the data given in the above figure,
q° + 80° = 180°
q° = 180° – 80°
q° = 100°
4r° + 100 = 180°
4r° = 180° – 100°
4r° = 80°
r° = 20°

Question 25.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 266

Solution:
d° = 5°

Explanation:
With reference to the data given in the above figure,
14d° = 70°
d° = 70/14
d° = 5°

Question 26.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 267

Solution:
y° = 30°, z° = 10°

Explanation:
With reference to the data given in the above figure,
3y° = 90°
y° = 90/3
y° = 30°
50° + 90° + 4z° = 180°
4z° = 180° – 140°
4z° = 40°
z° = 40/4
z° = 10°

Question 27.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 268

Solution:
m° = 44°
n° = 39°

Explanation:
With reference to the data given in the above figure,
m° = 44°
n° = 39°

Question 28.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 269

Solution:
c° = 28°

Explanation:
With reference to the data given in the above figure,
c° = ½ • 56
c° = 28°

10.5 Angle Relationships in Circles

Find the value of x.

Question 29.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 270

Solution:
x° = 250°

Explanation:
With reference to the data given in the above figure,
x° = 250°

Question 30.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 271

Solution:
x° = 106°

Explanation:
With reference to the data given in the above figure,
x° = ½(152 + 60)
x° = ½(212)
x° = 106°

Question 31.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 272

Solution:
x° = 28°

Explanation:
With reference to the data given in the above figure,
x° = ½(96 – 40)
x° = ½(56)
x° = 28°

Question 32.
Line l is tangent to the circle. Find m\(\widehat{X Y Z}\).
Big Ideas Math Geometry Answer Key Chapter 10 Circles 273

Solution:
m\(\widehat{X Y Z}\) = 240°

Explanation:
Given,
Line l is tangent to the circle.
With reference to the data given in the above figure,
m\(\widehat{X Y Z}\) = 2(120)
m\(\widehat{X Y Z}\) = 240°

10.6 Segment Relationships in Circles

Find the value of x.

Question 33.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 274

Solution:
x = 8

Explanation:
With reference to the data given in the above figure,
3 • x = 4 • 6
x = 24/3
x = 8

Question 34.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 275

Solution:
x = 3

Explanation:
With reference to the data given in the above figure,
(x + 3) • x = (6 – x) • 2x
(x + 3) • x = 12 – 2x
3x  = 9
x = 9/3
x = 3

Question 35.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 276

Solution:
x = 18

Explanation:
With reference to the data given in the above figure,
12² = 8 • x
144 = 8x
x = 144/8
x = 18

Question 36.
A local park has a circular ice skating rink. You are standing at point A, about 12 feet from the edge of the rink. The distance from you to a point of tangency on the rink is about 20 feet. Estimate the radius of the rink.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 277

Solution:
Estimated radius of the rink is 10 ft.

Explanation:
Given,
You are standing at point A, about 12 feet from the edge of the rink.
The distance from you to a point of tangency on the rink is about 20 feet.
With reference to the data given in the above figure,
20² = 12 • (2r + 12)
400 = 24r + 144
400 – 144 = 24r
256 = 24r
r = 256/24
r = 10.66
Therefore, estimated radius of the rink is 10 ft.

10.7 Circles in the Coordinate Plane

Write the standard equation of the circle shown.

Question 37.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 278

Solution:
(x – 4)² + (y + 1)² = 12.25

Explanation:
With reference to the data given in the above figure,
(x – 4)² + (y + 1)² = 3.5²
(x – 4)² + (y + 1)² = 12.25

Question 38.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 279

Solution:
(x – 8)² + (y – 5)² = 36

Explanation:
With reference to the data given in the above figure,
(x – 8)² + (y – 5)² = 6²
(x – 8)² + (y – 5)² = 36

Question 39.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 280

Solution:
x² + y² = 4

Explanation:
With reference to the data given in the above figure,
We know that,
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 2²
x² + y² = 4

Write the standard equation of the circle with the given center and radius.

Question 40.
center: (0,0), radius: 9

Solution:
x² + y² = 81

Explanation:
Given,
center: (0,0), radius: 9
We know that,
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 9²
x² + y² = 81

Question 41.
center: (- 5, 2), radius: 1.3

Solution:
(x + 5)² + (y – 2)² = 1.69

Explanation:
Given,
center: (- 5, 2), radius: 1.3
We know that,
(x – h)² + (y – k)² = r²
(x + 5)² + (y – 2)² = 1.3²
(x + 5)² + (y – 2)² = 1.69

Question 42.
center: (6, 21), radius: 4

Solution:
(x – 6)² + (y – 21)² = 16

Explanation:
Given,
center: (6, 21), radius: 4
We know that,
(x – h)² + (y – k)² = r²
(x – 6)² + (y – 21)² = 4²
(x – 6)² + (y – 21)² = 16

Question 43.
center: (- 3, 2), radius: 16

Solution:
(x + 3)² + (y – 2)² = 256

Explanation:
Given,
center: (- 3, 2), radius: 16
We know that,
(x – h)² + (y – k)² = r²
(x + 3)² + (y – 2)² = 16²
(x + 3)² + (y – 2)² = 256

Question 44.
center: (10, 7), radius: 3.5

Solution:
(x – 10)² + (y – 7)² = 12.25

Explanation:
Given,
center: (10, 7), radius: 3.5
We know that,
(x – h)² + (y – k)² = r²
(x – 10)² + (y – 7)² = 3.5²
(x – 10)² + (y – 7)² = 12.25

Question 45.
center: (0, 0), radius: 5.2

Solution:
x² + y² = 27.04

Explanation:
Given,
center: (0, 0), radius: 5.2
We know that,
(x – h)² + (y – k)² = r²
(x – 0)² + (y – 0)² = 5.2²
x² + y² = 27.04

Question 46.
The point (- 7, 1) is on a circle with center (- 7, 6). Write the standard equation of the circle.

Solution:
(x + 7)² + (y – 6)² = 25

Explanation:
Given,
The point (- 7, 1) is on a circle with center (- 7, 6).
We know that,
(x – h)² + (y – k)² = r²
r² = (-7 + 7)² + (6 – 1)²
r² = 5²
r = 5
So, radius is 5 and center is (-7, 6)
The standard equation of a circle is (x – (-7))² + (y – 6)² = 5²
(x + 7)² + (y – 6)² = 25

Question 47.
The equation of a circle is x2 + y2 – 12x + 8y + 48 = 0. Find the center and the radius of the circle. Then graph the circle.

Solution:
The radius of the circle is 2, the centre is (6, -4)

Explanation:
Given,
x2 + y2 – 12x + 8y + 48 = 0
x² – 12x + 36 + y² + 8y + 16 = 4
(x – 6)² + (y + 4)² = 2²
So, the radius of the circle is 2, the centre is (6, -4)
Big Ideas Math Geometry Answers Chapter 10 Circles 23

Question 48.
Prove or disprove that the point (4, – 3) lies on the circle centred at the origin and containing
the point (- 5, 0).

Solution:
The point (4, – 3) lies on the circle.

Explanation:
Given that the point (4, – 3) lies on the circle centered at the origin and containing
the point (- 5, 0).
By use the distance formula to find the radius of the circle with center (0, 0) and a point (-5, 0)
r = √(-5 – 0)² + (0 – 0)² = 5
The distance from the point (4, -3) to the center (0, 0)
d = √(4 – 0)² + (-3 – 0)² = √(16 +9) = 5
Since the radius of the circle is 5, the point lies on the circle.

Circles Chapter Test

Find the measure of each numbered angle in ⊙P. Justify your answer.

Question 1.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 281

Solution:
m∠1 = 72.5°
m∠2 = 145°

Explanation:
Based on the information given in the above figure,
m∠1 = \(\frac { 145 }{ 2 } \)
m∠1 = 72.5°
m∠2 = 145°

Question 2.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 282

Solution:
m∠1 = 60°, m∠2 = 90°

Explanation:
Based on the information given in the above figure,
A tangent is perpendicular to diameter.
So, m∠2 = 90°
m∠1 = 60°

Question 3.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 283

Solution:
m∠1 = 48°

Explanation:
Based on the information given in the above figure,
m∠1 = \(\frac { 96 }{ 2 } \)
m∠1 = 48°

Question 4.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 284

Answer:

Use the diagram.

Big Ideas Math Geometry Answer Key Chapter 10 Circles 285

Question 5.
AG = 2, GD = 9, and BG = 3. Find GF.

Answer:

Question 6.
CF = 12, CB = 3, and CD = 9. Find CE.

Answer:

Question 7.
BF = 9 and CB = 3. Find CA

Answer:

Question 8.
Sketch a pentagon inscribed in a circle. Label the pentagon ABCDE. Describe the relationship between each pair of angles. Explain your reasoning.

a. ∠CDE and ∠CAE

Answer:

b. ∠CBE and ∠CAE

Answer:

Find the value of the variable. Justify your answer.

Question 9.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 286

Solution:
x = 5

Explanation:
Based on the information given in the above figure,
5x – 4 = 3x + 6
5x – 3x = 6 + 4
2x = 10
x = 10/2
x = 5

Question 10.
Big Ideas Math Geometry Answer Key Chapter 10 Circles 287

Solution:
r = 9

Explanation:
Based on the information given in the above figure,
(6 + r)² = 12² + r²
(6 + r)² is in the form of (a + b)²
36 + 12r + r² = 144 + r²
12r = 144 – 36
12r = 108
r = 108/12
r = 9

Question 11.
Prove or disprove that the point (2√2, – 1) lies on the circle centered at (0, 2) and containing the point (- 1, 4).

Solution:
Disproved.

Explanation:
Given that the point (2√2, – 1) lies on the circle centered at (0, 2) and containing the point (- 1, 4).
We consider the circle centered at the A(0, 2) and containing the point B(-1, 4).
AB = √(-1 – 0)² + (4 – 2)² = √1 + 4 = √5
The distance between centre A(0, 2) and P(2√2, – 1) is
AP = √(2√2 – 0)² + (-1 – 2)² = √8 + 9 = √17
AB ≠ AP
So, the point (2√2, – 1) dies not lie on the circle.

Prove the given statement.

Question 12.
\(\widehat{S T} \cong \widehat{R Q}\)
Big Ideas Math Geometry Answer Key Chapter 10 Circles 288

Answer:

Question 13.
\(\widehat{J M} \cong \widehat{L M}\)
Big Ideas Math Geometry Answer Key Chapter 10 Circles 289

Answer:

Question 14.
\(\widehat{D G} \cong \widehat{F G}\)
Big Ideas Math Geometry Answer Key Chapter 10 Circles 290

Answer:

Question 15.
A bank of lighting hangs over a stage. Each light illuminates a circular region on the stage. A coordinate plane is used to arrange the lights, using a corner of the stage as the origin. The equation (x – 13)2 + (y – 4)2 = 16 represents the boundary of the region illuminated by one of the lights. Three actors stand at the points A(11, 4), B(8, 5), and C(15, 5). Graph the given equation. Then determine which actors are illuminated by the light.

Solution:
The actors who stand at points A and C are illuminated by the light.

Explanation:
The equation (x – 13)² + (y – 4)²= 16 represents the standard equation of the circle with center (13, 4) and radius 4.
Graph the circle with center S(13, 4), radius 4.
Then graph the points A(11,4), B (8, 5), C(15,5) which represents the places where the actors stand.
Big Ideas Math Geometry Answers Chapter 10 Circles 24
From the graph, we can observe that points A and C inside the circle and point B is outside the circle.
Therefore, actors who stand at points A and C are illuminated by the light.

Question 16.
If a car goes around a turn too quickly, it can leave tracks that form an arc of a circle. By finding the radius of the circle, accident investigators can estimate the speed of the car.
Answer:

Big Ideas Math Geometry Answer Key Chapter 10 Circles 291

a. To find the radius, accident investigators choose points A and B on the tire marks. Then the investigators find the midpoint C of \(\overline{A B}\). Use the diagram to find the radius r of the circle. Explain why this method works.

Solution:
The radius r of the circle = 155.71 ft.

Explanation:
With reference to the above given figure,
AC = 130 ft, CD = 70 ft
CE = (r – 70) ft
r² = a² + b²
r²= 130²+ (r – 70)²
(r – 70)² is in the form of (a – b)²
r² = 16900 + r² – 140r + 4900
140r = 21,800
r = 21800/140
r = 155.71 ft

b. The formula S = 3.87√fr can be used to estimate a car’s speed in miles per hour, where f is the coefficient of friction and r is the radius of the circle in feet. If f = 0.7, estimate the car’s speed in part (a).

Solution:
The estimated car’s speed is 39.67 miles per hour.

Explanation:
Given,
f = 0.7,
With reference to the above given data,
S = 3.87√fr
S = 3.8 √(0.7 x 155.71)
S = 3.8 √108.997
S = 3.8 x 10.44
S = 39.67

Circles Cumulative Assessment

Question 1.
Classify each segment as specifically as possible.
Big Ideas Math Geometry Solutions Chapter 10 Circles 292
a. \(\overline{B G}\)

Solution:
\(\overline{B G}\) is a chord.

Explanation:
Based on the information given in the above figure,
\(\overline{B G}\) is a chord.
We know that,
The chord is a line segment that joins two points on the circumference of the circle.

b. \(\overline{C D}\)

Solution:
\(\overline{C D}\) is radius.

Explanation:
Based on the information given in the above figure,
\(\overline{C D}\) is radius.
We know that,
Radius of a circle is the distance from the center of the circle to any point on it’s circumference.

c. \(\overline{A D}\)

Solution:
\(\overline{A D}\) is diameter.

Explanation:
Based on the information given in the above figure,
\(\overline{AD}\) is diameter.
We know that,
The diameter is the length of the line through the center that touches two points on the edge of the circle.

d. \(\overline{F E}\)

Solution:
\(\overline{F E}\) is a chord.

Explanation:
Based on the information given in the above figure,
\(\overline{F E}\) is a chord.
We know that,
The chord is a line segment that joins two points on the circumference of the circle.

Question 2.
Copy and complete the paragraph proof.

Big Ideas Math Geometry Solutions Chapter 10 Circles 293

Given Circle C with center (2, 1) and radius 1,
Circle D with center (0, 3) and radius 4
Prove Circle C is similar to Circle D.

Map Circle C to Circle C’ by using the _________ (x, y) → _________ so that Circle C’ and Circle D have the same center at (____, _____). Dilate Circle C’ using a cellIer of dilation (_____, _____) and a scale factor of _____ . Because there is a _________ transformation that maps Circle C to Circle D, Circle C is __________ Circle D.

Solution:
Based on the information given in the above figure,
Map Circle C to Circle C’ by using the scale factor (x, y) → (0, 3)
So, that Circle C’ and Circle D have the same center at (2, 1).
Dilate Circle C’ using a celler of dilation (2, 1) and a scale factor of circles.
Because there is a transformation that maps Circle C to Circle D, Circle C is similar to Circle D.

Question 3.
Use the diagram to write a proof.
Big Ideas Math Geometry Solutions Chapter 10 Circles 294
Given ∆JPL ≅ ∆NPL
\(\overline{P K}\) is an altitude of ∆JPL
\(\overline{P M}\) is an altitude ∆NPL
Prove ∆PKL ~ ∆NMP

Solution:
∆PKL ~ ∆NMP

Explanation:
With reference to the data given in the above figure,
∆JPL is similar to ∆NPL.
PK is the altitude of ∆JPL and PM is an altitude of ∆NPL.
We know that,
Altitude is a line drawn from one vertex to the opposite site.
It is perpendicular to the side.
So, ∆PKL is similar to ∆NMP.

Question 4.
The equation of a circle is x² + y² + 14x – 16y + 77 = 0. What are the center and radius of the circle?
(A) center: (14, – 16). radius: 8.8
(B) center: (- 7, 8), radius: 6
(C) center (- 14, 16), radius: 8.8
(D) center: (7, – 8), radius: 5.2

Solution:
(B) center: (- 7, 8), radius: 6

Explanation:
With reference to the data given in the above figure,
x² + y² + 14x – 16y + 77 = 0
x² + 14x + 49 + y² – 16y + 64 = 36
(x + 7)² + (y – 8)² = 6²
So, the center is (-7, 8) and radius is 6.

Question 5.
The coordinates of the vertices of a quadrilateral are W(- 7, – 6), X(1, – 2), Y(3, – 6) and Z(- 5, – 10). Prove that quadrilateral WXYZ is a rectangle.

Solution:
Proved.

Explanation:
Given,
The coordinates of the vertices of a quadrilateral are W(- 7, – 6), X(1, – 2), Y(3, – 6) and Z(- 5, – 10).
WY = √(-7 – 3)² + (-6 + 6)² = √(-10)² = 10
ZX = √(1 + 5)² + (-2 + 10)² = √6² + 8² = 10
WY = ZX,
Hence, the diagonals are congruent.
By using the slope formula to find the slopes of diagonals,
Slope of WY = \(\frac { -6 + 6 }{ -7 – 3 } \) = 0
Slope of ZX = \(\frac { -2 + 10 }{ 1 + 5 } \) = \(\frac { 4 }{ 3 } \)
Hence, the product of slopes of diagonals is 0 and the diagonals are not perpendicular.
Therefore, the quadrilateral WXYZ is a rectangle.

Question 6.
Which angles have the same measure as ∠ACB? Select all that apply.
im – 295
∠DEF ∠JGK ∠KGL ∠LGM ∠MGJ
∠QNR ∠STV ∠SWV ∠VWU ∠XYZ

Solution:
∠VWU

Question 7.
Classify each related conditional statement based on the conditional statement
“If you are a soccer player. then you are an athlete.”

a. If you are not a soccer player, then you are not an athlete.

Solution:
False

Explanation:
A statement written in the if-then form is known as conditional statement.
Given,
“If you are a soccer player. then you are an athlete.”
So, If you are not a soccer player, then you are not an athlete condition does not apply.

b. If you are an athlete, then you are a soccer player.

Solution:
False

Explanation:
A statement written in the if-then form is known as conditional statement.
Given,
“If you are a soccer player. then you are an athlete.”
So, If you are an athlete, then you are a soccer player condition does not apply.

c. You are a soccer player if and only if you are an athlete.

Solution:
True

Explanation:
A statement written in the if-then form is known as conditional statement.
Given,
“If you are a soccer player. then you are an athlete.”
So, you are a soccer player if and only if you are an athlete condition apply based on given statement.

d. If you are not an athlete, then you are not a soccer player.

Solution:
False

Explanation:
A statement written in the if-then form is known as conditional statement.
Given,
“If you are a soccer player. then you are an athlete.”
So, If you are not an athlete, then you are not a soccer player condition does not apply.

Question 8.
Your friend claims that the quadrilateral shown can be inscribed in a circle. Is your friend correct? Explain our reasoning.
Big Ideas Math Geometry Solutions Chapter 10 Circles 296

Solution:
My friend is correct.

Explanation:
We know that,
If the sum of any two angles is 180°, then the quadrilateral is inscribed in a circle.
With reference to the information given in the above figure,
70° + 110° = 180°, 110° + 70°= 180°
So, my friend is correct.

Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals

Big Ideas Math Book 6th Grade Advanced Answer Key Chapter 2 Fractions and Decimals

Fractions and Decimals STEAM Video/ Performance Task

STEAM Video

Big Ideas Math Answer Key Grade 6 Advanced Chapter 2 Fractions and Decimals 1

Space is Big
An astronomical unit (AU) is the average distance between Earth and the Sun, about 93 million miles. Why do asn000mers use astronomical units to measure distances in space? In what different ways can you compare the distances between objects and the locations of objects using the four mathematical operations?

Watch the STEAM Video ‘Space is Big: Then answer the following questions.

Question 1.
You know the distances between the Sun and each planet. How can you find the minimum and maximum distances between two planets as they rotate around the Sun?
Answer:

Question 2.
‘The table shows the distances of three celestial bodies from Earth. It takes about three days to travel from Earth to the Moon. I low can you estimate the amount of Lime it would take to travel from Earth to the Sun or to Venus?
Big Ideas Math Answer Key Grade 6 Advanced Chapter 2 Fractions and Decimals 2
Answer:

Performance Task

Space Explorers

Big Ideas Math Answer Key Grade 6 Advanced Chapter 2 Fractions and Decimals 3

After completing this Chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task.

You will use a table that shows the average distances between the Sun and each planet in our solar system to find several distances in space. Then you will use the speed of the Orion spacecraft to answer questions about time and distance.

Is it realistic for a manned spacecraft to travel to each planet in our solar system? Explain why or why not.
Answer:

Getting Ready for Chapter Fractions and Decimals

Chapter Exploration

Work with a partner. ‘the area model represents the multiplication of two fractions. Copy and complete the statement.

Question 1.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 2 Fractions and Decimals 4
Answer:

Question 2.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 2 Fractions and Decimals 5
Answer:

Question 3.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 2 Fractions and Decimals 6
Answer:

Question 4.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 2 Fractions and Decimals 7
Answer:

Work with a partner. use an area model to find the product.

Question 5.
\(\frac{1}{2} \times \frac{1}{3}\)
Answer:

Question 6.
\(\frac{4}{5} \times \frac{1}{4}\)
Answer:

Question 7.
\(\frac{1}{6} \times \frac{3}{4}\)
Answer:

Question 8.
\(\frac{3}{5} \times \frac{1}{4}\)
Answer:

Question 9.
MODELING REAL LIFE
You have a recipe that serves 6 people. The recipe uses three-fourths of a cup of milk.
Big Ideas Math Answer Key Grade 6 Advanced Chapter 2 Fractions and Decimals 8
a. How can you use the recipe to serve more people? How much milk would you need? Give 2 examples.
Answer:

b. How can you use the recipe to serve fewer people? How much milk would you need? Give 2 examples.
Answer:

Vocabulary

The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.
reciprocals               multiplicative inverses
Answer:

Section 2.1 Multiplying Fractions

Exploration 1

Using Models to Solve a Problem

Work with a partner. A bottle of water is \(\frac { 1 }{ 2 }\) full. You drink \(\frac { 2 }{ 3 }\) the water. Use one of the models to find the portion of the bottle of water that you drink. Explain your steps.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 9

number line
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 10

area model
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 11

tape diagram
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 12
Answer:

Exploration 2

Solving a Problem Involving Fractions

Work with a partner. A park has a playground that is \(\frac { 3 }{ 4 }\) of its width and \(\frac { 4 }{ 5 }\) of its length.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 13
a. use a model to find the portion of the park that is covered by the playground. Explain your steps.
Answer:

b. How can you find the solution of part (a) without using a model?
Answer:

Math Practice
Find General Methods
How can you use your answer to find a method for multiplying fractions?
Answer:

Lesson 2.1 Multiplying Fractions

Try It MuItiply

Question 1.
\(\frac{1}{3} \times \frac{1}{5}\)
Answer:

Question 2.
\(\frac{2}{3} \times \frac{3}{4}\)
Answer:

Question 3.
\(\frac{1}{2} \cdot \frac{5}{6}\)
Answer:

Try It Multiply. Write the answer in simplest form.

Question 4.
\(\frac{3}{7} \times \frac{2}{3}\)
Answer:

Question 5.
\(\frac{4}{9} \cdot \frac{3}{10}\)
Answer:

Question 6.
\(\frac{6}{5} \cdot \frac{5}{8}\)
Answer:

Question 7.
WHAT IF?
You use \(\frac{1}{4}\) of the flour to make the dough. How much of the entire bag do you use to make the dough?
Answer:

Try It Multiply. Write the answer in simplest form.

Question 8.
\(\frac{1}{3} \times 1 \frac{1}{6}\)
Answer:

Question 9.
\(3 \frac{1}{2} \times \frac{4}{9}\)
Answer:

Question 10.
\(4 \frac{2}{3} \cdot \frac{3}{4}\)
Answer:

Try It Multiply. Write the answer in simplest form.

Question 11.
\(1 \frac{7}{8} \cdot 2 \frac{2}{5}\)
Answer:

Question 12.
\(5 \frac{5}{7} \times 2 \frac{1}{10}\)
Answer:

Question 13.
\(2 \frac{1}{3} \cdot 7 \frac{2}{3}\)
Answer:

MULTIPLYING FRACTIONS AND MIXED NUMBERS
Multiply. Write the answer in simplest form.

Question 14.
\(\frac{1}{8} \times \frac{1}{6}\)
Answer:

Question 15.
\(\frac{3}{8} \cdot \frac{2}{3}\)
Answer:

Question 16.
\(2 \frac{1}{6} \cdot 4 \frac{2}{5}\)
Answer:

Question 17.
MP REASONING
What is the missing denominator?
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 14
Answer:

Question 18.
USING TOOLS
Write a multiplication problem involving fractions that is represented by the model. Explain your reasoning.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 15
Answer:

Question 19.
USING TOOLS
Use the number line to find \(\frac{3}{4} \times \frac{1}{2}\) Explain your reasoning.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 16
Answer:

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 20.
You spend \(\frac{5}{12}\) of a day at an amusement park. You spend \(\frac{2}{5}\) of that time riding waterslides. How many hours do you spend riding waterslides? Draw a model to show why your answer makes sense.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 17
Answer:

Question 21.
A venue is preparing for a concert on the floor shown. The width of the red carpet is \(\frac{1}{6}\) of the width of the floor. What is the area of the red carpet?
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 17
Answer:

Question 22.
You travel 9\(\frac{3}{8}\) miles from your house to a shopping mall. You travel \(\frac{2}{3}\) of that distance on an interstate. The only road construction you encounter is on the first \(\frac{2}{5}\) of the interstate. On how many miles of your trip do you encounter construction?
Answer:

Multiplying Fractions Practice 2.1

Review & Refresh

Find the LCM of the numbers.

Question 1.
8, 10
Answer:

Question 2.
5, 7
Answer:

Question 3.
2, 5, 7
Answer:

Question 4.
6, 7, 10
Answer:

Divide. Use a diagram to justify your answer.

Question 5.
6 ÷ \(\frac{1}{2}\)
Answer:

Question 6.
\(\frac{1}{4}\) ÷ 8
Answer:

Question 7.
4 ÷ \(\frac{1}{3}\)
Answer:

Question 8.
\(\frac{1}{5}\) ÷ 4
Answer:

Write the product as a power.
Question 9.
10 × 10 × 10
Answer:

Question 10.
5 × 5 × 5 × 5
Answer:

Question 11.
How many inches arc in 5\(\frac{1}{2}\) yards?
A. 15\(\frac{1}{2}\)
B. 16\(\frac{1}{2}\)
C. 66
D. 198
Answer:

Concepts, Skills, & Problem Solving

MP CHOOSE TOOLS
A bottle of water is \(\frac{2}{3}\) full. You drink the given portion of the water. Use a model to find the portion of the bottle of water that you drink.
Question 12.
\(\frac{1}{2}\)
Answer:

Question 13.
\(\frac{1}{4}\)
Answer:

Question 14.
\(\frac{3}{4}\)
Answer:

MULTIPLYING FRACTIONS
Multiply. Write the answer in simplest form.

Question 15.
\(\frac{1}{7} \times \frac{2}{3}\)
Answer:

Question 16.
\(\frac{5}{8} \cdot \frac{1}{2}\)
Answer:

Question 17.
\(\frac{1}{4} \times \frac{2}{5}\)
Answer:

Question 18.
\(\frac{3}{7} \times \frac{1}{4}\)
Answer:

Question 19.
\(\frac{2}{3} \times \frac{4}{7}\)
Answer:

Question 20.
\(\frac{5}{7} \times \frac{7}{8}\)
Answer:

Question 21.
\(\frac{3}{8} \cdot \frac{1}{9}\)
Answer:

Question 22.
\(\frac{5}{6} \cdot \frac{2}{5}\)
Answer:

Question 23.
\(\frac{5}{12}\) × 10
Answer:

Question 24.
6 • \(\frac{7}{8}\)
Answer:

Question 25.
\(\frac{3}{4} \times \frac{8}{15}\)
Answer:

Question 26.
\(\frac{4}{9} \times \frac{4}{5}\)
Answer:

Question 27.
\(\frac{3}{7} \cdot \frac{3}{7}\)
Answer:

Question 28.
\(\frac{5}{6} \times \frac{2}{9}\)
Answer:

Question 29.
\(\frac{13}{18} \times \frac{6}{7}\)
Answer:

Question 30.
\(\frac{7}{9} \cdot \frac{21}{10}\)
Answer:

Question 31.
MODELING REAL LIFE
In an aquarium, \(\frac{2}{5}\) of the fish are surgeonfish. of these, \(\frac{3}{4}\) are yellow tangs. What portion of all fish in the aquarium are yellow tangs?
Answer:

Question 32.
MODELING REAL LIFE
You exercise for \(\frac{3}{4}\) of an hour. You jump rope for \(\frac{1}{3}\) of that time. What portion of the hour do you spend jumping rope?
Answer:

MP REASONING
Without finding the products copy and complete the statement using <, >, or =. Explain your reasoning.
Question 33.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 18
Answer:

Question 34.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 19
Answer:

Question 35.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 20
Answer:

MULTIPLYING FRACTIONS AND MIXED NUMBERS
Multiply. Write the answer in simplest form.

Question 36.
\(1 \frac{1}{3} \cdot \frac{2}{3}\)
Answer:

Question 37.
\(6 \frac{2}{3} \times \frac{3}{10}\)
Answer:

Question 38.
\(2 \frac{1}{2} \cdot \frac{4}{5}\)
Answer:

Question 39.
\(\frac{3}{5} \cdot 3 \frac{1}{3}\)
Answer:

Question 40.
\(7 \frac{1}{2} \times \frac{2}{3}\)
Answer:

Question 41.
\(\frac{5}{9} \times 3 \frac{3}{5}\)
Answer:

Question 42.
\(\frac{3}{4} \cdot 1 \frac{1}{3}\)
Answer:

Question 43.
\(3 \frac{3}{4} \times \frac{2}{5}\)
Answer:

Question 44.
\(4 \frac{3}{8} \cdot \frac{4}{5}\)
Answer:

Question 45.
\(\frac{3}{7} \times 2 \frac{5}{6}\)
Answer:

Question 46.
\(1 \frac{3}{10} \times 18\)
Answer:

Question 47.
\(15 \cdot 2 \frac{4}{9}\)
Answer:

Question 48.
\(1 \frac{1}{6} \times 6 \frac{3}{4}\)
Answer:

Question 49.
\(2 \frac{5}{12} \cdot 2 \frac{2}{3}\)
Answer:

Question 50.
\(5 \frac{5}{7} \cdot 3 \frac{1}{8}\)
Answer:

Question 51.
\(2 \frac{4}{5} \times 4 \frac{1}{16}\)
Answer:

YOU BE THE TEACHER
Your friend finds the product. Is your friend correct? Explain your reasoning.

Question 52.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 21

Question 53.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 22

Question 54.
MODELING REAL LIFE
A vitamin C tablet contains \(\frac{1}{4}\) of a gram of vitamin C. You take 1\(\frac{1}{2}\) every day. How many grams of vitamin C do you take every day?

Question 55.
MP PROBLEM SOLVING
You make a banner for a football ralIy.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 23
a. What is the area of the banner?
Answer:

b. You add a\(\frac{1}{4}\) foot border on each side. What is the area of the new banner?
Answer:

MULTIPLYING FRACTIONS AND MIXED NUMBERS
Multiply. Write the answer in simplest form.
Question 56.
\(\frac{1}{2} \times \frac{3}{5} \times \frac{4}{9}\)
Answer:

Question 57.
\(\frac{4}{7} \cdot 4 \frac{3}{8} \cdot \frac{5}{6}\)
Answer:

Question 58.
\(1 \frac{1}{15} \times 5 \frac{2}{5} \times 4 \frac{7}{12}\)
Answer:

Question 59.
\(\left(\frac{3}{5}\right)^{3}\)
Answer:

Question 60.
\(\left(\frac{4}{5}\right)^{2} \times\left(\frac{3}{4}\right)^{2}\)
Answer:

Question 61.
\(\left(\frac{5}{6}\right)^{2} \cdot\left(1 \frac{1}{10}\right)^{2}\)
Answer:

Question 62.
OPEN-ENDED
Find a fraction that, when multiplied by \(\frac{1}{2}\), is less than \(\frac{1}{4}\).
Answer:

Question 63.
MP LOGIC
You are in a bike race. When you get to the first checkpoint, you arc \(\frac{2}{5}\) of the distance to the second checkpoint. When you get to the second checkpoint, you are \(\frac{1}{4}\) of the distance to the finish. What is the distance from the start to the first checkpoint?
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 24
Answer:

Question 64.
MP NUMBER SENSE
Is the product of two positive mixed numbers ever less than 1? Explain.
Answer:

Question 65.
MP REASONING
You plan to add a fountain to your garden.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 25
a. Draw a diagram o[thc fountain in the garden. Label the dimensions.
Answer:

b. Describe two methods for finding the area of the garden that surrounds the fountain.
Answer:

c. find the area. Which method did you use, and why?
Answer:

Question 66.
MP PROBLEM SOLVING
The cooking time for a ham is \(\frac{2}{5}\) of an hour for each pound. What time should you start cooking a ham that weighs 12\(\frac{3}{4}\) pounds so that it is done at 4:45 P.M. ?
Answer:

Question 67.
MP PRECISION
Complete the Four Square for \(\frac{7}{8} \times \frac{1}{3}\)
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 26
Answer:

Question 68.
DIG DEEPER!
You ask 150 people about their pets. The results show that \(\frac{9}{25}\) of the people own a dog. of the people who own a dog, \(\frac{1}{6}\) of them also own a cat.
a. What portion of the people own a dog and a cat?
Answer:

b. How many people own a dog but not a cat? Explain.
Answer:

Question 69.
MP NUMBER SENSE
Use each of the numbers from 1 to 9 exactly once to create three mixed numbers with the greatest possible product. then use each of the numbers exactly once to create three mixed numbers with the least possible product. Find each product. Explain your reasoning. The fraction portion of each mixed number should be proper.
Big Ideas Math Answers 6th Grade 6 Advanced Chapter 2 Fractions and Decimals 27
Answer:

Section 2.2 Dividing Fractions

Exploration 1

Dividing by Fractions
Work with a partner. Answer each question using a model.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 28
a. How many two-thirds arc in four?
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 29
Answer:

b. How many three-fourths arc in three?
Answer:

c. How many two-fifths arc in four-fifths?
Answer:

d. How many two-thirds arc in three?
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 30
Answer:

e. How many one-thirds arc in five-sixths?

Exploration 2

Finding a Pattern

Work with a partner. The table shows the division expressions from Exploration 1. Complete each multiplication expression so that it has the same value as the division expression above it. What can you conclude about dividing by fractions?
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 32
Answer:

Math Practice
Look for Structure
Can the pattern you found be applied to division by a whole number? Why or why not?
Answer:

Lesson 2.2 Dividing Fractions

Try It Write hie reciprocal of the number.

Question 1.
\(\frac{3}{4}\)
Answer:

Question 2.
5

Question 3.
\(\frac{7}{2}\)
Answer:

Question 4.
\(\frac{4}{9}\)
Answer:

Try It Divide. Write the answer in simplest form. Use a model to justify your answer.

Question 5.
\(\frac{1}{2} \div \frac{1}{8}\)
Answer:

Question 6.
\(\frac{2}{5} \div \frac{3}{10}\)
Answer:

Question 7.
\(\frac{3}{8} \div \frac{3}{4}\)
Answer:

Question 8.
\(\frac{2}{7} \div \frac{9}{14}\)
Answer:

Try It Divide. Write the answer in simplest form.

Question 9.
\(\frac{1}{3}\) ÷ 3
Answer:

Question 10.
\(\frac{2}{3}\) ÷ 10
Answer:

Question 11.
\(\frac{5}{8}\) ÷ 4
Answer:

Question 12.
\(\frac{6}{7}\) ÷ 4
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

DIVIDING FRACTIONS
Divide. Write the answer in simplest form. Draw a model to justify your answer.

Question 13.
\(\frac{2}{3} \div \frac{5}{6}\)
Answer:

Question 14.
\(\frac{6}{7}\) ÷ 3
Answer:

Question 15.
WHICH ONE DOESN’T BELONG?
Which of the following does not belong with the other three? Explain your reasoning.
\(\frac{2}{3} \div \frac{4}{5}\)     \(\frac{3}{2} \cdot \frac{4}{5}\)     \(\frac{5}{4} \times \frac{2}{3}\)       \(\frac{5}{4} \div \frac{3}{2}\)
Answer:

MATCHING
Match the expression with its value.

16. \( \frac{2}{5} \div \frac{8}{15}\) (A) \( \frac{1}{12}\)
17. \( \frac{8}{15} \div \frac{2}{5}\) (B) \( \frac{3}{4}\)
18. \( \frac{2}{15} \div \frac{8}{5}\) (c) 12
19. \( \frac{8}{5} \div \frac{2}{15}\) (D) 1\( \frac{1}{3}\)

Answer:

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding, of the success criteria in your journal.

Question 20.
You have 5 cups of rice to make bibimbap, a popular Korean meal. The recipe calls for \(\frac{4}{5}\) cup of rice per serving. H0w many full servings of bibimbap can you make? How much rice is left over?
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 33
Answer:

Question 21.
A band earns \(\frac{2}{3}\) of their profit from selling concert tickets and \(\frac{1}{5}\) of their profit from selling merchandise. The band earns a profit of $1500 from selling concert tickets. How much profit does the band earn from selling merchandise?
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 33
Answer:

Dividing Fractions Practice 2.2

Review & Refresh

Multiply. Write the answer in simplest form.

Question 1.
\(\frac{7}{10} \cdot \frac{3}{4}\)
Answer:

Question 2.
\(\frac{5}{6} \times 2 \frac{1}{3}\)
Answer:

Question 3.
\(\frac{4}{9} \times \frac{3}{8}\)
Answer:

Question 4.
\(2 \frac{2}{5} \cdot 6 \frac{2}{3}\)
Answer:

Match the expression with its value.

5. 3 + 2 × 42 A. 22
6. (3 + 2) × 42 B. 35
7. 2 + 3 × 42 C. 50
8. 42 + 2 × 3 D. 80

Answer:

Find the area of the rectangle.

Question 9.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 34
Answer:

Question 10.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 35
Answer:

Question 11.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 36
Answer:

Concepts, skills, & Problem Solving

CHOOSE TOOLS
Answer the question using a model. (See Exploration 1, Page.No 53.)

Question 12.
How many three-fifths are in three?
Answer:

Question 13.
How many two-ninths are in eight-ninths?
Answer:

Question 14.
How many three-fourths are in seven-eighths?
Answer:

WRITING RECIPROCALS
Write the reciprocal of the number.

Question 15.
8
Answer:

Question 16.
\(\frac{6}{7}\)
Answer:

Question 17.
\(\frac{2}{5}\)
Answer:

Question 18.
\(\frac{11}{8}\)
Answer:

DIVIDING FRACTIONS
Divide. Write the answer in simplest form.

Question 19.
\(\frac{1}{3} \div \frac{1}{2}\)
Answer:

Question 20.
\(\frac{1}{8} \div \frac{1}{4}\)
Answer:

Question 21.
\(\frac{2}{7} \div 2\)
Answer:

Question 22.
\(\frac{6}{5} \div 3\)
Answer:

Question 23.
\(\frac{2}{3} \div \frac{4}{9}\)
Answer:

Question 24.
\(\frac{5}{6} \div \frac{2}{7}\)
Answer:

Question 25.
\(12 \div \frac{3}{4}\)
Answer:

Question 26.
\(8 \div \frac{2}{5}\)
Answer:

Question 27.
\(\frac{3}{7} \div 6\)
Answer:

Question 28.
\(\frac{12}{25} \div 4\)
Answer:

Question 29.
\(\frac{2}{9} \div \frac{2}{3}\)
Answer:

Question 30.
\(\frac{8}{15} \div \frac{4}{5}\)
Answer:

Question 31.
\(\frac{1}{3} \div \frac{1}{9}\)
Answer:

Question 32.
\(\frac{7}{10} \div \frac{3}{8}\)
Answer:

Question 33.
\(\frac{14}{27} \div 7\)
Answer:

Question 34.
\(\frac{5}{8} \div 15\)
Answer:

Question 35.
\(\frac{27}{32} \div \frac{7}{8}\)
Answer:

Question 36.
\(\frac{4}{15} \div \frac{10}{13}\)
Answer:

Question 37.
\(9 \div \frac{4}{9}\)
Answer:

Question 38.
\(10 \div \frac{5}{12}\)
Answer:

YOU BE THE TEACHER
Your friend finds the quotient. Is your friend correct? Explain your reasoning.

Question 39.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 37
Answer:

Question 40.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 38
Answer:

Question 41.
MP REASONING
You have \(\frac{3}{5}\) of an apple pie. You divide the remaining pie into 5 equal slices. What portion of the original pie is each slice?
Answer:

Question 42.
MP PROBLEM SOLVING
How many times longer is the baby alligator than the baby gecko?
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 39
Answer:

OPEN-ENDED
Write a real-life problem for the expression. Then solve the problem.
Question 47.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 40
Answer:

Question 48.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 41
Answer:

Question 49.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 42
Answer:

MP REASONING
Without finding the quotient, copy and complete the statement using <, >, or =. Explain your reasoning.

Question 50.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 43
Answer:

Question 51.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 44
Answer:

Question 52.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 45
Answer:

Question 53.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 46
Answer:

ORDER OF OPERATIONS
Evaluate the expression. Write the answer in simplest form.

Question 54.
\(\frac{1}{6}\) ÷ 6 ÷ 6
Answer:

Question 55.
\(\frac{7}{12}\) ÷ 14 ÷ 6
Answer:

Question 56.
\(\frac{3}{5} \div \frac{4}{7} \div \frac{9}{10}\)
Answer:

Question 57.
\(4 \div \frac{8}{9}-\frac{1}{2}\)
Answer:

Question 58.
\(\frac{3}{4}+\frac{5}{6} \div \frac{2}{3}\)
Answer:

Question 59.
\(\frac{7}{8}-\frac{3}{8} \div 9\)
Answer:

Question 60.
\(\frac{9}{16} \div \frac{3}{4} \cdot \frac{2}{13}\)
Answer:

Question 61.
\(\frac{3}{14} \cdot \frac{2}{5} \div \frac{6}{7}\)
Answer:

Question 62.
\(\frac{10}{27} \cdot\left(\frac{3}{8} \div \frac{5}{24}\right)\)
Answer:

Question 63.
MP NUMBER SENSE
When is the reciprocal of a fraction a whole number? Explain.
Answer:

Question 64.
MODELING REAL LIFE
You use \(\frac{1}{8}\) of your battery for every \(\frac{2}{5}\) of an hour that you video chat. You use \(\frac{3}{4}\) of your battery video chatting. How long did you video chat?
Answer:

Question 65.
MP PROBLEM SOLVING
The table shows the portions of a family budget that are spent on several expenses.

Expense Portion of Budget
Housing \( \frac{2}{5}\)
Food \( \frac{4}{9}\)
Automobiles \( \frac{1}{15}\)
Recreation \( \frac{1}{40}\)

a. How many times more is the expense for housing than for automobiles?
Answer:

b. How many times more is the expense for food than for recreation?
Answer:

c. The expense for automobile fuel is \(\frac{1}{60} 0\) of the total expenses. What portion of the automobile expense is spent on fuel?
Answer:

Question 66.
CRITICAL THINKING
A bottle of juice is \(\frac{2}{3}\) full. the bottle contains \(\frac{4}{5}\) of a cup of juice.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 47
a. Write a division expression that represents the capacity of the bottle.
Answer:

b. Write a related multiplication expression that represents the capacity of the bottle.
Answer:

e. Explain how you can use the diagram to verify the expression In part (b).
Answer:

d. Find the capacity of the bade.
Answer:

Question 67.
DIG DEEPER!
You have 6 pints of glaze. It takes \(\frac{7}{8}\) of a pint to glaze a bowl and \(\frac{9}{16}\) of a pint to glaze a plate.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 48
a. how many howls can you completely glaze? How many plates can you completely glaze?
Answer:

b. You want to glaze 5 bowls, and then use the rest for plates. How many plates can you completely glaze? How much glaze will be left over?
Answer:

c. How many of each object can you completely glaze so that there is no glaze left over? Explain how you found your answer.
Answer:

Question 68.
MP REASONING
A water tank is \(\frac{1}{8}\) full. The tank is \(\frac{3}{4}\) full when 42 gallons of water are added to the tank.
Big Ideas Math Answers Grade 6 Advanced Chapter 2 Fractions and Decimals 49
a. How much water can the tank hold?
Answer:

b. How much water was originally in the tank?
Answer:

c. How much water is in the tank when it is \(\frac{1}{2}\) full?
Answer:

Section 2.3 Dividing Mixed Numbers

Exploration 1

Dividing Mixed Numbers

Work with a partner. Write a real-life problem that represents each division expression described. Then solve each problem using a model. Check your answers.

a. How many three-fourths arc in four and one-half?
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 50
Answer:

b. How many three-eighths arc in two and one-fourth?
Answer:

c. How many one and one-halves arc in six?
Answer:

d. How many seven-sixths arc in three and one-third?
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 51
Answer:

e. How many one and one-fifths are in five?
Answer:

f. How many three and one-halves arc in two and one-half?
Answer:
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 52

g. How many four and one-halves arc in one and one-half?
Answer:

Lesson 2.3 Dividing Mixed Numbers

Try It Divide. Write the answer in simplest form.

Question 1.
\(3 \frac{2}{3} \div \frac{1}{3}\)
Answer:

Question 2.
\(1 \frac{3}{7} \div \frac{2}{3}\)
Answer:

Question 3.
\(2 \frac{1}{6} \div \frac{3}{4}\)
Answer:

Question 4.
\(6 \frac{1}{2} \div 2\)
Answer:

Question 5.
\(10 \frac{2}{3} \div 2 \frac{2}{3}\)
Answer:

Question 6.
\(8 \frac{1}{4} \div 1 \frac{1}{2}\)
Answer:

Question 7.
\(3 \div 1 \frac{3}{4}\)
Answer:

Question 8.
\(\frac{3}{4} \div 2 \frac{1}{2}\)
Answer:

Try It Evaluate the expression. Write the answer in simplest form.

Question 9.
\(1 \frac{1}{2} \div \frac{1}{6}-\frac{7}{8}\)
Answer:

Question 10.
\(3 \frac{1}{3} \div \frac{5}{6}+\frac{8}{9}\)
Answer:

Question 11.
\(\frac{2}{5}+2 \frac{4}{5} \div 2\)
Answer:

Question 12.
\(\frac{2}{5}+2 \frac{4}{5} \div 2\)
Answer:

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

EVALUATING EXPRESSIONS
Evaluate the expression. Write the answer in simplest form.

Question 13.
\(4 \frac{4}{7} \div \frac{4}{7}\)
Answer:

Question 14.
\(\frac{1}{2} \div 5 \frac{1}{4}\)
Answer:

Question 15.
\(\frac{3}{4}+6 \frac{2}{5} \div 1 \frac{3}{5}\)
Answer:

Question 16.
MP NUMBER SENSE
Is \(2 \frac{1}{2} \div 1 \frac{1}{4}\) the sense as \(1 \frac{1}{4} \div 2 \frac{1}{2}\)? Use models to justify your answer.
Answer:

Question 17.
DIFFERENT WORDS. SAME QUESTION
Which is different? Find “b0th” answers.
What is 5\(\frac{1}{2}\) divided by \(\frac{1}{8}\)?
What is the quotient of 5\(\frac{1}{2}\) and \(\frac{1}{8}\)?
What is 5\(\frac{1}{2}\) times 8?
What is \(\frac{1}{8}\) of 5\(\frac{1}{2}\)?
Answer:

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.
A water cooler contains 160 cups of water. During practice, each person on a team fills a water bottle with 3\(\frac{1}{3}\) cups of water from the cooler. Is there enough water for all 45 people on the team to fill their water bottles? Explain.
Answer:

Question 19.
A cyclist is 7\(\frac{3}{4}\) kilometers from the finish line of a race. The cyclist rides at a rate of 25\(\frac{5}{6}\) kilometers per hour. How many minutes will it take the cyclist to finish the race?
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 53
Answer:

Dividing Mixed Numbers Practice 2.3

Review & Refresh

Divide. Write the answer in simplest form.

Question 1.
\(\frac{1}{8} \div \frac{1}{7}\)
Answer:

Question 2.
\(\frac{7}{9} \div \frac{2}{3}\)
Answer:

Question 3.
\(\frac{5}{6}\) ÷ 10
Answer:

Question 4.
12 ÷ \(\frac{3}{8}\)
Answer:

Find the LCM of the numbers.

Question 5.
8, 14
Answer:

Question 6.
9, 11, 12
Answer:

Question 7.
12, 27, 30
Answer:

Find the volume of the rectangular prism.

Question 8.
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 54
Answer:

Question 9.
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 55
Answer:

Question 10.
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 56
Answer:

Question 11.
Which number is not a prime factor of 286?
A. 2
B. 7
C. 11
D. 13
Answer:

Concepts, Skills & Problem Solving

MP CHOOSE TOOLS
Write a real-life problem that represents the division expression described. Then solve the problem using a model. Check your answer algebraically. (See Exploration 1, Page.No 61.)
Question 12.
How many two-thirds are in three and one-third?
Answer:

Question 13.
How many one and one-sixths are in five and five-sixths?
Answer:

Question 14.
How man two and one-halves are in eight and three-fourths?
Answer:

DIVIDING WITH MIXED NUMBERS
Divide. Write the answer in simplest form.

Question 15.
\(2 \frac{1}{4} \div \frac{3}{4}\)
Answer:

Question 16.
\(3 \frac{4}{5} \div \frac{2}{5}\)
Answer:

Question 17.
\(8 \frac{1}{8} \div \frac{5}{6}\)
Answer:

Question 18.
\(7 \frac{5}{9} \div \frac{4}{7}\)
Answer:

Question 19.
\(7 \frac{1}{2} \div 1 \frac{9}{10}\)
Answer:

Question 20.
\(3 \frac{3}{4} \div 2 \frac{1}{12}\)
Answer:

Question 21.
\(7 \frac{1}{5} \div 8\)
Answer:

Question 22.
\(8 \frac{4}{7} \div 15\)
Answer:

Question 23.
\(8 \frac{1}{3} \div \frac{2}{3}\)
Answer:

Question 24.
\(9 \frac{1}{6} \div \frac{5}{6}\)
Answer:

Question 25.
\(13 \div 10 \frac{5}{6}\)
Answer:

Question 26.
\(12 \div 5 \frac{9}{11}\)
Answer:

Question 27.
\(\frac{7}{8} \div 3 \frac{1}{16}\)
Answer:

Question 28.
\(\frac{4}{9} \div 1 \frac{7}{15}\)
Answer:

Question 29.
\(4 \frac{5}{16} \div 3 \frac{3}{8}\)
Answer:

Question 30.
\(6 \frac{2}{9} \div 5 \frac{5}{6}\)
Answer:

Question 31.
YOU BE THE TEACHER
Your friend finds the quotient of 3\(\frac{1}{2}\) and 1\(\frac{2}{3}\). Is your friend correct? Explain your reasoning.
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 57
Answer:

Question 32.
MP PROBLEM SOLVING
A platinum nugget weighs 3\(\frac{1}{2}\) ounces. How many \(\frac{1}{4}\) ounce pieces can he cut from the nugget?
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 58
Answer:

ORDER OF OPERATIONS
Evaluate the expression. Write the answer in simplest form.

Question 33.
\(3 \div 1 \frac{1}{5}+\frac{1}{2}\)
Answer:

Question 34.
\(4 \frac{2}{3}-1 \frac{1}{3} \div 2\)
Answer:

Question 35.
\(\frac{2}{5}+2 \frac{1}{6} \div \frac{5}{6}\)
Answer:

Question 36.
\(5 \frac{5}{6} \div 3 \frac{3}{4}-\frac{2}{9}\)
Answer:

Question 37.
\(6 \frac{1}{2}-\frac{7}{8} \div 5 \frac{11}{16}\)
Answer:

Question 38.
\(9 \frac{1}{6} \div 5+3 \frac{1}{3}\)
Answer:

Question 39.
\(3 \frac{3}{5}+4 \frac{4}{15} \div \frac{4}{9}\)
Answer:

Question 40.
\(\frac{3}{5} \times \frac{7}{12} \div 2 \frac{7}{10}\)
Answer:

Question 41.
\(4 \frac{3}{8} \div \frac{3}{4} \cdot \frac{4}{7}\)
Answer:

Question 42.
\(1 \frac{9}{11} \times 4 \frac{7}{12} \div \frac{2}{3}\)
Answer:

Question 43.
\(3 \frac{4}{15} \div\left(8 \cdot 6 \frac{3}{10}\right)\)
Answer:

Question 44.
\(2 \frac{5}{14} \div\left(2 \frac{5}{8} \times 1 \frac{3}{7}\right)\)
Answer:

Question 45.
MP LOGIC
Your friend uses the model shown to state that \(2 \frac{1}{2} \div 1 \frac{1}{6}=2 \frac{1}{6}\). Is your friend correct? Justify your answer using the model.
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 59
Answer:

Question 46.
MODELING REAL LIFE
A bag contains 42 cups of dog food. Your dog cats 2\(\frac{1}{3}\) cups of dog food each day. Is there enough food to last 3 weeks? Explain.
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 60
Answer:

Question 47.
DIG DEEPER!
You have 12 cups of granola and 8\(\frac{1}{2}\) cups of peanuts to make trail mix. What is the greatest number of full batches of trail mix you can make? Explain how you found your answer.
Big Ideas Math Solutions Grade 6 Advanced Chapter 2 Fractions and Decimals 61
Answer:

Question 48.
MP REASONING
At a track and field meet, the longest shot-put throw by a boy is 25 feet 8 inches. the longest shot-put throw by a girl is 19 feet 3 inches. How many times greater is the longest shot-put throw by the boy than by the girl?
Answer: