Go through the Math in Focus Grade 8 Workbook Answer Key Cumulative Review Chapters 1-2 to finish your assignments.

## Math in Focus Grade 8 Course 3 A Cumulative Review Chapters 1-2 Answer Key

**Concepts and Skills**

**Write the prime factorization of each number in exponential notation. (Lesson 1.1)**

Question 1.

16,807

Answer:

Prime factorization of 16,807 is 7×7×7×7×7×7

The number in the exponential notation is 7⁶.

Question 2.

25,920

Answer:

Prime factorization of 25,920 is 2×2×2×2×2×2×3×3×3×3×5.

The number in the exponential form is 2⁶ × 3⁴ × 5.

**Simplify each expression. Write your answer in exponential notation. (Lessons 1.2, 1.3)**

Question 3.

\(\frac{\left[\left(\frac{3}{5}\right) \cdot\left(\frac{3}{5}\right)^{3}\right]^{4}}{\left[\left(\frac{3}{5}\right)^{2}\right]^{2}}\)

Answer:

((3/5) • (3/5)³)⁴/((3/5)²)²

⅗ • ( ⅗)³)⁴ • ((⅗)²)²

⅗ • (⅗)¹² • (⅗)⁴

Bases are equal powers should be added.

⅗ • (⅗)¹⁶

(⅗)¹⁷

Question 4.

(a^{6} • a^{7})^{3} ÷ (4a^{3})^{2}

Answer:

Given (a6 • a7)3 ÷ (4a3)2

(a⁶)³ • (a⁷)³ ÷ (4a)⁶

a¹⁸ • a²¹ ÷ 4a⁶

Bases are equal powers should be added.

a³⁹/4a⁶

¼ × a³⁹/a⁶

¼ × a³³

**Simplify each expression. Write your answer using a positive exponent. (Lessons 1.2, 1.3, 1.4, 1.5)**

Question 5.

\(\frac{6^{3} \cdot 15^{3}}{\left(7^{0}\right)^{3}}\)

Answer:

6³ • 15³/(7⁰)³

= 90³/7⁰

= 90³/1

= 90³

Question 6.

\(\frac{2^{8} \cdot(-3)^{8} \cdot 3^{0}}{5^{-8}}\)

Answer:

Given 2⁸ • (-3)⁸ • 3⁰/5^{-8}

= -6⁸ • 1/5^{-8}

= -6⁸/5^{-8}

Question 7.

[12^{2} • 3^{2}]^{3} ÷ 3^{6}

Answer:

Given [122 • 32]3 ÷ 36

12⁶ • 3⁶ ÷ 3⁶

36⁶/3⁶

12⁶/1⁶

Question 8.

(16^{7} ÷ 16^{4}) • \(\frac{\left(5^{0}\right)^{3}}{2^{3} \cdot 4^{3}}\)

Answer:

Given (167 ÷ 164) • (5⁰)³/2³ • 4³

16⁷/16⁴ • 5⁰/2³ • 4³

16³ • 1/2³ • 4³

64³ • (½)³

Question 9.

8^{-2} • \(\frac{3^{0} \cdot 8^{-3}}{4^{-5}}\) .

Answer:

Given 8-2 • 3⁰ • 8-3/4^{-5}

8-2 • 1 • 8-3/4^{-5}

8-2 • 8-3/4^{-5}

8-5/4^{-5}

2-5/1^{-5}

Question 10.

6^{-4} • (5^{0})^{-4} • (\(\frac{1}{3}\))^{-4} ÷ 3^{-4}

Answer:

6^{-4} (5^{0})^{-4} ∙ (\(\frac{1}{3}\))^{-4} ÷ 3^{-4} We are given the expression:

6^{-4} ∙ (5^{0})^{-4} ∙ (\(\frac{1}{3}\))^{-4} ÷ 3^{-4} Simplify:

= 6^{-4} ∙ \(\frac{1}{3^{-4}} \cdot \frac{1}{3^{-4}}\)

= \(\frac{6^{-4}}{3^{-4}} \cdot \frac{1}{\frac{1}{3^{4}}}\)

= \(\left(\frac{6}{3}\right)^{-4} \cdot 3^{4}\)

= 2^{-4} ∙ 3^{4}

= \(\frac{3^{4}}{2^{4}}\)

= \(\left(\frac{3}{2}\right)^{4}\)

**Evaluate the square roots of each number. Round your answer to the nearest tenth when you can. (Lesson 1.6)**

Question 11.

576

Answer:

576 = (24)²

Square of 576 is 24

24 rounded to the nearest tenth is 20

Question 12.

1,003.4

Answer:

1,003.4 = (31.6764897046)²

Square of 1,003.4 is 31.6764897046

31.6764897046 rounded to the nearest tenth is 30

**Evaluate the cube root of each number. Round your answer to the nearest tenth when you can. (Lesson 1.6)**

Question 13.

\(\frac{27}{216}\)

Answer:

27/216

0.125

0.125 = (0.5)³

Cubic root of 27/216 is 0.5

Question 14.

-629.5

Answer:

-629.5 = (-8.57035039349)³

Cubic root of -629.5 is -8.57035039349

**Evaluate each expression and write your answer in scientific notation. Identify the greater number. (Lessons 2.1, 2.2, 2.3)**

Question 15.

3.27 • 10^{11} + 3.13 • 10^{11} and 9.28 • 10^{11} – 4.15 • 10^{11}

Answer:

3.27 • 1011 + 3.13 • 1011

3.27 + 3.13 • 10¹¹

6.4 • 10¹¹

And 9.28 • 1011 – 4.15 • 1011

9.28 – 4.15 • 10¹¹

5.13 • 10¹¹

6.4 • 10¹¹ is the greater number

Question 16.

9.1 • 10^{-5} – 8.2 • 10^{-6} and 1.2 • 10^{-6} – 5.5 • 10^{-7}

Answer:

9.1 • 10^{-5} – 8.2 • 10^{-6}

9.1 • 10^{-5} – 0.82 • 10^{-5}

9.1 – 0.82 • 10^{-5}

8.28 • 10^{-5}

And 1.2 • 10-6 – 5.5 • 10^{-7}

1.2 • 10-6 – 0.55 • 10^{-6}

1.2 – 0.55 • 10^{-6}

0.65 • 10^{-6}

8.28 • 10^{-5} is the greater number.

Question 17.

8.4 • 10^{5} • 2 • 10^{5} and 3.2 • 10^{-7} • 2 • 10^{-5}

Answer:

8.4 • 10^{5} • 2 • 10^{5}

8.4 • 10⁵ • 2 • 10^{5}

16.8 • 10¹⁰

And 3.2 • 10^{-7} • 2 • 10^{-5}

6.4 • 10^{-12}

16.8 • 10¹⁰ is the greater number.

Question 18.

9.1 • 10^{3} ÷ (7 • 10^{5}) and 7.2 • 10^{-4} ÷ (1.2 • 10^{-4})

Answer:

9.1 • 10^{3} ÷ (7 • 10^{5})

9.1/7 • 10³/10⁵

1.3 • 1/10²

1.3 • 10-2

And 7.2 • 10^{-4} ÷ (1.2 • 10^{-4})

7.2/1.2 • 10^{-4}/10^{-4}

6 • 1

6 = 0.6 • 10^{1}

0.6 • 10^{1} is the greater number.

**Write each measurement in the appropriate unit in prefix form. (Lesson 2.2)**

Question 19.

0.000020 meter

Answer:

0.02 • 10^{3} meter

0.02 millimeter

Question 20.

0.070 gram

Answer:

Given 0.070 gram

0.07 •10^{-3} gram

Question 21.

35,000,000 bytes

Answer:

Given 35000000 bytes

35 • 10⁶ bytes

0.035 kilobytes

Question 22.

42,000 volts

Answer:

Given 42,000 volts

42 • 10³ volts

42 millivolts.

**Problem Solving**

**Solve. Show your work.**

Question 23.

The total surface area of a cube is 4,704 square inches. What is the length of each side? (Chapter 1)

Answer:

From the given question

Total surface area = 4,704 in².

We know that

The total surface area of a cube is 6a²

Where a = side of the cube

6a² = 4704

a² = 4704/6

a² = 784

a = √784

a = 28 in

Question 24.

The volume of a spherical balloon is 12.348π cubic feet. (Chapter 1)

a) Find its radius. Round to the nearest tenth.

Answer:

Given that the volume of a spherical balloon is 12.348π cubic feet.

We know that

The volume of the spherical balloon is 4/3 × πr³

4/3 × πr³ = 12.348π cubic feet.

r³ = 12.348π/4/3× π

r³ = 12.348/ 1.33

r³ = 9.261

r = 2.1

2.1 rounded to the nearest tenth is 2.1

b) Air is pumped into the balloon, so that its radius doubles every 10 seconds. Using 3.14 as an approximation for n, find its surface area after 30 seconds. Round to the nearest tenth.

Answer:

Let r be the radius of the balloon

The radius of the balloon doubles for every 10 seconds.

For 10 seconds radius = r²

For 20 seconds radius = r³

For 30 seconds radius = r⁴

Therefore r⁴ = 30

r = ± 2.340

2.340 rounded to the nearest tenth is 2.3

Question 25.

An oxygen atom has a total of 8 protons. If the mass of one proton is 1.67 • 10^{-24} gram, find the total mass of the protons in the oxygen atom. Write your answer in scientific notation. Round the coefficient to 3 significant digits. (Chapters 1, 2)

Answer:

Given that the oxygen atom has a total of 8 protons

Mass of one proton = 1.67 • 10-24

Mass of 8 protons = 8

Total mass of a protons = 8 × 1.67 • 10-24

= 13.36• 10-24

13.36• 10-24 Round the coefficient to 3 significant digits is 13.4

Question 26.

The table lists the energy in Calories contained in 100 grams of fruits. (Chapter 2)

a) Calculate the total energy of the three fruits. Write your answer in scientific notation.

Answer:

Given that the energy in calories contained in 100 grams of apples = 4.9 • 10⁴

The energy in calories contained in 100 grams of oranges = 6.2 • 10⁴

The energy in calories contained in 100 grams of pear = 3.5 • 10⁴

Total energy in all the three fruits = 4.9 • 10⁴ + 6.2 • 10⁴ + 3.5 • 10⁴

= 4.9 + 6.2 + 3.5 • 10⁴

= 14.6 • 10⁴

b) Find the difference in energy contained between 100 grams of apple and 100 grams of pear.

Answer:

Given that the energy in calories contained in 100 grams of apples = 4.9 • 10⁴

The energy in calories contained in 100 grams of pear = 3.5 • 10⁴

Difference = 4.9 • 10⁴ – 3.5 • 10⁴

= 1.4 • 10⁴

The difference in the energy contained between 100 grams of apple and 100 grams of pear is 1.4 • 10⁴

c) How many times more energy does 100 grams .of orange have compared to 100 grams of apple? Round to the nearest tenth.

Answer:

Given that the energy in calories contained in 100 grams of oranges = 6.2 • 10⁴

The energy in calories contained in 100 grams of apples = 4.9 • 10⁴

6.2 • 10⁴ – 4.9 • 10⁴

6.2 – 4.9 • 10⁴

1.3 • 10⁴

The energy in calories contained in 100 grams of oranges is 1.3 • 10⁴ times more than the 100 grams of apple

Question 27.

Jim deposits $2,000 in a bank, which gives 6% interest, compounded yearly. Use the formula A = P (1 + r )^{n} to find the amount of money in his account after 15 years. A represents the final amount of investment, P is the original principal, r is the interest rate, and n is the number of years it was invested. (Chapter 1)

Answer:

Given that

Jim deposits $2,000 in a bank

It gives 6% interest

Using the formula A = P (1 + r )n

P is the original principal

r is the interest rate

n is the number of years

A = 2000(1+6)¹⁵

A = 2000(7)¹⁵

A = 2000 • 7¹⁵