# Eureka Math Algebra 2 Module 3 Lesson 2 Answer Key

## Engage NY Eureka Math Algebra 2 Module 3 Lesson 2 Answer Key

### Eureka Math Algebra 2 Module 3 Lesson 2 Opening Exercise Answer Key

Opening Exercise:
In the last lesson, you worked with the thickness of a sheet of gold foil (a very small number) and some very large numbers that gave the size of a piece of paper that actually could be folded in half more than 13 times.

a. Convert 0.28 millionth of a meter to centimeters, and express your answer as a decimal number.
$$\frac{0.28}{1,000,000} \mathrm{~m} \cdot \frac{100 \mathrm{~cm}}{1 \mathrm{~m}}=\frac{28}{1,000,000} \mathrm{~cm}$$ = 0.000028 cm

b. The length of a piece of square notebook paper that can be folded in half 13 times is 3,294. 2 in. Use this number to calculate the area of a square piece of paper that can be folded in half 14 times. Round your answer to the nearest million.
(2,3294.2)2 = 43,407,014.56
Rounded to the nearest million, the area is 43,000,000 square inches.

c. Match the equivalent expressions without using a calculator.
3.5 × 105
3,500,000
-6
350,000
-6 × 100
6 × 10-1
0.6
0.0000035
3.5 × 10-6
3.5 × 106
3.5 × 105 is equal to 350,000.
-6 × 100 is equal to -6.
6 × 10-1 is equal to 0.6.
3.5 × 106 is equal to 3,500,000.
3.5 × 10-6 is equal to 0.0000035.

### Eureka Math Algebra 2 Module 3 Lesson 2 Example Answer Key

Example 1:

Write each number as a product of a decimal number between 1 and 10 and a power of 10.

a. 234,000
2.34 . 100,000 = 2.34 × 105

b. 0.0035
$$\frac{35}{10000}=\frac{3.5}{1000}$$ = 3.5 . $$\frac{1}{1000}$$ = 3.5 × 10-3

c. 532,100,000
5.321.100,000,000 = 5.321 × 108

d. 0.0000000012
$$\frac{12}{10,000,000,000}=\frac{1.2}{1,000,000,000}$$ = 1.2 . $$\frac{1}{1,000,000,000}$$ = 1.2 × 10-9

e. 3.331
3.331 . 1 = 3.331 × 100

Example 2: Arithmetic Operations with Numbers Written Using Scientific Notation

a. (2.4 × 1020) + (4.5 × 1021)
(2.4 × 1020) + (45 × 1020) = 47.4 × 1020 = 4.74 × 1021

b. (7 × 10-9)(5 × 105)
(7 . 5) × (10-9 . 105) = 35 × 10-4 = 3. 5 × 10-3

c. $$\frac{1.2 \times 10^{15}}{3 \times 10^{7}}$$
$$\frac{1.2}{3}$$ × 1015 – 7 = 0.4 × 108 = 4 × 107

### Eureka Math Algebra 2 Module 3 Lesson 2 Exercise Answer Key

Exercises 1 – 6:

For Exercises 1 – 6, write each number in scientific notation.

Exercise 1.
532,000,000
5.32 × 108

Exercise 2.
0.0000000000000000123 (16 zeros after the decimal place)
1.23 × 10-17

Exercise 3.
8,900,000,000,000,000(14 zeros after the 9)
8.9 × 1015

Exercise 4.
0.00003382
3.382 × 10-5

Exercise 5.
34, 000,000,000, 000, 000, 000,000,000 (24 zeros after the 4)
3.4 × 1025

Exercise 6.
0. 0000000000000000000004 (21 zeros after the decimal place)
4 × 10-22

Exercises 7 – 8:

Exercise 7.
Use the fact that the average distance between the sun and Earth is 151, 268,468 km, the average distance between the sun and Jupiter is 780, 179,470 km, and the average distance between the sun and Pluto is 5,908, 039, 124 km to approximate the following distances. Express your answers in scientific notation (d × 10n), where d is rounded to the nearest tenth.

a. Distance from the sun to Earth:
1.5 × 108 km

b. Distance from the sun to Jupiter:
7.8 × 108 km

c. Distance from the sun to Pluto:
5.9 × 109 km

d. How much farther Jupiter is from the sun than Earth Is from the sun:
780, 179,470 km – 151,268,468 km = 628,911,002 km
6.3 × 108 km

e. How much farther Pluto is from the sun than Jupiter is from the sun:
5,908,039,124 km – 780,179,470 km = 5,127,859,654 km
5.1 × 109 km

Exercise 8.
Order the numbers in Exercise 7 from smallest to largest. Explain how writing the numbers in scientific notation
helps you to quickly compare and order them.
The numbers from smallest to largest are 1. 5 × 108, 6.3 × 108, 7.8 × 108 5.1 × 109, and 5.9 × 109. The power of 10 helps to quickly sort the numbers by their order of magnitude, and then it is easy to quickly compare the numbers with the same order of magnitude because they are only written as a number between one and ten.

Exercises 9 – 10:

Exercise 9.
Perform the following calculations without rewriting the numbers in decimal form.

a. (1.42 × 1015) – (2 × 1013)
142 × 1013 – 2 × 1013 = (142 – 2) × 1013 = 140 × 1013 = 1.4 × 1015

b. (1.42 × 1015) (24 × 1013)
(1.42 . 2.4) × (1015 . 1013) = 3.408 × 1028

c. $$\frac{1.42 \times 10^{-5}}{2 \times 10^{13}}$$
$$\frac{1.42 \times 10^{-5}}{2 \times 10^{13}}$$ = 0.71 × 10-5-13 = 0.71 × 10-18 = 7.1× 10-19

Exercise 10.
Estimate how many times farther Jupiter is from the sun than Earth is from the sun. Estimate how many times farther Pluto is from the sun than Earth is from the sun.
Earth is approximately 1.5 × 108 km from the sun, and Jupiter is approximately 7.8 × 108 km from the sun. Therefore, Jupiter is about 5 times as far from the sun as Earth is from the sun. Pluto is approximately 5.9 × 109 km from the sun. Therefore, since $$\frac{59 \times 10^{8}}{1.5 \times 10^{8}}=\frac{59}{1.5}$$ ≈ 39.33

Pluto is approximately 39 times as far from the sun as Earth is from the sun.

### Eureka Math Algebra 2 Module 3 Lesson 2 Problem Set Answer Key

Question 1.
Write the following numbers used in these statements in scientific notation. (Note: Some of these numbers have been rounded.)

a. The density of helium is 0.0001785 gram per cubic centimeter.
1.785 × 10-4

b. The boiling point of gold is 5,200°F.
5.2 × 103

c. The speed of light is 186,000 miles per second.
1.86 × 105

d. One second is 0.000278 hour.
2.78 × 10-4

e. The acceleration due to gravity on the sun is 900 f t/s.
9 × 102

f. One cubic inch is 0.0000214 cubic yard.
2.14 × 10-5

g. Earth’s population in 2012 was 7,046,000,000 people.
7.046 × 109

h. Earth’s distance from the sun is 93,000,000 miles.
9.3 × 107

i. Earth’s radius is 4,000 miles.
4 × 103

j. The diameter of a water molecule is 0.000000028 cm.
2.8 × 10-8

Question 2.
Write the following numbers in decimal form. (Note: Some of these numbers have been rounded.)

a. A light year is 9.46 × 1015 m.
9,460,000,000,000,000

b. Avogadro’s number is 6.02 × 1023 mol-1.
602, 000,000,000,000,000, 000,000

c. The universal gravitational constant is 6.674 × 10-11 N$$\left(\frac{\mathrm{m}}{\mathrm{kg}}\right)^{2}$$.
0.00000000006674

d. Earth’s age is 4.54 × 109 years.
4,540,000,000

e. Earth’s mass is 5.97 × 1024 kg.
5,970,000,000,000,000,000,000,000

f. A foot is 1.9 × 10-4 mile.
0.00019

g. The population of C0hina in 2014 was 1.354 × 109 people.
1,354,000,000

h. The density of oxygen is 1.429 × 10-4 grams per liter.
0.0001429

i. The width of a pixel on a smartphone is 7.8 × 10-2 mm.
0.078

j. The wavelength of light used in optic fibers is 1.55 × 10-6 m.
0.00000155

Question 3.
State the necessary value of n that will make each statement true.

a. 0 .000027=27 × 10n
-5

b. -3.125 = -3.125 × 10n
o

c. 7,540,000,000 = 7.54 × 10n
9

d. 0.033 = 3.3 × 10n
-2

e. 15 = 1.5 × 10n
1

f. 26,000 × 200 = 5.2 × 10n
6

g. 3000 × 0.0003 = 9 × 10n
-1

h. 0.0004 × 0.002.=8 × 10n
-7

i. $$\frac{16000}{80}$$ = 2 × 10n
2

j. $$\frac{500}{0.002}$$ = 2.5 × 10n
5

Question 4.
Perform the following calculations without rewriting the numbers in decimal form.

a. (2.5 × 104) + (3.7 × 103)
2.87 × 104

b. (6.9 × 10-3) – (8.1 × 10-3)
-1.2 × 10-3

c. (6 × 1011) (2.5 × 10-5)
1.5 × 107

d. $$\frac{4.5 \times 10^{8}}{2 \times 10^{10}}$$
2.25 × 10-2

Question 5.
The wavelength of visible light ranges from 650 nanometers to 850 nanometers, where 1 nm = 1 × 10-7 cm. Express the range of wavelengths of visible light in centimeters.
Convert 650 nanometers to centimeters: (6.5 × 102) (1 × 10-7) = 6.5 × 10-5
Convert 850 nanometers to centimeters: (8.5 × 102) (1 × 10-7) = 8.5 × 10-5
The wavelength of visible light in centimeters is 6. 5 × 10-5 cm to 8.5 × 10-5 cm.

Question 6.
In 1694, the Dutch scientist Antonie van Leeuwenhoek was one of the first scientists to see a red blood cell in a microscope. He approximated that a red blood cell was “25,000 times as small as a grain of sand.” Assume a grain of sand is $$\frac{1}{2}$$ mm wide, and a red blood cell is approximately 7 micrometers wide. One micrometer is 1 × 10-6 m. Support or refute Leeuwenhoek’s claim. Use scientific notation in your calculations.
Convert millimeters to meters: (5 × -1)(1 × 10-3) = 5 × 10-4. A medium-size grain of sand measures 5 × 10-4 m across. Similarly, a red blood cell is approximately 7 × 10-6 m across. Dividing these numbers produces

$$\frac{5 \times 10^{-4}}{7 \times 10^{-6}}$$ = 0.714 × 102 = 7.14 × 101.

So, a red blood cell is 71.4 times as small as a grain of sand. Leeuwenhoek’s claim was off by approximately a factor of 350.

Question 7.
When the Mars Curiosity Rover entered the atmosphere of Mars on its descent in 2012, it was traveling roughly 13,200 mph. On the surface of Mars, its speed averaged 0.00073 mph. How many times faster was the speed when it entered the atmosphere than its typical speed on the planet’s surface? Use scientific notation in your calculations.
$$\frac{1.32 \times 10^{4}}{7.3 \times 10^{-4}}$$ = 0.18 × 108 = 1.8 × 107
The speed when it entered the atmosphere is greater than its surface speed by an order of magnitude of 7.

Question 8.
Earth’s surface is approximately 70% water. There is no water on the surface of Mars, and its diameter is roughly half of Earth’s diameter. Assume both planets are spherical. The radius of Earth is approximately 4,000 miles. The surface area of a sphere is given by the formula SA = 4πr2, where r is the radius of the sphere. Which has more land mass, Earth or Mars? Use scientific notation in your calculations.
The surface area of Earth: 4πr(4000 mi)2 ≈ 2 × 108 mi2
The surface area of Mors: 4πr(2000 mi)2 ≈ 5 × 107 mi2
Thirty percent of Earth’s surface area is approximately 6 × 107 square miles. Earth has more land mass by approximately 20%.

Question 9.
There are approximately 25 trillion (2.5 × 1013) red blood cells in the human body at any one time. A red blood cell is approximately 7 × 10-6 m wide. Imagine If you could line up all your red blood cells end to end. How long would the line of cells be? Use scientific notation in your calculations.
Because (2.5 × 1013)(7 × 10-6) = 1.75 × 108, the line of cells would be 1. 75 × 108 m long, which is 1.75 × 105 km. One mile is equivalent to 1.6 km, so the line of blood cells measures $$\frac{1.75 \times 10^{5}}{1.6}$$ km, which is approximately 109.375 mi which is almost halfway to the moon!

Question 10.
Assume each person needs approximately 100 square feet of living space. Now imagine that we are going to build a giant apartment building that will be 1 mile wide and 1 mile long to house all the people in the United States, estimated to be 313.9 million people in 2012. If each floor of the apartment building is 10 feet high, how tall will the apartment building be?
Since (3.139 × 108)(100) = 3.139 × 1010 we need 3.139 × 1010 ft2 of living space.
Next, divide the total number of square feet by the number of square feet per floor to get the number of needed floors. Remember that 1 mi2 = 52802 ft2 = 27878400 ft2.
$$\frac{3.139 \times 10^{10} \mathrm{ft}^{2}}{2.78784 \times 10^{7} \mathrm{ft}^{2}}$$ ≈ 1.126 × 103

Multiplying the number of floors by 10 feet per floor gives a height of 11,260 feet, which is approximately 2.13 miles.

### Eureka Math Algebra 2 Module 3 Lesson 2 Exit Ticket Answer Key

Question 1.
A sheet of gold foil is 0.28 millionth of a meter thick. Write the thickness of a gold foil sheet measured in centimeters using scientific notation.
The thickness is 0.28 × 10-6 m. In scientific notation, the thickness of a gold foil sheet is 2.8 × 10-7 m, which is 2.8 × 10-5 cm.

Question 2.
Without performing the calculation, estimate which expression is larger. Explain how you know.
(4 × 1010) (2 × 105) and $$\frac{4 \times 10^{12}}{2 \times 10^{-4}}$$