## Engage NY Eureka Math 8th Grade Module 2 Lesson 10 Answer Key

### Eureka Math Grade 8 Module 2 Lesson 10 Example Answer Key

Example 1.

The world population is about 7 billion. There are 4.6× 10^{7} ants for every human on the planet. About how many ants are there in the world?

First, write 7 billion in scientific notation: (7× 10^{9}).

To find the number of ants in the world, we need to multiply the world population by the known number of ants for each person: (7× 10^{9})(4.6× 10^{7} ).

(7× 10^{9})(4.6× 10^{7} )=(7×4.6)( 10^{9}× 10^{7}) By repeated use of the associative and commutative properties

=32.2× 10^{16} By the first law of exponents

=3.22×10× 10^{16}

=3.22× 10^{17} By the first law of exponents

There are about 3.22× 10^{17} ants in the world!

Example 2.

A certain social media company processes about 990 billion likes per year. If the company has approximately 8.9× 10^{8} users of the social media, about how many likes is each user responsible for per year? Write your answer in scientific and standard notation.

First, write 990 billion in scientific notation: 9.9× 10^{11}.

To find the number of likes per person, divide the total number of likes by the total number of users: \(\frac{9.9 \times 10^{11}}{8.9 \times 10^{8}}\)

\(\frac{9.9 \times 10^{11}}{8.9 \times 10^{8}}\) = \(\frac{9.9}{8.9}\)×\(\frac{10^{11}}{10^{8}}\) By the product formula

=1.11235…× 10^{3} By the first law of exponents

≈1.1× 10^{3}

≈1100

Each user is responsible for about 1.1× 10^{3}, or 1,100, likes per year.

Example 3.

In 2010, Americans generated 2.5× 10^{8} tons of garbage. There are about 2,000 landfills in the United States. Assuming that each landfill is the same size and that trash is divided equally among them, determine how many tons of garbage were sent to each landfill in 2010.

First, write 2,000 in scientific notation: 2× 10^{3}.

To find the number of tons of garbage sent to each landfill, divide the total weight of the garbage by the number of landfills: \(\frac{2.5 \times 10^{8}}{2 \times 10^{3}}\).

\(\frac{2.5 \times 10^{8}}{2 \times 10^{3}}\) = \(\frac{2.5}{2}\)× \(\frac{10^{8}}{10^{3}}\) By the product formula

=1.25× 10^{5} By the first law of exponents

Each landfill received 1.25× 10^{5} tons of garbage in 2010.

Actually, not all garbage went to landfills. Some of it was recycled and composted. The amount of recycled and composted material accounted for about 85 million tons of the 2.5× 10^{8} tons of garbage. Given this new information, how much garbage was actually sent to each landfill?

First, write 85 million in scientific notation: 8.5× 10^{7}.

Next, subtract the amount of recycled and composted material from the garbage: 2.5× 10^{8}-8.5× 10^{7}. To subtract, we must give each number the same order of magnitude and then use the distributive property.

2.5× 10^{8}-8.5× 10^{7}=(2.5×10)× 10^{7}-8.5× 10^{7} By the first law of exponents

=(2.5×10)-8.5))× 10^{7} By the distributive property

=(25-8.5)× 10^{7}

=16.5× 10^{7}

=1.65×10× 10^{7}

=1.65× 10^{8} By the first law of exponents

Now, divide the new amount of garbage by the number of landfills: \(\frac{1.65 \times 10^{8}}{2 \times 10^{3}}\).

\(\frac{1.65 \times 10^{8}}{2 \times 10^{3}}\) = \(\frac{1.65}{2}\)× \(\frac{10^{8}}{10^{3}}\) By the product formula

=0.825× 10^{5} By the first law of exponents

=0.825×10× 10^{4} By the first law of exponents

=8.25× 10^{4}

Each landfill actually received 8.25× 10^{4} tons of garbage in 2010.

### Eureka Math Grade 8 Module 2 Lesson 10 Exercise Answer Key

Exercise 1.

The speed of light is 300,000,000 meters per second. The sun is approximately 1.5× 10^{11} meters from Earth. How many seconds does it take for sunlight to reach Earth?

Answer:

300 000 000=3× 10^{8}

\(\frac{1.5 \times 10^{11}}{3 \times 10^{8}}\) = \(\frac{1.5}{3}\)×\(\frac{10^{11}}{10^{8}}\)

=0.5× 10^{3}

=0.5×10× 10^{2}

=5× 10^{2}

It takes 500 seconds for sunlight to reach Earth.

Exercise 2.

The mass of the moon is about 7.3× 10^{22} kg. It would take approximately 26,000,000 moons to equal the mass of the sun. Determine the mass of the sun.

Answer:

26 000 000=2.6× 10^{7}

(2.6× 10^{7} )(7.3× 10^{22})=(2.6×7.3)( 10^{7}× 10^{22})

=18.98× 10^{29}

=1.898×10× 10^{29}

=1.898× 10^{30}

The mass of the sun is 1.898× 10^{30} kg.

Exercise 3.

The mass of Earth is 5.9× 10^{2}4 kg. The mass of Pluto is 13,000,000,000,000,000,000,000 kg. Compared to Pluto, how much greater is Earth’s mass than Pluto’s mass?

Answer:

13 000 000 000 000 000 000 000=1.3× 10^{22}

5.9× 10^{24}-1.3× 10^{22}=(5.9× 10^{2} )× 10^{22}-1.3× 10^{22}

=(590-1.3)× 10^{22}

=588.7× 10^{22}

=5.887× 10^{2}× 10^{22}

=5.887× 10^{24}

The mass of Earth is 5.887× 10^{24} kg greater than the mass of Pluto.

Exercise 4.

Using the information in Exercises 2 and 3, find the combined mass of the moon, Earth, and Pluto.

Answer:

7.3× 10^{22}+1.3× 10^{22}+5.9× 10^{24}=(7.3× 10^{22}+1.3× 10^{22})+5.9× 10^{24}

=8.6× 10^{22}+5.9× 10^{24}

=(8.6+590)× 10^{22}

=598.6× 10^{22}

=5.986× 10^{2}× 10^{22}

=5.986× 10^{24}

The combined mass of the moon, Earth, and Pluto is 5.986× 10^{24} kg.

Exercise 5.

How many combined moon, Earth, and Pluto masses (i.e., the answer to Exercise 4) are needed to equal the mass of the sun (i.e., the answer to Exercise 2)?

Answer:

\(\frac{1.898 \times 10^{30}}{5.986 \times 10^{24}}\)=\(\frac{1.898}{5.986}\) × \(\frac{10^{30}}{10^{24}}\)

≈0.32× 10^{6}

=0.32×10× 10^{5}

=3.2× 10^{5}

It would take 3.2× 10^{5} combined masses of the moon, Earth, and Pluto to equal the mass of the sun.

### Eureka Math Grade 8 Module 2 Lesson 10 Problem Set Answer Key

Have students practice operations with numbers written in scientific notation and standard notation.

Question 1.

The sun produces 3.8× 10^{27} joules of energy per second. How much energy is produced in a year? (Note: a year is approximately 31,000,000 seconds).

Answer:

31 000 000=3.1× 10^{7}

(3.8× 10^{2}7 )(3.1× 10^{7} )=(3.8×3.1)( 10^{2}7× 10^{7} )

=11.78× 10^{34}

=1.178×10× 10^{34}

=1.178× 10^{35}

The sun produces 1.178× 10^{35} joules of energy in a year.

Question 2.

On average, Mercury is about 57,000,000 km from the sun, whereas Neptune is about 4.5× 10^{9} km from the sun. What is the difference between Mercury’s and Neptune’s distances from the sun?

Answer:

57 000 000=5.7× 10^{7}

4.5× 10^{9}-5.7× 10^{7}=(4.5× 10^{2} )× 10^{7}-5.7× 10^{7}

=450× 10^{7}-5.7× 10^{7}

=(450-5.7)× 10^{7}

=444.3× 10^{7}

=4.443× 10^{2}× 10^{7}

=4.443× 10^{9}

The difference in the distance of Mercury and Neptune from the sun is 4.443× 10^{9} km.

Question 3.

The mass of Earth is approximately 5.9× 10^{24} kg, and the mass of Venus is approximately 4.9× 10^{24} kg.

a. Find their combined mass.

Answer:

5.9× 10^{24}+4.9× 10^{24}=(5.9+4.9)× 10^{24}

=10.8× 10^{24}

=1.08×10× 10^{24}

=1.08× 10^{25}

The combined mass of Earth and Venus is 1.08× 10^{25} kg.

b. Given that the mass of the sun is approximately 1.9× 10^{30} kg, how many Venuses and Earths would it take to equal the mass of the sun?

Answer:

\(\frac{1.9 \times 10^{30}}{1.08 \times 10^{25}}\) = \(\frac{1.9}{1.08}\) × \(\frac{10^{30}}{10^{25}}\)

=1.75925…× 10^{5}

≈1.8× 10^{5}

It would take approximately 1.8× 10^{5} Venuses and Earths to equal the mass of the sun.

### Eureka Math Grade 8 Module 2 Lesson 10 Exit Ticket Answer Key

Question 1.

The speed of light is 3× 10^{8} meters per second. The sun is approximately 230,000,000,000 meters from Mars. How many seconds does it take for sunlight to reach Mars?

Answer:

230 000 000 000=2.3× 10^{11}

\(\frac{2.3 \times 10^{11}}{3 \times 10^{8}}\) =\(\frac{2.3}{3}\)×\(\frac{10^{11}}{10^{8}}\)

=0.7666″…”× 10^{3}

≈0.77×10× 10^{2}

≈7.7× 10^{2}

It takes approximately 770 seconds for sunlight to reach Mars.

Question 2.

If the sun is approximately 1.5× 10^{11} meters from Earth, what is the approximate distance from Earth to Mars?

Answer:

(2.3× 10^{11} )-(1.5× 10^{11} )=(2.3-1.5)× 10^{11}

=0.8× 10^{11}

=0.8×10× 10^{10}

=8× 10^{10}

The distance from Earth to Mars is 8× 10^{10} meters.