Eureka Math

Engage NY Eureka Math 5th Grade Module 2 Lesson 9 Answer Key

Eureka Math Grade 5 Module 2 Lesson 9 Problem Set Answer Key

Solve.

Question 1.
An office space in New York City measures 48 feet by 56 feet. If it sells for $565 per square foot, what is the total cost of the office space?

Answer:
The total cost of the office space = $565.

Explanation:
In the above-given question,
given that,
An office space in New York City measures 48 feet by 56 feet.
if it sells for $565 per square foot.
48 x 56 =
$ 565

Question 2.
Gemma and Leah are both jewelry makers. Gemma made 106 beaded necklaces. Leah made 39 more necklaces than Gemma.
a. Each necklace they make has exactly 104 beads on it. How many beads did both girls use altogether while making their necklaces?

Answer:
The beads did both girls use altogether while making their necklaces = 26104.

Explanation:
In the above-given question,
given that,
Each necklace they make has exactly 104 beads on it.
106 + 39 = 145.
145 + 106 = 251.
251 x 104 = 26104.

b. At a recent craft fair, Gemma sold each of her necklaces for $14. Leah sold each of her necklaces for $10 more. Who made more money at the craft fair? How much more?

Answer:
The money at the craft fair = $24.

Explanation:
In the above-given question,
given that,
At a recent craft fair, Gemma sold each of her necklaces for $14.
Leah sold each of her necklaces for $10 more.
Leah made more money at the craft fair.

Question 3.
Peng bought 26 treadmills for her new fitness center at $1,334 each. Then, she bought 19 stationary bikes for $749 each. How much did she spend on her new equipment? Write an expression, and then solve.

Answer:
The money she spends on her new equipment = 34684.

Explanation:
In the above-given question,
given that,
peng bought 26 treadmills for her new fitness center at $1,334 each.
Then, she bought 19 stationary bikes for $749 each.
26 x $ 1334.
34684.

Question 4.
A Hudson Valley farmer has 26 employees. He pays each employee $410 per week. After paying his workers for one week, the farmer has $162 left in his bank account. How much money did he have at first?

Answer:
The money did he have at first = 10660.

Explanation:
In the above-given question,
given that,
A Hudson Valley farmer has 26 employees.
He pays each employee $410 per week.
After paying his workers for one week, the farmer has $162 left in his bank account.
26 x 410 = 10660.

Question 5.
Frances is sewing a border around 2 rectangular tablecloths that each measure 9 feet long by 6 feet wide. If it takes her 3 minutes to sew on 1 inch of border, how many minutes will it take her to complete her sewing project? Write an expression, and then solve.

Answer:
The time it takes her to complete the sewing project = 162.

Explanation:
In the above-given question,
given that,
Frances is sewing a border around 2 rectangular tablecloths that each measure 9 feet long by 6 feet wide.
If it takes her 3 minutes to sew on 1 inch of border.
9 x 6 = 54.
54 + 54 + 54 = 162.

Question 6.
Each grade level at Hooperville Schools has 298 students.
a. If there are 13-grade levels, how many students attend Hooperville Schools?

Answer:
The students attend Hooperville Schools = 3874.

Explanation:
In the above-given question,
given that,
If there are 13-grade levels, each grade level at Hooperville Schools has 298 students.
13 x 298 = 3874.

b. A nearby district, Willington, is much larger. They have 12 times as many students. How many students attend schools in Willington?

Answer:
The students attend Schools in Willington = 46488.

Explanation:
In the above-given question,
given that,
A nearby district, Willington, is much larger.
They have 12 times as many students.
3874 x 12 = 46488.

Eureka Math Grade 5 Module 2 Lesson 9 Exit Ticket Answer Key

Juwad picked 30 bags of apples on Monday and sold them at his fruit stand for $3.45 each. The following week he picked and sold 26 bags.
a. How much money did Juwad earn in the first week?

Answer:
The money Juwad earns in the first week = $103.5.

Explanation:
In the above-given question,
given that,
Juwad picked 30 bags of apples on Monday and sold them at his fruit stand for $3.45 each.
30 x $3.45 = 103.5.

b. How much money did he earn in the second week?

Answer:
The money did he earn in the second week = $89.7.

Explanation:
In the above-given question,
given that,
Juwad sold 26 bags of apples.
26 x $3.45.
89.7

c. How much did Juwad earn selling bags of apples these two weeks?

Answer:
The money did he earn in these two weeks = $193.2.

Explanation:
In the above-given question,
given that,
103.5 + 89.7.
193.2.

d. Extension: Each bag Juwad picked holds 15 apples. How many apples did he pick in two weeks? Write an expression to represent this problem.

Answer:
The money did he earn in these two weeks = 51.75 .

Explanation:
In the above-given question,
given that,
Each bag Juwad picked holds 15 apples.
15 x $3.45.
51.75.

Eureka Math Grade 5 Module 2 Lesson 9 Homework Answer Key

Solve.

Question 1.
Jeffery bought 203 sheets of stickers. Each sheet has a dozen stickers. He gave away 907 stickers to his family and friends on Valentine’s Day. How many stickers does Jeffery have remaining?

Answer:
The stickers do Jeffery has = 1529.

Explanation:
In the above-given question,
given that,
Jeffery bought 203 sheets of stickers.
Each sheet has a dozen stickers.
He gave away 907 stickers to his family and friends on Valentine’s Day.
203 x 12 = 2436.
2436 – 907 = 1529.

Question 2.
During the 2011 season, a quarterback passed for 302 yards per game. He played in all 16 regular season games that year.
a. For how many total yards did the quarterback pass?

Answer:
The total yards did the quarterback pass = 4832.

Explanation:
In the above-given question,
given that,
During the 2011 season, a quarterback passed for 302 yards per game.
He played in all 16 regular-season games that year.
302 x 16 = 4832.

b. If he matches this passing total for each of the next 13 seasons, how many yards will he pass for in his career?

Answer:
The many yards will he pass for in his career = 3926.

Explanation:
In the above-given question,
given that,
If he matches this passing total for each of the next 13 seasons.
302 x 13 = 3926.

Question 3.
Bao saved $179 a month. He saved $145 less than Ada each month. How much would Ada save in three and a half years?

Answer:
The Ada save in three and a half years = $324.

Explanation:
In the above-given question,
given that,
Bao saved $179 a month.
He saved $145 less than Ada each month.
$179 + $145 = $324.

Question 4.
Mrs. Williams is knitting a blanket for her newborn granddaughter. The blanket is 2.25 meters long and 1.8 meters wide. What is the area of the blanket? Write the answer in centimeters.

Answer:
The area of the blanket = 405 cms.

Explanation:
In the above-given question,
given that,
Mrs. Williams is knitting a blanket for her newborn granddaughter.
The blanket is 2.25 meters long and 1.8 meters wide.
the area of the blanket = 2.25 x 1.8 meters.
2.25 x 1.8 = 4.05 meters.
1 meter = 100 cms.
4.05 x 100 = 405 cms

Question 5.
Use the chart to solve.
Soccer Field Dimensions

	FIFA Regulation (in yards)	New York State High Schools (in yards)
Minimum Length	110	100
Maximum Length	120	120
Minimum Width	70	55
Maximum Width	80	80

a. Write an expression to find the difference in the maximum area and minimum area of a NYS high school soccer field. Then, evaluate your expression.

Answer:
The minimum area of NYS high school soccer field =

Explanation:
In the above-given question,
given that,
The maximum length = 110.
minimum length = 100.
110 – 100 = 10.

b. Would a field with a width of 75 yards and an area of 7,500 square yards be within FIFA regulation? Why or why not?

Answer:
No, there does not have a field with a width of 75 yards and an area of 7,500 square yards.

Explanation:
In the above-given question,
given that,
Would a field with a width of 75 yards and an area of 7500 square yards.
70 x 55 = 3850.
80 x 80 = 6400.
120 x 120 = 14,400.
no there is no field with a width of 75 yards and an area of 7500 square yards.

c. It costs $26 to fertilize, water, mow, and maintain each square yard of a full size FIFA field (with maximum dimensions) before each game. How much will it cost to prepare the field for next week’s match?

Answer:
The cost prepares the field for next week’s match = $3120.

Explanation:
In the above-given question,
given that,
It costs $26 to fertilize, water, mow, and maintain each square yard of a full size.
120 x $26.
3120.

Engage NY Eureka Math Kindergarten Module 1 Lesson 4 Answer Key

Eureka Math Kindergarten Module 1 Lesson 4 Problem Set Answer Key

Use the cutouts. Glue the pictures to show where to keep each thing.

Answer:

Cutouts for the Problem Set

Eureka Math Kindergarten Module 1 Lesson 4 Exit Ticket Answer Key

Circle the animals that belong to one group, and underline the animals that belong to the other group.

Answer:

What is the same about the animals in each group? (Discuss with a friend.)
Answer:

Eureka Math Kindergarten Module 1 Lesson 4 Homework Answer Key

Circle the things that belong to one group, and underline the things that belong to the other group. Tell an adult why the items in each group belong together.

Answer:

Engage NY Eureka Math 1st Grade Module 1 Lesson 37 Answer Key

Eureka Math Grade 1 Module 1 Lesson 37 Sprint Answer Key

A
*Write the missing number for each number sentence. Pay attention to the + and – signs.

Question 1.
9 + 1 = ☐
Answer:10
Explination:
9+1=10

Question 2.
1 + 9 = ☐
Answer:10
Explination:
1+9=10

Question 3.
10 – 1 = ☐
Answer:9
Explination:
10-1=9

Question 4.
10 – 9 = ☐
Answer:1
Explination:
10+9=1

Question 5.
10 + 0 = ☐
Answer:10
Explination:
10+0=10

Question 6.
0 + 10 = ☐
Answer:10
Explination:
0+10=10

Question 7.
10 – 0 = ☐
Answer:10
Explination:
10-0=10

Question 8.
10 – 10 = ☐
Answer:0
Explination:
10-10=0

Question 9.
8 + 2 = ☐
Answer:10
Explination:
8+2=10

Question 10.
2 + 8 = ☐
Answer:10
Explination:
2+8=10

Question 11.
10 – 2 = ☐
Answer:8
Explination:
10-2=8

Question 12.
10 – 8 = ☐
Answer:2
Explination:
10-8=2

Question 13.
7 + 3 = ☐
Answer:10
Explination:
7+3=10

Question 14.
3 + 7 = ☐
Answer:
Explination:
3+7=10

Question 15.
10 – 3 = ☐
Answer:7
Explination:
10-3=7

Question 16.
10 – 7 = ☐
Answer:3
Explination:
10-7=3

Question 17.
10 = 7 + ☐
Answer:3
Explination:
10=7+3

Question 18.
10 = 3 + ☐
Answer:7
Explination:
10=3+7

Question 19.
10 = 6 + ☐
Answer:4
Explination:
10=6+4

Question 20.
10 = 4 + ☐
Answer:6
Explination:
10=4+6

Question 21.
10 = 5 + ☐
Answer:5
Explination:
10=5+5

Question 22.
10 – ☐ = 5
Answer:5
Explination:
10-5=5

Question 23.
5 = 10 – ☐
Answer:5
Explination:
5=10-5

Question 24.
6 = 10 – ☐
Answer:4
Explination:
6=10-4

Question 25.
7 = 10 – ☐
Answer:3
Explination:
7=10-3

Question 26.
7 = ☐ – 3
Answer:10
Explination:
7=10-3

Question 27.
4 = 10 – ☐
Answer:6
Explination:
4=10-6

Question 28.
5 = ☐ – 5
Answer:5
Explination:
5=10-5

Question 29.
6 = 10 – ☐
Answer:
Explination:
6=10-4

Question 30.
7 = ☐ – 3
Answer:10
Explination:
7=10-3

B
*Write the missing number for each number sentence. Pay attention to the + and – signs.

Question 1.
8 + 2 = ☐
Answer:10
Explination:
8+2=10

Question 2.
2 + 8 = ☐
Answer:10
Explination:
2+8=10

Question 3.
10 – 2 = ☐
Answer:8
Explination:
10-2=8

Question 4.
10 – 8 = ☐
Answer:2
Explination:
10-8=2

Question 5.
9 + 1 = ☐
Answer:10
Explination:
9+1=10

Question 6.
1 + 9 = ☐
Answer:10
Explination:
1+9=10

Question 7.
10 – 1 = ☐
Answer:9
Explination:
10-1=9

Question 8.
10 – 9 = ☐
Answer:9
Explination:
10-9=1

Question 9.
10 + 0 = ☐
Answer:10
Explination:
10+0=10

Question 10.
0 + 10 = ☐
Answer:10
Explination:
0+10=10

Question 11.
10 – 0 = ☐

Answer:10
Explination:
10+0=10

Question 12.
10 – 10 = ☐
Answer:0
Explination:
10+10=0

Question 13.
6 + 4 = ☐
Answer:10
Explination:
6+4=10

Question 14.
4 + 6 = ☐
Answer:10
Explination:
4+6=10

Question 15.
10 – 4 = ☐
Answer:6
Explination:
10-4=6

Question 16.
10 – 6 = ☐
Answer:4
Explination:
10-6=4

Question 17.
10 = 8 + ☐
Answer:2
Explination:
10=8+2

Question 18.
10 = 7 + ☐
Answer:3
Explination:
10=7+3

Question 19.
10 = 3 + ☐
Answer:7
Explination:
10=3+7

Question 20.
10 = 4 + ☐
Answer:6
Explination:
10=4+6

Question 21.
10 = 5 + ☐
Answer:5
Explination:
10=5+5

Question 22.
10 – ☐ = 5
Answer:5
Explination:
10-5=5

Question 23.
6 = 10 – ☐
Answer:4
Explination:
6=10-4

Question 24.
7 = 10 – ☐
Answer:3
Explination:
7=10-3

Question 25.
8 = 10 – ☐
Answer:2
Explination:
8=10-2

Question 26.
7 = ☐ – 3
Answer:10
Explination:
7=10-3

Question 27.
2 = 10 – ☐
Answer:8
Explination:
2=10-8

Question 28.
4 = ☐ – 6
Answer:2
Explination:
4=2-6

Question 29.
3 = 10 – ☐
Answer:7
Explination:
3=10-7

Question 30.
7 = ☐ – 3
Answer:10
Explination:
7=10-3

Eureka Math Grade 1 Module 1 Lesson 37 Problem Set Answer Key

Solve the sets. Cross off on the 5-groups. Write the related subtraction sentence that would have the same number bond.

Question 1.

9 – 8 = __
9 – 1 = ___
Answer: 9- 8 = 1
9-1=8

Question 2.

9 – 7 = __
________
Answer:9-7=2
9-2=7

Question 3.

9 – 9 = __
_____
Answer:9-9=0
9-0=9

Make a 5-group drawing. Solve, and write a related subtraction sentence that would have the same number bond. Cross off to show.

Question 4.

9 – 6 = ___

___________
Answer:9-6=3
9-3=6

Question 5.

9 – 4 = __
___________
Answer: 9-4=5
9-4=5

Question 6.

9 – 3 = __
___________
Answer:9-3=6
9-6=3

Subtract. Then, write the related subtraction sentence.
Make a math drawing if needed, and complete a number bond.

Question 7.

9 – 5 = ___
______________
Answer:9-5=4
9-4=5

Question 8.

9 – 8 = ___
______________
Answer:9-8=1
9-1=8

Question 9.

9 – 7 = ___
________
Answer:9-7=2
9-2=7

Question 10.

9 – 3 = ___
______________
Answer:9-3=6
9-6=3

Question 11.
Fill in the missing part. Write the 2 matching subtraction sentences.
a.
Answer:

b.
Answer:

c.
Answer:

d.
Answer:

e.
Answer:

Eureka Math Grade 1 Module 1 Lesson 37 Exit Ticket Answer Key

Fill in the missing part. Draw a math picture if needed. Write the 2 matching subtraction sentences.

Question 1.

_________________
_________________
Answer: 9-3=6
9-3

Question 2.

_________________
_______________
Answer:9-7=2
9-7

Question 3.

_________________
_______________
Answer:9-4=5
9-5=4

Eureka Math Grade 1 Module 1 Lesson 37 Homework Answer Key

Make 5-group drawings and solve. Use the first number sentence to help you write a related number sentence that matches your picture.

Question 1.

9 – 2 = ____
__ – __ = __
Answer: 9 – 2 = 7
9 – 7 = 2

Question 2.

9 – 8 = __
__ – __ = __
Answer: 9 – 8 = 1
9 – 1 = 8

Question 3.

9 – 4 = __
__ – __ = __
Answer: 9 – 4 = 5
9 – 5 = 4

Subtract. Then, write the related subtraction sentence. Make a math drawing if needed, and complete a number bond for each.

Question 4.

9 – 7 = ___
_________
Answer: 9 – 7 = 2
9 – 2 = 7

Question 5.

9 – 2 = __
___
Answer: 9-2=7
9-7=2

Question 6.

9 – __ = 6
___;
Answer: 9-3=6
9-6=3

Question 7.

9 – __ = 1
____
Anwer:9-8=1
9-1=8

Question 8.

__ = 9 – 5
_____
Answer:4=9+5
9=5+4

Question 9.
Use 5-group drawings to help you complete the number bond. Match the number bond to the related subtraction sentence. Write the other related subtraction number sentence.

a.
Answer:

b.
Answer:

c.
Answer:

d.
Answer:

e.
Answer:

Engage NY Eureka Math Algebra 1 Module 1 Lesson 18 Answer Key

Eureka Math Algebra 1 Module 1 Lesson 18 Example Answer Key

Examples 1–2
Work through the examples as a whole class.

Example 1.
Consider the equation \(\frac{1}{x}\)=\(\frac{3}{x-2}\).
a. Rewrite the equation into a system of equations.
Answer:
\(\frac{1}{x}\)=\(\frac{3}{x-2}\) and x≠0 and x≠2

b. Solve the equation for x, excluding the value(s) of x that lead to a denominator of zero.
Answer:
x=-1 and x≠0 and x≠2 Solution set: {-1}

Example 2.
Consider the equation \(\frac{x+3}{x-2}\)=\(\frac{5}{x-2}\).
a. Rewrite the equation into a system of equations.
Answer:
\(\frac{x+3}{x-2}\)=\(\frac{5}{x-2}\) and x≠2

b. Solve the equation for x, excluding the value(s) of x that lead to a denominator of zero.
Answer:
x=2 and x≠2 Solution set: ɸ

Emphasize the process of recognizing a rational equation as a system composed of the equation itself and the excluded value(s) of x. For Example 1, this is really the compound statement:
\(\frac{1}{x}\)=\(\frac{3}{x-2}\) and x≠0 and x-2≠0
→ By the properties of equality, we can multiply through by nonzero quantities. Within this compound statement, x and x-2 are nonzero, so we may write x-2=3x and x≠0 and x-2≠0, which is equivalent to -2=2x and x≠0 and x≠2.
→ All three declarations in this compound statement are true for x=-1. This is the solution set.
In Example 2, remind students of the previous lesson on solving equations involving factored expressions. Students will need to factor out the common factor and then apply the zero-product property.
→ What happens in Example 2 when we have x=2 and x≠2? Both declarations cannot be true. What can we say about the solution set of the equation?
→ There is no solution.

Eureka Math Algebra 1 Module 1 Lesson 18 Exercise Answer Key

Opening Exercise
Nolan says that he checks the answer to a division problem by performing multiplication. For example, he says that
20÷4=5 is correct because 5×4 is 20, and \(\frac{3}{\frac{1}{2}}\)=6 is correct because 6×\(\frac{1}{2}\) is 3.
a. Using Nolan’s reasoning, explain why there is no real number that is the answer to the division problem
5÷0.
Answer:
There is no number n such that 0×n=5.

b. Quentin says that \(\frac{0}{0}\)=17. What do you think?
Answer:
While it is true that 0×17=0, the problem is that by that principle \(\frac{0}{0}\) could equal any number.

c. Mavis says that the expression \(\frac{3 x-6}{x-2}\) has a meaningful value for whatever value one chooses to assign to x. Do you agree?
Answer:
No. The expression does not have a meaningful value when x=-2.

d. Bernoit says that the expression \(\frac{3 x-6}{x-2}\) always has the value 3 for whichever value one assigns to x. Do you agree?
Answer:
The expression does equal 3 for all values of x except x=2.

Note that the problem with \(\frac{0}{0}\) is that too many numbers pass Nolan’s criterion! Have students change 17 to a different number. It still passes Nolan’s multiplication check. Like \(\frac{5}{0}\), it is a problematic notion. For this reason, we want to disallow the possibility of ever dividing by zero.
Point out that an expression like \(\frac{5}{x+2}\) is really accompanied with the clause “under the assumption the denominator is not zero.” So, \(\frac{5}{x+2}\) should be read as a compound statement:
\(\frac{5}{x+2}\) and x+2≠0 OR \(\frac{5}{x+2}\) and x≠-2

Exercises 1–2
Give students a few minutes to complete the problems individually. Then, elicit answers from students.

Exercise 1.
Rewrite \(\frac{10}{x+5}\) as a compound statement.
Answer:
\(\frac{10}{x+5}\) and x≠-5

Exercise 2.
Consider \(\frac{x^{2}-25}{\left(x^{2}-9\right)(x+4)}\).

a. Is it permissible to let x=5 in this expression?
Answer:
Yes. \(\frac{0}{144}\)=0

b. Is it permissible to let x=3 in this expression?
Answer:
No. –\(\frac{16}{0}\) is not permissible.

c. Give all the values of x that are not permissible in this expression.
Answer:
x≠-3, 3, -4

Exercises 3–11
Allow students time to complete the problems individually. Then, have students compare their work with another student. Make sure that students are setting up a system of equations as part of their solution.
Work through Exercises 11 as a class.

Exercises 3–11
Rewrite each equation into a system of equations excluding the value(s) of x that lead to a denominator of zero; then, solve the equation for x.

Exercise 3.
\(\frac{5}{x}\)=1
Answer:
\(\frac{5}{x}\)=1 and x ≠ 0
{5}

Exercise 4.
\(\frac{1}{x-5}\)=3
Answer:
\(\frac{1}{x-5}\)=3 and x ≠ 5
{\(\frac{16}{3}\)}

Exercise 5.
\(\frac{x}{x+1}\)=4
Answer:
\(\frac{x}{x+1}\)=4 and x ≠ -1
{-\(\frac{4}{3}\)}

Exercise 6.
\(\frac{2}{x}\)=\(\frac{3}{x-4}\)
Answer:
\(\frac{2}{x}\)=\(\frac{3}{x-4}\) and x≠0 and x≠4
{-8}

Exercise 7.
\(\frac{x}{x+6}\)=-\(\frac{6}{x+6}\)
Answer:
\(\frac{x}{x+6}\)=-\(\frac{6}{x+6}\) and x≠-6
No solution

Exercise 8.
\(\frac{x-3}{x+2}\)=0
Answer:
\(\frac{x-3}{x+2}\)=0 and x≠-2
{3}

Exercise 9.
\(\frac{x+3}{x+3}\)=5
Answer:
\(\frac{x+3}{x+3}\)=5 and x≠-3
No solution

Exercise 10.
\(\frac{x+3}{x+3}\)=1
Answer:
\(\frac{x+3}{x+3}\)=1 and x≠-3
All real numbers except -3

Exercise 11.
A baseball player’s batting average is calculated by dividing the number of times a player got a hit by the total number of times the player was at bat. It is expressed as a decimal rounded to three places. After the first 10 games of the season, Samuel had 12 hits off of 33 at bats.
a. What is his batting average after the first 10 games?
Answer:
\(\frac{12}{33}\)≈0 .364

b. How many hits in a row would he need to get to raise his batting average to above 0.500?
Answer:
\(\frac{12+x}{33+x}\)>0.500
x>9
He would need 10 hits in a row to be above 0.500.

c. How many at bats in a row without a hit would result in his batting average dropping below 0.300?
Answer:
\(\frac{12}{33+x}\)<0.300 x > 7
If he went 8 at bats in a row without a hit, he would be below 0.300.

Ask:
→ What was the difference between Exercises 9 and 10? How did that affect the solution set?

Eureka Math Algebra 1 Module 1 Lesson 18 Problem Set Answer Key

Question 1.
Consider the equation \(\frac{10\left(x^{2}-49\right)}{3 x\left(x^{2}-4\right)(x+1)}\) =0. Is x=7 permissible? Which values of x are excluded? Rewrite as a system of equations.
Answer:
Yes, x=7 is permissible. The excluded values are 0, ±2, and -1. The system is \(\frac{10\left(x^{2}-49\right)}{3 x\left(x^{2}-4\right)(x+1)}\) =0 and x≠0 and x≠-2 and x≠-1 and x≠2.

Question 2.
Rewrite each equation as a system of equations excluding the value(s) of x that lead to a denominator of zero. Then, solve the equation for x.
a. 25x=\(\frac{1}{x}\)
Answer:
System: 25x=\(\frac{1}{x}\) and x≠0; solution set: {±\(\frac{1}{5}\)}

b. \(\frac{1}{5x}\)=10
Answer:
System: \(\frac{1}{5}\)x=10 and x≠0; solution set: {\(\frac{1}{5}\)0}

c. \(\frac{x}{7-x}\)=2x
Answer:
System: \(\frac{x}{7-x}\)=2x and x≠7; solution set: {0,\(\frac{13}{2}\)}

d. \(\frac{2}{x}\)=\(\frac{5}{x+1}\)
Answer:
System: \(\frac{2}{x}\)=\(\frac{5}{x+1}\) and x≠-1 and x≠0; solution set: {\(\frac{2}{3}\)}

e. \(\frac{3+x}{3-x}\)=\(\frac{3+2x}{3-2x}\)
Answer:
System: \(\frac{3+x}{3-x}\)=\(\frac{3+2x}{3-2x}\) and x≠3/2 and x≠3; solution set: {0}

Question 3.
Ross wants to cut a 40-foot rope into two pieces so that the length of the first piece divided by the length of the second piece is 2.
a. Let x represent the length of the first piece. Write an equation that represents the relationship between the pieces as stated above.
Answer:
\(\frac{x}{40-x}\)=2

b. What values of x are not permissible in this equation? Describe within the context of the problem what situation is occurring if x were to equal this value(s). Rewrite as a system of equations to exclude the value(s).
Answer:
40 is not a permissible value because it would mean the rope is still intact. System: \(\frac{x}{40-x}\)=2 and x≠40

c. Solve the equation to obtain the lengths of the two pieces of rope. (Round to the nearest tenth if necessary.)
Answer:
First piece is \(\frac{80}{3}\)≈26.7 feet long; second piece is 40/3≈13.3 feet long.

Question 4.
Write an equation with the restrictions x≠14, x≠2, and x≠0.
Answer:
Answers will vary. Sample equation: \(\frac{1}{x}\)(x-2)(x-14) =0

Question 5.
Write an equation that has no solution.
Answer:
Answers will vary. Sample equation: \(\frac{1}{x}\)=0

Eureka Math Algebra 1 Module 1 Lesson 18 Exit Ticket Answer Key

Question 1.
Rewrite the equation \(\frac{x-2}{x-9}\)=2 as a system of equations. Then, solve for x.
Answer:
\(\frac{x-2}{x-9}\)=2 and x ≠ 9
x-2=2(x-9)
x-2=2x-18
16=x
{16}

Question 2.
Write an equation that would have the restriction x≠-3.
Answer:
Sample answer: \(\frac{4}{x+3}\)=2

Engage NY Eureka Math 5th Grade Module 5 Lesson 5 Answer Key

Eureka Math Grade 5 Module 5 Lesson 5 Problem Set Answer Key

Question 1.
Determine the volume of two boxes on the table using cubes, and then confirm by measuring and multiplying.

Box Number	Number of Cubes Packed	Measurements Length               Width              Height			Volume

Answer:

Question 2.
Using the same boxes from Problem 1, record the amount of liquid that your box can hold.

Box Number	Liquid the Box Can Hold
	mL
	mL

Answer:

Question 3.
Shade to show the water in the beaker.

Answer:

Question 4.
What conclusion can you draw about 1 cubic centimeter and 1 mL?
Answer:

When 1 cubic centimeter is added the water level raised by 1 millimetres.

Therefore, 1 cubic cm. is equal to 1 ml

Question 5.
The tank, shaped like a rectangular prism, is filled to the top with water.

Will the graduated cylinder hold all the water in the tank? If yes, how much more will the beaker hold? If no, how much more will the tank hold than the beaker? Explain how you know.
Answer:

No, The beaker cannot hold all the water in the tank

Explanation :

Given, the measurements of the tank are

length = 8 cm, width = 13 cm and height = 10 cm

Now, the volume of tank = length x width x height

V= 8 x 13 x 10

V = 1,040 cubic centimetres.

Given, the capacity of the beaker = 1 liter

1 liter = 1000 millilitres

1,040 – 1000 = 40

Therefore, the  tank can hold 40 ml more than the beaker.

Question 6.
A rectangular fish tank measures 26 cm by 20 cm by 18 cm. The tank is filled with water to a depth of 15 cm.

a. What is the volume of the water in mL?
b. How many liters is that?
c. How many more mL of water will be needed to fill the tank to the top? Explain how you know.
Answer:

a.

Given, the measurements of the fish tank  are 26 cm 20 cm and 18

Volume = length x breadth x height of water

V = 26 x 20 x 15

V = 7,800 cubic cm.

V = 7800 ml

Therefore, the volume of water = 7800 ml

b.

We know that, 1 litre = 1000 millilitres

Now, 7,800/ 100

= 7.8 litres

Therefore, the amount of water = 7.8 litres

c.

Volume = length x width x height

V = 26 x 20 x 3

V  = 1560 ml

Therefore, the remaining part = 26 x 20 x 3

Question 7.
A rectangular container is 25 cm long and 20 cm wide. If it holds 1 liter of water when full, what is its height?
Answer:

Volume =

25 x 20 = 500 sq. cm

The quantity of container = 1 litre

1 litre = 1000 millilitres

1000 / 500 = 2

Therefore, the height of the container = 2 cm

Eureka Math Grade 5 Module 5 Lesson 5 Exit Ticket Answer Key

a. Find the volume of the prism.
b. Shade the beaker to show how much liquid would fill the box.
Answer:

Volume = length x width x height

V= 15 x 5 x 3

V = 225 cubic centimetres.

Eureka Math Grade 5 Module 5 Lesson 5 Homework Answer Key

Question 1.
Johnny filled a container with 30 centimeter cubes. Shade the beaker to show how much water the container will hold. Explain how you know.

Answer:

Question 2.
A beaker contains 250 mL of water. Jack wants to pour the water into a container that will hold the water. Which of the containers pictured below could he use? Explain your choices.

Answer:

Volume of container A= 12 x 12 x 6 = 864 cubic. cm

Volume of container B = 20 sq. cm x 12 = 240 cubic. cm

Volume of container C = 25 cm x 5 cm x 2 cm = 250 cubic cm.

Volume of container E = 75 sq. cm x 3 = 225 cubic cm.

Therefore, both A and C can hold 250 ml of water.

Question 3.
On the back of this paper, describe the details of the activities you did in class today. Include what you learned about cubic centimeters and milliliters. Give an example of a problem you solved with an illustration.
Answer:

In today,s class, I learned that, cubic centimetres and millilitres are same

1 cubic cm. = 1 ml

 

Engage NY Eureka Math 3rd Grade Module 1 Lesson 4 Answer Key

Eureka Math Grade 3 Module 1 Answer Key

Eureka Math Grade 3 Module 1 Lesson 4 Sprint Answer Key

A
Repeated Addition as Multiplication

Answer:

Question 1.
5 + 5 + 5 =

Answer:
5 + 5 + 5 = 15,

Explanation:
Given 5 + 5 + 5 we add 5 by 3 times
we get 15, So 5 + 5 + 5 = 15.

Question 2.
3 × 5 =

Answer:
3 × 5 = 15,

Explanation:
Given 3 × 5 we multiply 3 with 5,
we get 15 as 3 × 5 = 15.

Question 3.
5 × 3 =

Answer:
5 × 3 = 15,

Explanation:
Given 5 × 3 we multiply 5 with 3,
we get 15 as 5 × 3 = 15.

Question 4.
2 + 2 + 2 =

Answer:
2 + 2 + 2 = 6,

Explanation:
Given 2 + 2 + 2 we add 2 by 3 times
we get 6, So 2 + 2 + 2 = 6.

Question 5.
3 × 2 =

Answer:
3 × 2 = 6,

Explanation:
Given 3 × 2 we multiply 3 with 2,
we get 6 as 3 × 2 = 6.

Question 6.
2 × 3 =

Answer:
2 × 3 = 6,

Explanation:
Given 2 × 3 we multiply 2 with 3,
we get 6 as 2 × 3 = 6.

Question 7.
5 + 5 =

Answer:
5 + 5 = 10,

Explanation:
Given 5 + 5 we add 5 by 2 times
we get 10, So 5 + 5 = 10.

Question 8.
2 × 5 =

Answer:
2 X 5 = 10,

Explanation:
Given 2 × 5 we multiply 2 with 5,
we get 10 as 2 × 5 = 10.

Question 9.
5 × 2 =

Answer:
5 × 2 = 10,

Explanation:
Given 5 × 2 we multiply 5 with 2,
we get 10 as 5 × 2 = 10.

Question 10.
2 + 2 + 2 + 2 =

Answer:
2 + 2 + 2 + 2 = 8,

Explanation:
Given 2 + 2 + 2 + 2 we add 2 by 4 times
we get 8, So 2 + 2 + 2 + 2= 8.

Question 11.
4 × 2 =

Answer:
4 × 2 = 8,

Explanation:
Given 4 × 2 we multiply 4 with 2,
we get 8 as 4 × 2 = 8.

Question 12.
2 × 4 =

Answer:
2 × 4 = 8,

Explanation:
Given 2 × 4 we multiply 2 with 4,
we get 8 as 2 × 4 = 8.

Question 13.
2 + 2 + 2 + 2 + 2 =

Answer:
2 + 2 + 2 + 2 + 2 = 10,

Explanation:
Given 2 + 2 + 2 + 2 + 2 we add 2 by 5 times
we get 10, So 2+ 2 + 2 + 2 + 2= 10.

Question 14.
5 × 2 =

Answer:
5 × 2 = 10,

Explanation:
Given 5 × 2 we multiply 5 with 2,
we get 10 as 5 × 2 = 10.

Question 15.
2 × 5 =

Answer:
2 × 5 = 10,

Explanation:
Given 2 × 5 we multiply 2 with 5,
we get 10 as 2 × 5 = 10.

Question 16.
3 + 3 =

Answer:
3 + 3 = 6,

Explanation:
Given 3 + 3 we add 3 by 2 times
we get 6, So 3 + 3 = 6.

Question 17.
2 × 3 =

Answer:
2 × 3 = 6,

Explanation:
Given 2 × 3 we multiply 2 with 3,
we get 6 as 2 × 3 = 6.

Question 18.
3 × 2 =

Answer:
3 × 2 = 6,

Explanation:
Given 3 × 2 we multiply 3 with 2,
we get 6 as 3 × 2 = 6.

Question 19.
5 + 5 + 5+ 5 =

Answer:
5 + 5 + 5 + 5 = 20,

Explanation:
Given 5 + 5 + 5 + 5 we add 5 by 4 times
we get 20, So 5 + 5 + 5 + 5 = 20.

Question 20.
4 × 5 =

Answer:
4 × 5 = 20,

Explanation:
Given 4 × 5 we multiply 4 with 5,
we get 20 as 4 × 5 = 20.

Question 21.
5 × 4 =

Answer:
5 × 4 = 20,

Explanation:
Given 5 × 4 we multiply 5 with 4,
we get 20 as 5 × 4 = 20.

Question 22.
2 × 2 =

Answer:
2 × 2 = 4,

Explanation:
Given 2 × 2 we multiply 2 with 2,
we get 4 as 2 × 2 = 4.

Question 23.
3 + 3 + 3 + 3 =

Answer:
3 + 3 + 3 + 3 = 12,

Explanation:
Given 3 + 3 + 3 + 3 we add 3 by 4 times
we get 12, So 3 + 3 + 3 + 3 = 12.

Question 24.
4 × 3 =

Answer:
4 × 3 = 12,

Explanation:
Given 4 × 3 we multiply 4 with 3,
we get 12 as 4 × 3 = 12.

Question 25.
3 × 4 =

Answer:
3 × 4 = 12,

Explanation:
Given 3 × 4 we multiply 3 with 4,
we get 12 as 3 × 4 = 12.

Question 26.
3 + 3 + 3 =

Answer:
3 + 3 + 3 = 9,

Explanation:
Given 3 + 3 + 3 we add 3 by 3 times
we get 9, So 3 + 3 + 3 = 9.

Question 27.
3 × 3 =

Answer:
3 × 3 = 9,

Explanation:
Given 3 × 3 we multiply 3 with 3,
we get 9 as 3 × 3 = 9.

Question 28.
3 + 3 + 3 + 3 + 3 =

Answer:
3 + 3 + 3 + 3 + 3 = 15,

Explanation:
Given 3 + 3 + 3 + 3 + 3 we add 3 by 5 times
we get 15, So 3 + 3 + 3 + 3 + 3 = 15.

Question 29.
5 × 3 =

Answer:
5 × 3 = 15,

Explanation:
Given 5 × 3 we multiply 5 with 3,
we get 15 as 5 × 3 = 15.

Question 30.
3 × 5 =

Answer:
3 × 5 = 15,

Explanation:
Given 3 × 5 we multiply 3 with 5,
we get 15 as 3 × 5 = 15.

Question 31.
7 + 7 =

Answer:
7 + 7 = 14,

Explanation:
Given 7 + 7 we add 7 by 2 times
we get 14, So 7 + 7 = 14.

Question 32.
2 × 7 =

Answer:
2 × 7 = 14,

Explanation:
Given 2 × 7 we multiply 2 with 7,
we get 14 as 2 × 7 = 14.

Question 33.
7 × 2 =

Answer:
7 × 2 = 14,

Explanation:
Given 7 × 2 we multiply 7 with 2,
we get 14 as 7 × 2 = 14.

Question 34.
9 + 9 =

Answer:
9 + 9 = 18,

Explanation:
Given 9 + 9 we add 9 by 2 times
we get 18, So 9 + 9 = 18.

Question 35.
2 × 9 =

Answer:
2 × 9 = 18,

Explanation:
Given 2 × 9 we multiply 2 with 9,
we get 18 as 2 × 9 = 18.

Question 36.
9 × 2 =

Answer:
9 × 2 = 18,

Explanation:
Given 9 × 2 we multiply 9 with 2,
we get 18 as 9 × 2 = 18.

Question 37.
6 + 6 =

Answer:
6 + 6 = 12,

Explanation:
Given 6 + 6 we add 6 by 2 times
we get 12, So 6 + 6 = 12.

Question 38.
6 × 2 =

Answer:
6 × 2 = 12,

Explanation:
Given 6 × 2 we multiply 6 with 2,
we get 12 as 6 × 2 = 12.

Question 39.
2 × 6 =

Answer:
2 × 6 = 12,

Explanation:
Given 2 × 6 we multiply 2 with 6,
we get 12 as 2 × 6 = 12.

Question 40.
8 + 8 =

Answer:
8 + 8 = 16,

Explanation:
Given 8 + 8 we add 8 by 2 times
we get 16, So 8 + 8 = 16.

Question 41.
2 × 8 =

Answer:
2 × 8 = 16,

Explanation:
Given 2 × 8 we multiply 2 with 8,
we get 16 as 2 × 8 = 16.

Question 42.
8 × 2 =

Answer:
8 × 2 = 16,

Explanation:
Given 8 × 2 we multiply 8 with 2,
we get 16 as 8 X 2 = 16.

Question 43.
7 + 7 + 7 + 7 =

Answer:
7 + 7 + 7 + 7 = 28,

Explanation:
Given 7 + 7 + 7 + 7 we add 7 by 4 times
we get 28, So 7 + 7 + 7 + 7 = 28.

Question 44.
4 × 7 =

Answer:
4 × 7 = 28,

Explanation:
Given 4 × 7 we multiply 4 with 7,
we get 28 as 4 × 7 = 28.

B
Repeated Addition as Multiplication

Answer:

Question 1.
2 + 2 + 2 =

Answer:
2 + 2 + 2 = 6,

Explanation:
Given 2 + 2 + 2 we add 2 by 3 times
we get 6, So 2 + 2 + 2 = 6.

Question 2.
3 × 2 =

Answer:
3 × 2 = 6,
Explanation:
Given 3 × 2 we multiply 3 with 2,
we get 6 as 3 × 2 = 6.

Question 3.
2 × 3 =

Answer:
2 × 3 = 6,
Explanation:
Given 2 × 3 we multiply 2 with 3,
we get 6 as 2 × 3 = 6.

Question 4.
5 + 5 + 5 =

Answer:
5 + 5 + 5 = 15,
Explanation:
Given 5 + 5 + 5 we add 5 by 3 times
we get 15, So 5 + 5 + 5 = 15.

Question 5.
3 × 5 =

Answer:
3 × 5 = 15,
Explanation:
Given 3 × 5 we multiply 3 with 5,
we get 15 as 3 × 5 = 15.

Question 6.
5 × 3 =

Answer:
5 × 3 = 15,
Explanation:
Given 5 × 3 we multiply 5 with 3,
we get 15 as 5 × 3 = 15.

Question 7.
2 + 2 + 2 + 2 =

Answer:
2 + 2 + 2 + 2 = 8,

Explanation:
Given 2 + 2 + 2 + 2 we add 2 by 4 times
we get 8, So 2 + 2 + 2 + 2= 8.

Question 8.
4 × 2 =

Answer:
4 × 2 = 8,
Explanation:
Given 4 × 2 we multiply 4 with 2,
we get 8 as 4 × 2 = 8.

Question 9.
2 × 4 =

Answer:
2 X 4 = 8,
Explanation:
Given 2 × 4 we multiply 2 with 4,
we get 8 as 2 × 4 = 8.

Question 10.
5 + 5 =

Answer:
5 + 5 = 10,
Explanation:
Given 5 + 5 we add 5 by 2 times
we get 10, So 5 + 5 = 10.

Question 11.
2 × 5 =

Answer:
2 × 5 = 10,
Explanation:
Given 2 × 5 we multiply 2 with 5,
we get 10 as 2 × 5 = 10.

Question 12.
5 × 2 =

Answer:
5 × 2 = 10,
Explanation:
Given 5 × 2 we multiply 5 with 2,
we get 10 as 5 × 2 = 10.

Question 13.
3 + 3 =

Answer:
3 + 3 = 6,

Explanation:
Given 3 + 3 we add 3 by 2 times
we get 6, So 3 + 3 = 6.

Question 14.
2 × 3 =

Answer:
2 X 3 = 6,

Explanation:
Given 2 × 3 we multiply 2 with 3,
we get 6 as 2 × 3 = 6.

Question 15.
3 × 2 =

Answer:
3 × 2 = 6,

Explanation:
Given 3 × 2 we multiply 3 with 2,
we get 6 as 3 × 2 = 6.

Question 16.
2 + 2 + 2 + 2 + 2 =

Answer:
2 + 2 + 2 + 2 + 2 = 10,
Explanation:
Given 2 + 2 + 2 + 2 + 2 we add 2 by 5 times
we get 10, So 2+ 2 + 2 + 2 + 2= 10.

Question 17.
5 × 2 =

Answer:
5 × 2 = 10,

Explanation:
Given 5 × 2 we multiply 5 with 2,
we get 10 as 5 × 2 = 10.

Question 18.
2 × 5 =

Answer:
2 × 5 = 10,
Explanation:
Given 2 × 5 we multiply 2 with 5,
we get 10 as 2 × 5 = 10.

Question 19.
5 + 5 + 5 + 5 =

Answer:
5 + 5 + 5 + 5 = 20,

Explanation:
Given 5 + 5 + 5 + 5 we add 5 by 4 times
we get 20, So 5 + 5 + 5 + 5 = 20.

Question 20.
4 × 5 =

Answer:
4 × 5 = 20,
Explanation:
Given 4 × 5 we multiply 4 with 5,
we get 20 as 4 × 5 = 20.

Question 21.
5 × 4 =

Answer:
5 × 4 = 20,

Explanation:
Given 5 × 4 we multiply 5 with 4,
we get 20 as 5 × 4 = 20.

Question 22.
2 × 2 =

Answer:
2 X 2 = 4,
Explanation:
Given 2 × 2 we multiply 2 with 2,
we get 4 as 2 × 2 = 4.

Question 23.
4 + 4 + 4 =

Answer:
4 + 4 + 4 = 12,

Explanation:
Given 4 + 4 + 4 we add 4 by 3 times
we get 12, So 4 + 4 + 4 = 12.

Question 24.
3 × 4 =

Answer:
3 × 4 = 12,

Explanation:
Given 3 × 4 we multiply 3 with 4,
we get 12 as 3 × 4 = 12.

Question 25.
4 × 3 =

Answer:
4 × 3 = 12,

Explanation:
Given 4 × 3 we multiply 4 with 3,
we get 12 as 4 × 3 = 12.

Question 26.
4 + 4 + 4 + 4 =

Answer:
4 + 4 + 4 + 4 = 16,

Explanation:
Given 4 + 4 + 4 + 4 we add 4 by 4 times
we get 16, So 4 + 4 + 4 +4 = 16.

Question 27.
4 × 4 =

Answer:
4 × 4 = 16,

Explanation:
Given 4 × 4 we multiply 4 with 4,
we get 14 as 4 × 4 = 16.

Question 28.
4 + 4 + 4 + 4 + 4 =

Answer:
4 + 4 + 4 + 4 + 4 = 20,

Explanation:
Given 4 + 4 + 4 + 4 + 4 we add 4 by 5 times
we get 20, So 4 + 4 + 4 + 4 + 4 = 20.

Question 29.
4 × 5 =

Answer:
4 × 5 = 20,

Explanation:
Given 4 × 5 we multiply 4 with 5,
we get 20 as 4 × 5 = 20.

Question 30.
5 × 4 =

Answer:
5 × 4 = 20,

Explanation:
Given 5 × 4 we multiply 5 with 4,
we get 20 as 5 × 4 = 20.

Question 31.
6 + 6 =

Answer:
6 + 6 = 12,

Explanation:
Given 6 + 6 we add 6 by 2 times
we get 12, So 6 + 6 = 12.

Question 32.
6 × 2 =

Answer:
6 × 2 = 12,

Explanation:
Given 6 × 2 we multiply 6 with 2,
we get 12 as 6 × 2 = 12.

Question 33.
2 × 6 =

Answer:
2 × 6 = 12,

Explanation:
Given 2 × 6 we multiply 2 with 6,
we get 12 as 2 × 6 = 12.

Question 34.
8 + 8 =

Answer:
8 + 8 = 16,

Explanation:
Given 8 + 8 we add 8 by 2 times
we get 16, So 8 + 8 = 16.

Question 35.
2 × 8 =

Answer:
2 × 8 = 16,

Explanation:
Given 2 × 8 we multiply 2 with 8,
we get 16 as 2 × 8 = 16.

Question 36.
8 × 2 =

Answer:
8 × 2 = 16,

Explanation:
Given 8 × 2 we multiply 8 with 2,
we get 16 as 8 × 2 = 16.

Question 37.
7 + 7 =

Answer:
7 + 7 = 14,

Explanation:
Given 7 + 7 we add 7 by 2 times
we get 14, So 7 + 7 = 14.

Question 38.
2 × 7 =

Answer:
2 × 7 = 14,

Explanation:
Given 2 × 7 we multiply 2 with 7,
we get 14 as 2 X 7 = 14.

Question 39.
7 × 2 =

Answer:
7 × 2 = 14,

Explanation:
Given 7 × 2 we multiply 7 with 2,
we get 14 as 7 × 2 = 14.

Question 40.
9 + 9 =

Answer:
9 + 9 = 18,

Explanation:
Given 9 + 9 we add 9 by 2 times
we get 18, So 9 + 9 = 18.

Question 41.
2 × 9 =

Answer:
2 × 9 = 18,

Explanation:
Given 2 × 9 we multiply 2 with 9,
we get 18 as 2 × 9 = 18.

Question 42.
9 × 2 =

Answer:
9 × 2 = 18,

Explanation:
Given 9 × 2 we multiply 9 with 2,
we get 18 as 9 × 2 = 18.

Question 43.
6 + 6 + 6 + 6 =

Answer:
6 + 6 + 6 + 6 = 24,

Explanation:
Given 6 + 6 + 6 + 6 we add 6 by 4 times
we get 24, So 6 + 6 + 6 + 6 = 24.

Question 44.
4 × 6 =

Answer:
4 × 6 = 24,

Explanation:
Given 4 × 6 we multiply 4 with 6,
we get 24 as 4 × 6 = 24.

Eureka Math Grade 3 Module 1 Lesson 4 Problem Set Answer Key

Question 1.

14 flowers are divided into 2 equal groups.
There are ____7_____ flowers in each group.

Answer:
There are 7 flowers in each group,

Explanation:
Given 14 flowers are divided into 2 equal groups,
So there are 14 ÷ 2 = 7 flowers in 2 equal groups.

Question 2.

28 books are divided into 4 equal groups.
There are _____7____ books in each group.

Answer:
There are 7 books in each group.

Explanation:
Given 28 books are divided into 4 equal groups,
So there are 28 ÷ 4 = 7 books in 4 equal groups.

Question 3.

30 apples are divided into ___3___ equal groups.
There are ____10_____ apples in each group.

Answer:
30 apples are divided into 3 equal groups.
There are 10 apples in each group.

Explanation:
Given in the picture there are30 apples divided into
3 equal groups. So there are 30 ÷ 3 = 10 apples
in each group.

Question 4.

___12____ cups are divided into ___2____ equal groups.
There are ____6_____ cups in each group.
12 ÷ 2 = ___6______

Answer:
12 cups are divided into 2 equal groups.
There are 6 cups in each group. 12 ÷ 2 = 6 cups,

Explanation:
As given in the picture there are 12 cups divided
into 2 equal groups, There are 6 cups  in each group
as 12 ÷ 2 = 6 cups.

Question 5.

There are ____15_____ toys in each group.
15 ÷ 3 = ____5_____

There are ____15_____ toys in each group,

Explanation:
As given in the picture there are 15 toys divided
as 15 ÷ 3 = 5 toys in each group, So, there are
15 toys in 3 equal groups.

Question 6.

9 ÷ 3 = ____3______

Answer:
There are 3 cars in each group,

Explanation:
As given in the picture there are 9 cars divided as 9 ÷ 3 = 3 cars in each group.
So, there are 3 cars in 3 equal groups.

Question 7.
Audrina has 24 colored pencils. She puts them in 4 equal groups. How many colored pencils are in each group?

There are ___6____ colored pencils in each group.
24 ÷ 4 = ___6____

Answer:
There are 6 colored pencils in each group,

Explanation:
Given Audrina has 24 colored pencils. She puts them in 4 equal groups. So number of  colored pencils in each group are 24 ÷ 4 = 6 pencils in 4 equal groups.

Question 8.
Charlie picks 20 apples. He divides them equally between 5 baskets. Draw the apples in each basket.

There are _____4______ apples in each basket.
20 ÷ ____5____ = ____4______

Answer:
There are 4 apples in each basket,

Explanation:
Given Charlie picks 20 apples. He divides them equally between 5 baskets. Drawn the apples in each basket as 20 ÷ 5 = 4 apples in 5 equal groups.

Question 9.
Chelsea collects butterfly stickers. The picture shows how she placed them in her book. Write a division sentence to show how she equally grouped her stickers.
There are ______3______ butterflies in each row.
____15______ ÷ _____5_____ = ____3______

Answer:

Division sentence : 15 ÷ 5 = 3, Chelsea equally grouped 3 butterflies in her stickers.

Explanation:
Given Chelsea collects butterfly stickers.
The picture is showing 15 butterflies she placed them in her book. Wrote a division sentence as 15 ÷ 5 = 3 butterflies to show how she equally grouped 3 butterflies in her stickers.

Eureka Math Grade 3 Module 1 Lesson 4 Exit Ticket Answer Key

Question 1.
There are 16 glue sticks for the class. The teacher divides them into 4 equal groups. Draw the number of glue sticks in each group.

There are _____16______ glue sticks in each group.
16 ÷ ___4_____ = ____4______

Answer:
There are 16 glue sticks in each group.

Explanation:
Given there are 16 glue sticks for the class. The teacher divides them into 4 equal groups.
Drawn the number of glue sticks in each group as 16 ÷ 4 =  4 glue sticks in 4 equal groups.

Question 2.
Draw a picture to show 15 ÷ 3. Then, fill in the blank to make a true division sentence.
15 ÷ 3 = ____5______

Drawn a picture to show division sentence as
15 ÷ 3 = 5,
Filled in the blank to make a true division sentence as
15 ÷ 3 = ____5____,

Explanation:
Drawn 15 dogs and wrote division sentence as 15 ÷ 3 = 5 as shown above and filled in the blank to make a true division sentence as 15 ÷ 3 = ____5__ or 5 × 3 = 15.

Eureka Math Grade 3 Module 1 Lesson 4 Homework Answer Key

Question 1.

12 chairs are divided into 2 equal groups.
There are ____6_____ chairs in each group.

Answer:
There are 6 chairs in each group,

Explanation:
As given in the picture there are 12 chairs divided as 12 ÷ 2 = 6 chairs in each group, So, there are 6 chairs in 2 equal groups.

Question 2.

21 triangles are divided into 3 equal groups.
There are ____7_____ triangles in each group.

Answer:
There are 7 triangles in each group,

Explanation:
As given in the picture there are 21 triangles divided as 21 ÷ 3 =7 triangles in each group, So, there are 7 triangles in 3 equal groups.

Question 3.

25 erasers are divided into ___5___ equal groups.
There are ____5_____ erasers in each group.

Answer:
25 erasers are divided into 5 equal groups.
There are 5 erasers in each group as 25 ÷ 5 = 5 erasers,

Explanation:
As given in the picture there are 25 erasers divided into 5 equal groups.
There are 5 erasers in each group as 25 ÷ 5 = 25 erasers.

Question 4.

___9____ chickens are divided into ___3____ equal groups.
There are ____3_____ chickens in each group.
9 ÷ 3 = ____3______

Answer:
9 chickens are divided into 3 equal groups.
There are 3 chickens in each group as 9 ÷ 3 = 3 chickens,

Explanation:
As given in the picture there are 9 chickens divided into 3 equal groups, There are 3 chickens in each group as 9 ÷ 3 = 3 chickens.

Question 5.

There are ____3_____ buckets in each group.
12 ÷ 4 = ____3____

Answer:
12 buckets are divided into 4 equal groups.
There are 3 buckets in each group as 12 ÷ 4 = 3 buckets,

Explanation:
As given in the picture there are 12 buckets divided into 4 equal groups, There are 3 buckets in each group as 12 ÷ 4 = 3 buckets.

Question 6.

16 ÷ 4 = _4_

Answer:
16 bricks are divided into 4 equal groups.
There are 4 bricks in each group as 16 ÷ 4 = 4 bricks,

Explanation:
As given in the picture there are 16 bricks divided into 4 equal groups, There are 4 bricks in each group as 16 ÷ 4 = 4 bricks.

Question 7.
Andrew has 21 keys. He puts them in 3 equal groups.
How many keys are in each group?

There are ___7____ keys in each group.
21 ÷ 3 = ____7______

Answer:
21 keys are divided into 3 equal groups.
There are 7 keys in each group.

Explanation:
Given Andrew has 21 keys. He puts them in 3 equal groups. So, number of keys in each group are 7 keys as 21 ÷ 3 = 7 keys in each group.

Question 8.
Mr. Doyle has 20 pencils. He divides them equally between 4 tables. Draw the pencils on each table.

There are _____5_____ pencils on each table.
20 ÷ ___4_____ = ____5______

Answer:
There are 5 pencils on each table,

Explanation:
Given Mr. Doyle has 20 pencils and he divides them equally between 4 tables. Drawn the pencils on each table as 20 ÷ 4= 5 pencils on each table as shown above.

Question 9.
Jenna has markers. The picture shows how she placed them on her desk. Write a division sentence to represent how she equally grouped her markers.
There are ______4______ markers in each row.

____20______ ÷ ___5_______ = ____4____

Answer:
Division sentence to represent Jenna is 20 ÷ 5 = 4 markers equally grouped in each row,

Explanation:
Given Jenna has markers and the picture shows how she placed them on her desk. Wrote a division sentence 20 ÷ 5 = 4 markers to represent how Jenna equally grouped 4 markers in each row.

Eureka Math Grade 3 Module 7 Answer Key provided lays a deeper understanding of concepts. Practice using the Engage NY Math 3rd Grade Module 7 Solutions and develop the Knowledge that builds upon itself throughout the learning process. Avail the Problem-Solving Methods used in Eureka Math Grade 3 Answer Key Module 7 and solve related problems in no time. Detailed Solutions provided in the 3rd Grade Engage NY Module 7 Answer Key make it easy for you to retain the concepts for a long time.

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EngageNY Math Grade 3 Module 7 Answer Key | Eureka Math 3rd Grade Module 7 Answer Key

Engage NY Math Grade 3 module 7 Geometry and Measurement Answer Key provides a flexible, knowledge-building curriculum along with any time learning environment. Third Grade Eureka Math Module 7 Answer Key has questions from Lessons, Extra Practice, Mid Module, End Module Assessments, Review Tests, Assessments, etc. You can access the Topicwise Engage NY Eureka Math 3rd Grade Module 7 Solutions via the direct links available. Just tap on them and learn the underlying concepts within them efficiently and score better grades in exams.

Eureka Math Grade 3 Module 7 Geometry and Measurement Word Problems

Eureka Math Grade 3 Module 7 Topic A Solving Word Problems

• Eureka Math Grade 3 Module 7 Lesson 1 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 2 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 3 Answer Key

Eureka Math 3rd Grade Module 7 Topic B Attributes of Two-Dimensional Figures

• Eureka Math Grade 3 Module 7 Lesson 4 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 5 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 6 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 7 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 8 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 9 Answer Key

Engage NY Math 3rd Grade Module 7 Topic C Problem Solving with Perimeter

• Eureka Math Grade 3 Module 7 Lesson 10 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 11 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 12 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 13 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 14 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 15 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 16 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 17 Answer Key

Eureka Math Grade 3 Module 7 Mid Module Assessment Answer Key

EngageNY Math Grade 3 Module 7 Topic D Recording Perimeter and Area Data on Line Plots

• Eureka Math Grade 3 Module 7 Lesson 18 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 19 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 20 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 21 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 22 Answer Key

3rd Grade Eureka Math Module 7 Topic E Problem Solving with Perimeter and Area

• Eureka Math Grade 3 Module 7 Lesson 23 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 24 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 25 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 26 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 27 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 28 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 29 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 30 Answer Key

Eureka Math Grade 3 Module 7 End of Module Assessment Answer Key

Engage NY Grade 3 Module 7 Topic F Year in Review

• Eureka Math Grade 3 Module 7 Lesson 31 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 32 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 33 Answer Key
• Eureka Math Grade 3 Module 7 Lesson 34 Answer Key

Wrapping Up

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Engage NY Eureka Math 1st Grade Module 2 Lesson 9 Answer Key

Eureka Math Grade 1 Module 2 Lesson 9 Problem Set Answer Key

Make ten to solve. Use a number bond to show how you took 2 out to make ten.
Ben has 8 green grapes and 3 purple grapes. How many grapes does he have?

Ben has ___ grapes.
Answer:

Ben has 11 grapes.
Explanation:
Ben has 8 green grapes and 3 purple grapes. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe three purple grapes can be divided into two parts as one and two. ADD two purple grapes with eight green grapes in order to get ten. ADD ten grapes with one purple grape then we got eleven grapes. Ben has eleven grapes.

Question 2.

Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe four can be divided into two parts as two and two. ADD two with eight in order to get ten. ADD ten with two then we got twelve.

Use number bonds to show your thinking. Write the 10+ fact.

Question 3.
8 + 5 = ____                                                     ____ + ____ = ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe five can be divided into two parts as two and three. ADD two with eight in order to get ten. ADD ten with three then we got thirteen.

Question 4.
8 + 7 = ____                                                     ____ + ____ = ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe seven can be divided into two parts as two and five. ADD two with eight in order to get ten. ADD ten with five then we got fifteen.

Question 5.
4 + 8 = ____                                                     ____ + ____ = ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe four can be divided into two parts as two and two. ADD two with eight in order to get ten. ADD ten with two then we got twelve.

Question 6.
7 + 8 = ____                                                     ____ + ____ = ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe seven can be divided into two parts as five and two. ADD two with eight in order to get ten. ADD ten with five then we got fifteen.

Question 7.
8 + ____ = 17                                                   ____ + ____ = ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe nine can be divided into two parts as two and seven. ADD two with eight in order to get ten. ADD ten with seven then we got seventeen.

Complete the addition sentences and number bonds.

Question 8.
a. 10 + 1 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD ten with one which results eleven. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of ten and one which results eleven.

b. 8 + 3 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe three can be divided into two parts as two and one. ADD eight with two which results ten. ADD ten with one which results eleven. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of eight and three which results eleven.

Question 9.
a. 10 + 5 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD ten with five which results fifteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of ten and five which results fifteen.

b. 8 + 7 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe seven can be divided into two parts as two and five. ADD eight with two which results ten. ADD ten with five which results fifteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of eight and seven which results fifteen.

Question 10.
a. 10 + 6 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD ten with six which results sixteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of ten and six which results sixteen.

b. 8 + 8 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe eight can be divided into two parts as two and six. ADD eight with two which results ten. ADD ten with six which results sixteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of eight and eight which results sixteen.

Question 11.
a. 2 + 10 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD two with ten which results twelve. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of two and ten which results twelve.

b. 4 + 8 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe four can be divided into two parts as two and two. ADD two with eight which results ten. ADD ten with two which results twelve. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of four and eight which results twelve.

Question 12.
a. 4 + 10 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD four with ten which results fourteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of four and ten which results fourteen.

b. 6 + 8 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe six can be divided into two parts as four and two. ADD two with eight which results ten. ADD ten with four which results fourteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of six and eight which results fourteen.

Eureka Math Grade 1 Module 2 Lesson 9 Exit Ticket Answer Key

Question 1.
Seyla has 3 stamps in her collection. Her father gives her 8 more stamps. How many stamps does she have now? Show how you make ten, and write the 10+ fact.

Answer:

Explanation:
Seyla has 3 stamps in her collection. Her father gives her 8 more stamps. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe three can be divided into two parts as one and two. ADD two with eight which results ten. ADD ten with one then we got eleven. Seyla have eleven stamps.

Question 2.
Complete the addition sentences and the number bonds.

a. 8 + 6 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe six can be divided into two parts as two and four. ADD eight with two which results ten. ADD ten with four which results fourteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of eight and six which results fourteen.

b. 10 + ___ = 14
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD ten with four which results fourteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of ten and four which results fourteen.

Eureka Math Grade 1 Module 2 Lesson 9 Homework Answer Key

Use number bonds to show your thinking. Write the 10+ fact.

Question 1.
8 + 3 = ____ 10 + ____ = ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe three can be divided into two parts as two and one. ADD eight with two which results ten. ADD ten with one which results eleven.

Question 2.
6 + 8 = ____ ____ + 10 = ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe six can be divided into two parts as four and two. ADD eight with two which results ten. ADD four with ten which results fourteen.

Question 3.
____ = 8 + 8 ____ = 10 + ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe eight can be divided into two parts as two and six. ADD eight with two which results ten. ADD ten with six which results sixteen.

Question 4.
____ = 5 + 8 ____ = 10 + ____
Answer:

Explanation:
A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe five can be divided into two parts as three and two. ADD two with eight which results ten. ADD ten with three which results thirteen.

Complete the addition sentences and the number bonds.

Question 5.
a. 7 + 8 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe seven can be divided into two parts as five and two. ADD eight with two which results ten. ADD ten with five which results fifteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of seven and eight which results fifteen.

b. 10 + 5 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD ten with five which results fifteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of ten and five which results fifteen.

Question 6.
a. 16 = ___ + 8
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe eight can be divided into two parts as six and two. ADD eight with two which results ten. ADD ten with six which results sixteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of eight and eight which results sixteen.

b. 10 + 6 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD ten with six which results sixteen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of ten and six which results sixteen.

Question 7.
a. ___ = 9 + 8
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. In the above image we can observe nine can be divided into two parts as seven and two. ADD eight with two which results ten. ADD ten with seven which results seventeen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of nine and eight which results seventeen.

b. 10 + 7 = ___
Answer:

Explanation:
An addition sentence is a mathematical expression that shows two or more values added together. ADD ten with seven which results seventeen. A number bond is a simple addition of two numbers that add up to give the sum. In the above image we can observe a number bond of ten and seven which results seventeen.

Draw a line to the matching number sentence. You may use a number bond or 5-group drawing to help you

Answer:

Explanation:
8. In the above image we can observe a number sentence 11 = 8 + 3. A number bond is a simple addition of two numbers that add up to give the sum. In the above number sentence three can be divided into two parts as two and one. ADD eight with two which results ten. ADD ten with one then we got eleven. The number sentence 11= 8 + 3 is equal to the 10 + 1 = 11 number sentence.
9. Lisa had 5 red rocks and 8 white rocks. A number bond is a simple addition of two numbers that add up to give the sum. A number sentence 5 + 8 = 13. In this five can be divided into two parts as three and two. ADD two with eight results ten. ADD ten with three results thirteen. The number sentence 5 + 8 = 13 is equal to 13 = 10 + 3 number sentence.
10. A number bond is a simple addition of two numbers that add up to give the sum. In the above number bond ten and four results fourteen. A number sentence 8 + 6 = 14. In this six can be divided into two parts as two and four. ADD two with eight results ten. ADD ten with four results fourteen. The number bond ten and four equal to the 8 + 6 = 14 number sentence.

Engage NY Eureka Math 8th Grade Module 1 Lesson 8 Answer Key

Eureka Math Grade 8 Module 1 Lesson 8 Example Answer Key

Example 1.
In 1723, the population of New York City was approximately 7,248. By 1870, almost 150 years later, the population had grown to 942,292. We want to determine approximately how many times greater the population was in 1870 compared to 1723.
The word approximately in the question lets us know that we do not need to find a precise answer, so we approximate both populations as powers of 10.
→ Population in 1723: 7248<9999<10000=10⁴
→ Population in 1870: 942 292<999 999<1 000 000=10⁶
We want to compare the population in 1870 to the population in 1723:
\(\frac{10^{6}}{10^{4}}\)
Now we can use what we know about the laws of exponents to simplify the expression and answer the question:
\(\frac{10^{6}}{10^{4}}\) =10².
Therefore, there were approximately 100 times more people in New York City in 1870 compared to 1723.

Example 2.
Let’s compare the population of New York City to the population of New York State. Specifically, let’s find out how many times greater the population of New York State is compared to that of New York City.
The population of New York City is 8,336,697. Let’s round this number to the nearest million; this gives us 8,000,000. Written as single-digit integer times a power of 10:
8 000 000=8×10⁶.
The population of New York State is 19,570,261. Rounding to the nearest million gives us 20,000,000. Written as a single-digit integer times a power of 10:
20 000 000=2×10⁷.
To estimate the difference in size we compare state population to city population:
\(\frac{2 \times 10^{7}}{8 \times 10^{6}}\)
Now we simplify the expression to find the answer:
\(\frac{2 \times 10^{7}}{8 \times 10^{6}}\) = \(\frac{2}{8}\) × \(\frac{10^{7}}{10^{6}}\)By the product formula
= \(\frac{1}{4}\)×10 By equivalent fractions and the first law of exponents
=0.25×10
=2.5
Therefore, the population of the state is 2.5 times that of the city.

Example 3.
There are about 9 billion devices connected to the Internet. If a wireless router can support 300 devices, about how many wireless routers are necessary to connect all 9 billion devices wirelessly?
Because 9 billion is a very large number, we should express it as a single-digit integer times a power of 10.
9 000 000 000=9×10⁹
The laws of exponents tells us that our calculations will be easier if we also express 300 as a single-digit integer times a power of 10, even though 300 is much smaller.
300=3×10²
We want to know how many wireless routers are necessary to support 9 billion devices, so we must divide
\(\frac{9 \times 10^{9}}{3 \times 10^{2}}\)
Now, we can simplify the expression to find the answer:
\(\frac{9 \times 10^{9}}{3 \times 10^{2}}\) = \(\frac{9}{3}\) × \(\frac{10^{9}}{10^{2}}\) By the product formula
=3×10⁷ By equivalent fractions and the first law of exponents
=30 000 000
About 30 million routers are necessary to connect all devices wirelessly.

Example 4.
The average American household spends about $40,000 each year. If there are about 1×〖10〗⁸ households, what is the total amount of money spent by American households in one year?
Let’s express $40,000 as a single-digit integer times a power of 10.
40000=4×〖10〗⁴
The question asks us how much money all American households spend in one year, which means that we need to multiply the amount spent by one household by the total number of households:
(4×10⁴ )(1×10⁸ )=(4×1)(10⁴×10⁸ ) By repeated use of associative and commutative properties
=4×10¹² By the first law of exponents
Therefore, American households spend about $4,000,000,000,000 each year altogether!

Eureka Math Grade 8 Module 1 Lesson 8 Exercise Answer Key

Exercise 1.
The Federal Reserve states that the average household in January of 2013 had $7,122 in credit card debt. About how many times greater is the U.S. national debt, which is $16,755,133,009,522? Rewrite each number to the nearest power of 10 that exceeds it, and then compare.
Answer:
Household debt=7122<9999<10000=10⁴.
U.S.debt =16 755 133 009 522<99 999 999 999 999<100 000 000 000 000=10¹⁴.
\(\frac{10^{14}}{10^{4}}\) = 10^14-4=10¹⁰. The U.S. national debt is 10¹⁰ times greater than the average household’s credit card debt.

Exercise 2.
There are about 3,000,000 students attending school, kindergarten through Grade 12, in New York. Express the number of students as a single-digit integer times a power of 10.
Answer:
3 000 000=3×10⁶

The average number of students attending a middle school in New York is 8×〖10〗². How many times greater is the overall number of K–12 students compared to the average number of middle school students?
Answer:
\(\frac{3 \times 10^{6}}{8 \times 10^{2}}\)= \(\frac{3}{8}\)×\(\frac{10^{6}}{10^{2}}\)
= \(\frac{3}{8}\)×\(\frac{10^{6}}{10^{2}}\)
=0.375×〖10〗⁴
=3750
There are about 3,750 times more students in K–12 compared to the number of students in middle school.

Exercise 3.
A conservative estimate of the number of stars in the universe is 6×10²². The average human can see about 3,000 stars at night with his naked eye. About how many times more stars are there in the universe compared to the stars a human can actually see?

Answer:
\(\frac{6 \times 10^{22}}{3 \times 10^{3}}\) = \(\frac{6}{3}\) ×\(\frac{10^{22}}{10^{3}}\)= 2×10^22-3 = 2×10¹⁹
There are about 2×10¹⁹ times more stars in the universe compared to the number we can actually see.

Exercise 4.
The estimated world population in 2011 was 7×10⁹. Of the total population, 682 million of those people were left-handed. Approximately what percentage of the world population is left-handed according to the 2011 estimation?
Answer:
682 000 000≈700 000 000=7×10⁸
\(\frac{7 \times 10^{8}}{7 \times 10^{9}}\)=\(\frac{7}{7}\)×\(\frac{10^{8}}{10^{9}}\)
= 1×\(\frac{1}{10}\)
=\(\frac{1}{10}\)
About one-tenth of the population is left-handed, which is equal to 10%.

Exercise 5
The average person takes about 30,000 breaths per day. Express this number as a single-digit integer times a power of 10.
Answer:
30000=3×10⁴

If the average American lives about 80 years (or about 30,000 days), how many total breaths will a person take in her lifetime?
Answer:
(3×10⁴ )×(3×10⁴ )=9×10⁸
The average American takes about 900,000,000 breaths in a lifetime.

Eureka Math Grade 8 Module 1 Lesson 8 Problem Set Answer Key

Students practice estimating size of quantities and performing operations on numbers written in the form of a single-digit integer times a power of 10.

Question 1.
The Atlantic Ocean region contains approximately 2×10¹⁶ gallons of water. Lake Ontario has approximately 8,000,000,000,000 gallons of water. How many Lake Ontarios would it take to fill the Atlantic Ocean region in terms of gallons of water?
Answer:
8 000 000 000 000=8×10¹²
\(\frac{2 \times 10^{16}}{8 \times 10^{12}}\)=\(\frac{2}{8}\)×\(\frac{10^{16}}{10^{12}}\)
=\(\frac{1}{4}\)×10⁴
=0.25×10⁴
=2500
2,500 Lake Ontario’s would be needed to fill the Atlantic Ocean region.

Question 2.
U.S. national forests cover approximately 300,000 square miles. Conservationists want the total square footage of forests to be 300,000〗² square miles. When Ivanna used her phone to do the calculation, her screen showed the following:

a. What does the answer on her screen mean? Explain how you know.
Answer:
The answer means 9×10¹⁰. This is because:
(300 000)²=(3×10⁵)²
=3²×(10⁵ )²
=9×10¹⁰

b. Given that the U.S. has approximately 4 million square miles of land, is this a reasonable goal for conservationists? Explain.
Answer:
4 000 000=4×10⁶. It is unreasonable for conservationists to think the current square mileage of forests could increase that much because that number is greater than the number that represents the total number of square miles in the U.S,
9×10¹⁰>4×10⁶.

Question 3.
The average American is responsible for about 20,000 kilograms of carbon emission pollution each year. Express this number as a single-digit integer times a power of 10.
Answer:
20 000=2×10⁴

Question 4.
The United Kingdom is responsible for about 1× 10⁴ kilograms of carbon emission pollution each year. Which country is responsible for greater carbon emission pollution each year? By how much?
Answer:
2× 10⁴>1×10⁴
America is responsible for greater carbon emission pollution each year. America produces twice the amount of the U.K. pollution.

Eureka Math Grade 8 Module 1 Lesson 8 Exit Ticket Answer Key

Most English-speaking countries use the short-scale naming system, in which a trillion is expressed as 1,000,000,000,000. Some other countries use the long-scale naming system, in which a trillion is expressed as 1,000,000,000,000,000,000,000. Express each number as a single-digit integer times a power of ten. How many times greater is the long-scale naming system than the short-scale?
Answer:
1 000 000 000 000=10¹²
1 000 000 000 000 000 000 000= 10²¹
\(\frac{10^{21}}{10^{12}}\) = 10⁹. The long-scale is about 10⁹ times greater than the short-scale.

Eureka Math Grade 8 Module 1 Lesson 8 Sprint Answer Key

Applying Properties of Exponents to Generate Equivalent Expressions—Round 1
Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. When appropriate, express answers without parentheses or as equal to 1. All letters denote numbers.

Answer:

Applying Properties of Exponents to Generate Equivalent Expressions—Round 2

Directions: Simplify each expression using the laws of exponents. Use the least number of bases possible and only positive exponents. When appropriate, express answers without parentheses or as equal to 1. All letters denote numbers.

Answer:

Engage NY Eureka Math 5th Grade Module 2 Lesson 23 Answer Key

Eureka Math Grade 5 Module 2 Lesson 23 Problem Set Answer Key

Question 1.
Divide. Then, check using multiplication.

a. 4,859 ÷ 23

Answer:
4859/23 = 6.

Explanation:
In the above-given question,
given that,
Divide, and then check.
23 x 2 = 46.
48 – 46 = 2.
23 x 1 = 23.
29-23 = 6.

b. 4,368 ÷ 52

Answer:
4368/52 = 0.

Explanation:
In the above-given question,
given that,
Divide, and then check.
52 x 8 = 416.
436 – 416 = 20.
52 x 4 = 208.
208-208 = 0.

c. 7,242 ÷ 34

Answer:
7242/34 = 0.

Explanation:
In the above-given question,
given that,
Divide, and then check.
34 x 2 = 68.
72 – 68 = 4.
34 x 1 = 34.
102-102 = 0.

d. 3,164 ÷ 45

Answer:
3164/45 = 5.

Explanation:
In the above-given question,
given that,
Divide, and then check.
45 x 7 = 315.
316 – 315 = 1.
45 x 3 = 140.
140-135 = 5.

e. 9,152 ÷ 29

Answer:
9152/29 = 17.

Explanation:
In the above-given question,
given that,
Divide, and then check.
29 x 2 = 87.
91 – 87 = 4.
29 x 5 = 145.
162-145 = 17.

f. 4,424 ÷ 63

Answer:
4424/63 = 22.

Explanation:
In the above-given question,
given that,
Divide, and then check.
63 x 5 = 315.
442 – 315 = 27.
63 x 4 = 279.
279-252 = 22.

Question 2.
Mr. Riley baked 1,692 chocolate cookies. He sold them in boxes of 36 cookies each. How much money did he collect if he sold them all at $8 per box?

Answer:
The money did she collect = $5.9.

Explanation:
In the above-given question,
given that,
Mr. Riley baked 1692 chocolate cookies.
he sold them in boxes of 36 cookies each.
36x 8 = 288.
1692/288.
5.9$.

Question 3.
1,092 flowers are arranged into 26 vases, with the same number of flowers in each vase. How many flowers would be needed to fill 130 such vases?

Answer:
The number of flowers needed to fill 130 such vases = 8 flowers.

Explanation:
In the above-given question,
given that,
1092 flowers are arranged into 26 vases,
with the same number of flowers in each vase.
1092/26 = 42.
1092/130 = 8.4.

Question 4.
The elephant’s water tank holds 2,560 gallons of water. After two weeks, the zookeeper measures and finds that the tank has 1,944 gallons of water left. If the elephant drinks the same amount of water each day, how many days will a full tank of water last?

Answer:
The number of days will a full tank of water last = 4.15 gallons.

Explanation:
In the above-given question,
given that,
The elephant’s water tank holds 2560 gallons of water.
after two weeks, the zookeeper measures and finds that the tank has 1944 gallons of water left.
if the elephant drinks the same amount of water each day.
2560 – 1944.
616.
2560/616.
4.15 gallons.

Eureka Math Grade 5 Module 2 Lesson 23 Exit Ticket Answer Key

Divide. Then, check using multiplication.
a. 8,283 ÷ 19

Answer:
8283/41 = 1.

Explanation:
In the above-given question,
given that,
Divide, and then check.
41 x 2 = 82.
82 – 82 = 0.
41 x 2 = 82.
83-82 = 1.

b. 1,056 ÷ 37

Answer:
1056/37 = 20.

Explanation:
In the above-given question,
given that,
Divide, and then check.
37 x 2 = 74.
105 – 74 =31.
37 x 8 = 296.
316-296 = 20.

Eureka Math Grade 5 Module 2 Lesson 23 Homework Answer Key

Question 1.
Divide. Then, check using multiplication.
a. 9,962 ÷ 41

Answer:
9962/41 = 40.

Explanation:
In the above-given question,
given that,
Divide, and then check.
41 x 2 = 82.
99 – 82 = 17.
41 x 2 = 122.
122-82 = 40.

b. 1,495 ÷ 45

Answer:
1495/45 = 10.

Explanation:
In the above-given question,
given that,
Divide, and then check.
45 x 3 = 135.
149 – 135 = 14.
45 x 3 = 135.
145-135 = 10.

c. 6,691 ÷ 28

Answer:
6691/28 = 27.

Explanation:
In the above-given question,
given that,
Divide, and then check.
28 x 2 = 56.
66 – 56 = 10.
23 x 8 = 251.
251-224 = 27.

d. 2,625 ÷ 32

Answer:
2625/32 = 33.

Explanation:
In the above-given question,
given that,
Divide, and then check.
32 x 8 = 256.
262 – 256 = 6.
32 x 1 = 32.
65-32 = 33.

e. 2,409 ÷ 19

Answer:
2409/19 = 15.

Explanation:
In the above-given question,
given that,
Divide, and then check.
19 x 1 = 19.
24 – 19 = 5.
19 x 6 = 126.
126-114 = 15.

f. 5,821 ÷ 62

Answer:
5821/62 = 55.

Explanation:
In the above-given question,
given that,
Divide, and then check.
62 x 9 = 558.
582 – 558 = 24.
62 x 3 = 241.
241-186 = 55.

Question 2.
A political gathering in South America was attended by 7,910 people. Each of South America’s 14 countries was equally represented. How many representatives attended from each country?

Answer:
The representatives attend from each country = 565.

Explanation:
In the above-given question,
given that,
A political gathering in south America was attended by 7910 people.
each of south America’s 14 countries was equally represented.
7910/14 = 565.

Question 3.
A candy company packages caramel into containers that hold 32 fluid ounces. In the last batch, 1,848 fluid ounces of caramel were made. How many containers were needed for this batch?

Answer:
The containers were needed for this batch = 58 containers.

Explanation:
In the above-given question,
given that,
A candy company packages caramel into containers that hold 32 fluid ounces.
In the last batch, 1,848 fluid ounces of caramel were made.
1848/32.
57.75.
58.

Engage NY Eureka Math Algebra 1 Module 1 Lesson 21 Answer Key

Eureka Math Algebra 1 Module 1 Lesson 21 Example Answer Key

Example
The solution to x+y=20 is shown on the graph below.

a. Graph the solution to x+y≤20.
Answer:

b. Graph the solution to x+y≥20.
Answer:

c. Graph the solution to x+y<20.
Answer:

d. Graph the solution to x+y>20.
Answer:

Exercises 3–5
Students will need graph paper for this portion of the lesson. Have students work individually to complete as much of Exercise 3 as they can in 8 minutes, reserving the final 7 minutes for comparing with a neighbor and debating any conflicting answers. Alternatively, differentiate by assigning only a subset of the problems most appropriate for each student or group of students. In any case, make the assignments in pairs so that students have someone with whom to compare answers. Students may struggle as they work on parts (f)–(j), graphing solutions to equations like y=5; allow the students to struggle and discuss with each other. (Exercise 4 will revisit this idea with the entire class.) Students will rely on their experiences in Grade 8 as well as their explorations in Lessons 1–5 of this module to distinguish between the linear and nonlinear inequalities and answer the question that concludes Exercise 3.

Allow students to debate and discuss. Guide them to the correct conclusion, and then review the definition of a half-plane that follows Exercise 3, clarifying for students that a strict inequality does not include the or equal to option. It must be either strictly less than or greater than.

Exercises 3–5

Exercise 3.
Using a separate sheet of graph paper, plot the solution sets to the following equations and inequalities:
a. x-y=10
b. x-y<10 c. y>x-10
d. y≥x
e. x≥y
f. y=5
g. y<5 h. x≥5 i. y≠1 j. x=0 k. x>0
l. y<0
m. x² -y=0
n. x² +y² >0
o. xy≤0
Answer:

Which of the inequalities in this exercise are linear inequalities?
Answer:
Parts (a)–(l) are linear. Parts (m)–(o) are not.

Exercise 4.
Describe in words the half-plane that is the solution to each inequality.
a. y≥0
Answer:
The half-plane lying above the x-axis and including the x-axis.

b. x<-5
Answer:
The half plane to the left of the vertical line x=-5, not including the line x=-5.

c. y≥2x-5
Answer:
The line y=2x-5 and the half-plane lying above it.

d. y<2x-5
Answer:
The half-plane lying below the line y=2x-5.

Exercise 5.
Graph the solution set to x<-5, reading it as an inequality in one variable, and describe the solution set in words. Then graph the solution set to x<-5 again, this time reading it as an inequality in two variables, and describe the solution set in words.
Answer:
Read in one variable: All real numbers less than -5. The graph will have an open circle at the endpoint -5 and extend as a ray to the left of -5 on the number line.

Read in two variables: All ordered pairs (x,y) such that x is less than -5. The graph will be a dashed vertical line through x=-5, and all points to the left of the line will be shaded.

Eureka Math Algebra 1 Module 1 Lesson 21 Exercise Answer Key

Exercises 1–2

Exercise 1.
Circle each ordered pair (x,y) that is a solution to the equation 4x-y≤10.

i. (3,2) (2,3) (-1,-14) (0,0) (1,-6)
Answer:

ii. (5,10) (0,-10) (3,4) (6,0) (4,-1)
Answer:

b. Plot each solution as a point (x,y) in the coordinate plane.
Answer:

c. Describe the location of the solutions in the coordinate plane.
Answer:
(Students may struggle to describe the points. Here is one possible description.) The points do not all fall on any one line, but if I drew a line through any two of the points, the others are not too far away from that line.

Exercise 2.
Discover as many additional solutions to the inequality 4x-y≤10 as possible. Organize solutions by plotting each as a point (x,y) in the coordinate plane. Be prepared to share the strategies used to find the solutions.
Answer:
(There are an infinite number of correct answers, as well as an infinite number of incorrect answers. Some sample correct answers are shown.)
(1,1), (1,-3), (-2,2), (-5,4)

b. Graph the line y=4x-10. What do you notice about the solutions to the inequality 4x-y≤10 and the graph of the line y=4x-10?
Answer:
All of the points are either on the line, to the left of the line, or above the line.

c. Solve the inequality for y.
Answer:
y≥4x-10

d. Complete the following sentence.
If an ordered pair is a solution to 4x-y≤10, then it will be located _____ the line y = 4x – 10
Answer:
If an ordered pair is a solution to 4x-y≤10, then it will be located on, above, or to the left of the line y=4x-10.
Explain how you arrived at your conclusion.
Answer:
I observed that all the points were on one side of the line, and then I tested some points on the other side of the line and found that none of the points I tested from that side of the line were solutions to the inequality.

Next have the groups complete parts (b)–(d). As they work, circulate around the room answering questions and providing support. Make sure that students reversed the inequality symbol when solving for y in part (c). Discuss the following:
→ I noticed some of you wrote that all the points are on the left side of the line and others wrote that all the points are above the line. Are both of those descriptions correct?
→ Now, look at your answer to part (c). When you solved the inequality for y, what does that statement seem to tell you?
→ It tells me that all the y-values have to be greater than or equal to something related to x.
→ Then which description would you say best correlates to the inequality we wrote in part (c)? Points to the left of the line or points above the line? Why?
→ Points above the line because when we solve for y, we are describing where the y-values are in relation to the line, and y-values are plotted on the vertical axis; therefore, the words above and below are the accurate descriptors.
→ How can we depict the entire solution set of ALL the points above the line? When we worked with equations in one variable and graphed our solution set on the number line, how did we show what the solution set was?
→ We colored it darker or shaded it. So we can just shade in the entire area above the line.
→ What about the line itself, is it part of the solution set?
→ Yes.
→ What if it wasn’t? What if the inequality was y>4x-10? How could we show that it is all the points except that line?
→ We traditionally make the line a dashed line instead of a solid line to indicate that the points on the line are not part of the solution set.
Before moving on, make sure students understand that any ordered pair in the solution set will be a point (x,y) that is located on or above the line because that is the portion of the coordinate plane where y is greater than or equal to the difference of 4x and 10.

Eureka Math Algebra 1 Module 1 Lesson 21 Problem Set Answer Key

Question 1.
Match each inequality with its graph. Explain your reasoning.

a. 2x-y>6
Answer:
Graph 2

b. y≤2x-6
Answer:
Graph 4

c. 2x<y+6
Answer:
Graph 3

d. 2x-6≤y
Answer:
Graph 1
Answer:
Student explanations will vary. Sample response: I rearranged each equation and found that they were all the same except for the inequality symbol. The strict inequalities are the dashed lines, and the others are solid lines. When solving for y, you can decide the shading. Greater than is shaded above the line, and less than is shaded below the line.

Question 2.
Graph the solution set in the coordinate plane. Support your answer by selecting two ordered pairs in the solution set and verifying that they make the inequality true.
a. -10x+y>25
b. -6≤y
c. y≤-7.5x+15
d. 2x-8y≤24
e. 3x<y
f. 2x>0
Answer:

Question 3.
Marti sells tacos and burritos from a food truck at the farmers market. She sells burritos for $3.50 each and tacos for $2.00 each. She hopes to earn at least $120 at the farmers market this Saturday.
a. Identify three combinations of tacos and burritos that will earn Marti more than $120.
Answer:
Answers will vary but should be solutions to the inequality 3.5x+2y>120.

b. Identify three combinations of tacos and burritos that will earn Marti exactly $120.
Answer:
Answers will vary but should be solutions to the equation 3.5x+2y=120.

c. Identify three combinations of tacos and burritos that will not earn Marti at least $120.
Answer:
Answers will vary but should not be solutions to the inequality or equation.

d. Graph your answers to parts (a)–(c) in the coordinate plane and then shade a half-plane that contains all possible solutions to this problem.
Answer:
The graph shown for part (d) is shown to the right. Answers to part (a) should lie in the shaded half-plane. Answers to part (b) should lie on the line, and answers to part (c) should lie in the un-shaded half-plane.

e. Create a linear inequality that represents the solution to this problem. Let x equal the number of burritos that Marti sells, and let y equal the number of tacos that Marti sells.
Answer:
3.5x+2y≥120

f. Are the points (10,49.5) a solution to the inequality you created in part (e)? Explain your reasoning.
Answer:
The point would not be valid because it would not make sense in this situation to sell a fractional amount of tacos or burritos.

Eureka Math Algebra 1 Module 1 Lesson 21 Exit Ticket Answer Key

What pairs of numbers satisfy the statement: The sum of two numbers is less than 10?
Create an inequality with two variables to represent this situation, and graph the solution set.
Answer:
Let x= one number, and let y= a second number.
Inequality: x+y<10
Graph the line y=-x+10 using a dashed line and shade below the line.