## Big Ideas Math Book 8th Grade Answer Key Chapter 8 Exponents and Scientific Notation

### Exponents and Scientific Notation STEAM Video/ Performance Task

**STEAM Video**

Carbon Atoms

Carbon is one of the four main elements of life. The number of carbon atoms in a compound can be represented using exponents. In what other real-life situations are exponents used?

Watch the STEAM Video “Carbon Atoms.” Then answer the following questions.

1. The table shows the percent carbon by weight for humans and plants. How many pounds of carbon are in a 130-pound person? a 25-pound plant?

a. Pounds of Carbon in 130-pound person is 2.57 X 10^{21}.

b. Pounds of Carbon in 25 -pound plant is 1.23975 X 10^{21}.

Explanation:

Given 1 carbon atoms consists of 5 X 10^{22},

a. So 1 person has 18% of carbon means 18 ÷ 100 X 5 × 10^{22} =

18 X 5 X 10^{22-2} = 90 X 10^{20 }= 9 X 10 X 10^{20 }= 9 X 10^{21 }now,

We know 1 gram is equal to 0.00220462 pound

So 0.00220462 X 9 X 10^{21 }as 0.00220462 approximately equal to ≈

2.204 X 10^{-3 }X 9 X 10^{21 }= 19.836 X 10^{21-3 }= 19.836 X 10^{18 }= 1.9836 X 10^{19 }now

in 130-pound person is 130 X 1.9836 X 10^{19 }= 257.868 X 10^{19 }= 2.57 X 10^{21}.

b. So 1 plant has 45% of carbon means 45 ÷ 100 X 5 × 10^{22} =

45 X 5 X 10^{22-2} = 225 X 10^{20 }= 2.25 X 100 X 10^{20 }= 2.25 X 10^{22 }now,

We know 1 gram is equal to 0.00220462 pound

So 0.00220462 X 2.25 X 10^{22 }as 0.00220462 approximately equal to ≈

2.204 X 10^{-3 }X 2.25 X 10^{22 }= 4.959 X 10^{22-3 }= 4.959 X 10^{19 }= 4.959 X 10^{19 }now

in 25-pound plant is 25 X 4.959 X 10^{19 }=123.975 X 10^{19 }= 1.23975 X 10^{21}.

2. Steven says 5 × 10^{22}, carbon atoms are in 1 gram of carbon. How many carbon atoms are in 3 grams of carbon?

In 3 grams of carbon = 3 X (5 x10^{22}) = ^{ }15 x 10^{22 }carbon atoms are available

Explanation:

Given Steven says 5 x10^{22 },carbon atoms are in 1 gram of carbon,

in 3 grams of carbon it will be (5 x 10^{22}) X 3 = 5 X 3 X (10^{22})=

15 x 10^{22 }carbon atoms are available.

**Performance Task**

Elements in the Universe

After completing this chapter, you will be able to use the concepts you learned to answer the questions in the STEAM Video Performance Task. You will be given information about the atomic masses of the four most common elements in the universe: oxygen, hydrogen, helium, and carbon.

You will be asked to solve problems about the amounts of carbon dioxide in Earth’s atmosphere for several years. What might cause the amount of carbon dioxide in the atmosphere to increase over time?

### Exponents and Scientific Notation Getting Ready for Chapter 8

**Chapter Exploration**

1. Work with a partner. Write each distance as a whole number. Which numbers do you know how to write in words? For instance, in words, 10^{2} is equal to one hundred.

a. 10^{27 }meters diameter of the observable universe is

10000000 0000000000 0000000000 is equal to

octillion ( quadrilliard )

b. 10^{21 }meters diameter of the Milky Way galaxy is

10 0000000000 0000000000 is equal to

sextillion (trilliard)

c. 10^{16 }meters diameter of the solar system is

10 0000000000 00000 is equal to

10 quadrillion or 10 thousand trillion

d. 10^{7 }meters diameter of Earth is 10000000

is equal to ten million (crore (India))

e. 10^{4 }meters diameter of Halley’s Comet is 10000

is equal to ten thousand

f. 10^{3 }meters diameter of a meteor crater is

1000 is equal to thousand.

Explanation:

a. 10^{27 }meters diameter of the observable universe,

we call the number 10 is called the base

and the number 27 is called the exponent, we multiply 10 by 27 times,

we write 10 to the 27th power as 1,000,000,000,000,000,000,000,000,000

is equal to octillion ( quadrilliard )

b. 10^{21 }meters diameter of the Milky Way galaxy,

we call the number 10 is called the base

and the number 21 is called the exponent, we multiply 10 by 21 times,

we write 10 to the 21th power as 1,000,000,000,000,000,000,000

is equal to sextillion (trilliard)

c. 10^{16 }meters diameter of the solar system

we call the number 10 is called the base

and the number 16 is called the exponent, we multiply 10 by 16 times,

we write 10 to the 16th power as 10,000,000,000,000,000 is equal to

10 quadrillion or 10 thousand trillion

d. 10^{7 }meters diameter of Earth

we call the number 10 is called the base

and the number 7 is called the exponent, we multiply 10 by 7 times,

we write 10 to the 7th power as 10,000,000 is equal to ten million (crore (India))

e. 10^{4 }meters diameter of Halley’s Comet we call the number 10 is called the base

and the number 4 is called the exponent, we multiply 10 by 4 times,

we write 10 to the 4th power as 10,000 is equal to ten thousand

f. 10^{3 }meters diameter of a meteor crater we call the number 10 is called the base

and the number 27 is called the exponent, we multiply 10 by 3 times

we write 10 to the 3th power as 1,000 is equal to thousand.

2. Work with a partner. Write the numbers of wives, sacks, cats, and kits as powers.

Man ,wives are 7^{0 } + 7^{1 }= 1 + 7 = 8,^{
}Man, wives, sacks are 7^{0 }+ 7^{2 }= 1 + 49 = 50

Man, wives, sacks, cats are 7^{0 } + 7^{3 }= 1 + 343 = 344,

Man, wives, sacks, cats , kits are 7^{0 } + 7^{4 }= 1 + 2401 = 2402

Total 2402 are going to St. Ives.

Explanation:

Given I met a man with seven wives so

writing as powers man and seven wives means 1 + 7^{1 }= 8,

man, each wives had seven sacks means 1 + 7^{2 }= 49,

man, wives, each sack had seven cats are 1+ 7^{3 }= 1 + 343 = 344,

man, wives, sacks , Each cat had seven kits are 1+ 7^{4 }= 1 + 2401 = 2402,

So in total kits, cats, sacks, wives and man are 2402 are going to St. Ives.

**Vocabulary**

The following vocabulary terms are defined in this chapter. Think about what each term might mean and record your thoughts.

power

exponent of a power

base of a power

scientific notation

### Lesson 8.1 Exponents

The expression 3^{5} is called a power. The base is 3. The exponent is 5.

**EXPLORATION 1**

Using Exponent Notation

Work with a partner.

a. Copy and complete the table.

Explanation:

Given Powers wrote repeated multiplication form and value as

(-3)^{3}= (-3).(-3).(-3)= -27, multiplied -3 three times as power is 3

(-3)^{4}= (-3).(-3).(-3).(-3)= 81,multiplied -3 four times as power is 4

(-3)^{5}= (-3).(-3).(-3).(-3).(-3)= -243,multiplied -3 five times as power is 5

(-3)^{6}= (-3).(-3).(-3).(-3).(-3).(-3)= 729,multiplied -3 five times as power is 6

(-3)^{7}= (-3).(-3).(-3).(-3).(-3).(-3).(-3)= 2187,multiplied -3 five times as power is 7

b. Describe what is meant by the expression (- 3)^{n}. How can you find the value of (- 3)^{n}?

Answer:

The expression (-3)^{n} is called a power. The base is -3. The exponent is n.

We find the value of (- 3)^{n }we multiply -3 with n number of times.

Explanation:

An expression that represents repeated multiplication

of the same factor is called a power. Here the expression

(-3)^{n} is called a power of n and the number -3 is called the base,

and the number n is called the exponent. The exponent corresponds

to the number of times the base is used as a factor.

**EXPLORATION 2**

Using Exponent Notation

Work with a partner. On a game show, each small cube is worth $3. The small cubes are arranged to form a large cube. Show how you can use a power to find the total value of the large cube. Then write an explanation to convince a friend that your answer is correct.

Answer:

The total value of the large cube is =$3 X (320)^{320 }

My answer is correct as given base as 320 and exponent is 320.

Explanation:

Given small cubes are arranged to form large cube,

and power is 320, base is 320 and exponent is 320,

each small cube is worth $3 so the total value of the large

cube is $3 multiplied by 320 and multiplied $3 with 320 by 320 times is

$3 X (320)^{320 }as my answer is correct I say my friend because

given base is 320 and exponent is 320 we write as (320)^{320 }and

multiply by $3 to get the value of the large cube.

**Try It**

**Write the product using exponents.**

Question 1.

\(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\)

Answer:

\(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\) = (\(\frac{1}{4}\))^{5
}

Explanation:

As \(\frac{1}{4}\) is multiplied by 5 times,

the expression is (\(\frac{1}{4}\))^{5}.

The base is \(\frac{1}{4}\) and the exponent is 5.

Question 2.

0.3 • 0.3 • 0.3 • 0.3 • x • x

Answer:

0.3 • 0.3 • 0.3 • 0.3 • x • x = (0.3)^{4 }X (x)^{2 }

Explanation:

As 0.3 is multiplied 4 times we write as (0.3)^{4 }and x

is multiplied twice so (x)^{2
}so the expression is (0.3)^{4 }X (x)^{2}.

Evaluate the expression.

Question 3.

12^{2}

Answer:

12^{2 }= 144

Explanation:

Given 12^{2 }means 12 is multiplied twice

as 12 X 12 we get 144.

here base is 12 and exponent is 2.

Question 4.

(- 2)^{6}

Answer:

(- 2)^{6 }= 64

Explanation:

Given (- 2)^{6 }we multiply -2 by 6 times as

-2 X -2 X -2 X -2 X -2 X -2 we get 64

here base is -2 and exponent is 6.

Question 5.

– 5^{4}

Answer:

– 5^{4 }=625

Explanation:

Given – 5^{4 }we multiply -5 by 4 times as

-5 X -5 X -5 X -5 = 625,

here base is -5 and 4 is exponent.

Question 6.

\(\left(-\frac{1}{6}\right)^{3}\)

Answer:

\(\left(-\frac{1}{6}\right)^{3}\) = –\(\frac{1}{216}\)

Explanation:

Given \(\left(-\frac{1}{6}\right)^{3}\) we multiply

– \(\frac{1}{6}\) by 3 times as –\(\frac{1}{6}\) X

– \(\frac{1}{6}\) X –\(\frac{1}{6}\) we get

– \(\frac{1}{216}\) here base is –\(\frac{1}{6}\)

and 3 is exponent.

**Evaluate the expression.**

Question 7.

9 – 2^{5} . 0.5

Answer:

9 – 2^{5} . 0.5 = -7

Explanation:

Given 9 – 2^{5} . 0.5 = First we solve 2^{5} . 0.5

2^{5} X 0.5 as 0.5 can be written as \(\frac{1}{2}\),

= 32 X \(\frac{1}{2}\) = 16 now we subtract 16 from 9

9-16 = -7

Question 8.

|- 3^{3} ÷ 27|

Answer:

|- 3^{3} ÷ 27| = -1

Explanation:

First we calculate – 3^{3} we multiply -3 by 3 times as

-3 X -3 X -3 = -27 now we divide -27 by 27 we get -1.

Question 9.

(7 . 4 – 4^{3}) ÷ 6

Answer:

(7 . 4 – 4^{3}) ÷ 6 = -6

Explanation:

First we calculate the value of 4^{3}

we multiply 4 by 3 times as 4 X 4 X 4 = 64,

Now we multiply 7 X 4 = 28 now we subtract 64 from 28

we get (7 . 4 – 4^{3}) =(28 – 64) = -36 now we divide this by 6 we get

-36 ÷ 6 = -6, So (7 . 4 – 4^{3}) ÷ 6 = -6.

**Self-Assessment for Concepts & Skills**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

**WRITING EXPRESSIONS USING EXPONENTS** Write the product using exponents.

Question 10.

(- 0.9) • (- 0.9) • (- 0.9)

Answer:

(- 0.9) • (- 0.9) • (- 0.9 )= (-0.9)^{3}

Explanation:

As -0.9 is multiplied by 3 times we write as

(-0.9)^{3 }here -0.9 is base and 3 is exponent.

Question 11.

\(\frac{1}{8}\) • \(\frac{1}{8}\) • y • y • y

Answer:

\(\frac{1}{8}\) • \(\frac{1}{8}\) • y • y • y =

(\(\frac{1}{8}\))^{2 }X (y)^{3
}

Explanation:

As \(\frac{1}{8}\) is multiplied by 2 times we write as

(\(\frac{1}{8}\))^{2}

and y is multiplied 3 times we write as (y)^{3},

So \(\frac{1}{8}\) • \(\frac{1}{8}\) • y • y • y =

(\(\frac{1}{8}\))^{2 }X (y)^{3}.

**EVALUATING EXPRESSIONS** Evaluate the expression.

Question 12.

11^{2}

Answer:

11^{2 }= 121

Explanation:

Given 11^{2 }means 11 is multiplied twice

as 11 X 11 = 121.

Question 13.

– 6^{3}

Answer:

– 6^{3 }= – 216

Explanation:

Given – 6 is multiplied by 3 times as

– 6 X -6 X -6 = -216

Question 14.

(- 0.3)^{4}

Answer:

(- 0.3)^{4 }= 0.0081

Explanation:

As (-0.3) is multiplied by 4 times we get

-0.3 X -0.3 X -0.3 X -0.3 = 0.0081

**USING ORDER OF OPERATIONS** Evaluate the expression.

Question 15.

|- 24 ÷ 2^{2}|

Answer:

|- 24 ÷ 2^{2}|= -6

Explanation:

First we calculate 2^{2 }we get 4,

now we divide -24 by 4 we get -6.

Question 16.

(3^{3} – 6 • 8) ÷ 7

Answer:

(3^{3} – 6 • 8) ÷ 7 = -3

Explanation:

First we calculate 6 X 8 we get 48 Now we subtract

48 from 3^{3} as 3^{3} is 3 X 3 X 3 = 27 we get (27 – 48 )= -21

now we divide -21 by 7 we get -3 therefore

(3^{3} – 6 • 8) ÷ 7 = -3 .

Question 17.

**WHICH ONE DOESN’T BELONG?**

Which expression does not belong with the other three? Explain your reasoning.

Answer:

The expression – 8^{2 }does not belongs to other three.

Explanation:

Given expressions (-2)^{6},- 8^{2 },-8^{2 }and 2^{6 }**
**the values are (-2)

^{6}= -2 X -2 X -2 X -2 X -2 X -2 = 64,

– 8

^{2 }=- ( 8 X 8) = -64,

8

^{2 }= 8 X 8 = 64 and 2

^{6 }= 2 X 2 X 2 X 2 X 2 X 2 = 64

as (-2)

^{6},-8

^{2 }and 2

^{6 }have same value 64 only – 8

^{2 }=-64 is different,

so the expression – 8

^{2 }does not belongs to other three.

**Self-Assessment for Problem Solving**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.

**DIG DEEPER!**

Consider the diameters of three planets.

Planet A: 10^{9}m

Planet B: 10^{7}n

Planet C: 10^{8}m

a. Write each diameter as a whole number.

b. A dwarf planet is discovered with a radius that is \(\frac{1}{100}\) the radius Planet C. Write the diameter of the dwarf planet as a power.

Answer:

a. Planet A: 10^{9}m = 1000000000m

Planet B: 10^{7}n = 10000000n

Planet C: 10^{8}m = 100000000m

b. diameter = 2 X 10^{6}m

Explanation:

The diameters of three planets are given as

Planet A: 10^{9}m, Planet B: 10^{7}n, Planet C: 10^{8}m

in part a,we write whole for Planet A as 10 is multiplied by 9 times

so 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 =1000000000m

Planet B as 10 is multiplied by 7 times therefore it is

10 X 10 X 10 X 10 X 10 X 10 X 10 =10000000n and Planet C

as 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 =100000000m and

in part b, Given a dwarf planet is discovered with a radius that is

\(\frac{1}{100}\) the radius Planet C and

we know diameter = 2 X radius so diameter is

2 X 100000000 X \(\frac{1}{100}\) = 2 X 1000000,

therfore the diameter of the dwarf planet as a power is 2 X 10^{6}m.

Question 19.

A fish jumps out of the water at a speed of 12 feet per second. The height y (in feet) of the fish above the surface of the water is represented by the equation y = – 16x^{2} + 12x, where x is the time (in seconds) since the jump began. The fish reaches its highest point above the surface of the water after 0.375 second. How far above the surface is the fish at this time?

Answer:

The fish is 2.25 feet above the surface at 0.375 second.

Explanation:

The height y (in feet) of the fish above the surface of the water

is represented by the equation y = – 16x^{2} + 12x,

where x is the time (in seconds) since the jump began.

The fish reaches its highest point above the surface

of the water after 0.375 second, So X = 0.375 second

we substitute In the equation as

y=- 16 (0.375 X 0.375 ) + 12 (0.375)= – 16 ( 0.140625) + 4.5

= -2.25 + 4.5 = 2.25

The fish is 2.25 feet above the surface at 0.375 second.

### Exponents Homework & Practice 8.1

**Review & Refresh**

**Sketch a graph that represents the situation.**

Question 1.

A trading card becomes more valuable over time. The value increases at a constant rate, and then at a faster and faster rate.

Answer:

Sketch is

Explanation:

We represent the graph Trading card as

on X axis Time and on Y axis Value,

Given the value increases at a constant rate,

so first we draw a straight Line with positive slope,

and increase rate is represented by exponential growth as

shown in the figure above.

Question 2.

The water level of a river remains constant, and then decreases at a constant rate.

Answer:

Explanation:

Given the water level of a river remains constant

and then decreases at a constant rate so in the graph

we draw a straight Line with positive slope as constant,

and decrease rate is represented as decay shown in the figure above.

The vertices of a figure are given. Rotate the figure as described. Find the coordinates of the image.

Question 3.

A(0, – 4), B(0, – 1), C(2, – 1)

90° clockwise about the origin

Answer:

A^{/}(-4,0)

B^{/}(-1,0)

C^{/}(-1,-2)

Explanation:

When we rotate a figure of 90 degrees clockwise about the origin,

each point of the given figure has to be changed from (x, y) to (y, -x).

So A(0,-4) becomes A^{/}(-4,-0)= A^{/}(-4,0) ,

B(0,-1) becomes B^{/}(-1,-0)= B^{/}(-1,0) and

C(2, – 1) becomes C^{/}(-1,-2)

Question 4.

E(1, 2), F(1, 3), G(4, 3), H(4, 2)

180° about the origin

Answer:

E^{/}(-1,-2)

F^{/}(-1,-3)

G^{/}(-4,-3)

H^{/}(-4,-2)

Explanation:

Rotation of a point through 180°, about the origin when a point A (x, y)

is rotated about the origin O through 180° in anticlockwise

or clockwise direction, it takes the new position A’ (-x, -y)

So E(1,2) becomes E^{/}(-1,-2), F(1,3) becomes F^{/}(-1,-3), G(4,3)

becomes G^{/}(-4,-3) and H(4,2) becomes H^{/}(-4,-2).

**Concepts, Skills, & Problem Solving**

**USING EXPONENT NOTATION** Write the power in repeated multiplication form. Then find the value of the power.(See Exploration 1, p. 319.)

Question 5.

4^{4}

Answer:

4 X 4 X 4 X 4, 256

Explanation:

First we write 4^{4 }in repeated multiplication form as

power is 4 times 4 X 4 X 4 X 4 and the value is 256.

Question 6.

(- 8)^{2}

Answer:

-8 X -8 , 64

Explanation:

First we write (- 8)^{2 }in repeated multiplication form as

– 8 X -8 as power is 2 times and the value is 64.

Question 7.

(- 2)^{3}

Answer:

– 2 X -2 X -2 , -8

Explanation:

First we write (- 2)^{3} ^{ }in repeated multiplication form as

– 2 X -2 X -2 as power is 3 times and the value is -8.

**WRITING EXPRESSIONS USING EXPONENTS** Write the product using exponents.

Question 8.

3 • 3 • 3 • 3

Answer:

3 • 3 • 3 • 3 = (3)^{4} ^{ }

Explanation:

We write the product 3 • 3 • 3 • 3 in exponents as (3)^{4} ^{
}because 3 is multiplied by 4 times.

Question 9.

(- 6) • (- 6)

Answer:

(- 6) • (- 6) = (-6)^{2} ^{ }

Explanation:

We write the product (-6) X (-6) in exponents as (-6)^{2} ^{
}because -6 is multiplied by 2 times.

Question 10.

(- \(\frac{1}{2}\)) • (- \(\frac{1}{2}\)) • (- \(\frac{1}{2}\))

Answer:

(- \(\frac{1}{2}\)) • (- \(\frac{1}{2}\)) • (- \(\frac{1}{2}\))=

(- \(\frac{1}{2}\))^{3} ^{ }

Explanation:

We write the product (- \(\frac{1}{2}\)) X(- \(\frac{1}{2}\))

X (- \(\frac{1}{2}\)) in exponent as (- \(\frac{1}{2}\))^{3} ^{
}here – \(\frac{1}{2}\) is multiplied by 3 times.

Question 11.

\(\frac{1}{3}\) • \(\frac{1}{3}\) • \(\frac{1}{3}\)

Answer:

\(\frac{1}{3}\) • \(\frac{1}{3}\) • \(\frac{1}{3}\)=

(\(\frac{1}{3}\))^{3} ^{
}

Explanation:

We write the product ( \(\frac{1}{3}\)) X ( \(\frac{1}{3}\)) X

(\(\frac{1}{3}\)) in exponent as (\(\frac{1}{3}\))^{3} ^{
}here \(\frac{1}{3}\) is multiplied by 3 times.

Question 12.

π • π • π • x • x • x • x

Answer:

π • π • π • x • x • x • x = (π)^{3} X (x)^{4}

Explanation:

We write the product π • π • π • x • x • x • x

in exponent as (π)^{3} X (x)^{4 }here π is multiplied by

3 times and X is multiplied 4 times.

Question 13.

(- 4) • (- 4) • (- 4) • y • y

Answer:

(- 4) • (- 4) • (- 4) • y • y = (-4)^{3 }X (y)^{2}

Explanation:

Here we write the product (- 4) • (- 4) • (- 4) • y • y

as (-4)^{3 }X (y)^{2},-4 is multiplied three times and y is

multiplied by 2 times

Question 14.

6.4 • 6.4 • 6.4 • 6.4 • b • b • b

Answer:

6.4 • 6.4 • 6.4 • 6.4 • b • b • b = (6.4)^{4 }X (b)^{3
}

Explanation:

Here we write the product 6.4 • 6.4 • 6.4 • 6.4 • b • b • b

as (6.4)^{4 }X (b)^{3},6.4 is multiplied four times and b is

multiplied by 3 times.

Question 15.

(- t) • (- t) • (- t) • (- t) • (- t)

Answer:

(- t) • (- t) • (- t) • (- t) • (- t ) = (-t)^{5}

Explanation:

Here we write the given product (- t) • (- t) • (- t) • (- t) • (- t)

as (-t)^{5 }because – t is multiplied by 5 times.

Question 16.

– (7 • 7 • 7 • 7 • 7)

Answer:

– (7 • 7 • 7 • 7 • 7) = -(7)^{5}

Explanation:

We write the given product as – (7 • 7 • 7 • 7 • 7)

= -(7)^{5 }here -(7) is multiplied by 5 times.

Question 17.

\(-\left(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\right)\)

Answer:

\(-\left(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\right)\)= – (\(\frac{1}{4}\))^{4}

Explanation:

We write the given product as

\(-\left(\frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4} \cdot \frac{1}{4}\right)\)

as – (\(\frac{1}{4}\))^{4} because \(\frac{1}{4}\)

is multiplied by 4 times.

**EVALUATING EXPRESSIONS** Evaluate the expression.

Question 18.

5^{2}

Answer:

5^{2 }= 5 X 5 = 25

Explanation:

Given 5^{2 }we write the expression as 5 X 5

and the value is 25

Question 19.

– 11^{3}

Answer:

– 11^{3 }= (-11 X -11 X -11) = -1331

Explanation:

Given – 11^{3 } we write it as (-11 X -11 X -11)

we get -1331.

Question 20.

(- 1)^{6}

Answer:

(- 1)^{6}= (-1 X -1 X -1 X -1 X -1 X -1) = 1

Explanation:

Given (- 1)^{6 }as power is 6 we multiply -1 by

6 times as (-1 X -1 X -1 X -1 X -1 X -1) we get 1.

Question 21.

(\(\frac{1}{6}\))^{6}

Answer:

(\(\frac{1}{6}\))^{6}

= \(\frac{1}{6}\) ^{ }X \(\frac{1}{6}\) X

\(\frac{1}{6}\) X \(\frac{1}{6}\) X

\(\frac{1}{6}\) X \(\frac{1}{6}\) = \(\frac{1}{46656}\)

Explanation:

Given (\(\frac{1}{6}\))^{6 }here power is 6 we multiply

\(\frac{1}{6}\) by 6 times as \(\frac{1}{6}\) ^{ }X \(\frac{1}{6}\) X

\(\frac{1}{6}\) X \(\frac{1}{6}\) X

\(\frac{1}{6}\) X \(\frac{1}{6}\) we get \(\frac{1}{46656}\)

Question 22.

(- \(\frac{1}{12}\))^{2}

Answer:

(- \(\frac{1}{12}\))^{2 }= –\(\frac{1}{12}\) X –\(\frac{1}{12}\) = \(\frac{1}{144}\)

Explanation:

Given (- \(\frac{1}{12}\))^{2
}the expression has power 2 we write it as –\(\frac{1}{12}\) X –\(\frac{1}{12}\)

we get \(\frac{1}{144}\)

Question 23.

– (\(\frac{1}{9}\))^{3}

Answer:

– (\(\frac{1}{9}\))^{3 }= – \(\frac{1}{9}\) X – \(\frac{1}{9}\) X –\(\frac{1}{9}\)=

– \(\frac{1}{729}\)

Explanation:

Given – (\(\frac{1}{9}\))^{3 }as in the expression we have power 3 we multiply

– \(\frac{1}{9}\) by three times as – \(\frac{1}{9}\) X – \(\frac{1}{9}\) X –\(\frac{1}{9}\) we get –\(\frac{1}{729}\)

Question 24.

**YOU BE THE TEACHER**

Your friend evaluates the power – 6^{2}. Is your friend correct? Explain your reasoning.

Answer:

Yes friend is correct,

Explanation:

Given friend evaluates the power – 6^{2 }as -6 X -6 = 36,

as given -6 has power 2 we multiply -6 twice so we get 36,

which is similar to what friend has evaluated so friend is correct.

**STRUCTURE** Write the prime factorization of the number using exponents.

Question 25.

675

Answer:

675 = 3^{3 }X 5^{2}

Explanation:

The number 675 is a composite number so, it is possible to factorize it.

In other words, 675 can be divided by 1, by itself and at least by 3 and 5.

A composite number is a positive integer that has at least one positive divisor

other than one or the number itself.

In other words, a composite number is any integer greater than one that is not a prime number.

The prime factorization of 675 = 3^{3 }X 5^{2}.

The prime factors of 675 are 3 and 5.

Question 26.

280

Answer:

280 = 2^{3 }X 5 X 7

Explanation:

The number 280 is a composite number so, it is possible to factorize it.

In other words, 280 can be divided by 1, by itself and at least by 2, 5 and 7.

A composite number is a positive integer that has at least one

positive divisor other than one or the number itself.

In other words, a composite number is any integer greater than

one that is not a prime number.

The prime factorization of 280 = 2^{3 }X 5 X 7.

The prime factors of 280 are 2, 5 and 7.

Question 27.

363

Answer:

363 = 3 X 11^{2}

Explanation:

The number 363 is a composite number so, it is possible to factorize it.

In other words, 363 can be divided by 1, by itself and at least by 3 and 11.

A composite number is a positive integer that has at least one

positive divisor other than one or the number itself.

In other words, a composite number is any integer greater than

one that is not a prime number.

The prime factorization of 363 = 3 X 11^{2}.

The prime factors of 363 are 3 and 11.

Question 28.

**PATTERNS**

The largest doll is 12 inches tall. The height of each of the other dolls is \(\frac{7}{10}\) the height of the next larger doll. Write an expression involving a power that represents the height of the smallest doll. What is the height of the smallest doll?

Answer:

The height of the smallest doll is 4.116 inches.

Explanation:

Given the largest doll is 12 inches tall, The height of each of the other dolls is

\(\frac{7}{10}\) the height of the next larger doll, There are 4 dolls

So 12 X \(\frac{7}{10}\) X \(\frac{7}{10}\) X \(\frac{7}{10}\) =

\(\frac{4116}{1000}\) = 4.116 inches, therefore the height of the smallest doll is 4.116 inches.

**USING ORDER OF OPERATIONS** Evaluate the expression.

Question 29.

5 + 2 • 2^{3}

Answer:

5 + 2 • 2^{3 }= 21

Explanation:

Given 5 + 2 • 2^{3 }we first simplify 2^{3 }and multiply with 2

we get 2 X 2 X 2 X 2 = 16 and add 5, 16 + 5 = 21,

therefore 5 + 2 • 2^{3 }= 21

Question 30.

2 + 7 • (- 3)^{2}

Answer:

2 + 7 • (- 3)^{2 }= 65

Explanation:

Given expression as 2 + 7 • (- 3)^{2 }we first simplify 7 • (- 3)^{2}

we multiply – 3 twice as – 3 X -3 = 9 and multiply with 7 we get

7 X 9 = 63 now we add 2 to 63 now we get 2 + 63 = 65,

therefore 2 + 7 • (- 3)^{2 }= 65.

Question 31.

(13^{2} – 12^{2}) ÷ 5

Answer:

(13^{2} – 12^{2}) ÷ 5= 5.

Explanation:

We have expression as (13^{2} – 12^{2}) ÷ 5 first

we calculate (13^{2} – 12^{2}) so 13 X 13 = 169 and

12 X 12 = 144 we subtract 144 from 169 we get

169 – 144 = 25 now we divide 25 by 5 we get 5,

therefore (13^{2} – 12^{2}) ÷ 5= 5.

Question 32.

\(\frac{1}{2}\)(4^{3} – 6 • 3^{2})

Answer:

\(\frac{1}{2}\)(4^{3} – 6 • 3^{2}) = 5

Explanation:

Given expression as \(\frac{1}{2}\)(4^{3} – 6 • 3^{2}) First we evaluate

(4^{3} – 6 • 3^{2}) as 3^{2}) as 3 X 3 = 9 now multiply by 6 we get 6 X 9 = 54,

4^{3} = 4 x 4 X 4 = 64, So 64 – 54 =10 Now we multiply 10 with \(\frac{1}{2}\)

we get 5, therefore \(\frac{1}{2}\)(4^{3} – 6 • 3^{2}) = 5.

Question 33.

|\(\frac{1}{2}\)(7 + 5^{3})|

Answer:

|\(\frac{1}{2}\)(7 + 5^{3})| = 66

Explanation:

given expression is |\(\frac{1}{2}\)(7 + 5^{3})| so first we evaluate

(7 + 5^{3}) = 7 + 5 X 5 X 5 = 7 + 125 = 132 now we multiply 132 with \(\frac{1}{2}\)

we get \(\frac{1}{2}\) X 132 = 66. So |\(\frac{1}{2}\)(7 + 5^{3})| = 66.

Question 34.

|(- \(\frac{1}{2}\))^{3} ÷ (\(\frac{1}{4}\))^{2}|

Answer:

|(- \(\frac{1}{2}\))^{3} ÷ (\(\frac{1}{4}\))^{2}| = -2

Explanation:

Given expression as |(- \(\frac{1}{2}\))^{3} ÷ (\(\frac{1}{4}\))^{2}|

first we evaluate (- \(\frac{1}{2}\))^{3}= – \(\frac{1}{2}\) X – \(\frac{1}{2}\) X

– \(\frac{1}{2}\) = – \(\frac{1}{8}\), Now (\(\frac{1}{4}\))^{2}

= \(\frac{1}{4}\) X \(\frac{1}{4}\) = \(\frac{1}{16}\),

now we multiply – \(\frac{1}{8}\) with \(\frac{1}{16}\) = -2.

Question 35.

(9^{2} – 15 • 2) ÷ 17

Answer:

(9^{2} – 15 • 2) ÷ 17 = 3

Explanation:

The expression is (9^{2} – 15 • 2) ÷ 17 we evaluate first (9^{2} – 15 • 2) as

9 X 9 = 81 and 15 X 2 = 30 so 81 – 30 = 51 now we divide 51 by 17

we get 3 as 17 x 3 = 51 therefore (9^{2} – 15 • 2) ÷ 17 = 3.

Question 36.

– 6 • (- 5^{2} + 20)

Answer:

– 6 • (- 5^{2} + 20) = 30

Explanation:

The given expression is – 6 • (- 5^{2} + 20) we first find

(- 5^{2} + 20) = – 5 X -5 = -25 + 20 = -5 now we multiply -5 with -6

we get – 5 X -6 = 30, So – 6 • (- 5^{2} + 20) = 30.

Question 37.

(- 4 + 12 – 6^{2}) ÷ 7

Answer:

(- 4 + 12 – 6^{2}) ÷ 7 = – 4

Explanation:

Given expression is (- 4 + 12 – 6^{2}) ÷ 7 we calculate first (- 4 + 12 – 6^{2}) as

-4 +12 – (6 X 6)= -4 +12 -36 = -40 + 12 = – 28 now we divide -28 by 7 we

get -4, therefore (- 4 + 12 – 6^{2}) ÷ 7 = – 4

Question 38.

**STRUCTURE**

Copy and complete the table. Compare the values of 2^{h} – 1 with the values of 2^{h – 1}. When are the values the same?

Answer:

Comparing the values of 2^{h} – 1 with the values of 2^{h – 1 }both do not

have any same values.

Explanation:

First we calculate 2^{h }-1 we substitute h as 1,2,3,4,5

we get If h is 1 , 2^{1 }-1= 2 – 1 = 2, if h is 2 we get 2^{2 }-1= 4 -1 = 3,

now h = 3, 2^{3 }-1 = 8 – 1 = 7, if h =4 , 2^{4 }-1= 16 – 1 = 15 and

h=5, 2^{5 }-1= 32 – 1 = 31. So for h =1,2,3,4,5 we get 2^{h }-1 = 2,3,7,15,31 respectively

Now we substitute for h= 1, 2^{h-1 }= 2^{1-1}= 2^{0}= 1, for h = 2 it is 2^{2-1 }= 2^{1 }= 2, now h = 3

2^{3-1 }= 2^{2 }= 2 X 2 = 4 , if h is 4 we get 2^{4-1 }= 2^{3 }= 2 X 2 X = 8 and if h is 5

we get 2^{5-1 }= 2^{4 }= 2 X 2 X 2 X 2 = 16, So for h = 1,2,3,4,5, we get 2^{h-1 }= 1,

2,4,8,16 respectively. As comparing the values of 2^{h} – 1 with the values of 2^{h – 1
}both do not have any same values. Hence no common values.

Question 39.

**MODELING REAL LIFE**

Scientists use carbon-14 dating to determine the age of a sample of organic material.

a. The amount C(in grams) of carbon-14 remaining after t years of a

sample of organic material is represented by the equation C = 100(0.99988)^{t}. Find the amount of carbon-14 remaining after 4 years.

b. What percent of the carbon-14 remains after 4 years?

Answer:

a. The amount of carbon – 14 remaining after 4 years is 99.95 grams.

b. The percent of the carbon – 14 remains after 4 years is 99.95%.

Explanation:

a. Given the amount C(in grams) of carbon-14 remaining

after t years of a sample of organic material is represented

by the equation C = 100(0.99988)^{t }we the amount of carbon-14 remaining

after 4 years as t = 4 we substitute C = 100(0.99988)^{4 }we get C =

100 X 0.99988 X 0.99988 X 0.99988 X 0.99988 = 99.95 grams,

therefore The amount of carbon – 14 remaining after 4 years is 99.95 grams.

b. Now the percent of the carbon – 14 remains after 4 years , we have the amount

of carbon after 4 years is 99.95 grams,

So Percentage is 100 X 99.95 by 100 = 99.95 %.

Question 40.^{
}

**DIG DEEPER!**

The frequency (in vibrations per second) of a note on a piano is represented by the equation F = 440(1.0595)^{n}, where n is the number of notes above A440. Each black or white key represents one note.

a. How many notes do you take to travel from A440 to A?

b. What is the frequency of A?

c. Compare the frequency of A to the frequency of A440.

Answer:

a. There are 12 notes to travel from A440 to A.

b. The frequency of A is 880 vibrations.

c. The frequency of A is twice to the frequency of A440.

Explanation:

a. As each black or white key represents one note, to reach

from A 440 to A if we count there are 12 notes to travel.

b. The frequency (in vibrations per second) of a note on a piano

is represented by the equation F = 440(1.0595)^{n}, For note A, n =12

F = 440(1.0595)^{12}, F = 440 X 1.0595 X 1.0595 X 1.0595 X 1.0595 X

1.0595 X 1.0595 X 1.0595 X 1.0595 X 1.0595 X 1.0595 X 1.0595 X 1.0595 =

F=880.37 therefore the frequency of A is 880 vibrations.

c. The frequency of A to the frequency of A440 is 880 by 440(1.0595)^{0},

we get approximately 2, So the frequency of A is twice to the frequency of A440.

### Lesson 8.2 Product of Powers Property

**EXPLORATION 1**

Finding Products of Powers

Work with a partner.

a. Copy and complete the table. Use your results to write a general rule for finding a^{m} • a^{n}, a product of two powers with the same base.

b. Show how to use your rule in part(a) to write each expression below as a single power. Then write a general rule for finding (a^{m})^{n}, a power of a power.

Answer:

a.

General rule for a^{m} • a^{n }= a^{m+n }a product of two powers with the same base

then powers are added.

b.

(7^{3})^{2 }= 7^{3×2} = 7^{6},

(6^{2})^{2} = 6^{2×2} = 6^{4},

(3^{2})^{3} = 3^{2×3} = 3^{6},

(2^{2} )^{4}= 2^{2×4 }= 2^{8},

((\(\frac{1}{2}\))^{2})^{5 }= (\(\frac{1}{2}\))^{2 x 5 }= (\(\frac{1}{2}\))^{10
}General rule for finding (a^{m})^{n }power of a power,

If two powers have the same base then

we can multiply the powers as (a^{m})^{n }= a^{m x }^{n} .

**
**Explanation:

a. Completed the table as shown above as Product,

Repeated Multiplication Form and Power as

(2

^{2 }X 2

^{4}) = 2

^{2+4 }= 2 X 2 X 2 X 2 X 2 X 2 = 2

^{6 }

(-3)

^{2 }X (-3)

^{4 }= (-3)

^{2+4 }= -3 X -3 X -3 X -3 X -3 X -3 = (-3)

^{6 }

7

^{3 }X 7

^{2 }= (7)

^{3+2 }= 7 X 7 X 7 X 7 X 7 = 7

^{5 }

5.1

^{1 }X 5.1

^{6 }= (5.1)

^{1+6 }= 5.1 X 5.1 X 5.1 X 5.1 X 5.1 X 5.1 X5.1 =(5.1)

^{7 }

(-4)

^{2 }X (-4)

^{2 }= (-4)

^{2+2 }= -4 X -4 X -4 X -4 = (-4)

^{4 }

10

^{3 }X 10

^{5}= 10

^{3+5 }= 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 = 10

^{8}

(\(\frac{1}{2}\))

^{5 }X (\(\frac{1}{2}\))

^{5}=(\(\frac{1}{2}\))

^{5+5 }=\(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\)

^{ }X \(\frac{1}{2}\) = (\(\frac{1}{2}\))

^{10 }General rule for a

^{m}• a

^{n }= a

^{m+n }a product of two powers with the same base,

powers are added. If two powers have the same base

then we can multiply the powers.

When we multiply two powers we add their exponents.

b. We write (7

^{3})

^{2 }as 7

^{3×2}= 7

^{6},

(6

^{2})

^{2}as 6

^{2×2}= 6

^{4},

(3

^{2})

^{3}as 3

^{2×3}= 3

^{6},

(2

^{2})

^{4 }as 2

^{2×4 }= 2

^{8},

((\(\frac{1}{2}\))

^{2})

^{5 }= (\(\frac{1}{2}\))

^{2 x 5 }= (\(\frac{1}{2}\))

^{10 }wrote each expression as a single power above,

General rule for finding (a

^{m})

^{n}a power of a power,

If two powers have the same base then

we can multiply the powers as (a

^{m})

^{n }= a

^{m x }

^{n}.

**EXPLORATION 2**

Finding Powers of Products

Work with a partner. Copy and complete the table. Use your results to general rule write a for finding (ab)^{m}, a power of a product.

Answer:

Explanation:

Completed the table as shown above as first

Repeated Multiplication Form and Product of Powers as

(2 X 3)^{3} = 2 X 2 X 2 X 3 X 3 X 3 = 2^{3} X 3^{3
}(2 X 5)^{2 }= 2 X 2 X 5 X 5 = 2^{2} X 5^{2
}(5 X 4)^{3} = 5 X 5 X 5 X 4 X 4 X 4 = 5^{3} X 4^{3
}(-2 X 4)^{2 }= -2 X -2 X 4 X 4 = -2^{2 }X 4^{2
}(-3 X 2)^{4 }= -3 X -3 X -3 X -3 X 2 X 2 X 2 X 2 = -3^{4} X 2^{4
}We know general rule to write (ab)^{m } power of a product is a^{m }X b^{m }

**Try It**

**Simplify the expression. Write your answer as a power.**

Question 1.

6^{2} • 6^{4}

Answer:

6^{2} • 6^{4 }= 6^{6}

Explanation:

Given 6^{2} • 6^{4 }we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added. As both base are 6 so 6^{2} • 6^{4 }= 6^{2+4 }= 6^{6}

Question 2.

(- \(\frac{1}{2}\))^{3} • (- \(\frac{1}{2}\))^{6}

Answer:

(- \(\frac{1}{2}\))^{3} • (- \(\frac{1}{2}\))^{6 }=(- \(\frac{1}{2}\))^{9
}Given (- \(\frac{1}{2}\))^{3} • (- \(\frac{1}{2}\))^{6 }we have general rule for

a^{m} • a^{n }= a^{m+n }If product of two powers with the same base

then powers are added as both bases are same so (- \(\frac{1}{2}\))^{3} • (- \(\frac{1}{2}\))^{6 }=

(- \(\frac{1}{2}\))^{3+6 }=(- \(\frac{1}{2}\))^{9}

Question 3.

z • z^{12}

Answer:

z • z^{12 }= z^{13}

Explanation:

Given z • z^{12 }we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added, here both bases are z so z • z^{12 }= z^{1+12}= z^{13}

**Simplify the expression. Write your answer as a power.**

Question 4.

(4^{3})^{5}

Answer:

(4^{3})^{5 }=(4)^{15}

Explanation:

Given (4^{3})^{5 }we have general rule for finding (a^{m})^{n }a power of a power,

If two powers have the same base then we can multiply the

powers as (a^{m})^{n }= a^{m x }^{n }. Here both base is 4 so (4^{3})^{5 }=(4^{3x}^{5}) = (4)^{15}

Question 5.

(y^{2})^{4}

Answer:

(y^{2})^{4 }=(y)^{8}

Explanation:

Given (y^{2})^{4 }we have general rule for finding (a^{m})^{n }a power of a power,

If two powers have the same base then we can multiply the

powers as (a^{m})^{n }= a^{m x }^{n }. Here both base is y so (y^{2})^{4 }=(y^{2x}^{4}) = (y)^{8}

Question 6.

((- 4)^{3})^{2}

Answer:

((- 4)^{3})^{2 }= (-4)^{6}

Explanation:**
**Given ((- 4)

^{3})

^{2 }we use general rule for finding (a

^{m})

^{n }a power of a power,

If two powers have the same base then we can multiply the

powers as (a

^{m})

^{n }= a

^{m x }

^{n }. Here both base is 4 so ((- 4)

^{3})

^{2 }=(- 4)

^{3x}

^{2 }= (-4)

^{6}

Simplify the expression.

Question 7.

(5y)^{4}

Answer:

(5y)^{4 }= 5^{4} X y^{4}

Explanation:

Given (5y)^{4 }to simplify the expression we use general rule to

write (ab)^{m } power of a product as a^{m }X b^{m },So (5y)^{4 }= 5^{4} X y^{4}

Question 8.

(ab)^{5}

Answer:

(ab)^{5}= a^{5 }X b^{5
}

Explanation:

Given (ab)^{5 }to simplify the expression we use general rule to

write (ab)^{m } power of a product as a^{m }X b^{m },So (ab)^{5}= a^{5 }X b^{5}

Question 9.

(0.5 mn)^{2}

Answer:

(0.5 mn)^{2}= 0.5^{2 }X m^{2 }X n^{2}

Explanation:

Given (0.5 mn)^{2} to simplify the expression we use general rule to

write (abc)^{m } power of a product as a^{m }X b^{m }X c^{m }, So (0.5 mn)^{2}= 0.5^{2 }X m^{2 }X n^{2}

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

**FINDING POWERS** Simplify the expression. Write your answer as a power.

Question 10.

4^{7} • 4^{4}

Answer:

4^{7} • 4^{4}= 4^{11}

Explanation:

Given 4^{7} • 4^{4 }we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added. So 4^{7} • 4^{4 }= 4^{7+4 }= 4^{11}

Question 11.

(g^{6})^{3}

Answer:

(g^{6})^{3}=g^{18}

Explanation:

Given (g^{6})^{3 }we have general rule for finding (a^{m})^{n }a power of a power,

If two powers have the same base then we can multiply the

powers as (a^{m})^{n }= a^{m x }^{n }. So (g^{6})^{3}= g^{6x}^{3}= g^{18}

Question 12.

(- \(\frac{1}{3}\))^{5} • (- \(\frac{1}{3}\))^{7}

Answer:

(- \(\frac{1}{3}\))^{5} • (- \(\frac{1}{3}\))^{7}= (- \(\frac{1}{3}\))^{12}

Explanation:

Given (- \(\frac{1}{3}\))^{5} • (- \(\frac{1}{3}\))^{7}**
**we have general rule for a

^{m}• a

^{n }= a

^{m+n }If product of two powers with the same base then

powers are added, here both bases are – \(\frac{1}{3}\) so

(- \(\frac{1}{3}\))

^{5}• (- \(\frac{1}{3}\))

^{7}= (- \(\frac{1}{3}\))

^{5+7}= (- \(\frac{1}{3}\))

^{12}

**FINDING A POWER OF A PRODUCT Simplify the expression.**

Question 13.

(8t)^{4}

Answer:

(8t)^{4 }= 8^{4 }X t^{4}

Explanation:

Given (8t)^{4 }we have general rule to write (ab)^{m }

power of a product as a^{m }X b^{m }therefore (8t)^{4 }= 8^{4 }X t^{4}

Question 14.

(yz)^{6}

Answer:

(yz)^{6}= y^{6} X z^{6}

Explanation:

Given (yz)^{6 }we have general rule to write (ab)^{m }

power of a product as a^{m }X b^{m }therefore (yz)^{6}= y^{6} X z^{6
}

Question 15.

(\(\frac{1}{4}\)gh)^{3}

Answer:

(\(\frac{1}{4}\)gh)^{3}= (\(\frac{1}{4}\))^{3 }X g^{3 }X h^{3}

Explanation:

Given (\(\frac{1}{4}\)gh)^{3 }to simplify the expression we use general rule to

write (abc)^{m } power of a product as a^{m }X b^{m }X c^{m }, So (\(\frac{1}{4}\)gh)^{3}= (\(\frac{1}{4}\))^{3 }X g^{3 }X h^{3}

Question 16.

**CRITICAL THINKING**

Can you use the Product of Powers Property to simplify 5^{2} • 6^{4}? Explain.

Answer:

No, we can not use the Product of Powers Property to simplify 5^{2} • 6^{4}

Explanation:

Given to simplify 5^{2} • 6^{4 }as both bases are different

and Product of Powers Property is in general,

for all real numbers we multiply two powers having

the same base we add the exponents but here the bases are different

5,6 so no, we can not use the Product of Powers Property to simplify 5^{2} • 6^{4}

Question 17.

**OPEN-ENDED**

Write an expression that simplifies to x^{12} using the Product of Powers Property.

Answer:

The expression is x^{2} X x^{10} simplifies to x^{12}

Explanation:

Given an expression that simplifies to x^{12} by

using the Product of Powers Property we write as x^{12} = x^{2} X x^{10} **
**as we know Product of Powers Property is for a

^{m}• a

^{n }= a

^{m+n }If product of two powers with the same base then

powers are added. As base is x here we have m + n = 12,

so we can take powers (m , n) as (2,10) or (6,6) or (1,11) or (3,9) or

(4,8) or (5,7) or (7,5) or (8,4) or (9,3) or (10,2) or (11,1). So I took (2,10)

as m, n and wrote expression as x

^{2}X x

^{10}= x

^{12}.

Self-Assessment for Problem Solving

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.

A newborn blue whale weighs 3^{7} kilograms. An adult blue whale weighs 81 times the weight of the newborn. How many kilograms does the adult blue whale weigh?

Answer:

The adult blue whale weighs 81 X 3^{7} kilograms = 3^{11} kilograms

Explanation:

Given A newborn blue whale weighs 3^{7} kilograms.

An adult blue whale weighs 81 times the weight of the newborn.

So the adult blue whale weighs 81 X 3^{7} kilograms , we write 81 as

multiple of 3 we get 81 = 3 X 3 X 3 X 3, 81 = 3^{4} therefore

81 X 3^{7} kilograms = 3^{4} X 3^{7} kilograms,

when bases are same powers are added

3^{4} X 3^{7} kilograms = 3^{4+7} = 3^{11} kilograms,

therefore the adult blue whale weighs

81 X 3^{7} kilograms = 3^{11} kilograms.

Question 19.

One megabyte of cell phone storage space is 2^{20} bytes. An app uses 4^{4} megabytes of storage space. How many bytes of storage space does the app use?

Answer:

The app used 2^{28 }bytes of storage space.

Explanation:

Given One megabyte of cell phone storage space is 2^{20} bytes.

and an app uses 4^{4} megabytes of storage space.

the number of bytes of storage space does the app use is

4^{4 }X 2^{20} bytes as 4 we can write as 2 X 2 = 2^{2} (2^{2})^{4} X 2^{20}

4^{4 }X 2^{20 }= (2^{2})^{4} X 2^{20} = 2^{8 }X 2^{20}= 2^{8+20 }= 2^{28 }bytes.

Question 19.

**DIG DEEPER!**

The diagram shows the area of a small circular rug. The radius of a large circular rug is 3 times the radius of the small rug. Write an expression for the area of the large rug in terms of x. Justify your answer.

Answer:

The area of the large rug in terms of x is 9πx^{2 }X \(\frac{1}{4}\)

Explanation:

We have area of small circular rug as

A= \(\frac{1}{4}\) X πx^{2 }

A = π X (\(\frac{x}{2}\))^{2
}hence the radius of small rug is \(\frac{x}{2}\) and radius

of large rug is \(\frac{3x}{2}\), therefore the area of the

large rug is π^{ }X (\(\frac{3x}{2}\))^{2}= 9πx^{2 }X \(\frac{1}{4}\),

therefore the area of the large rug in terms of x is 9πx^{2 }X \(\frac{1}{4}\).

### Product of Powers Property Homework & Practice 8.2

**Review & Refresh**

**Write the product using exponents.**

Question 1.

11 • 11 • 11 • 11 • 11

Answer:

11 • 11 • 11 • 11 • 11 = 11^{5}

Explanation:

Given 11 • 11 • 11 • 11 • 11 as 11 is multiplied by 5 times

we write as 11 • 11 • 11 • 11 • 11 = 11^{5}

Question 2.

(- 6) • (- 6) • (- 6) • z • z

Answer:

(- 6) • (- 6) • (- 6) • z • z = (-6)^{3 }X z^{2}

Explanation:**
**Given (- 6) • (- 6) • (- 6) • z • z we have -6 multiplied

by 3 times and z twice so write the product as

(- 6) • (- 6) • (- 6) • z • z = (-6)

^{3 }X z

^{2}

Find the value of for the given value of x.

Question 3.

y = – 4x; x = 7

Answer:

y= -28

Explanation:

Given y = -4x and value of x as 7,

we substitute x as y = -4 X 7, So y = -28.

Question 4.

y = 5x + 6; x = – 2

Answer:

y= -4

Explanation:

Given y= 5x + 6 and value of x as -2,

we substitute x as y= 5 X -2 + 6 = -10 + 6 = -4,

therefore y = -4.

Question 5.

y = 10 – 3x ; x = 3

Answer:

y=1

Explanation:

Given y= 10 – 3x and value of x as 3,

we substitute x as y= 10 – 3 X 3 = 10 – 9 = 1,

therefore y =1.

Question 6.

What is the measure of each interior angle of the regular polygon?

A. 45°

B. 135°

C. 1080°

D. 1440°

Answer:

The measure of each interior angle of the

regular polygon given is B.135°.

Explanation:

We know each Angle (of a Regular Polygon) =

(n−2) × 180° / n where n is number of sides

in the given figure we 8 sides so n is 8,

the measure of each interior angle is (8-2) X 180° / 8=

6 X 180° / 8 = 135° . So the measure of each

interior angle of the regular polygon given is B.135°.

**Concepts, Skills, &Problem Solving**

**FINDING PRODUCTS OF POWERS** Write the expression in repeated multiplication form. Then write the expression as a power. (See Exploration 1, p. 325.)

Question 7.

5^{6} • 5^{3}

Answer:

5^{6} • 5^{3 }= 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 = 5^{9}

Explanation:

Given 5^{6} • 5^{3 }the expression in repeated multiplication form is

5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 X 5 and expression as power

is 5^{6} • 5^{3}= 5^{6+3} = 5^{9}

Question 8.

(6^{4})^{2}

Answer:

(6^{4})^{2 }= 6 X 6 X 6 X 6 X 6 X 6 X 6X 6 = 6^{8}

Explanation:

Given (6^{4})^{2} the expression in repeated multiplication form is

6 X 6 X 6 X 6 X 6 X 6 X 6 X 6 and expression as power

is (6^{4})^{2 }= 6^{4×2} = 6^{8}

Question 9.

(- 8)^{3} • (- 8)^{4}

Answer:

(- 8)^{3} • (- 8)^{4}= (- 8)^{7}

Explanation:

Given (- 8)^{3} • (- 8)^{4 }the expression in repeated multiplication form is

-8 X -8 X -8 X -8 X -8 X -8 X -8 and expression as power

is (- 8)^{3} • (- 8)^{4}= (- 8)^{3+4} = (- 8)^{7}

**FINDING POWERS Simplify the expression. Write your answer as a power.**

Question 10.

3^{2} • 3^{2}

Answer:

3^{2} • 3^{2 }= 3^{4}

Explanation:

We write the given 3^{2} • 3^{2 }expression as a power,

so 3^{2} • 3^{2 }as bases are same 3 powers are added 3^{2+2} = 3^{4}

Question 11.

8^{10} • 8^{4}

Answer:

8^{10} • 8^{4 }= 8^{14}

Explanation:

We write the given 8^{10} • 8^{4 }expression as a power,

here 8^{10} • 8^{4}^{ }has same bases 8 so same powers are

added as 8^{10+4} = 8^{14}

Question 12.

(5^{4})^{3}

Answer:

(5^{4})^{3 }= (5)^{12}

Explanation:

We write the given expression (5^{4})^{3 }as a power,

so (5^{4})^{3 } has powers of powers therefore powers

are multiplied as (5)^{4 x 3 }= 5^{12}

Question 13.

((- 3)^{2})^{4}

Answer:

((- 3)^{2})^{4 }= (- 3)^{8}

Explanation:

We write the given expression ((- 3)^{2})^{4 }as a power,

so ((- 3^{2})^{4 } has powers of powers therefore powers

are multiplied as (-3)^{2 x 4 }= (-3)^{8}

Question 14.

(- 4)^{5} • (- 4)^{7}

Answer:

(- 4)^{5} • (- 4)^{7}= (- 4)^{12}

Explanation:

We write the given expression (- 4)^{5} • (- 4)^{7} as a power,

here (- 4)^{5} • (- 4)^{7}^{ }has same bases -4 so same powers are

added as (- 4)^{5+7} = (- 4)^{12}.

Question 15.

h^{6} • h

Answer:

h^{6} • h = h^{7}

Explanation:

We write the given expression h^{6} • h as a power,

here h^{6} • h^{ }has same bases h so same powers are

added as (h)^{6+1} = (h)^{7}.

Question 16.

(b^{12})^{3}

Answer:

(b^{12})^{3 }= (b)^{36}

Explanation:

We write the given expression (b^{12})^{3 }as a power,

so (b^{12})^{3 } has powers of powers therefore powers

are multiplied as (b)^{12 x 3 }= (b)^{36}

Question 17.

(\(\frac{2}{3}\))^{2} • (\(\frac{2}{3}\))^{6}

Answer:

(\(\frac{2}{3}\))^{2} • (\(\frac{2}{3}\))^{6}= (\(\frac{2}{3}\))^{8}

Explanation:

We write the given expression (\(\frac{2}{3}\))^{2} • (\(\frac{2}{3}\))^{6
}as a power, here (\(\frac{2}{3}\))^{2} • (\(\frac{2}{3}\))^{6}^{
}has same bases so same powers are added as (\(\frac{2}{3}\))^{2+6}

= (\(\frac{2}{3}\))^{8}^{ }

Question 18.

(3.8^{3})^{4}

Answer:

(3.8^{3})^{4}= (3.8)^{12}

Explanation:

We write the given expression (3.8^{3})^{4}^{ }as a power,

so (3.8^{3})^{4}^{ } has powers of powers therefore powers

are multiplied as (3.8)^{3 x 4 }= (3.8)^{12}

Question 19.

(n^{3})^{5}

Answer:

(n^{3})^{5 }= (n)^{15}

Explanation:

We write the given expression (n^{3})^{5}^{ }as a power,

so (n^{3})^{5 }has powers of powers therefore powers

are multiplied as (n)^{3 X 5}= (n)^{15}

Question 20.

((- \(\frac{3}{4}\))^{5})^{2}

Answer:

((- \(\frac{3}{4}\))^{5})^{2}= (- \(\frac{3}{4}\))^{10}

Explanation:

We write the given expression ((- \(\frac{3}{4}\))^{5})^{2 }as a power,

so ((- \(\frac{3}{4}\))^{5})^{2}^{ }has powers of powers therefore powers

are multiplied as (- \(\frac{3}{4}\))^{5 x 2}= (- \(\frac{3}{4}\))^{10}

Question 21.

(- \(\frac{1}{2}\))^{8} • (- \(\frac{1}{2}\))^{9}

Answer:

(- \(\frac{1}{2}\))^{8} • (- \(\frac{1}{2}\))^{9}= (- \(\frac{1}{2}\))^{17}

Explanation:

We write the given expression (-\(\frac{1}{2}\))^{8} • (- \(\frac{1}{2}\))^{9
}as a power, here (-\(\frac{1}{2}\))^{8} • (- \(\frac{1}{2}\))^{9}^{
}has same bases so same powers are added as (-\(\frac{1}{2}\))^{8+9}

= (-\(\frac{1}{2}\))^{17}^{ }

**YOU BE THE TEACHER** Your friend simplifies the expression.

Is your friend correct? Explain your reasoning.

Question 22.

Answer:

No, Friend is in correct as 5^{2 }X 5^{9 }= 5^{11 }not 25^{11}

Explanation:

Given expression is 5^{2 }X 5^{2
}here as base is same 5 we add powers

as 5^{2+9 }we get 5^{11 }not 25^{11}

Question 23.

Answer:

No, friend is In correct as (r^{6})^{4}= (r)^{6 x 4 }= (r)^{24 }not r^{10}

Explanation:**
**Given expression is (r

^{6})

^{4 }here we have

powers of powers therefore powers

are multiplied not added so (r

^{6})

^{4}= (r)

^{6 x 4 }= (r)

^{24 }not r

^{10 }therefore friend is incorrect.

**Simplify the expression.**

FINDING A POWER OF A PRODUCT

FINDING A POWER OF A PRODUCT

Question 24.

(6g)

^{3}

Answer:

(6g)

^{3 }= 216 g

^{3}

Explanation:

Given (6g)^{3 }we simplify as 6^{3 }X g^{3}= 6 X 6 X 6 X g X g X g

= 216 g^{3 }here we multiplied 6 by 3 times and g by

3 times as both has power 3.

Question 25.

(- 3v)^{5
}Answer:

(- 3v)^{5 }= -243v^{5}

Explanation:

Given (-3v)^{5 }we simplify as

– 3 X -3 X -3 X -3 X -3 X v X v X v X v X v = -243v^{5},

here we multiplied -3 and v by

5 times as power for both is 5.

Question 26.

(\(\frac{1}{5}\)k)^{2}

Answer:

(\(\frac{1}{5}\)k)^{2 }= \(\frac{1}{25}\)k^{2
}

Explanation:

Given expression (\(\frac{1}{5}\)k)^{2 }we simplify as

\(\frac{1}{5}\) X \(\frac{1}{5}\) X k X k = \(\frac{1}{25}\)k^{2}

here we multiply \(\frac{1}{5}\) and k

by 2 times as power for both is 2.

Question 27.

(1.2 m)^{4}

Answer:

(1.2 m)^{4 }= 2.0736m^{4}

Explanation:

Given expression as (1.2 m)^{4 }we simplify as

1.2 X 1.2 X 1.2 X m X m X m X m is 2.0736m^{4}

here we multiply 1.2 and m by 4 times as

power for both is 4.

Question 28.

(rt)^{12}

Answer:

(rt)^{12 }= r^{12 }X t^{12}

Explanation:

Given expression as (rt)^{12 }we simplify as

r X r X r X r X r X r X r X r X r X r X r X r X

t X t X t X t X t X t X t X t X t X t X t X t is r^{12 }X t^{12}

here we multiply r and t by 12 times as

power for both is 12.

Question 29.

(- \(\frac{3}{4}\)p)^{3}

Answer:

(- \(\frac{3}{4}\)p)^{3 }= –\(\frac{27}{64}\)p^{3}

Explanation:

Given expression as (- \(\frac{3}{4}\)p)^{3 } we simplify as

– \(\frac{3}{4}\) X – \(\frac{3}{4}\) X – \(\frac{3}{4}\) X p X p X p is

–\(\frac{27}{64}\)p^{3}

here we multiply –\(\frac{3}{4}\) and p by 3 times as

power for both is 3.

Question 30.

**PRECISION**

Is 3^{2} + 3^{3} equal to 3^{5}? Explain.

Answer:

No,3^{2} + 3^{3} is not equal to 3^{5
}

Explanation:

Given expression is 3^{2} + 3^{3} first we solve

3^{2} as 3 X 3 = 9 and 3^{3} = 3 X 3 X 3 = 27 so

3^{2} + 3^{3}= 9 + 27 = 36 now we have 3^{5} which is

equal to 3 X 3 X 3 X 3 X 3 = 243 as 36 ≠ 243 so

no, 3^{2} + 3^{3 }≠ 3^{5}

Question 31.

**PROBLEM SOLVING**

A display case for the artifact shown is in the shape of a cube. Each side of the display case is three times longer than the width w of the artifact.

a. Write a power that represents the volume of the case.

b. Simplify your expression in part(a).

Answer:

a. The volume of the case is (3w)^{3}.

b. Simplified form is 27w^{3}.

Explanation:

Given a display case for the artifact shown

is in the shape of a cube. Each side of the display

case is three times longer than the width w of the artifact.

a. we have volume of cube as (edge)^{3
}So the volume of the case = 3w X 3w X 3w = (3w)^{3}.

b. Simplified form of (3w)^{3}= 3 X 3 X 3 X w X w X w = 27w^{3}.

Question 32.

**LOGIC**

Show that (3 • 8 • x)^{7} = 6^{7} • 4^{7} • x^{7}.

Answer:

(3 • 8 • x)^{7} = ( 3^{7} X 2^{7}) X 4^{7} X x^{7} = 6^{7} x 4^{7} x x^{7}

Explanation:

Given expression is (3 • 8 • x)^{7} we write 8 as

multiple of 2 X 4 so (3 X 2 X 4 X x)^{7}

and now we multiply first 3 and 2 as (6 X 4 X x)^{7}

we simplify we get 6^{7} x 4^{7} x x^{7}

So (3 • 8 • x)^{7} = 6^{7} • 4^{7} • x^{7}.

Question 33.

**MODELING REAL LIFE**

The lowest altitude of an altocumulus cloud is about 3^{8 }feet. The highest altitude of an altocumulus cloud is about 3 times the lowest altitude. What is the highest altitude of an altocumulus cloud? Write your answer as a power.

Answer:

The highest altitude of an altocumulus cloud is

3 X 3^{8}= 3^{9 }feet.

Explanation:

Given The lowest altitude of an altocumulus cloud is

about 3^{8 }feet. The highest altitude of an altocumulus cloud is

about 3 times the lowest altitude. So the highest altitude

of an altocumulus cloud is 3 X 3^{8 }as bases are same

powers are added 3^{1+8 }= 3^{9 }feet.

Question 34.

**GEOMETRY**

A square pyramid has a height h and a base with side lengths. The side lengths of the base increase by 50%. Write a formula for the volume of the new pyramid in terms of s and h.

Answer:

The formula for the volume of the

new pyramid is terms of s and h is 3s^{2}h by 4.

Explanation:

Given a square pyramid has a height h and

a base with side lengths. The side lengths

of the base increase by 50%. So volume is

s^{2}\(\frac{h}{3}\) and side length s

increases by 50%, so s + 50%s = s + \(\frac{50}{100}\)s

= s + \(\frac{1}{2}\)s = \(\frac{3s}{2}\).

The new volume is (\(\frac{3s}{2}\))^{2 }X \(\frac{h}{3}\)

= 9s^{2 }X \(\frac{h}{12}\) = 3s^{2}h by 4

therefore the formula for the volume of the

new pyramid is terms of s and h is 3s^{2}h by 4.

Question 35.

**MODELING REAL LIFE**

The United States Postal Service delivers about 2^{4} • 3 • 5^{3} pieces of mail each second. There are 2^{8} • 3^{4} • 5^{2} seconds in 6 days. How many pieces of mail does the United States Postal Service deliver in 6 days? Write your answer as an expression involving three powers.

Answer:

Number of pieces of mails does the United States Postal Service

deliver in 6 days is 2^{12} X 3^{5} X 5^{5} .

Explanation:

Given the United States Postal Service delivers about

2^{4} • 3 • 5^{3} pieces of mail each second. There are 2^{8} • 3^{4} • 5^{2}

seconds in 6 days. so number of pieces of mail does

the United States Postal Service deliver in 6 days is

2^{4} X 3 X 5^{3 }X 2^{8} X 3^{4} X 5^{2 }we add the same bases power we get

2^{4 }X 2^{8} X 3 X 3^{4} X 5^{3 }X 5^{2 }as 2^{4+8} X 3^{1+4} X 5^{3+2
}= 2^{12} X 3^{5} X 5^{5 }therefore number of pieces of mails

does the United States Postal Service

deliver in 6 days is 2^{12} X 3^{5} X 5^{5} .

Question 36.

**REASONING**

The row numbers y and column numbers x of a chessboard are shown. Each position on the chessboard has a stack of pennies. (Only the first row is shown.) The number of pennies in each stack is 2^{x} • 2^{y}.

a. Which locations have 32 pennies in their stacks?

b. How much money (in dollars) is in the location with the tallest stack?

c. A penny is about 0.06 inch thick. About how tall is the tallest stack?

Answer:

a. The locations are the combinations of (x,y)=

(1,4),(2,3),(3,2),(4,1).

b. $655.36 is the money in the location with the tallest stack.

c. The tallest stack is 3932.16 inches.

Explanation:

The row numbers y and column numbers x,

The number of pennies in each stack is 2^{x} • 2^{y}

a. The locations that have 32 pennies in their stacks is

2^{x} • 2^{y }= 32 we write 32 as power of 2 we get

2^{x} • 2^{y }= 2^{5} we write 2^{x} • 2^{y }(bases are same powers are added)

as 2^{x+y} = 2^{5 }therefore x + y = 5, So the locations

are the combinations of (x, y)= (1,4),(2,3),(3,2),(4,1).

b. The tallest stack will be in the location of (8,8)

so the maximum combination is 2^{8} X 2^{8 }= 2^{8+8} = 2^{16
}= 65536 pennies. Now converting pennies into dollar

1 penny = 0.01 dollar so 65536 X 0.01 = $655.36,

therefore $655.36 is the money in the location

with the tallest stack.

c. Given each penny is about 0.06 inch thick

the tallest stack is 65536 X 0.06 = 3932.16 inches.

Question 37.

**CRITICAL THINKING**

Find the value of x in the equation without evaluating the power.

Answer:

a. x = 3

b. x = 4

Explanation:

a. Given 2^{5} X 2^{x }= 256, we write 256 as 2^{8},

so 2^{5} X 2^{x }= 2^{5 + x} = 2^{8} , 5 + x = 8 so x = 8 – 5,

therefore x = 3.

b. Given (\(\frac{1}{3}\))^{2 }X (\(\frac{1}{3}\))^{x }= \(\frac{1}{729}\)

\(\frac{1}{729}\) as multiple of \(\frac{1}{3}\) we get

\(\frac{1}{729}\) as (\(\frac{1}{3}\)))^{6 } so (\(\frac{1}{3}\))^{2 }X (\(\frac{1}{3}\))^{x }= (\(\frac{1}{3}\)))^{6 } as bases are same

we equate powers as 2 + x = 6 therefore x = 6 – 2 = 4, So X = 4.

**Lesson 8.3 Quotient of Powers Property**

**EXPLORATION 1**

Finding Quotients of Powers

Work with a partner.

a. Copy and complete the table. Use your results to write a general rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base.

b. Use your rule in part(a) to simplify the quotients in the first column of the table above. Does your rule give the results in the third column?

Answer:

a.

The general rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base is a^{m-n}

b. Yes my rule gives the result in third column.

Explanation:

a. Completed the given table as shown above.

We use the general rule for finding \(\frac{a^{m}}{a^{n}}\),

a quotient of two powers with the same base is a^{m-n .}

b. We use part(a) rule to simplify the quotients in

the first column of the table above

and rule gives the results in the third column as

\(\frac{2^{4}}{2^{2}}\) =2^{4-2} = 2^{2
}\(\frac{-4^{5}}{-4^{2}}\) = (-4)^{5-2} = (-4)^{3}

\(\frac{7^{7}}{7^{3}}\) = 7^{7-3} = 7^{4}

\(\frac{8.5^{9}}{8.5^{6}}\) = (8.5)^{9-6} = (8.5)^{3}

\(\frac{10^{8}}{10^{5}}\) = 10^{8-5 }= 10^{3}

\(\frac{3^{12}}{3^{4}}\) = 3^{12-4} = 3^{8}

\(\frac{-5^{7}}{-5^{5}}\) = (-5)^{7-2 }=(-5)^{2}

\(\frac{11^{4}}{11^{1}}\) = 11^{4-1} =11^{3}

\(\frac{x^{6}}{x^{2}}\) = x^{6 – 2}= x^{4}

**Try It**

**Simplify the expression. Write your answer as a power.**

**Question 1.**

Answer:

\(\frac{9^{7}}{9^{4}}\) = 9^{7-4 }= 9^{3}

Explanation:

Given \(\frac{9^{7}}{9^{4}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\),

a quotient of two powers with the same base as a^{m-n}

so \(\frac{9^{7}}{9^{4}}\) = 9^{7-4 }= 9^{3}

Question 2.

Answer:

\(\frac{4.2^{6}}{4.2^{5}}\) = 4.2^{6-5 }= 4.2^{1 }= 4.2

Explanation:

Given \(\frac{4.2^{6}}{4.2^{5}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n}

so \(\frac{4.2^{6}}{4.2^{5}}\) = 4.2^{6-5 }= 4.2^{1 }= 4.2

Question 3.

Answer:

\(\frac{-8^{8}}{-8^{4}}\) = (-8)^{8-4 }= (-8)^{4 }

Explanation:

Given \(\frac{-8^{8}}{-8^{4}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n}

so \(\frac{-8^{8}}{-8^{4}}\) = (-8)^{8-4 }= (-8)^{4 }

Question 4.

Answer:

\(\frac{x^{8}}{x^{3}}\) = (x)^{8-3 }= (x)^{5 }

Explanation:

Given \(\frac{x^{8}}{x^{3}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\),

a quotient of two powers with the same base as a^{m-n}

so \(\frac{x^{8}}{x^{3}}\) = (x)^{8-3 }= (x)^{5 }

Simplify the expression. Write your answer as a power.

Question 5.

Answer:

\(\frac{6^{7}}{6^{5}}\) X 6^{3 }= 6^{5}

Explanation:

Given \(\frac{6^{7}}{6^{5}}\) X 6^{3 }we use rule for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n}

first we solve \(\frac{6^{7}}{6^{5}}\) = = 6^{7-5 }= 6^{2
}and now we multiply by = 6^{2 }X 6^{3 }as bases are same 6 now

we add powers as = 6^{2+3 }= 6^{5 }

therefore \(\frac{6^{7}}{6^{5}}\) X 6^{3 }= 6^{5}

Question 6.

Answer:

= 2^{7}

Explanation:

First we multiply denominators as bases are same

we add powers so 2^{3} X 2^{5 }= 2^{3+5 }= 2^{8}

now we solve \(\frac{2^{15}}{2^{8}}\) we use rule for finding

\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n
}\(\frac{2^{15}}{2^{8}}\) = 2^{15-8 }= 2^{7}

therefore = 2^{7}

Question 7.

Answer:

= m^{9}

Explanation:

First we multiply numerators as bases are same

we add powers so m^{8} X m^{6 }= m^{8+6 }= m^{14}

now we solve \(\frac{m^{14}}{m^{5}}\) we use rule for finding

\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n
} \(\frac{m^{14}}{m^{5}}\) = m^{14-5 }= m^{9}

**Simplify the expression. Write your answer as a power.**

Question 8.

Answer:

= (-5)^{6}

Explanation:

First we calculate separately values of numerators

and denominators then divide , we have numerator (-5)^{7 }X (-5)^{6
}we have same bases as -5 so we add powers as (-5)^{7+6 }= (-5)^{13
}we have denominator (-5)^{5 }X (-5)^{2
}we have same base as -5 so we add powers as (-5)^{5+2 }= (-5)^{7}

Now we have \(\frac{-5^{13}}{-5^{7}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\), a quotient of

two powers with the same base as a^{m-n
} \(\frac{-5^{13}}{-5^{7}}\) = (-5)^{13-7 }= (-5)^{6}

Question 9.

Answer:

= d^{5}

Explanation:

First we calculate separately values of numerators

and denominators then divide , we have numerator (d)^{5 }X (d)^{9
}we have same bases as d so we add powers as (d)^{5+9 }= (d)^{14
}we have denominator (d)^{ }X (d)^{8
}we have same base as d so we add powers as (d)^{1+8 }= (d)^{9}

Now we have \(\frac{d^{14}}{d^{9}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\), a quotient of

two powers with the same base as a^{m-n
} \(\frac{d^{14}}{d^{9}}\) = (d)^{14-9 }= d^{5}

Question 10.

Answer:

= p^{10}

Explanation:

First we calculate separately values of numerators

and denominators then divide , we have numerator (p)^{3 }X (p)^{6} X (p)^{4
}we have same bases as p so we add powers as (p)^{3+6+4 }= (d)^{13
}we have denominator (p)^{2 }X (p)

we have same bases as p so we add powers as (p)^{2+1 }= (p)^{3}

Now we have \(\frac{p^{13}}{p^{3}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{p^{13}}{p^{3}}\) = (p)^{13-3 }= p^{10}

**Self-Assessment for Concepts & Skills**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

**SIMPLIFYING EXPRESSIONS** Simplify the expression. Write your answer as a power.

Question 11.

Answer:

= (-3)^{7 }

Explanation:

Given \(\frac{-3^{9}}{-3^{2}}\) we use rule for

finding \(\frac{a^{m}}{a^{n}}\), a quotient of two

powers with the same base as a^{m-n}

so \(\frac{-3^{9}}{-3^{2}}\) = (-3)^{9-2 }= (-3)^{7 }

Question 12.

Answer:

= (8)^{3}

Explanation:

First we calculate separately values of numerators

and denominators then divide , we have numerator (8)^{6 }X (8)^{2
}we have same bases as d so we add powers as (8)^{6+2 }= (8)^{8
}\(\frac{8^{8}}{8^{5}}\) we use rule for

finding \(\frac{a^{m}}{a^{n}}\), a quotient of two

powers with the same base as a^{m-n}

so \(\frac{8^{8}}{8^{5}}\) = (8)^{8-5 }= (8)^{3}

Question 13.

Answer:

= x

Explanation:

First we multiply denominators as bases are same

we add powers so x^{4} X x^{6 }= x^{4+6 }= x^{10}

now we solve \(\frac{x^{11}}{x^{10}}\) we use rule for finding

\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n
}\(\frac{x^{11}}{x^{10}}\) = x^{11-10 }= x

Question 14.

Answer:

= 5^{6}

Explanation:

First we calculate separately values of numerators

and denominators then divide , we have numerator (5)^{6 }X (5)^{3} ^{
}we have same bases as 5 so we add powers as (5)^{6+3 }= (5)^{9
}we have denominator 5 X (5)^{2 }

we have same bases as 5 so we add powers as (5)^{1+2 }= (5)^{3}

Now we have \(\frac{5^{9}}{5^{3}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{5^{9}}{5^{3}}\) = (5)^{9-3 }= 5^{6}

Question 15.

Answer:

= (-2)^{5}

Explanation:

First we calculate separately values of numerators

and denominators then divide , we have numerator (-2)^{9 }X (-2)^{4
}we have same bases as -2 so we add powers as (-2)^{9+4 }= (-2)^{13
}we have denominator (-2)^{4}^{ }X (-2)^{4
}we have same base as -2 so we add powers as (-2)^{4+4 }= (-2)^{8}

Now we have \(\frac{-2^{13}}{-2^{8}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\), a quotient of two

powers with the same base as a^{m-n
} \(\frac{-2^{13}}{-2^{8}}\) = (-2)^{13-8 }= (-2)^{5}

Question 16.

Answer:

= b^{13}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (b)^{10 }X (b)^{3 }X (b)^{5}

we have same bases as b so we add powers as (b)^{10+3+5 }= (b)^{18
}we have denominator (b)^{2}^{ }X (b)^{3
}we have same base as b so we add powers as (b)^{2+3 }= (b)^{5}

Now we have \(\frac{b^{18}}{b^{5}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{b^{18}}{b^{5}}\) = (b)^{18-5 }= b^{13}

Question 17.

**WHICH ONE DOESN’T BELONG?**

Which quotient does not belong with the other three? Explain your reasoning.

Answer:

\(\frac{-4^{8}}{-3^{4}}\) does not belongs with the other three.

Explanation:

Given \(\frac{-10^{7}}{-10^{2}}\) have same base -10,

\(\frac{6^{3}}{6^{2}}\) have same base 6,

\(\frac{5^{6}}{5^{3}}\) have same base 5 but

\(\frac{-4^{8}}{-3^{4}}\) has different bases -4 and -3,

So, \(\frac{-4^{8}}{-3^{4}}\) does not belongs with the other three.

**Self-Assessment for Problem Solving**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 18.

You want to purchase a cat tracker. Tracker A detects your cat within a radius of 4 • 10^{2} feet of your home. Tracker B detects your cat within a radius of 10^{4} feet of your home. Which tracker has a greater radius? How many times greater?

Answer:

Tracker B has greater radius than Tracker A and

Tracker B is greater by radius of 9,600 feet.

Explanation:

Given Tracker A detects your cat within a radius of 4 • 10^{2} feet of your home.

Tracker B detects your cat within a radius of 10^{4} feet of your home,

Tracker A = 4 X 10^{2} = 4 X 100 = 400 feet

Tracker B = 10^{4} feet = 10 X 10 X 10 X 10 = 10,000 feet as

comparing Tracker A and Tracker B,Tracker B is greater than Tracker A,

by 10,000 – 400 = 9,600 feet, therfore Tracker B

has greater radius than Tracker A and

Tracker B is greater by radius of 9,600 feet.

Question 19.

**DIG DEEPER!**

An earthquake of magnitude 3.0 is 10^{2} times stronger than an earthquake of magnitude 1.0. An earthquake of magnitude 8.0 is 10^{7} times stronger than an earthquake of magnitude 1.0. How many times stronger is an earthquake of magnitude 8.0 than an earthquake of magnitude 3.0?

Answer:

10^{5} times stronger is an earthquake of magnitude 8.0

more than an earthquake of magnitude 3.0.

Explanation:

Given an earthquake of magnitude 3.0 is 10^{2} times

stronger than an earthquake of magnitude 1.0 and

an earthquake of magnitude 8.0 is 10^{7} times stronger

than an earthquake of magnitude 1.0. So how many times

stronger is an earthquake of magnitude 8.0 than an

earthquake of magnitude 3.0 is \(\frac{10^{7}}{10^{2}}\) ,

so we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}as \(\frac{10^{7}}{10^{2}}\) = 10^{7-2} = 10^{5}

therefore 10^{5} times stronger is an earthquake of magnitude 8.0

more than an earthquake of magnitude 3.0.

Question 20.

The edge length of a cube-shaped crate is the square of the edge length of a cube-shaped box. Write an expression for the number of boxes that can fit in the crate. Justify your answer.

Answer:

The number of boxes that fit in the crate is x^{3}

Explanation:

Let the edge be x, So the volume of box is V = x^{3}

Given the edge length of a cube-shaped crate is the square of the

edge length of a cube-shaped box so

the volume crate with side equal to the side of the box x^{2} = ( x^{2})^{3 }

we use general rule for finding (a^{m})^{n} a power of a power,

If two powers have the same base then

we can multiply the powers as (a^{m})^{n }= a^{m x }^{n} so ( x^{2})^{3 }= x^{2 x }^{3}=( x)^{6
}The number of boxes that fit in crate is

\(\frac{x^{6}}{x^{3}}\) now we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}therefore \(\frac{x^{6}}{x^{3}}\) = x^{6-3}^{ }= x^{3
}so the number of boxes that fit in the crate is x^{3
}

### Quotient of Powers Property Homework & Practice 8.3

**Review & Refresh**

**Simplify the expression. Write your answer as a power.**

Question 1.

4^{2} • 4^{3}

Answer:

4^{2} • 4^{3 }= 4^{5}

Explanation:

Given expression as 4^{2} • 4^{3}

we use general rule for a^{m} • a^{n }= a^{m+n }a product of

two powers with the same base then powers are added.

So 4^{2} • 4^{3 }= 4^{2+3} = 4^{5}

Question 2.

(a^{5})^{5}

Answer:

(a^{5})^{5 }= a^{2}^{5}

Explanation:

Given expression is (a^{5})^{5 }we have general rule for

finding (a^{m})^{n }a power of a power, If two powers have the

same base then we can multiply the

powers as (a^{m})^{n }= a^{m x }^{n }. So (a^{5})^{5 }= a^{5X}^{5 }= a^{2}^{5 }.

Question 3.

(xy)^{7}

Answer:

(xy)^{7}= x^{7 }X y^{7} = x^{7 }y^{7}

Explanation:

Given expression as (xy)^{7}as both as same

power 7 we write as x^{7 }X y^{7} = x^{7 }y^{7}

^{
}**The red figure is similar to the blue figure. Describe a similarity transformation between the figures.**

Question 4.

Answer:

Similarity between two figures is

dilate the red figure using a scale factor of 2:3

and then reflect the image in the x- axis.

Explanation:

By comparing the side lengths, we can see that

the blue figure is 2:3 the size of red figure,

Similarity between two figures is dilate the red figure

using a scale factor of 2:3 and then reflect the image in the x- axis.

Question 5.

Answer:

Similarity between two figures is

dilate the red figure using a scale factor of 1/2

and then reflect the image in the x- axis.

Explanation:

By comparing the side lengths, we can see that

the blue figure is one-half the size of red figure,

Similarity between two figures is dilate the red figure

using a scale factor of 1/2 and then reflect the image in the x- axis.

**Concepts, Skills, & Problem Solving**

**FINDING QUOTIENTS OF POWERS** Write the quotient as repeated multiplication. Then write the quotient as a power.(See Exploration 1, p. 331.)

Question 6.

Answer:

= \(\frac{7 X 7 X 7 X 7 X 7 X 7 X 7 X 7 X 7}{7 X 7 X 7 X 7 X 7 X 7}\) = 7^{3}

Explanation:

Given \(\frac{7^{9}}{7^{6}}\) the repeated multiplication is

\(\frac{7 X 7 X 7 X 7 X 7 X 7 X 7 X 7 X 7}{7 X 7 X 7 X 7 X 7 X 7}\) and

the quotient as a power we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n},so \(\frac{7^{9}}{7^{6}}\) is 7^{9-6 }= 7^{3}

Question 7.

Answer:

= \(\frac{-4.5 X -4.5 X -4.5 X -4.5 X -4.5 X -4.5}{-4.5 X -4.5}\) = (-4.5)^{4}.

Explanation:

Given \(\frac{-4.5^{6}}{-4.5^{2}}\) the repeated multiplication is

\(\frac{-4.5 X -4.5 X -4.5 X -4.5 X -4.5 X -4.5}{-4.5 X -4.5}\) and

the quotient as a power we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} ,so \(\frac{-4.5^{6}}{-4.5^{2}}\)

is (-4.5)^{6-2 }= (-4.5)^{4}.

Question 8.

Answer:

= \(\frac{m X m X m X m X m X m X m X m X m X m}{m X m X m X m X m }\) = m^{5
}

Explanation:

Given \(\frac{m^{10}}{m^{5}}\) the repeated multiplication is

and \(\frac{m X m X m X m X m X m X m X m X m X m}{m X m X m X m X m }\)

the quotient as a power we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{m^{10}}{m^{5}}\)

is (m)^{10-5 } = m^{5}.

**DIVIDING POWERS WITH THE SAME BASE** Simplify the expression. Write your answer as a power.

Question 9.

Answer:

\(\frac{6^{10}}{6^{4}}\) = 6^{6}.

Explanation:

As given expression is \(\frac{6^{10}}{6^{4}}\) we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{6^{10}}{6^{4}}\) =

(6)^{10-4 }= 6^{6}.

Question 10.

Answer:

\(\frac{8^{9}}{8^{7}}\) = 8^{2}.

Explanation:

As given expression is \(\frac{8^{9}}{8^{7}}\) we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{8^{9}}{8^{7}}\) =

(8)^{9-7 }= 8^{2}.

Question 11.

Answer:

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

Explanation:

As given expression is \(\frac{-3^{4}}{-3^{1}}\) we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{-3^{4}}{-3^{1}}\) =

(-3)^{4-1 }= -3^{3}.

Question 12.

Answer:

\(\frac{4.5^{5}}{4.5^{3}}\) = 4.5^{2}.

Explanation:

As given expression is \(\frac{4.5^{5}}{4.5^{3}}\) we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{4.5^{5}}{4.5^{3}}\) =

(4.5)^{5-3 }= 4.5^{2}.

Question 13.

Answer:

\(\frac{64^{4}}{64^{3}}\) = 64.

Explanation:

As given expression is \(\frac{64^{4}}{64^{3}}\) we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{64^{4}}{64^{3}}\) =

(64)^{4-3 }= 64.

Question 14.

Answer:

\(\frac{-17^{5}}{-17^{2}}\) = (-17)^{3}.

Explanation:

As given expression is \(\frac{-17^{5}}{-17^{2}}\) we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{-17^{5}}{-17^{2}}\) =

(-17)^{5-2 }= (-17)^{3}.

Question 15.

Answer:

\(\frac{-6.4^{8}}{-6.4^{6}}\) = (-6.4)^{2}.

Explanation:

As given expression is \(\frac{-6.4^{8}}{-6.4^{6}}\) we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{-6.4^{8}}{-6.4^{6}}\) =

(-6.4)^{8-6 }= (-6.4)^{2}.

Question 16.

Answer:

\(\frac{π^{11}}{π^{7}}\) = π^{4}.

Explanation:

As given expression is \(\frac{π^{11}}{π^{7}}\) we use

rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{π^{11}}{π^{7}}\)=

(π)^{11-7 }= π^{4}.

Question 17.

**YOU BE THE TEACHER**

Your friend simplifies the quotient. Is your friend correct? Explain your reasoning.

Answer:

No, friend is incorrect as \(\frac{6^{15}}{6^{5}}\) = 6^{10 }≠ 6^{3
}

Explanation:

Given \(\frac{6^{15}}{6^{5}}\) as per rule finding

\(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{6^{15}}{6^{5}}\) = 6^{15-5}**
**= 6

^{10 }but friend says \(\frac{6^{15}}{6^{5}}\) = 6

^{15/5}= 6

^{3 }which is incorrect therefore \(\frac{6^{15}}{6^{5}}\) = 6

^{10 }≠ 6

^{3 }exponents should be subtracted not divided.

**SIMPLIFYING AN EXPRESSION Simplify the expression. Write your answer as a power.**

Question 18.

\(\frac{7^{5} \cdot 7^{3}}{7^{2}}\)

Answer:

\(\frac{7^{5} \cdot 7^{3}}{7^{2}}\) = 7^{6}.

Explanation:

Given Expression as \(\frac{7^{5} \cdot 7^{3}}{7^{2}}\)

first we calculate separately values of numerators

then divide with denominator, we have numerator (7)^{5 }X (7)^{3 }

we have same bases as 7 so we add powers as (7)^{5+3 }= (7)^{8
}Now we have \(\frac{7^{8}}{7^{2}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}So \(\frac{7^{8}}{7^{2}}\) = (7)^{8-2 }= 7^{6}

Question 19.

\(\frac{6^{13}}{6^{4} \cdot 6^{2}}\)

Answer:

\(\frac{6^{13}}{6^{4} \cdot 6^{2}}\) = 6^{7}

Explanation:

Given expression \(\frac{6^{13}}{6^{4} \cdot 6^{2}}\)

first we calculate separately values of denominators

then divide with numerator, we have denominator (6)^{4 }X (6)^{2 }

we have same bases as 6 so we add powers as (6)^{4+2 }= (6)^{6
}Now we have \(\frac{6^{13}}{6^{6}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n}

So \(\frac{6^{13}}{6^{6}}\) = (6)^{13-6 }= 6^{7}

Question 20.

Answer:

= = (-6.1)^{2}

Explanation:

Given expression \(\frac{-6.1^{11}}{-6.1^{7} \cdot -6.1^{2}}\)

first we calculate separately values of denominators

then divide with numerator, we have denominator (-6.1)^{7 }X (-6.1)^{2 }

we have same bases as -6.1 so we add powers as (-6.1)^{7+2 }= (-6.1)^{9
}Now we have \(\frac{-6.1^{11}}{-6.1^{9}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n}

So \(\frac{-6.1^{11}}{-6.1^{9}}\) = (-6.1)^{11-9 }= (-6.1)^{2}

Question 21.

\(\frac{\pi^{30}}{\pi^{18} \cdot \pi^{4}}\)

Answer:

\(\frac{\pi^{30}}{\pi^{18} \cdot \pi^{4}}\) = (π)^{8
}

Explanation:

Given expression is \(\frac{\pi^{30}}{\pi^{18} \cdot \pi^{4}}\)

first we calculate separately values of denominators

then divide with numerator, we have denominator (π)^{18 }X (π)^{4 }

we have same bases as π so we add powers as (π)^{18+4 }= (π)^{22
}Now we have \(\frac{\pi^{30}}{\pi^{22} \) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}So \(\frac{\pi^{30}}{\pi^{22} \) = (π)^{30-22 }= π^{8}

Question 22.

\(\frac{c^{22}}{c^{8} \cdot c^{9}}\)

Answer:

\(\frac{c^{22}}{c^{8} \cdot c^{9}}\) = c^{5}

Explanation:

Given expression is \(\frac{c^{22}}{c^{8} \cdot c^{9}}\)

first we calculate separately values of denominators

then divide with numerator, we have denominator c^{8 }X c^{9 }

we have same bases as c so we add powers as c^{8+9 }= c^{17
}Now we have \(\frac{c^{22}} {c^{17}} \) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}So \(\frac{c^{22}} {c^{17}} \) = c^{22-17 }= c^{5}

Question 23.

\(\frac{z^{8} \cdot z^{6}}{z^{8}}\)

Answer:

\(\frac{z^{8} \cdot z^{6}}{z^{8}}\) = z^{6}

Explanation:

Given expression is \(\frac{z^{8} \cdot z^{6}}{z^{8}}\)

first we calculate separately values of numerators

then divide with denominator, we have numerator z^{6 }X z^{8 }

we have same bases as z so we add powers as z^{6+8 }= z^{14
}Now we have \(\frac{z^{14}}{z^{8}} \) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}So \(\frac{z^{14}}{z^{8}} \) = z^{14-8 }= z^{6}

Question 24.

**MODELING REAL LIFE**

The sound intensity of a normal conversation is 10^{6} times greater than the quietest noise a person can hear. The sound intensity of a jet at takeoff is 10^{14} times greater than the quietest noise a person can hear. How many times more intense is the sound of a jet at takeoff than the sound of a normal conversation?

Answer:

10^{8} times more intense is the sound of a jet at takeoff than

the sound of a normal conversation.

Explanation:

Given the sound intensity of a normal conversation is 10^{6} times

greater than the quietest noise a person can hear.

The sound intensity of a jet at takeoff is 10^{14} times

greater than the quietest noise a person can hear.

Therefore more intense is the sound of a jet at takeoff than

the sound of a normal conversation is \(\frac{10^{14}}{10^{6}} \)

so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}So \(\frac{10^{14}}{10^{6}} \) = 10^{14-6}= 10^{8
}therefore 10^{8} times more intense is the sound of a jet at

takeoff than the sound of a normal conversation.

**SIMPLIFYING AN EXPRESSION** Simplify the expression. Write your answer as a power.

Question 25.

Answer:

= (-4)^{5}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (-4)^{8 }X (-4)^{3 }

we have same bases as -4 so we add powers as (-4)^{8+3 }= (-4)^{11
}we have denominator (-4)^{4}^{ }X (-4)^{2
}we have same base as -4 so we add powers as (-4)^{4+2 }= (-4)^{6}

Now we have \(\frac{-4^{11}}{-4^{6}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{-4^{11}}{-4^{6}}\) = (-4)^{11-6 }= (-4)^{5}

Question 26.

Answer:

= 6^{5
}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (6)^{2 }X (6)^{12 }

we have same bases as 6 so we add powers as (6)^{2+12 }= (6)^{14
}we have denominator (6)^{1}^{ }X (6)^{8
}we have same base as 6 so we add powers as (6)^{1+8 }= (6)^{9}

Now we have \(\frac{6^{14}}{6^{9}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{6^{14}}{6^{9}}\) = (6)^{14-9 }= 6^{5}

Question 27.

Answer:

= 3^{10}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (3)^{2 }X (3)^{6 }X (3)^{5
}we have same bases as 3 so we add powers as (3)^{2+6+5 }= (3)^{13
}we have denominator (3)^{2}^{ }X (3)^{1
}we have same base as 3 so we add powers as (3)^{2+1 }= (3)^{3}

Now we have \(\frac{3^{13}}{3^{3}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{3^{13}}{3^{3}}\) = (3)^{13-3 }= 3^{10}

Question 28.

Answer:

= z^{10}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (z)^{7 }X (z)^{6 }^{
}we have same bases as z so we add powers as z^{7+6 }= z^{13
}we have denominator (z)^{1}^{ }X (z)^{2
}we have same base as z so we add powers as (z)^{1+2 }= (z)^{3}

Now we have \(\frac{3^{13}}{3^{3}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{z^{13}}{z^{3}}\) = (z)^{13-3 }= z^{10}

Question 29.

Answer:

= x^{6}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (x)^{5 }X (x)^{13 }^{
}we have same bases as x so we add powers as x^{5+13 }= x^{18
}we have denominator (x)^{4}^{ }X (x)^{8
}we have same base as x so we add powers as (x)^{4+8 }= (x)^{12}

Now we have \(\frac{x^{18}}{x^{12}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{x^{18}}{x^{12}}\) = (x)^{18-12 }= x^{6}

Question 30.

Answer:

= y^{11}

Explanation:

First we calculate separately all multiple values of numerators

and denominators then divide, we have numerator (y)^{8 }X (y)^{2 }X (y)^{4 }X (y)^{7
}we have same bases as y so we add powers as y^{8+2+4+7 }= y^{21
}we have denominator (y)^{7}^{ }X (y)^{1 }X (y)^{2 }we have same base

as y so we add powers as (y)^{7+1+2 }= (y)^{10}

Now we have \(\frac{y^{21}}{y^{10}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{y^{21}}{y^{10}}\) = (y)^{21-10 }= y^{11}

Question 31.

**REASONING**

The storage capacities and prices of five devices are shown in the table.

a. How many times more storage does Device D have than Device B?

b. Do storage and price have a linear relationship? Explain.

Answer:

a. 4 times more storage capacity Device D have than Device B.

b. No, as the price increases by $20 storage capacity doubles.

Explanation:

Given the storage capacities and prices of five devices are

as shown above in the table,

a. Storage Device D has capacity of 2^{8 }GB and Device B has

capacity of 2^{8 }GB as Device D has more capacity by

\(\frac{2^{8}}{2^{6}}\) = (2)^{8-6 }= 2^{2}= 4,

therefore 4 times more storage capacity Device D

have than Device B.

b. As seen the price increases by $20 storage capacity doubles so

there is no linear relationship between storage and price.

Question 32.

**DIG DEEPER!**

Consider the equation \(\frac{9^{m}}{9^{n}}\) = 9^{2}

a. Find two numbers m and n that satisfy the equation.

b. Describe the number of solutions that satisfy the equation. Explain your reasoning.

Answer:

a. The two numbers m and n are 3,1.

b. We have more number of solutions that satisfy the equation as

explained below

Explanation:

Given the equation \(\frac{9^{m}}{9^{n}}\) = 9^{2}

a. We have rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n} so \(\frac{9^{m}}{9^{n}}\)= 9^{m-n}= 9^{2 }so

m-n=2, m=2+n we can take n=1 we get m= 2 + 1 = 3 so m,n is 3,1.

the equation is \(\frac{9^{3}}{9^{1}}\)= 9^{3-1}= 9^{2}

b. As m=2 + n we can take n any natural number from 1 to infinity,

as if n=1 m will be 2+1=3, if n=2 , m will be 2+2 = 4 and so on.

therefore (m, n)=(n+2,n) or (m,m-2)=(3,1)….(5,3),(∞,∞-2) are the

(infinity, infinity minus 2) numbers of solutions that satisfy the equation.

Question 33.

**MODELING REAL LIFE**

A scientist estimates that there are about 10^{24} stars in the universe and that each galaxy has, on average, approximately the same number of stars as the Milky Way galaxy. About how many galaxies are in the universe?

Answer:

10^{13 }galaxies are there in the universe.

Explanation:

A scientist estimates that there are about 10^{24} stars in the

universe and that each galaxy has, on average, approximately

the same number of stars as the Milky Way galaxy.

so number of galaxies in the universe are \(\frac{10^{24}}{10^{1} \cdot 10^{10}}\)

first we calculate separately values of denominators

then divide with numerator, we have denominator (10)^{1 }X (10)^{10 }

we have same bases as 10 so we add powers as (10)^{1+10 }= (10)^{11
}Now we have \(\frac{10^{24}}{10^{11}} \) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}\(\frac{10^{24}}{10^{11}} \) = (10)^{24-11 }= 10^{13}

therefore 10^{13 }galaxies are there in the universe.

Question 34.

**NUMBER SENSE**

Find the value of x that makes c = 8^{9} true. Explain how you found your answer.

Answer:

The value of x is 10,

Explanation:

Given \(\frac{8^{3 x}}{8^{2 x}+1}\) = 8^{9} is true,

so we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}as bases are same 8 so 8^{3x-(2x+1)} = 8^{9} now we equate powers we have

3x-(2x+1)= 9, means x-1=9 therefore x = 9 + 1 = 10.

### Lesson 8.4 Zero and Negative Exponents

**EXPLORATION 1**

Understanding Zero Exponents

Work with a partner.

a. Copy and complete the table.

b. Evaluate each expression in the first column of the table in part(a). How can you use these results to define a^{0}, where a ≠ 0?

Answer:

a.

b. We define a^{0}

Explanation:

a. We completed the table by using quotient of powers of property,

as \(\frac{a^{m}}{a^{n}}\) = a^{m-n
}then by the quotient rule for exponents we can write this as

a^{n-n }=\(\frac{a^{n}}{a^{n}}\) Then this becomes a problem about

dividing fractions. Since the numerator and denominator

are both the same this becomes .

EXPLORATION 2^{
}

Understanding Negative Exponents

Work with a partner.

a. Copy and complete the table.

b. How can you use the Multiplicative Inverse Property to rewrite the powers containing negative exponents in the first column of the table?

c. Use your results in parts (a) and (b) to define a^{-n}, where a ≠ 0 and n is an integer.

Answer:

a.

b. We rewrite the powers containing negative exponents in the first column of the table as

5^{-3 }X 5^{3 }as 6^{2}X 6^{-2 }as ^{4 }X 3^{-4 }as -4^{-5 }X -4^{5 }as =

c. a^{-n}, where a ≠ 0 we write a^{-n }as \(\frac{1}{a^{n}}\)

Explanation:

a. To complete the table first we write Product of Powers Property

then write their power and value.

As here we use Product of Powers Property

for a^{m} • a^{n }= a^{m+n }If product of two powers with the same base then

powers are added. So 5^{-3 }X 5^{3 }= 5^{-3+3 =} 5^{0 }= 1, 6^{2}X 6^{-2 }= 6^{2}^{-2 }= 6^{0} = 1,

^{4 }X 3^{-4 }= ^{4}^{-4 }=^{0 }= 1 and -4^{-5 }X -4^{5 }= -4^{-5+}^{5 }=-4^{0} = 1.

b. The inverse property of multiplication states that if you

multiply a number by its reciprocal, also called the multiplicative inverse,

the product will be 1. (a/b)*(b/a)=1,so we rewrite the powers

containing negative exponents in the first column of the table as

5^{-3 }X 5^{3 }as 6^{2}X 6^{-2 }as ^{4 }X 3^{-4 }as -4^{-5 }X -4^{5 }as =

**Try It**

**Evaluate the expression.**

Question 1.

4^{-2}

Answer:

4^{-2 }=

Explanation:

Given expression as 4^{-2 }so we write as

Question 2.

(- 2)^{– 5}

Answer:

(- 2)^{– 5}=

Explanation:

Given expression as (-2)^{-5 }so we write as

Question 3.

6^{-8} • 6^{8}

Answer:

6^{-8} • 6^{8 }= 1

Explanation:

we write the given expression 6^{-8} X 6^{8 }as Product of Powers Property

for a^{m} • a^{n }= a^{m+n }If product of two powers with the same base then

powers are added. So 6^{-8 }X 6^{8 }= 6^{-8+8 =} 6^{0 }= 1.

Question 4.

\(\frac{(-3)^{5}}{(-3)^{6}}\)

Answer:

\(\frac{(-3)^{5}}{(-3)^{6}}\) = – \(\frac{1}{3}\) or -3^{-1}

Explanation:

Given expression as \(\frac{(-3)^{5}}{(-3)^{6}}\) we use

the quotient rule for exponents we can write this as \(\frac{a^{n}}{a^{n}}\) = a^{m-n },

so (-3)^{5-6 },= (-3)^{-1 }or – \(\frac{1}{3}\).

Question 5.

\(\frac{1}{5^{7}} \cdot \frac{1}{5^{4}}\)

Answer:

\(\frac{1}{5^{7}} \cdot \frac{1}{5^{4}}\) = \(\frac{1}{5^{11}}\)

Explanation:

Given expression is \(\frac{1}{5^{7}} \cdot \frac{1}{5^{4}}\)

first we calculate separately values of denominators

then divide with numerator, we have denominator 5^{7 }X 5^{4}

we have same bases as 5 so we add powers as 5^{7+4 }= 5^{11
}as numerator is 1 we write as \(\frac{1}{5^{11}}\).

Question 6.

\(\frac{4^{5} \cdot 4^{3}}{4^{2}}\)

Answer:

\(\frac{4^{5} \cdot 4^{3}}{4^{2}}\) = 4^{6}

Explanation:

Given Expression as \(\frac{4^{5} \cdot 4^{3}}{4^{2}}\)

first we calculate separately values of numerators

then divide with denominator, we have numerator (4)^{5 }X (4)^{3 }

we have same bases as 4 so we add powers as (4)^{5+3 }= (4)^{8
}Now we have \(\frac{4^{8}}{4^{2}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}So \(\frac{4^{8}}{4^{2}}\) = (4)^{8-2 }= 4^{6}

**Simplify. Write the expression using only positive exponents.**

Question 7.

8x^{-2}

Answer:

8x^{-2 }= \(\frac{8}{x^{2}}\)

Explanation:

Given 8x^{-2 }we write the expression as positive exponents by using

so 8 X \(\frac{1}{x^{2}}\) or \(\frac{8}{x^{2}}\)

Question 8.

b^{0} • b^{-10}

Answer:

b^{0} • b^{-10 }= \(\frac{1}{b^{10}}\)

Explanation:

Given b^{0} • b^{-10 }we write the expression as positive exponents by using

so b^{0} X \(\frac{1}{b^{10}}\) as we know b^{0} =1,

b^{0} X \(\frac{1}{b^{10}}\)=1 X \(\frac{1}{b^{10}}\) or \(\frac{1}{b^{10}}\).

Question 9.

\(\frac{z^{6}}{15 z^{9}}\)

Answer:

\(\frac{z^{6}}{15 z^{9}}\) = \(\frac{1}{15 z^{3}}\)

Explanation:

Given expression as \(\frac{z^{6}}{15 z^{9}}\) first we write it as

\(\frac{1}{15}\) X \(\frac{z^{6}}{z^{9}}\) now we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n}

as \(\frac{1}{15}\) X z^{6-9 }we get \(\frac{1}{15}\) X z^{-3
}= \(\frac{1}{15 z^{3}}\).

Self-Assessment for Concepts & Skills

Solve each exercise. Then rate your understanding of the success criteria in your journal.

**EVALUATING EXPRESSIONS** Evaluate the expression.

Question 10.

7^{-2}

Answer:

7^{-2}= \(\frac{1}{49}\)

Explanation:

Given 7^{-2 }we write the expression as positive exponents by using

so \(\frac{1}{7^{2}}\)= \(\frac{1}{49}\)

Question 11.

4^{-3} • 4^{0}

Answer:

4^{-3} • 4^{0 }= \(\frac{1}{64}\)

Explanation:

Given 4^{-3 }X 4^{0 }we write the expression as positive exponents by using

so \(\frac{1}{4^{3}}\) X 4^{0} as we know 4^{0} =1,

\(\frac{1}{4^{3}}\) X 4^{0} =\(\frac{1}{4^{3}}\) x 1 = \(\frac{1}{64}\).

Question 12.

\(\frac{(-9)^{5}}{(-9)^{7}}\)

Answer:

\(\frac{(-9)^{5}}{(-9)^{7}}\) = \(\frac{1}{-9^{2}}\) = \(\frac{1}{81}\)

Explanation:

Given expression as \(\frac{-9^{5}}{-9^{7}}\) now we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n}

as -9^{5-7 }we get -9^{-2 }we write as \(\frac{1}{-9^{2}}\) = \(\frac{1}{81}\).

**SIMPLIFYING EXPRESSIONS** Simplify. Write the expression using only positive exponents.

Question 13.

10t^{-5}

Answer:

10t^{-5 }= \(\frac{10}{t^{5}}\)

Explanation:

Given 10t^{-5 }we write as 10 X t^{-5 }= now we write t^{-5 }as tSo 10 X t^{-5 }= 10 X

Question 14.

w^{3} • w^{-9}

Answer:

w^{3} • w^{-9 }=

Explanation:

Given expression is w^{3} • w^{-9 }first we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added, so we add powers as w^{3-9 }= w^{-6},

So w^{-6 }=

Question 15.

\(\frac{r^{8} \cdot r^{8}}{4}\)

Answer:

\(\frac{r^{8} \cdot r^{8}}{4}\) = \(\frac{r^{16}}{4}\)

Explanation:

Given expression is \(\frac{r^{8} \cdot r^{8}}{4}\) first we

solve numerator by using general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added, so we add powers as r^{8+8 }= r^{16},

now we write as r^{16 }X \(\frac{1}{4}\) = \(\frac{r^{16}}{4}\).

Question 16.

**DIFFERENT WORDS, SAME QUESTION**

Which is different? Find “both” answers.

Answer:

d bit is different because writing (-3) X (-3) X (-3) as a power

with an integer base is (-3)^{3 }no negative exponent.

one more answer is all a,b,c bits have fractions but only d bit

is not in fraction form.

Explanation:

a. Writing \(\frac{1}{3 X 3 X 3}\) using negative exponent is 3^{-3}.

b. Writing 3 to the negative third is 3^{-3}.

c. Writing \(\frac{1^{3}}{3^{3}}\) is 3^{-3}.

d. Writing (-3) X (-3) X (-3) as a power with an integer base is (-3)^{3}.

As a, b, c has value 3^{-3}only bit d has (-3)^{3}.

Therefore d bit is different because writing (-3) X (-3) X (-3) as a power

with an integer base is (-3)^{3 }we don not have a negative exponent.

one more answer is all a,b,c bits have fractions but only d bit

is not in fraction form.

Self-Assessment for Problem Solving

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.

The mass of a grain of sand is about 10^{-3} gram. About how many grains of sand are in a 10-kilogram bag of sand?

Answer:

There are about 10,000,000 grains of sand.

Explanation:

Given the mass of a grain of sand is about 10^{-3} gram.

We know 1 kg = 1000 grams ,10 kg = 10 X 1000 = 10,000 grams,

therefore one grain of sand is 10,000 X 1000 = 10,000,000 grains of sand.

Question 18.

A one-celled, aquatic organism called a dinoflagellate is 1000 micrometers long. A microscope magnifies the dinoflagellate 100 times. What is the magnified length of the dinoflagellate in meters? (1 micrometer is 10^{-6}; meter.)

Answer:

The magnified length of the dinoflagellate in meters10^{-1 }meters

Explanation:

Given one-celled, aquatic organism called a dinoflagellate is

1000 micrometers long. A microscope magnifies the

dinoflagellate 100 times, the magnified length in meters is

1 micrometer is 10^{-6 }meter now dinoflagellate is 1000 X 10^{-6 }X 100 =

10^{3 }X 10^{-6 }X 10^{2 }we use general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added so all have base 10 we add powers as

10^{3-6+2 }= 10^{-1 }meters therefore the magnified length of

the dinoflagellate in meters10^{-1 }meters.

Question 19.

**DIG DEEPER!**

A garden is 12 yards long. Assuming the snail moves at a constant speed, how many minutes does it take the snail to travel the length of the garden? Justify your answer.

Answer:

Snail takes 15 minutes to travel the length of the garden

Explanation:

Given a garden is 12 yards long and the constant speed of snail is

5^{-2 }foot per second. We know 1 yard is equal to 3 foot and

speed = distance by time ,So 5^{-2 }= 12 X 3 by time

therefore time = 12 X 3 X 5^{2 }seconds = 36 X 25 seconds = 900 seconds,

converting seconds in minutes 900 divide by 60 or \(\frac{900}{60}\)= 15 minutes, therefore snail takes 15 minutes to travel the length of the garden.

**Zero and Negative Exponents Homework & Practice 8.4**

**Review & Refresh**

**Simplify the expression. Write your answer as a power.**

Question 1.

\(\frac{10^{8}}{10^{4}}\)

Answer:

\(\frac{10^{8}}{10^{4}}\) = 10^{4}

Explanation:

Given expression \(\frac{10^{8}}{10^{4}}\) so

now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base has a^{m-n}

So \(\frac{10^{8}}{10^{4}}\) = 10^{8-4 }= 10^{4}

Question 2.

\(\frac{y^{9}}{y^{7}}\)

Answer:

\(\frac{y^{9}}{y^{7}}\) = y^{2}

Explanation:

Given expression \(\frac{y^{9}}{y^{7}}\) so

now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base has a^{m-n}

So \(\frac{y^{9}}{y^{7}}\) = y^{9-7 }= y^{2}

Question 3.

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

Answer:

\(\frac{(-3)^{8} \cdot(-3)^{3}}{(-3)^{2}}\) = (-3)^{9}

Explanation:

Given Expression as \(\frac{-3^{8} \cdot -3^{3}}{-3^{2}}\)

first we calculate separately values of numerators

then divide with denominator, we have numerator (-3)^{8 }X (-3)^{3 }

we have same bases as -3 so we add powers as (-3)^{8+3 }= (-3)^{11
}Now we have \(\frac{-3^{11}}{-3^{2}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}So \(\frac{-3^{11}}{-3^{2}}\) = (-3)^{11-2 }= (-3)^{9}

Tell whether the triangles are similar. Explain.

Question 4.

Answer:

Yes, The triangles are similar.

Explanation:

Given two triangles in figure to know if they are similar we have

two triangles are said to be similar if their corresponding angles

are congruent and the corresponding sides are in proportion.

In the above two figures the triangles have the same angle measures,

and the corresponding sides are in proportion. So triangles are similar.

Question 5.

Answer:

Yes, The triangles are similar.

Explanation:

Given two triangles in figure to know if they are similar we have

two triangles are said to be similar if their corresponding angles

are congruent and the corresponding sides are in proportion.

In the above two figures the triangles do not have the same angle measures,

So triangles are not similar.

Question 6.

Which data display best orders numerical data and shows how they are distributed?

A. bar graph

B. line graph

C. scatter plot

D. stem-and-leaf plot

Answer:

D. stem-and-leaf plot

Explanation:

A stem-and-leaf plot best orders numerical data and shows

how the data is distributed since it orders the values from

least to greatest and shows how many values lie under each “stem” in the stem-and leaf plot so you can see how the data is distributed.

**Concepts, Skills, &Problem Solving**

**UNDERSTANDING NEGATIVE EXPONENTS** Copy and complete the table. (See Exploration 2, p. 337.)

Answer:

Explanation:

To complete the table first we write Product of Powers Property

then write their power and value.

As here we use Product of Powers Property

for a^{m} • a^{n }= a^{m+n }If product of two powers with the same base then

powers are added. So 7. 7^{-4 }X 7^{4 }= 7^{-4+4 =} 7^{0 }= 1,

8. (-2)^{5}X (-2)^{-5 }= (-2)^{5}^{-5 }= (-2)^{0} = 1.

**EVALUATING EXPRESSIONS** Evaluate the expression.

Question 9.

\(\frac{8^{7}}{8^{7}}\)

Answer:

\(\frac{8^{7}}{8^{7}}\) = 1

Explanation:

Given expression is \(\frac{8^{7}}{8^{7}}\)= 8^{7 }X 8^{-7 },

Here we use Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added. So 8^{7 }X 8^{-7 }= 8^{7-7 =} 8^{0 }= 1.

Question 10.

5^{0} • 5^{3}

Answer:

5^{0} • 5^{3 }= 5^{3 }= 125.

Explanation:

Given expression is 5^{0 }X 5^{3 },

Here we use Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added. So 5^{0 }X 5^{3 }= 5^{0+3 =} 5^{3 }= 125.

Question 11.

(- 2)^{-8} • (- 2)^{8}

Answer:

(- 2)^{-8} • (- 2)^{8 }= 1

Explanation:

Given expression is -2^{-8 }X -2^{8 },

Here we use Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added. So (-2)^{-8 }X (-2)^{8 }= (-2)^{-8+8 =} (-2)^{0}= 1.

Question 12.

9^{4} • 9^{-4
}Answer:

9^{4} • 9^{-4 }= 1

Explanation:

Given expression is 9^{4 }X 9^{-4 },

Here we use Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added. So 9^{4 }X 9^{-4 }= 9^{4-4 =} 9^{0}= 1.

Question 13.

6^{-2}

Answer:

6^{-2}= \(\frac{1}{36}\)

Explanation:

Given expression as 6^{-2 }we write the expression as positive exponents by using

so \(\frac{1}{6^{2}}\) = \(\frac{1}{36}\).

Question 14.

158^{0}

Answer:

158^{0 }= 1

Explanation:

Given expression is 158^{0 }it is proven that any number or

expression raised to the power of zero is always equal to 1.

In other words, if the exponent is zero then the result is 1.

So 158^{0 }= 1.

Question 15.

\(\frac{4^{3}}{4^{5}}\)

Answer:

\(\frac{4^{3}}{4^{5}}\) = \(\frac{1}{4^{2}}\) = \(\frac{1}{16}\)

Explanation:

Given expression is \(\frac{4^{3}}{4^{5}}\)

now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base has a^{m-n }So \(\frac{4^{3}}{4^{5}}\) =

4^{3-5 }= 4^{-2 }= \(\frac{1}{4^{2}}\) = \(\frac{1}{16}\).

Question 16.

\(\frac{-3}{(-3)^{2}}\)

Answer:

\(\frac{-3}{(-3)^{2}}\) = –\(\frac{1}{3}\)

Explanation:

Given expression is \(\frac{-3}{(-3)^{2}}\)

now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base has a^{m-n },so \(\frac{-3}{(-3)^{2}}\) =

(-3)^{1-2 }= (-3)^{-1 }= –\(\frac{1}{3}\).

Question 17.

2^{2} • 2^{-4}

Answer:

2^{2} • 2^{-4 }= \(\frac{1}{4}\)

Explanation:

Given expression is 2^{2 }X 2^{-4 },

Here we use Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added so 2^{2-4 }^{ }= 2^{2 }X 2^{-4 }= 2^{-2 }= \(\frac{1}{2^{2}}\) = \(\frac{1}{4}\).

Question 18.

3^{-3} • 3^{-2}

Answer:

3^{-3} • 3^{-2 }= \(\frac{1}{243}\)

Explanation:

Given expression is 3^{-3} • 3^{-2}^{ },

Here we use Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added so 3^{-3-2 }^{ }= 3^{-5 }= \(\frac{1}{3^{5}}\) = \(\frac{1}{243}\).

Question 19.

\(\frac{1}{5^{3}} \cdot \frac{1}{5^{6}}\)

Answer:

\(\frac{1}{5^{3}} \cdot \frac{1}{5^{6}}\) = \(\frac{1}{5^{9}}\) =

\(\frac{1}{1953125}\)

Explanation:

Given expression is \(\frac{1}{5^{3}} \cdot \frac{1}{5^{6}}\)

now we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base has a^{m-n },so \(\frac{1}{5^{3}} \cdot \frac{1}{5^{6}}\) =

(5)^{-3-6 }= (5)^{-9 }= \(\frac{1}{5^{9}}\) =

\(\frac{1}{1953125}\).

Question 20.

\(\frac{(1.5)^{2}}{(1.5)^{2} \cdot(1.5)^{4}}\)

Answer:

\(\frac{(1.5)^{2}}{(1.5)^{2} \cdot(1.5)^{4}}\) = \(\frac{1}{5.0625}\)

Explanation:

Given expression is \(\frac{(1.5)^{2}}{(1.5)^{2} \cdot(1.5)^{4}}\)

First we multiply denominators as bases are same

we add powers so (1.5)^{2} X (1.5)^{4 }= (1.5)^{2+4 }= (1.5)^{6}

now we solve \(\frac{1.5^{2}}{1.5^{6}}\) we use rule for finding

\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n
}\(\frac{1.5^{2}}{1.5^{6}}\) = 1.5^{2-6 }= 1.5^{-4 }now we use =\(\frac{1}{1.5^{4}}\) =

\(\frac{1}{5.0625}\).

Question 21.

**YOU BE THE TEACHER**

Your friend evaluates 4^{-3}. Is your friend correct? Explain your reasoning.

Answer:

No, friend is incorrect as 4^{-3 }= \(\frac{1}{64}\) ≠ -64

Explanation:

Given expression is 4^{-3 }we have = \(\frac{1}{4^{3}}\) = \(\frac{1}{64}\),

but friend says = -64 which is in correct as 4^{-3 }= \(\frac{1}{64}\) ≠ -64.

Question 22.

**CRITICAL THINKING**

How can you write the number 1 as a power with base 2? a power with base 10?

Answer:

2^{1 }and 10^{1}

Explanation:

Given to write the number 1 as a power with base 2 is 2^{1 }and

a power with base 10 is 10^{1}.

Question 23.

**NUMBER SENSE**

Without evaluating, order 5^{0}, 5^{4}, and 5^{-5} from least to greatest. Explain your reasoning.

Answer:

Order 5^{0}, 5^{4}, and 5^{-5} from least to greatest is 5^{-5}, 5^{0} and 5^{4
}**
**Explanation:

Given order 5

^{0}, 5

^{4}, and 5

^{-5}from least to greatest as each has base 5,

we take consideration of orders as -5 < 0 < 4 so Order 5

^{0}, 5

^{4}, and 5

^{-5}

from least to greatest is 5

^{-5}, 5

^{0}and 5

^{4}.

**Simplify. Write the expression using only positive exponents.**

SIMPLIFYING EXPRESSIONS

SIMPLIFYING EXPRESSIONS

Question 24.

6y

^{-4}

Answer:

6y

^{-4 }= \(\frac{6}{y^{4}}\)

Explanation:

Given expression as 6y^{-4 }we write the expression using only

positive exponents for 6 X y^{-4 }as we write first y^{-4 }we have

=

\(\frac{1}{y^{4}}\) now 6 X \(\frac{1}{y^{4}}\) = \(\frac{6}{y^{4}}\).

Question 25.

8^{-2} • a^{7}

Answer:

8^{-2} • a^{7 }= \(\frac{a^{7}}{64}\)

Explanation:

Given expression as 8^{-2 }X a^{7 }we write the expression using only

positive exponents for 8^{-2 }X a^{7} as we write first 8^{-2 }we have

=

\(\frac{1}{8^{2}}\), So a^{7 }X \(\frac{1}{8^{2}}\) = \(\frac{a^{7}}{64}\).

Question 26.

\(\frac{9 c^{3}}{c^{4}}\)

Answer:

\(\frac{9 c^{3}}{c^{4}}\) = \(\frac{9}{c}\)

Explanation:

Given expression is \(\frac{9 c^{3}}{c^{4}}\) = 9 X \(\frac{c^{3}}{c^{4}}\)

First we solve \(\frac{c^{3}}{c^{4}}\) we use rule for finding

\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n
} \(\frac{c^{3}}{c^{4}}\) = c^{3-4 }= c^{-1
}now we use .

Question 27.

\(\frac{5 b^{2}}{b^{3}}\)

Answer:

\(\frac{5 b^{2}}{b^{3}}\) = \(\frac{5}{b}\)

Explanation:

Given expression is \(\frac{5 b^{2}}{b^{3}}\) = 5 X \(\frac{b^{2}}{b^{3}}\)

First we solve \(\frac{b^{2}}{b^{3}}\) we use rule for finding

\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n
} \(\frac{b^{2}}{b^{3}}\) = b^{2-3 }= b^{-1
}now we use .

Question 28.

\(\frac{8 x^{3}}{2 x^{9}}\)

Answer:

\(\frac{8 x^{3}}{2 x^{9}}\) = \(\frac{4}{x^{6}}\)

Explanation:

Given expression as \(\frac{8 x^{3}}{2 x^{9}}\) we use rule for finding

\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n
} \(\frac{x^{3}}{x^{9}}\) = x^{3-9 }= x^{-6 }for x^{-6 } we use

Question 29.

3d^{-4} • 4d^{4}

Answer:

3d^{-4} • 4d^{4 }= 12

Explanation:

Given expression 3d^{-4} • 4d^{4 }first we multiply d^{-4} X d^{4 },

here we use Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added, we have same base d so d^{-4+4} = d^{0}= 1,

Now we multiply 1 with 3,4 as 3 X 4 X 1 we get 12.

Question 30.

m^{-2} • n^{3}

Answer:

m^{-2} • n^{3 }=

Explanation:

Given expression is m^{-2} • n^{3 }so we write m^{-2} we use

Question 31.

\(\frac{3^{2} \cdot k^{0} \cdot w^{0}}{w^{6}}\)

Answer:

\(\frac{3^{2} \cdot k^{0} \cdot w^{0}}{w^{6}}\) =

Explanation:

Given expression \(\frac{3^{2} \cdot k^{0} \cdot w^{0}}{w^{6}}\) we know

k^{0 }and w^{0 }is 1 now we have \(\frac{3^{2}}{w^{6}}\) X 1 X 1 = \(\frac{9}{w^{6}}\).

Question 32.^{
}

**OPEN-ENDED**

Write two different powers with negative exponents that have the same value. Justify your answer.

Answer:

We write two different powers with negative exponents that

have the same value are 2and the value will be

Explanation:

Let us take two different powers with negative exponents that

have the same value are 2we have .

**
REASONING** In Exercises 33–36, use the table.

Question 33.

How many millimeters are in a decimeter?

Answer:

100 millimeters are in a decimeter.

Explanation:

Given to find millimeter are in a decimeter we have as per table

millimeters are in a decimeter we write as 10so 100 millimeters are in a decimeter.

Question 34.

How many micrometers are in a centimeter?

Answer:

10000 micrometers are in a centimeter

Explanation:

Given to find micrometers are in a centimeter we have as per table

micrometers are in a centimeter we write as 10so 10000 micrometers are in a centimeter.

Question 35.

How many nanometers are in a millimeter?

Answer:

1,000,000 nanometers are in a millimeter

Explanation:

Given to find nanometers are in a millimeter we have as per table

nanometers are in a millimeter we write as 10so 1,000,000 nanometers are in a millimeter.

Question 36.

How many micrometers are in a meter?

Answer:

1,000,000 micrometers are in a meter

Explanation:

Given to find micrometers are in a meter we have as per table

micrometers are in a meter we write as 10so 1,000,000 micrometers are in a meter.

Question 37.

**MODELING REAL LIFE**

A bacterium is 100 micrometers long. A virus is 1000 times smaller than the bacterium.

a. Using the table above, find the length of the virus in meters.

b. Is the answer to part (a) less than, greater than, or equal to 1 micrometer?

Answer:

a. The length of the virus in meters is

Explanation:

Given a bacterium is 100 micrometers long. A virus is 1000 times

smaller than the bacterium.

a. The length of the virus in meters is 1 micrometer = 10^{-6 }meters

= 100 X 10^{-6 }by 1000 = 10^{2 } X 10^{-6 }X 10^{-3
}we have Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added so = therefore the length of the virus in meters is b. We have 1 micrometer as length of the virus it is So t

Question 38.

**DIG DEEPER!**

Every 2 seconds, someone in the United States needs blood. A sample blood donation is shown.

a. One cubic millimeter of blood contains about 10^{4} white blood cells. How many white blood cells are in the donation? (1 mm^{3} = 10^{-3} mL)

b. One cubic millimeter of blood contains about 5 × 10^{6} red blood cells. How many red blood cells are in the donation?

c. Compare your answers for parts (a) and (b).

Answer:

a. The white blood cells in the donation are 10^{7 }white blood cells

and the blood donation contains 500 ml are 5 X 10^{9 }white blood cells.

b. The red blood cells in the donation are 5 X 10^{9 } red blood cells

and the blood donation contains 500 ml are 25 X 10^{11} red blood cells.

c. The white blood cells are 500 times more than the red blood cells.

Explanation:

a. One cubic millimeter of blood contains about 10^{4} white blood cells.

(1 mm^{3} = 10^{-3} mL),The white blood cells in the donation are

1 mm^{3} = 10^{-3} mL,

10^{4} = 10^{-3} mL,

1 mL= 10^{4+3}, therefore 1 mL is 10^{7} white blood cells,

Now the blood donation contains 500 ml = 5 X 10^{2 }X 10 ^{7 }=

5 X 10^{9 }white blood cells.

b.One cubic millimeter of blood contains about 5 × 10^{6} red blood cells,

The red blood cells in the donation are

1 mm^{3} = 10^{-3} mL so

5 X 10^{6} = 10^{-3} mL therefore 1 mL= 5 X 10^{6} X 10^{3}

= 5 X 10^{6+}^{3} = 5 X 10^{9} red blood cells

Now the blood donation contains 500 ml = 5 X 10^{2 }X 5 X 10^{9 }= 25 X 10^{2+9}

= 25 X 10^{11 }red blood cells.

c. Now comparison of the white blood cells and the red blood cells are

25 X 10^{11 }divide by 5 X 10^{9 }= 5 X 10^{11-9 }= 5 X 10^{2 }= 5 X 100 = 500 ,

So the white blood cells are 500 times more than the red blood cells.

Question 39.

**PRECISION**

Describe how to rewrite a power with a positive exponent as a fraction with a power in the denominator. Use the definition of negative exponents to justify your reasoning.

Answer:

Explanation:

To rewrite a power with a positive exponent as a fraction

with a power in the denominator by using the definition of

negative exponents we write the power as 1 divided by a power

with the same base and a negative exponents , Example

a^{n }= a^{-(-n) }=

Question 40.

**REASONING**

The definition of a negative exponent states that a^{-n} = \(\frac{1}{a^{n^{*}}}\).

Explain why this rule does not apply when a = 0.

Answer:

Explanation:

we have negative exponent states that a^{-n} = \(\frac{1}{a^{n^{*}}}\) but this

rule does not apply when a= 0 as 0^{-n} = 0 or

if we use negative exponent we get =

we could get 0 or 1 so it is undefined.

### Lesson 8.5 Estimating Quantities

**EXPLORATION 1**

Work with a partner. Match each picture with the most appropriate distance. Explain your reasoning.

Answer:

Explanation:

Here we match each picture with the most appropriate distance

a. If we look at the picture it is very away so we take the

least distance in all as 6 X 10^{-2} m .

b. If we look at the picture it is very far so we take the

far distance in all as 6 X 10^{3} m .

c. If we look at the picture it can jump very little distance so we take

as 2 X 10^{-1} m.

d. If we look at the picture it is at near distance so we take

as 1 X 10^{1} m.

**EXPLORATION 2**

Approximating Numbers

Work with a partner. Match each number in List 1 with its closest approximation in List 2. Explain your method.

Answer:

Explanation:

We matched each number in List 1 with its closest

approximation in List 2 as

a. 180,000,000,000,000 is nearly or approximately equal to ≈ 2,00,000,000,000,000,

we have after 2 followed by 14 zeros so we take as 2 X 10^{14} matches with C.

therefore we match a. to C.

b. 0.0000000011 is nearly or approximately equal to ≈ 1 X 10^{-9} matches with C,

we have divided 1 by 10 followed by 9 zeros.

therefore we match b. to F.

c. 302,000,000,000 is nearly or approximately equal to ≈ 300,000,000,000,

we have after 3 followed by 11 zeros so we take as 3 X 10^{11} matches with A.

therefore match c. to A.

d. 0.00000028 is nearly or approximately equal to ≈ 0.0000003,

so 3 X 10^{-7} matches with E, we have divided 3 by 10 followed by 7 zeros,

therefore we match d. to E.

e. 0.0000097 is nearly or approximately equal to ≈ 0.00001,

so 1 X 10^{-5} matches with B, we have divided 1 by 10 followed by 5 zeros,

therefore we match e. to B.

f. 330,000,000,000,000 is nearly or approximately equal to ≈ 3,00,000,000,000,000

we have after 3 followed by 14 zeros so we take as 3 X 10^{14} matches with H.

therefore we match f. to H.

g. 26,000,000,000,000 is nearly or approximately equal to ≈ 30,000,000,000,000

we have after 3 followed by 13 zeros so we take as 3 X 10^{13} matches with D.

therefore we match g. to D.

h. 0.000023 is nearly or approximately equal to ≈ 0.00002,

so 2 X 10^{-5} matches with G, we have divided 2 by 10 followed by 5 zeros,

therefore we match h. to G.

**Try It**

**Round the number. Write the result as the product of a single digit and a power of 10.**

Question 1.

8,031,426,100

Answer:

8,031,426,100 = 8 X 10^{9}

Explanation:

Given number is 8,031,426,100 is nearly or approximately

equal to ≈ 8,000,000,000 so we have 8 followed by 9 zeros,

so we write as 8 X 10^{9}.

Question 2.

98,247,836,218

Answer:

98,247,836,218 = 1 X 10^{11}

Explanation:

Given number is 98,247,836,218 is nearly or approximately

equal to ≈ 100,000,000,000 so we have 1 followed by 11 zeros,

so we write as 1 X 10^{11}.

**Round the number. Write the result as the product of a single digit and a power of 10.**

Question 3.

0.000384509

Answer:

0.000384509 = 4 X 10^{-4}

Explanation:

Given number is 0.000384509 is nearly or approximately

equal to ≈ 0.0004, we have divided 4 by 10 followed

by 4 zeros so we write as 4 X 10^{-4}.

Question 4.

0.00000726

Answer:

0.00000726 = 7 X 10^{-6}

Explanation:

Given number is 0.00000726 is nearly or approximately

equal to ≈ 0.000007, we have divided 7 by 10

followed by zeros so we write as 7 X 10^{-6}.

Question 5.

The distance from Mercury to Mars is about 105,651,744 miles. The distance from Saturn to Jupiter is about 4 times this distance. What is the approximate distance from Saturn to Jupiter?

Answer:

The distance from Saturn to Jupiter is (105,651,744)^{4 }miles

Explanation:

Given the distance from Mercury to Mars is about 105,651,744 miles,

and the distance from Saturn to Jupiter is about 4 times this distance,

so 4 times of 105,651,744 is 105,651,744 X 105,651,744 X

105,651,744 X 105,651,744 = (105,651,744)^{4 }miles.

Self-Assessment for Concepts & Skills^{
}

Solve each exercise. Then rate your understanding of the success criteria in your journal.

**APPROXIMATING A NUMBER** Round the number. Write the result as the product of a single digit and a power of 10.

Question 6.

899,032,878,300

Answer:

899,032,878,300 = 9 X 10^{11}

Explanation:

Given number is 899,032,878,300 is nearly or approximately

equal to ≈ 900,000,000,000 so we have 9 followed by 11 zeros,

so we write as 9 X 10^{11}.

Question 7.

62,322,118,987

Answer:

62,322,118,987= 6 X 10^{10
}

Explanation:

Given number is 62,322,118,987 is nearly or approximately

equal to ≈ 60,000,000,000 so we have 6 followed by 10 zeros,

so we write as 6 X 10^{10}.

Question 8.

0.00000278101

Answer:

0.00000278101 = 3 X 10^{-6
}

Explanation:

Given number is 0.00000278101 is nearly or approximately equal to ≈ 0.000003,

we have divided 3 by 10 followed by 6 zeros so we write as

3 X 10^{-6}.

Question 9.

0.000013094

Answer:

0.000013094 = 1 X 10^{-5}.

Explanation:

Given number is 0.000013094 is nearly or approximately

equal to ≈ 0.00001, we have divided 1 by 10

followed by 5 zeros so we write as 1 X 10^{-5}.

Question 10.

**APPROXIMATING A QUANTITY**

Lake A has a volume of 21,150,427,000 cubic meters. Lake B has a volume that is 2.5 times the volume of Lake A. What is the approximate volume of Lake B?

Answer:

Explanation:

Given Lake A has a volume of 21,150,427,000 cubic meters.

Lake B has a volume that is 2.5 times the volume of Lake A.

The approximate volume of Lake B is 2.5 X 21,150,427,000,

we take 21,150,427,000 is nearly or approximately

equal to ≈ 20,000,000,000 so 2.5 X 20,000,000,000

= 5 X 10,000,000,000 = 5 X 10^{10}.

**Self-Assessment for Problem Solving**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.

On average, a small dog’s heart beats about 530,000,000 times

during its lifetime, and a large dog’s heart beats about 1.4 times this amount.

What is the approximate number of heartbeats in the lifetime of a large dog?

Answer:

The approximate number of heartbeats in the lifetime of a

large dog is 7 X 10^{8 }times.

Explanation:

Given on average, a small dog’s heart beats about

530,000,000 times during its lifetime and a large dog’s

heart beats about 1.4 times this amount.

So number of heartbeats in the lifetime of a large dog

= 1.5 X 530,000,000 we take 530,000,000 is nearly or approximately

equal to ≈ 500,000,000 so 1.5 X 500,000,000 = 7.5 X 100,000,000

= 7 X 10^{8 }times, therefore the approximate number

of heartbeats in the lifetime of a large dog is 7 X 10^{8 }times.

Question 12.

**DIG DEEPER!**

A physicist observes a gamma ray with a wavelength of 0.00000000135 millimeter and an X-ray with a wavelength of 0.00000012 millimeter. (a) About how many times shorter is the wavelength of the gamma ray than the wavelength of the X-ray? (b) The diagram shows wavelengths of visible light. Which ray has a wavelength closer to the wavelength of dark blue light?

Answer:

a.10^{2 } or 100 times shorter is the wavelength of the

gamma ray than the wavelength of the X-ray.

b. X-ray has a wave length closer to the wavelength of dark blue light.^{ }

Explanation:

Given a physicist observes a gamma ray with a wavelength

of 0.00000000135 millimeter is nearly or approximately

equal to ≈ 0.000000001 millimeter = 1 X 10^{-9 } millimeters,

and an X-ray with a wavelength 0.00000012 millimeter

is nearly or approximately equal to ≈ 0.0000001 millimeter=

1 X 10^{-7 } millimeters so now comparing wavelength

of the gamma ray with the wavelength of the X-ray,

1 X 10^{-9 } millimeters with 1 X 10^{-7 } millimeters so

we have 10^{2 } or 100 times shorter is the wavelength of the

gamma ray than the wavelength of the X-ray.

b. We have dark blue wavelength as 4 X 10^{-4 } millimeters so

it is close to an X-ray with a wavelength 0.00000012 millimeter

is nearly or approximately equal to ≈ 0.0000001 millimeter =

1 X 10^{-7 } millimeters. So X-ray has a wave length closer to the

wavelength of dark blue light.^{ } ^{
}

### Estimating Quantities Homework & Practice 8.5

**Review & Refresh**

**Simplify. Write the expression using only positive exponents.**

Question 1.

3x^{-5}

Answer:

3x^{-5}= \(\frac{3}{x^{5}}\)

Explanation:

Given expression as 3x^{-5 }we write the expression using only

positive exponents for 3 X x^{-5 }as we write first x^{-5 }we have

=

\(\frac{1}{x^{5}}\) now 3 X \(\frac{1}{x^{5}}\) = \(\frac{3}{x^{5}}\).

Question 2.

d^{0} • d^{-4}

Answer:

d^{0} • d^{-4 }= \(\frac{1}{d^{4}}\)

Explanation:

Given d^{0} • d^{-4 }we write the expression as positive exponents by using

so d^{0} X \(\frac{1}{d^{4}}\) as we know d^{0} =1,

d^{0} X \(\frac{1}{d^{4}}\)=1 X \(\frac{1}{d^{4}}\) or \(\frac{1}{d^{4}}\).

Question 3.

\(\frac{a^{6}}{2 a^{11}}\)

Answer:

\(\frac{a^{6}}{2 a^{11}}\) = \(\frac{1}{2 a^{5}}\)

Explanation:

Given expression as \(\frac{a^{6}}{2 a^{11}}\) we use rule for finding

\(\frac{a^{m}}{a^{n}}\), a quotient of two powers with the same base as a^{m-n
} \(\frac{a^{6}}{a^{11}}\) = a^{6-11 }= a^{-5 }for a^{-5 }we use

Write an equation in point-slope form of the line that passes through the given point and has the given slope.

Question 4.

(- 1, 2); m = – \(\frac{1}{3}\)

Answer:

The equation in point-slope form is (y-2)= – \(\frac{1}{3}\) (x+1)

Explanation:

Given (- 1, 2); m = – \(\frac{1}{3}\) we know

for straight-line equations the “point-slope” form with

(x_{1}, y_{1}) and a slope m we have formula y – y_{1} = m(x – x_{1})

here we have (x_{1}, y_{1}) are (-1,2) and a slope m is – \(\frac{1}{3}\),

So the equation in point-slope form is (y-2)= – \(\frac{1}{3}\) (x+1).

Question 5.

(3, 4); m = \(\frac{3}{4}\)

Answer:

The equation in point-slope form is (y-4)= \(\frac{3}{4}\) (x-3)

Explanation:

Given (3, 4); m = \(\frac{3}{4}\) we know

for straight-line equations the “point-slope” form with

(x_{1}, y_{1}) and a slope m we have formula y – y_{1} = m(x – x_{1})

here we have (x_{1}, y_{1}) are (3,4) and a slope m is \(\frac{3}{4}\),

So the equation in point-slope form is (y-4)= \(\frac{3}{4}\) (x-3).

Question 6.

(1, – 4); m = – 2

Answer:

The equation in point-slope form is (y+4)=-2 (x-1)

Explanation:

Given (1, -4); m = -2 we know

for straight-line equations the “point-slope” form with

(x_{1}, y_{1}) and a slope m we have formula y – y_{1} = m(x – x_{1})

here we have (x_{1}, y_{1}) are (1,4) and a slope m is – 2,

So the equation in point-slope form is (y+4)= -2(x-1).

**Concepts, Skills, & Problem Solving**

**APPROXIMATING NUMBERS** Match the number with its closest approximation. (See Exploration 2, p. 343.)

Question 7.

0.000618

Answer:

0.000618 ≈ 0.0006 ≈ 6 X 10^{-4}

So we match with B with its closest approximation

Explanation:

0.000618 is nearly or approximately equal to ≈ 0.0006 ≈ 6 X 10^{-4} ,

we have divided 6 by 10 followed by 4 zeros,

therefore we match B with its closest approximation.

Question 8.

7,257,993,201

Answer:

7,257,993,201 ≈ 7,000,000,000 ≈ 7 X 10^{9}

therefore we match D with its closest approximation.

Explanation:

Given number is 7,257,993,201 nearly or approximately

equal to ≈ 7,000,000,000 so we have 7 followed by 9 zeros,

so we write as 7 X 10^{9 }therefore we match D with its closest approximation.

Question 9.

0.0006781004

Answer:

0.0006781004 ≈ 0.0007 ≈ 7 X 10^{-4}

So we match with C with its closest approximation

Explanation:

0.000618 is nearly or approximately equal to ≈ 0.0007 ≈ 7 X 10^{-4} ,

we have divided 7 by 10 followed by 4 zeros,

therefore we match C with its closest approximation.

Question 10.

782,309,441

Answer:

782,309,441≈ 800,000,000 ≈ 8 X 10^{8}

therefore we match A with its closest approximation.

Explanation:

Given number is 782,309,441 nearly or approximately

equal to ≈ 800,000,000 so we have 8 followed by 8 zeros,

so we write as 8 X 10^{8 }therefore we match A with its closest approximation.

A. 8 × 10^{8}

B. 6 × 10^{-4}

C. 7 × 10^{-4}

D. 7 × 10^{9}

**APPROXIMATING A LARGE NUMBER** Round the number. Write the result as a product of a single digit and a power of 10.

Question 11.

414,148,636,008

Answer:

414,148,636,008 ≈ 4 X 10^{11}

Explanation:

Given number is 414,148,636,008 nearly or approximately

equal to ≈ 400,000,000,000 so we have 4 followed by 11 zeros,

so we write as 4 X 10^{11}.

Question 12.

231,210

Answer:

231,210 ≈ 2 X 10^{5}

Explanation:

Given number is 231,210 nearly or approximately

equal to ≈ 200,000 so we have 2 followed by 5 zeros,

so we write as 2 X 10^{5}.

Question 13.

28,007,806,203

Answer:

28,007,806,203 ≈ 3 X 10^{10}

Explanation:

Given number is 28,007,806,203 nearly or approximately

equal to ≈ 30,000,000,000 so we have 3 followed by 10 zeros,

so we write as 3 X 10^{10}.

Question 14.

38,108,996,999

Answer:

38,108,996,999 ≈ 4 X 10^{10}

Explanation:

Given number is 38,108,996,999 nearly or approximately

equal to ≈ 40,000,000,000 so we have 4 followed by 10 zeros,

so we write as 4 X 10^{10}.

Question 15.

1,003,111,391,008

Answer:

1,003,111,391,008 ≈ 1 X 10^{12}

Explanation:

Given number is 1,003,111,391,008 nearly or approximately

equal to ≈ 1,000,000,000,000 so we have 1 followed by 12 zeros,

so we write as 1 X 10^{12}.

Question 16.

627,638,538

Answer:

627,638,538 ≈ 6 X 10^{8}

Explanation:

Given number is 627,638,538 nearly or approximately

equal to ≈ 6,00,000,000 so we have 6 followed by 8 zeros,

so we write as 6 X 10^{8}.

Question 17.

**APPROXIMATING A LARGE NUMBER**

A company earns $518,204,500. Round the number. Write the result as a product of a single digit and a power of 10.

Answer:

$518,204,500 ≈ 5 X 10^{8 }dollars

Explanation:

Given number is $518,204,500 nearly or approximately

equal to ≈ 500,000,000 so we have 5 followed by 8 zeros,

so we write as 5 X 10^{8 }dollars.

**APPROXIMATING A SMALL NUMBER** Round the number.

Write the result as a product of a single digit and a power of 10.

Question 18.

0.00000124

Answer:

0.00000124 ≈ 1 X 10^{-6}

Explanation:

As 0.00000124 is nearly or approximately equal to ≈ 0.000001 ≈

1 X 0.000001 = 1 X 10^{-6 }or we have divided 1 by 10 followed by 6 zeros.

Question 19.

0.00003946

Answer:

0.00003946 ≈ 4 X 10^{-5}

Explanation:

As 0.00003946 is nearly or approximately equal to ≈ 0.00004 ≈

4 X 0.00001= 4 X 10^{-5 }or we have divided 4 by 10 followed by 5 zeros.

Question 20.

0.00001726

Answer:

0.00001726 ≈ 2 X 10^{-5}

Explanation:

As 0.00001726 is nearly or approximately equal to ≈ 0.00002 ≈ 2 X 10^{-5},

or we have divided 2 by 10 followed by 5 zeros.

Question 21.

0.00063718

Answer:

0.00063718 ≈ 6 X 10^{-4}

Explanation:

As 0.00063718 is nearly or approximately equal to ≈ 0.0006 ≈ 6 X 10^{-4},

or we have divided 6 by 10 followed by 4 zeros.

Question 22.

0.00000000305

Answer:

0.00000000305 ≈ 3 X 10^{-9}

Explanation:

As 0.00000000305 is nearly or approximately

equal to ≈ 0.000000003 ≈ 3 X 10^{-9},

or we have divided 3 by 10 followed by 9 zeros.

Question 23.

0.000000000994

Answer:

0.000000000994 ≈ 1 X 10^{-9}

Explanation:

As 0.000000000994 is nearly or approximately

equal to ≈ 0.000000001 ≈ 1 X 10^{-9},

or we have divided 1 by 10 followed by 9 zeros.

Question 24.

**YOU BE THE TEACHER**

Your friend rounds 0.000468 to the nearest ten thousandth and writes the result as a product of a single digit and a power of 10. Is your friend correct? Explain your reasoning.

Answer:

Yes, Friend is correct as 0.000468 to the nearest ten thousandth is 5 X 10^{-4}

Explanation:

As 0.000468 is nearly or approximately to the nearest

ten thousandth is equal to ≈ 0.0005 ≈ 5 X 0.0001= 5 X 10^{-4},

which is equal to the value of friend, So Yes, Friend is correct

as 0.000468 to the nearest ten thousandth is 5 X 10^{-4}.

Question 25.

**APPROXIMATING A QUANTITY**

A series of mystery books contains 2,029,242 words. A series of science fiction books contains about 3.5 times the number of words as the mystery book series. What is the approximate number of words in the science fiction book series?

Answer:

The approximate number of words in the

science fiction book series are 7 X 10^{6 }words.

Explanation:

Given a series of mystery books contains 2,029,242 words,

a series of science fiction books contains about 3.5 times the

number of words as the mystery book series.

therefore approximate number of words in the science

fiction book series are 3.5 X 2,029,242 as 2,029,242 is

approximately equal to ≈ 2 X 1000,000 so 3.5 X 2 X 1,000,000

= 7 X 10^{6 }words.

Question 26.

**APPROXIMATING A QUANTITY**

A volcanic eruption ejects about 43,600,000,000 cubic feet of volcanic rock. A smaller volcanic eruption ejects about 75% of this amount. What is the approximate amount of volcanic rock that the smaller volcanic eruption ejects?

Answer:

The approximate amount of volcanic rock that the

smaller volcanic eruption ejects is 3 X 10^{10 }cubic feet.

Explanation:

Given a volcanic eruption ejects about 43,600,000,000 cubic feet

of volcanic rock. A smaller volcanic eruption ejects about 75% of

this amount, So the smaller volcanic eruption is 75% X 43,600,000,000

we get \(\frac{75}{100}\) X 43,600,000,000 = 75 X 43,6000000 =

32,700,000,000 is approximately equal to ≈ 3 X 10,000,000,000 = 3 X 10^{10 }cubic feet.

Question 27.

**STRUCTURE**

Find a number that is approximately 1.5 times 61,040,000,100.

Write the result as the product of a single digit and a power of 10.

Answer:

The number approximate value of 1.5 times 61,040,000,100 is 9 X 10^{10}

Explanation:

The number approximate value of 1.5 times 61,040,000,100 is

1.5 X 61,040,000,100 = 91,560,000,150 is approximately

equal to ≈ 9 X 10,000,000,000 = 9 X 10^{10 }.

Question 28.

**APPROXIMATING A QUANTITY**

A mitochondrion has a diameter of about 0.00000031 meter. The diameter of a chloroplast is about 3 times that of the mitochondrion. What is the approximate diameter of the chloroplast?

Answer:

The approximate diameter of the chloroplast is 9 X 10^{-7 }meters.

Explanation:

A mitochondrion has a diameter of about 0.00000031 meter.

The diameter of a chloroplast is about 3 times that of the

mitochondrion. So the approximate diameter of the chloroplast is

3 X 0.00000031 = 0.00000093 is approximately

equal to ≈ 9 X 0.0000001 = 9 X 10^{-7 }meters.

Question 29.

**MODELING REAL LIFE**

A photo taken with a smartphone has 1,227,104 pixels. A photo taken with a camera has 11,943,936 pixels. Approximately how many times more pixels are in the photo taken with the camera?

Answer:

10 times more pixels are in the photo taken with the camera.

Explanation:

Given a photo taken with a smartphone has 1,227,104 pixels and

A photo taken with a camera has 11,943,936 pixels.

Number of times the pixels are in the photo taken with the camera is

\(\frac{11943936}{1227104}\) = 9.7334341669 is approximately

equal to ≈ 10, So 10 times more pixels are in the photo taken with the camera.

Question 30.

**MODELING REAL LIFE**

A star has a core temperature of about 115,000,000°F. The temperature of a lightning strike is about 10,300°F. Approximately how many times hotter is the core temperature of the star than the temperature of the lightning strike?

Answer:

11165 times hotter is the core temperature of the star than the

temperature of the lightning strike

Explanation:

Given a star has a core temperature of about 115,000,000°F and

the temperature of a lightning strike is about 10,300°F.

Number of times hotter is the core temperature of the star

more than the temperature of the lightning strike is

\(\frac{115000000}{10300}\) = 11,165.048543689

is approximately equal to ≈ 11165, So 11165 times hotter is the

core temperature of the star than the temperature of the lightning strike.

Question 31.

**REASONING**

The table shows the diameters of five types of animal hairs.

a. Order the hair types from greatest to least diameter.

b. What unit should be used to represent these data? Explain your reasoning.

Answer:

a. The hair types from greatest to least diameter are

Cow, Buffalo, Camel, Donkey, Rat.

b. Millimeters unit should be used to represent these data.

Explanation:

a. Given the table shows the diameters of five types of animal hairs as

Buffalo = 0.00011 is approximately equal to ≈ 1 X 10^{-4
}Rat = 0.00004 is approximately equal to ≈ 4 X 10^{-5
}Camel = 0.00008 is approximately equal to ≈ 8 X 10^{-5}

Cow = 0.00016 is approximately equal to ≈ 1 X 10^{-4}

Donkey = 0.00005 is approximately equal to ≈ 5 X 10^{-5},

Now from greatest to least diameters are

1 X 10^{-4 }> 1 X 10^{-4 }> 8 X 10^{-5 }> 5 X 10^{-5}> 4 X 10^{-5},

therefore the hair types from greatest to least diameter are

Cow, Buffalo, Camel, Donkey, Rat.

b. As the smallest unit of length is millimeter,

Millimeter is the smallest common unit of length that is represented as

‘mm’. The relation between mm and m is 1 meter = 1000 millimeter.

So millimeters unit should be used to represent these data.

Question 32.

**PROBLEM SOLVING**

The distance between New York City and Princeton is about 68,500 meters. The distance between New York City and San Antonio is about 40 times this distance. What is the approximate distance between New York City and San Antonio? Write the result as the product of a single digit and a power of 10.

Answer:

The approximate distance between New York City and San Antonio is

3 X 10^{6 }meters.

Explanation:

Given the distance between New York City and Princeton

is about 68,500 meters. The distance between New York City

and San Antonio is about 40 times this distance.

So the approximate distance between New York City and San Antonio is

40 X 68500 = 2,740,000 is approximately equal to ≈ 3 X 1,000,000 = 3 X 10^{6 }meters.

Question 33.

**REASONING**

Is 5 × 10^{6} a better approximation of 5,447,040 or 5,305,004? Explain.

Answer:

5 × 10^{6} a better approximation of 5,305,004

Explanation:

We have 5,447,040 or 5,305,004 we take approximate values ,

now 5 × 10^{6} is approximately equal to ≈ 5,000,000 now if

we see for 5,000,000 the more near value is 5,305,004 than the

5,447,040. So 5 × 10^{6} a better approximation of 5,305,004.

Question 34.

**DIG DEEPER!**

A proton weighs 0.00000000000167 nanogram. About how much do 8 protons weigh? Write the result as the product of a single digit and a power of 10. Is your answer an overestimate or an underestimate?

Answer:

The weight of 8 protons is 1 × 10^{-11}

My answer is overestimate.

Explanation:

A proton weighs 0.00000000000167 nanogram.

So 8 protons weighs is 8 X 0.00000000000167 =1.336 X 10^{-11}

is approximately equal to ≈ 1 X 10^{-11}.

My answer is overestimate given proton weighs

0.00000000000167 nanogram is approximately equal to ≈ 1 X 10^{-12}

and the weight of 8 protons is 1 × 10^{-11 }if we see 1 X 10^{-11 }> 1 X 10^{-12 },

So my answer is overestimate.

### Lesson 8.6 Scientific Notation

**EXPLORATION 1**

Work with a partner. Use a graphing calculator.

a. Experiment with multiplying very large numbers until your calculator displays an answer that is not in standard form. What do you think the answer means?

b. Enter the function y = 10^{x} into your graphing calculator. Use the table feature to evaluate the function for positive integer values of x until the calculator displays a y-value that is not in standard form. Do the results support your answer in part(a)? Explain.

c. Repeat part(a) with very small numbers.

d. Enter the function y = (\(\frac{1}{10}\))^{x} into your graphing calculator. Use the table feature to evaluate the function for positive integer values of x until the calculator displays a y-value that is not in standard form. Do the results support your answer in part(c)? Explain.

a. Means the answer is big so answer is not in standard form it

will be in scientific notation.

b. Until at x =6 the calculator displays a y-value that is not in

standard form. Yes the results support my answer in part(a),

c. Means the answer is small so answer is not in standard form it

will be in scientific notation.

d. Until at x =6 the calculator displays a y-value that is not in

standard form. Yes the results support my answer in part(c).

Explanation:

a. If multiplying very large numbers until my calculator

displays an answer that is not in standard form. We use

scientific notation it is just a shorthand way of expressing

gigantic numbers also known as an exponential form,

scientific notation has been one of the oldest mathematical approaches.

If numbers are too big to be simply calculated we refer to

scientific notation to handle these circumstances.

For example 4.5 X 10^{9} years or (on a calculator)

4.5E9 years (1 billion in scientific notation means

10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10)

b. Entering the function y = 10^{x} into graphing calculator.

Using the table feature we evaluate the function for positive integer

values of until x =6 the calculator displays a y-value that is not in standard form.

Yes the results support my answer in part(a) as we multiplied

very large number, calculator displays an answer that is not in standard form.

c. If multiplying very small numbers until my calculator

displays an answer that is not in standard form. We use

scientific notation it is just a shorthand way of expressing

numbers also known as an exponential form,

For example 0.000003426

Step 1: Move the decimal so that there is only one digit in front of the decimal.

0.000003.426

Step 2: Count the number of moves from the original decimal to the new position.

0.000003.426, There are 6 moves

Step 3: Write the new number as a product with a power of ten.

3.426 x 10^{-6} , 3E-6 the number of moves becomes the exponent.

d. Entering the function y = (\(\frac{1}{10}\))^{x} into graphing calculator.

Using the table feature we evaluate the function for positive integer

values of until x =6 the calculator displays a y-value that is not in standard form.

Yes the results support my answer in part(c) as we multiplied

very small number, calculator displays an answer that is not in standard form.

**Try It**

**Write the number in scientific notation.**

Question 1.

50,000

Answer:

50,000 = 5 X 10^{4}

Explanation:

Given number is 50,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 50,000 = 5 X 10000 = 5 X 10^{4}.

Question 2.

25,000,000

Answer:

25,000,000 = 25 X 10^{6}

Explanation:

Given number is 25,000,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 25,000,000 = 25 X 1000000 = 25 X 10^{6}.

Question 3.

683

Answer:

683 = 6.83 X 10^{2}

Explanation:

Given number is 683 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 683 = 6.83 X 100 = 6.83 X 10^{2} .

Question 4.

0.005

Answer:

0.005 = 5 X 10^{-3}

Explanation:

Given number is 0.005 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.005 = 5 X 0.001 = 5 X 10^{-3}.

Question 5.

0.00000033

Answer:

0.00000033 = 3.3 X 10^{-7}

Explanation:

Given number is 0.00000033 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.00000033 = 3.3 X 0.0000001 = 3.3 X 10^{-7}.

Question 6.

0.000506

Answer:

0.000506 = 5.06 X 10^{-4}

Explanation:

Given number is 0.000506 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.000506 = 5.06 X 0.0001 = 5.06 X 10^{-4}.

**Write the number in standard form.**

Question 7.

6 × 10^{7}

Answer:

6 × 10^{7 }= 60,000,000

Explanation:

Given 6 × 10^{7 }the number in standard form is

6 X 10,000,000 , So 6 × 10^{7 }= 60,000,000.

Question 8.

9.9 × 10^{-5}

Answer:

9.9 × 10^{-5 }= 0.000099

Explanation:

Given 9.9 × 10^{-5 }^{ }the number in standard form is

1 × 0.000099^{ }= 0.000099.

Question 9.

1.285 × 10^{4}

Answer:

1.285 × 10^{4 }= 12850

Explanation:

Given 1.285 × 10^{4 }the number in standard form is

1.285 X 10,000 , So 1.285 × 10^{4 }= 12850.

**Self-Assessment for Concepts & Skills**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

**WRITING NUMBERS IN SCIENTIFIC NOTATION** Write the number in scientific notation.

Question 10.

675,000,000

Answer:

675,000,000 = 6.75 X 10^{8}

Explanation:

Given number is 675,000,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 675,000,000 = 6.75 X 1,00,000,000 = 6.75 X 10^{8} .

Question 11.

0.000000084

Answer:

0.000000084 = 8.4 X 10^{-8}

Explanation:

Given number is 0.000000084 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.000000084 = 8.4 X 0.00000001 = 8.4 X 10^{-8}.

Question 12.

0.000012001

Answer:

0.000012001 = 1 X 10^{-5}

Explanation:

Given number is 0.000012001 first we write

in approximately equal to ≈0.00001 now in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.00001= 1 X 0.00001 = 1 X 10^{-5}.

**WRITING NUMBERS IN STANDARD FORM** Write the number in standard form.

Question 13.

8 × 10^{-7}

Answer:

8 × 10^{-7 }= 0.0000008

Explanation:

Given 8 × 10^{-7 }^{ }the number in standard form is

8 × 0.0000001^{ }= 0.0000008.

Question 14.

3.876 × 10^{7}

Answer:

3.876 × 10^{7 }= 38,760,000

Explanation:

Given 3.876 × 10^{7 }the number in standard form is

3.876 X 10,000,000 = 38,760,000 so 3.876 × 10^{7 }= 38,760,000.

Question 15.

1.11 × 10^{-5}

Answer:

1.11 × 10^{-5 }= 0.0000111

Explanation:

Given 1.11 × 10^{-5 }^{ }the number in standard form is

1.11 × 0.00001^{ }= 0.0000111.

Question 16.

**WHICH ONE DOESN’T BELONG?**

Which number does not belong with the other three? Explain.

Answer:

d. 10 X 9.2^{-13 }does not belong with the other three

Explanation:

Given 2.8 X 10^{15 }, 4.3 X 10^{-30 }, 1.05 X 10^{28 }all are single digit to

the left of decimal sign and is multiplied by an integer power of 10,

but only 10 X 9.2^{-13 }^{ }has power for 9.2^{13 },so 10 X 9.2^{-13 }^{
}does not belong with the other three.

**Self-Assessment for Problem Solving**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.

A series of movies is about 3.285 × 10^{4} seconds long.

How long does it take to watch the series twice?

Express your answer using more-appropriate units.

Answer:

The time it takes to watch the series twice is 18.25 hours.

Explanation:

Given a series of movies is about 3.285 × 10^{4} seconds long

and the time it takes to watch the series twice is

2 X 3.285 × 10^{4} seconds = 6.57 X 10^{4} seconds,

now we convert seconds into hours as 1 hour = 3600 seconds,

so \(\frac{6.57}{3600}\) X 10^{4} = 18.25 hours, therefore

the time it takes to watch the series twice is 18.25 hours.

Question 18.

The total power of a space shuttle during launch is the sum of the power from its solid rocket boosters and the power from its main engines. e power from the solid rocket boosters is 9,750,000,000 watts. What is the power from the main engines?

Answer:

The power from the main engines is 1.99 X 10^{9} watts.

Explanation:

Given the power from the solid rocket boosters is

9,750,000,000 watts is approximately equal to ≈9.75 X10^{9} watts

Let the main engine power be x, Total Power =

power from the solid rocket boosters + power from the main engines,

1.174 X 10^{10} watts = 9.75 X 10^{9} watts + x, So x = (11.74 – 1.99) X 10^{9} watts

= 9.75 X 10^{9} watts, therefore the power from the main engines is 1.99 X 10^{9} watts.

Question 19.

The area of a trampoline is about 1.8 × 10^{4} square inches.

Write this number in standard form. Then represent the area of the

trampoline using more-appropriate units.

Answer:

The area of the trampoline is 125 square feet

Explanation:

Given the area of a trampoline is about 1.8 × 10^{4} square inches,

so we get 1.8 × 10^{4} as 18000 square inches, now we

represent the area of the trampoline using more-appropriate units

in square feet, We know 1 square feet = 144 square inches

= \(\frac{18000}{144}\) = 125 square feet,

so the area of the trampoline is 125 square feet.

Question 20.

**DIG DEEPER!**

The epidermis, dermis, and hypodermis are layers of your skin. The dermis is about 3.5 millimeters thick. The epidermis is about 1.25 × 10^{-3} meter thick. The hypodermis is about 0.15 centimeter thick. What is the difference in thickness of the thickest layer and the thinnest layer? Justify your answer.

Answer:

The difference in thickness of the thickest layer and

the thinnest layer is 0.002 meters

Explanation:

Given the epidermis, dermis, and hypodermis are layers of your skin.

The dermis is about 3.5 millimeters thick. The epidermis is about

1.25 × 10^{-3} meter thick. The hypodermis is about 0.15 centimeter thick.

So the difference in thickness of the thickest layer and the thinnest layer is

1 millimeter = 0.001 meters,

dermis = 3.5 X 0.001 = 0.0035 meters,

epidermis = 1.25 X 10^{-3} = 0.00125 meters,

hypodermis = 0.15 centimeter, we know 1 centimeter = 0.01 meter,

so 0.15 X 0.01 = 0.0015 meters, we have the thickest as 0.0035 meters and

thinnest is 0.00125 meters , So the difference in thickness of

the thickest layer and the thinnest layer is 0.0035 – 0.00125 =

0.00225 is approximately equal to ≈ 0.002 meters.

### Scientific Notation Homework & Practice 8.6

**Review & Refresh**

**Round the number. Write the result as the product of a single digit and a power of 10.**

Question 1.

0.00000129

Answer:

0.00000129 ≈ 1 X 10^{-6}

Explanation:

As 0.00000129 is nearly or approximately equal to ≈ 0.000001 ≈

1 X 0.000001 = 1 X 10^{-6 }or we have divided 1 by 10 followed by 6 zeros.

Question 2.

4,241,933,200

Answer:

4,241,933,200 ≈ 4 X 10^{9}

Explanation:

Given number is 4,241,933,200 nearly or approximately

equal to ≈ 2,000,000,000 so we have 4 followed by 9 zeros,

so we write as 4 X 10^{9}.

Question 3.

0.0000001801

Answer:

0.0000001801≈ 2 X 10^{-7}

Explanation:

As 0.0000001801 is nearly or approximately equal to ≈ 0.0000002 ≈

2 X 0.0000001 = 2 X 10^{-7 }or we have divided 2 by 10 followed by 7 zeros.

Question 4.

879,679,466

Answer:

879,679,466 ≈ 9 X 10^{8}

Explanation:

Given number is 879,679,466 nearly or approximately

equal to ≈ 9,00,000,000 so we have 9 followed by 8 zeros,

so we write as 9 X 10^{8}.

**Write the product using exponents.**

Question 5.

4 • 4 • 4 • 4 • 4

Answer:

4 • 4 • 4 • 4 • 4 = (4)^{5} ^{ }

Explanation:

We write the product 4 • 4 • 4 • 4 • 4 in exponents as (4)^{5}^{
}because 4 is multiplied by 5 times.

Question 6.

3 • 3 • 3 • y • y • y

Answer:

3 • 3 • 3 • y • y • y = 3^{3}y^{3}= (3y)^{3}

Explanation:

We write the product 3 • 3 • 3• y • y • y in exponents as (3y)^{3}^{
}because 3 is multiplied by 3 times and y is also multiplied by 3 times.

Question 7.

(- 2) • (- 2) • (- 2)

Answer:

(- 2) • (- 2) • (- 2) = (-2)^{3}

Explanation:

We write the product (-2) • (-2) • (-2) in exponents as (-2)^{3}^{
}because -2 is multiplied by 3 times.

**Concepts, Skills, &Problem Solving**

**USING A GRAPHING CALCULATOR** Use a graphing calculator to evaluate x the function when 10. Write the number in standard form. (See Exploration 1, p. 349.)

Question 8.

y = (\(\frac{1}{10}\))^{x}

Answer:

The number in standard form is 0.0000000001

Explanation:

Given y = (\(\frac{1}{10}\))^{x }and x = 10 so

y = (\(\frac{1}{10}\))^{10 }so 1 X 10-^{10 }= 1X 0.0000000001

= 0.0000000001 therefore the number in standard form is 0.0000000001.

Question 9.

y = 20^{x}

Answer:

The number in standard form is 10,240,000,000,000

Explanation:

Given y = 20^{x }and x = 10 so y= (20)^{10 }= 10,240,000,000,000,

therefore the number in standard form is 10,240,000,000,000.

**WRITING NUMBERS IN SCIENTIFIC NOTATION** Write the number in scientific notation.

Question 10.

0.0021

Answer:

0.0021 = 2.1 X 10^{-3}

Explanation:

Given number is 0.0021 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.0021 = 2.1 X 0.001 = 2.1 X 10^{-3}.

Question 11.

5,430,000

Answer:

5,430,000 = 5.43 X 10^{6}

Explanation:

Given number is 5,430,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 5,430,000 = 5.43 X 1,000,000 = 5.43 X 10^{6}.

Question 12.

321,000,000

Answer:

321,000,000 = 3.21 X 10^{8}

Explanation:

Given number is 321,000,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 321,000,000 = 3.21 X 1,00,000,000 = 3.21 X 10^{8}.

Question 13.

0.00000625

Answer:

0.00000625 = 6.25 X 10^{-6}

Explanation:

Given number is 0.00000625 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.00000625 = 6.25 X 0.000001 = 6.25 X 10^{-6}.

Question 14.

0.00004

Answer:

0.00004 = 4 X 10^{-5}

Explanation:

Given number is 0.00004 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.00004 = 4 X 0.00001 = 4 X 10^{-5}.

Question 15.

10,700,000

Answer:

10,700,000 = 1.07 X 10^{7}

Explanation:

Given number is 10,700,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 10,700,000 = 1.07 X 10,000,000 = 1.07 X 10^{7}.

Question 16.

45,600,000,000

Answer:

45,600,000,000 = 4.56 X 10^{10}

Explanation:

Given number is 45,600,000,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 45,600,000,000 = 4.56 X 10,000,000,000 = 4.56 X 10^{10}.

Question 17.

0.000000000009256

Answer:

0.000000000009256 = 9.256 X 10^{-12}

Explanation:

Given number 0.000000000009256 is in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.000000000009256 = 9.256 X 0.000000000001 = 9.256 X 10^{-12}.

Question 18.

840,000

Answer:

840,000 = 8.4 X 10^{5}

Explanation:

Given number is 840,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 840,000 = 8.4 X 100,000 = 8.4 X 10^{5}.

**WRITING NUMBERS IN STANDARD FORM** Write the number in standard form.

Question 19

7 × 10^{7}

Answer:

7 × 10^{7 }= 70,000,000

Explanation:

Given 7 × 10^{7 }the number in standard form is

7 X 10,000,000 =70,000,000.

Question 20.

8 × 10^{-3}

Answer:

8 X 10^{-3 }= 0.008

Explanation:

Given 8 × 10^{-3 }^{ }the number in standard form is

8 × 0.001^{ }= 0.008.

Question 21.

5 × 10^{2}

Answer:

5 × 10^{2 }= 500

Explanation:

Given 5 × 10^{2 }the number in standard form is

7 X 100 = 500.

Question 22.

2.7 × 10^{-4}

Answer:

2.7 × 10^{-4 }= 0.00027

Explanation:

Given 2.7 × 10^{-4 }^{ }the number in standard form is

2.7 × 0.0001^{ }= 0.00027.

Question 23.

4.4 × 10^{-5}

Answer:

4.4 × 10^{-5 }= 0.000044

Explanation:

Given 4.4 × 10^{-5 }^{ }the number in standard form is

4.4 × 0.00001^{ }= 0.000044.

Question 24.

2.1 × 10^{3}

Answer:

2.1 × 10^{3}= 2,100

Explanation:

Given 2.1 × 10^{3 }the number in standard form is

2.1 X 1000 = 2,100.

Question 25.

1.66 × 10^{9}

Answer:

1.66 × 10^{9 }= 1,660,000,000

Explanation:

Given 1.66 × 10^{9 }the number in standard form is

1.66 X 1,000,000,000 = 1,660,000,000.

Question 26.

3.85 × 10^{-8}

Answer:

3.85 × 10^{-8 }= 0.0000000385

Explanation:

Given 3.85 × 10^{-8 }^{ }the number in standard form is

3.85 × 0.00000001^{ }= 0.0000000385.

Question 27.

9.725 × 10^{6}

Answer:

9.725 × 10^{6 }= 9,725,000

Explanation:

Given 9.725 × 10^{6 }the number in standard form is

9.725 X 1,000,000 = 9,725,000.

Question 28.

**MODELING REAL LIFE**

The U.S.Brig Niagara, a warship from the Battle of Lake Erie in 1813, uses about 28,300 feet of rope to operate its sails and spars. Write this number in scientific notation.

Answer:

28,300 feet = 2.83 X 10^{4}

Explanation:

Given the U.S.Brig Niagara, a warship from the Battle of

Lake Erie in 1813, uses about 28,300 feet of rope to

operate its sails and spars this number in scientific notation as

we write a number in scientific notation has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 28,300 = 2.83 X 10,000 = 2.83 X 10^{4}.

Question 29.

**MODELING REAL LIFE**

The radius of a fishing line is 2.5 × 10^{-4} feet. Write this ×number in standard form. Then write your answer using inches.

Answer:

The number in standard form is 0.00025 feet,

In inches it is 0.003 inches.

Explanation:

The radius of a fishing line is 2.5 × 10^{-4} feet,

The number in standard form is 2.5 X 0.0001 = 0.00025 feet,

Now to convert feet into inches we know 1 foot is 12 inch so

0.00025 X 12 = 0.003 inches.

Question 30.

**MODELING REAL LIFE**

Platelets are cell-like particles in the blood that help form blood clots.

a. How many platelets are in 3 milliliters of blood? Write your answer in standard form.

b. An adult human body contains about 5 liters of blood. How many platelets are in an adult human body?

Answer:

a. There are 810,000,000 milliliters platelets in 3 milliliters of blood.

b. There are 1,350,000,000,000 milliliters platelets are in an adult human body.

Explanation:

a. Given blood has 2.7 X 10^{8} platelets per milliliters,

So the number of platelets in 3 milliliters of blood is

3 X 2.7 X 10^{8} = 8.1 X 10^{8} = 810,000,000 milliliters.

b. An adult human body contains about 5 liters of blood,

So the number of platelets in an human body are, we know

1 liter = 1000 milliliters so 5 liters is equal to 5,000 milliliters,

5000 X 2.7 X 10^{8} = 1,350,000,000,000 milliliters ,therefore

There are 1,350,000,000,000 milliliters platelets are in an adult human body.

**CHOOSING APPROPRIATE UNITS** Match each value with the most appropriate unit of measurement.

Answer:

We match each value with the most appropriate unit of measurement

as 31. D, 32. C, 33. A , 34. B

Explanation:

Given

a. Height of a skyscraper is : 2.6 X 10^{2} = 2.6 X 100 = 260,

so the most appropriate unit of measurement is meters as

it is used to measure big lengths.

b. Distance between two asteroids : 2.5 X 10^{5} = 2.5 X 100,000 = 250,000

so the most appropriate unit of measurement is miles as

it is used to measure long distances.

c. Depth of bathtub: 1.6 X 10^{1} = 16,

so the most appropriate unit of measurement is inches as

it is used to measure the length of small objects.

d. Length of memory chip : 7.8 X 10^{0} = 7.8 X 1 = 7.8

so the most appropriate unit of measurement is millimeters as

it is used to measure very short lengths or thicknesses.

Question 35.

**NUMBER SENSE**

Describe how the value of a number written in scientific notation changes when you increase the exponent by 1.

Answer:

The number increases by 10 times

Explanation:

Whenever we increase the exponent by 1

in scientific notation the number becomes 10 times of

previous number. Example we have number 280 we write

number in scientific notation as single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 280 = 2.8 X 100 = 2.8 X 10^{2 }now if we increase exponent by 1 we get

2.8 X 10^{3 }as 2.8 X 1000 = 2800 here the value has increase by 10 times from

280 to 2800.

Question 36.

**PROBLEM SOLVING**

The area of the Florida Keys National Marine Sanctuary is about 9600 square kilometers. The area of the Florida Reef Tract is about 16.2% of the area of the sanctuary. What is the area of the Florida Reef Tract? Write your answer in scientific notation.

Answer:

The area of the Florida Reef Tract is 1.5552 X 10^{3} square kilometers

Explanation:

Given the area of the Florida Keys National Marine Sanctuary

is about 9600 square kilometers. The area of the Florida Reef

Tract is about 16.2% of the area of the sanctuary.

So the area of the Florida Reef Tract is \(\frac{16.2}{100}\) X 9600 =

1555.2 square kilometers, Now we convert 1555.2 in scientific notation as

single digit to the left of decimal sign and is multiplied

by an integer power of 10, So 1.552 X 1000 = 1.552 X 10^{3 }square kilometers,

therefore the area of the Florida Reef Tract is 1.5552 X 10^{3} square kilometers.

Question 37.

**REASONING**

A gigameter is 1.0 × 10^{6} kilometers. How many square kilometers

are in 5 square gigameters?

Answer:

There are 5 X 10^{12} kilometers^{2 }are there in 5 square gigameters

Explanation:

Given a gigameter is 1.0 × 10^{6} kilometers, So 1 Square kilometer =

(1.0 X 10^{6})^{2 }kilometers^{2 }= 1.0 X 10^{12 }kilometers^{2}, So

5 gigameter^{2}= 5 X 10^{12 }kilometers^{2 }therefore there are

5 X 10^{12} kilometers^{2 }are there in 5 square gigameter.

Question 38.

**PROBLEM SOLVING**

There are about 1.4 × 10^{9} cubic kilometers of water on Earth.

About 2.5% of the water is freshwater. How much freshwater is on Earth?

Answer:

Fresh water available on earth is 3.5 X 10^{7} kilometers^{3
}

Explanation:

Given there are about 1.4 × 10^{9} cubic kilometers of water on Earth

and about 2.5% of the water is freshwater so the amount of

fresh water available is 2.5 % of 1.4 × 10^{9} cubic kilometers

= \(\frac{2.5}{100}\) X 1.4 × 10^{9} = 3.5 X 10^{7} kilometers^{3 },

therefore fresh water available on earth is 3.5 X 10^{7} kilometers^{3 }.

Question 39.

**CRITICAL THINKING**

The table shows the speed of light through each of five media. Determine in which media light travels the fastest and the slowest.

Answer:

Fastest is Vacuum 3.0 X 10^{8} m/sec and slowest is Glass 2.01 X 10^{8} m/sec,

Explanation:

Given the table shows the speed of light through each of five media

Air is 6.7 X 10^{8} mi/sec now convert into meters per second

6.7 X 10^{8} X \(\frac{1609}{3600}\) = 2.99 X 10^{8} m/sec,

Now speed in glass is 6.6 X 10^{8} ft/sec now convert into meters per second

6.6 X 10^{8} X 0.3048 = 2.01 X 10^{8} m/sec,

Now speed in Ice is 2.3 X 10^{5} km/sec now convert into meters per second

2.3 X 10^{5} X 1000 = 2.3 X 10^{8} m/sec, already Vacuum is in m/sec,

so now speed in Water is 2.3 X 10^{10} cm/sec now

convert into meters per second 2.3 X 10^{10} X \(\frac{1}{100}\) =

2.3 X 10^{8} m/sec, Now

Medium Speed in m/sec

Air 2.99 X 10^{8} m/sec

Glass 2.01 X 10^{8} m/sec

Ice 2.3 X 10^{8} m/sec

Vacuum 3.0 X 10^{8} m/sec

Water 2.3 X 10^{8} m/sec , So now if we see at speeds the

fastest is Vacuum 3.0 X 10^{8} m/sec and slowest is Glass 2.01 X 10^{8} m/sec.

Question 40.

**STRUCTURE**

The mass of an atom or molecule is measured in atomic mass units. Which is greater, a carat or a milligram? Explain.

Answer:

A carat is greater unit than milligram,

Explanation:

The mass of an atom or molecule is measured in atomic mass units.

One atomic mass unit(amu) is equivalent to

1 amu = 8.3 X 10^{-24} carat

1 amu = 1.66 X 10^{-21} milligram

now 8.3 X 10^{-24} carat = 1.66 X 10^{-21} milligram

8.3 carat = 1.66 X 10^{3} milligram

So 1 carat = \(\frac{1660}{8.3}\) = 200 milligrams,

therefore A carat is greater unit than milligram.

### Lesson 8.7 Operations in Scientific Notation

**EXPLORATION 1**

Adding and Subtracting in Scientific Notation

Work with a partner.

a. Complete the table by finding the sum and the difference of Expression 1 and Expression 2. Write your answers in scientific notation. Explain your method.

b. Use your results in part(a) to explain how to find (a × 10^{n}) + (b × 10^{n}) and (a × 10^{n}) – (b × 10^{n}).

Answer:

a.

b. (a × 10^{n}) + (b × 10^{n}) = (a + b ) X 10^{n
}(a × 10^{n}) – (b × 10^{n}) = (a-b) X 10^{n}

Explanation:

a. One of the properties of quantities with exponents is that

numbers with exponents can be added and subtracted

only when they have the same base and exponent.

Since all numbers in scientific notation have the same base (10),

we need only worry about the exponents.

To be added or subtracted, two numbers in scientific notation they

must be manipulated so that their bases have the same exponent,

this will ensure that corresponding digits in their coefficients

have the same place value.

So completed the table by finding

the sum and the difference of Expression 1 and Expression 2.

1. Sum = ( 3 X 10^{4}) + ( 1 X 10^{4}) = ( 3 + 1) X 10^{4 }= 4 X 10^{4},

Difference = ( 3 X 10^{4}) – ( 1 X 10^{4}) = ( 3 – 1) X 10^{4 }= 2 X 10^{4},

2. Sum = ( 4 X 10^{-3}) + ( 2 X 10^{-3}) = ( 4 + 2) X 10^{-3 }= 6 X 10^{-3},

Difference = ( 4 X 10^{-3}) – ( 2 X 10^{-3}) = ( 4 – 2) X 10^{-3 }= 2 X 10^{-3},

3. Sum=( 4.1 X 10^{-7}) + ( 1.5 X 10^{-7}) = ( 4.1 + 1.5) X 10^{-7 }= 5.6 X 10^{-7},

Difference = ( 4.1 X 10^{-7}) – ( 1.5 X 10^{-7}) = ( 4.1 – 1.5) X 10^{-7 }= 2.6 X 10^{-7},

4. Sum = ( 8.3 X 10^{6}) + ( 1.5 X 10^{6}) = ( 8.3 + 1.5) X 10^{6 }= 9.8 X 10^{6},

Difference = ( 8.3 X 10^{6}) – ( 1.5 X 10^{6}) = ( 8.3 – 1.5) X 10^{6 }= 6.8 X 10^{6.
}b. (a × 10^{n}) + (b × 10^{n}) can be completed using the distributive property

of multiplication over addition, i.e., factor out the common factor 10^{n
}as (a × 10^{n}) + (b × 10^{n}) = (a + b ) X 10^{n} and

(a × 10^{n}) – (b × 10^{n}) can be completed using the distributive property

of multiplication over subtraction, i.e., factor out the common factor 10^{n
}as (a × 10^{n}) – (b × 10^{n}) = (a-b) X 10^{n}.

EXPLORATION 2

Multiplying and Dividing in Scientific Notation

Work with a partner.

a. Complete the table by finding the product and the quotient of

Expression 1 and Expression 2. Write your answers in scientific notation.

Explain your method.

b. Use your results in part(a) to explain how to find (a × 10^{n}) × (b × 10^{m}) and

(a × 10^{n}) ÷ (b ÷ 10^{m}). Describe any properties that you use.

Answer:

b. (a × 10^{n}) × (b × 10^{m}) = (a × b) X (10^{n} × 10^{m}) = (a × b) X (10^{n+}^{m}) ,

Separately we multiply the coefficients and exponents and

we use Product of Powers Property for a^{n} • a^{m }= a^{n+m
}If product of two powers with the same base then

powers are added and (a × 10^{n}) ÷ (b ÷ 10^{m}) = a ÷ b X (10^{n–}^{m}),

Separately we divide the coefficients and exponents.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

We use the general rule for finding \(\frac{a^{m}}{a^{n}}\),

a quotient of two powers with the same base is a^{m-n }.

Explanation:

a. To multiply numbers in scientific notation these are the steps:

If the numbers are not in scientific notation, convert them.

Regroup the numbers using the commutative and

associative properties of exponents. Now multiply the two

numbers written in scientific notation, we work out

the coefficients and the exponents separately.

we use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

Join the new coefficient to the new power of 10 to get the answer.

If the product of the coefficients is greater than 9, convert it to

scientific notation and multiply by the new power of 10.

To divide two numbers written in scientific notation, follow the steps below:

Separately we divide the coefficients and exponents.

For the division of bases, use the division rule of exponents,

where the exponents are subtracted.

Combine the result of coefficients by the new power of 10.

If the quotient from division of coefficients is not

less than 10 and greater than 1, convert it to scientific notation

and multiply it by the new power of 10.

Noted that when you dividing exponential terms,

always subtract the denominator from the numerator.

So completed the table by finding the product and the

quotient of Expression 1 and Expression 2.

Now first we do write Products as

1.(3 X 10^{4}) X ( 1 X 10^{4}) = (3 X 1) X ( 10^{4+}^{4}) = 3 X 10^{8} .

2.(4 X 10^{3}) X (2 X 10^{2}) = (4 X 2 ) X ( 10^{3+2}) = 8 X 10^{5}.

3.(7.7 X 10^{-2}) X (1.1 X 10^{-3}) = (7.7 X 1.1 ) X (10^{-2+(-3)}) = 8.47 X 10^{-5}.

4.(4.5 X 10^{5}) X (3 X 10^{-1}) = (4.5 X 3) X ( 10^{5+(-1)}) = 13.5 X 10^{4}.

Now we write Quotients as separately and divide the

coefficients and exponents.

For the division of bases, use the division rule of exponents,

where the exponents are subtracted.

1. (3 X 10^{4}) ÷ ( 1 X 10^{4}) = (3 ÷ 1) X ( 10^{4-}^{4}) = 3 X 10^{0} = 3 X 1 = 3.

2. (4 X 10^{3}) ÷ (2 X 10^{2}) = (4 ÷ 2 ) X ( 10^{3-2}) = 2 X 10^{1 }= 2 X 10 = 20.

3. (7.7 X 10^{-2}) ÷ (1.1 X 10^{-3}) = (7.7 ÷ 1.1 ) X (10^{-2-(-3)}) =7 X 10^{1 }=7 X 10 = 70

4. (4.5 X 10^{5}) ÷ (3 X 10^{-1}) = (4.5 ÷ 3) X ( 10^{5-(-1)}) = 1.5 X 10^{6}.

b. Here we multiply the coefficients separately and

use Product of Powers Property for a^{n} • a^{m }= a^{n+m
}we have rule of product of two powers with the same base then

we add powers, so (a × 10^{n}) × (b × 10^{m}) =

(a × b) X (10^{n} × 10^{m}) = (a × b) X (10^{n+}^{m}).

Separately we divide the coefficients and exponents.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

We use the general rule for finding \(\frac{a^{m}}{a^{n}}\),

a quotient of two powers with the same base is a^{m-n }.

So first we divide divide a ÷ b and then divide exponential terms

therefore (a × 10^{n}) ÷ (b ÷ 10^{m}) = a ÷ b X (10^{n-}^{m}).

**Try It**

**Find the sum or difference.**

Question 1.

(8.2 × 10^{2}) + (3.41 × 10^{-1})

Answer:

(8.2 × 10^{2}) + (3.41 × 10^{-1}) =8203.41 X 10^{-1}= 820.341

Explanation:

Given Expressions as (8.2 × 10^{2}) + (3.41 × 10^{-1})

as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{2} as 10^{-1} ·So we take 10^{3} and then grouping the 10^{3 }with 8.2.

So 8.2 X 10^{2} becomes 8200 X 10^{-1} therefore

(8.2 × 10^{2}) + (3.41 × 10^{-1}) =(8.2 X 10^{3} × 10^{-1}) + (3.41 × 10^{-1}) =

(8200 X 10^{-1}) + (3.41 × 10^{-1})

using the distributive property of multiplication over addition,

we get (8200+3.41) X 10^{-1} =8203.41 X 10^{-1}= 820.341

Question 2.

(7.8 × 10^{-5}) – (4.5 × 10^{-5})

Answer:

(7.8 × 10^{-5}) – (4.5 × 10^{-5}) = 3.3 X 10^{-5}

Explanation:

Given expressions as (7.8 × 10^{-5}) – (4.5 × 10^{-5})

using the distributive property of multiplication over subtraction,

we get (7.8 – 4.5 ) X 10^{-5} = 3.3 X 10^{-5}.

**Find the product.**

Question 3.

6 × (8 × 10^{-5})

Answer:

6 × (8 × 10^{-5}) = 48 X 10^{-5}

Explanation:

As given expression is 6 × (8 × 10^{-5}) so we use

associative law of multiplication as a X (b X c) = (a X b ) X c,

So 6 × (8 × 10^{-5}) = (6 × 8) × 10^{-5} = 48 X 10^{-5}.

Question 4.

(7 × 10^{2}) × (3 × 10^{5})

Answer:

(7 × 10^{2}) × (3 × 10^{5}) = 21 X 10^{7}

Explanation:

Given expressions as (7 × 10^{2}) × (3 × 10^{5})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (7 × 10^{2}) × (3 × 10^{5}) = (7 X 3) X (10^{2} × 10^{5}) = 21 X (10^{2+5}) = 21 X 10^{7} .

Question 5.

(2 × 10^{4}) × (6 × 10^{-7})

Answer:

(2 × 10^{4}) × (6 × 10^{-7}) = 12 X 10^{-3
}

Explanation:

Given expressions as (2 × 10^{4}) × (6 × 10^{-7})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (2 × 10^{4}) × (6 × 10^{-7}) = (2 X 6) X (10^{4} × 10^{-7}) = 12 X (10^{4-7}) = 12 X 10^{-3} .

Question 6.

(3 × 10^{8}) × (9 × 10^{3})

Answer:

(3 × 10^{8}) × (9 × 10^{3}) = 27 X 10^{11}

Explanation:

Given expressions as (3 × 10^{8}) × (9 × 10^{3})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (3 × 10^{8}) × (9 × 10^{3}) = (3 X 9) X (10^{8} × 10^{3}) = 27 X (10^{8+3}) = 27 X 10^{11} .

Find the quotient.

Question 7.

(9.2 × 10^{12}) ÷ 4.6

Answer:

(9.2 × 10^{12}) ÷ 4.6 = 2 X 10^{12}

Explanation:

Given expression as (9.2 × 10^{12}) ÷ 4.6,

Separately we divide the coefficients

and multiply exponent with base

so (9.2 ÷ 4.6) X 10^{12 }= 2 X 10^{12}.

Question 8.

(1.5 × 10^{-3}) ÷ (7.5 × 10^{2})

Answer:

(1.5 × 10^{-3}) ÷ (7.5 × 10^{2}) = 0.2 X 10^{-5}

Explanation:

Given expressions as (1.5 × 10^{-3}) ÷ (7.5 × 10^{2}),

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (1.5 × 10^{-3}) ÷ (7.5 × 10^{2}) = (1.5 ÷ 7.5 ) X ( 10^{-3-2}) = 0.2 X 10^{-5}.

Question 9.

(3.75 × 10^{-8}) ÷ (1.25 × 10^{-7})

Answer:

(3.75 × 10^{-8}) ÷ (1.25 × 10^{-7}) = 3 X 10^{-1 }or 0.3

Explanation:

Given expressions as (3.75 × 10^{-8}) ÷ (1.25 × 10^{-7}),

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (3.75 × 10^{-8}) ÷ (1.25 × 10^{-7}) =

(3.75 ÷ 1.25 ) X ( 10^{-8-(-7)}) = 3 X 10^{-1}= 0.3.

Question 10.

(9.2 × 10^{6}) ÷ (2.3 × 10^{12})

Answer:

(9.2 × 10^{6}) ÷ (2.3 × 10^{12}) = 4 X 10^{-6}

Explanation:

Given expressions as (9.2 × 10^{6}) ÷ (2.3 × 10^{12}) ,

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (9.2 × 10^{6}) ÷ (2.3 × 10^{12}) =

(9.2 ÷ 2.3 ) X ( 10^{6-12}) = 4 X 10^{-6}.

**Self-Assessment for Concepts & Skills**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 11.

**WRITING**

Describe how to add or subtract two numbers written in scientific notation with

the same power of 10.

Answer:

Example (a × 10^{m}) + (b × 10^{m}) can be completed using the distributive property

of multiplication over addition, i.e., factor out the common factor 10^{m
}as (a × 10^{m}) + (b × 10^{m}) = (a + b ) X 10^{m} , Here we have the same power 10.

Explanation:

One of the properties of quantities with exponents is that

numbers with exponents can be added and subtracted

only when they have the same base and exponent.

Since all numbers in scientific notation have the same base (10),

we need only worry about the exponents.

To be added or subtracted, two numbers in scientific notation they

must be manipulated so that their bases have the same exponent,

this will ensure that corresponding digits in their coefficients

have the same place value.

Question 12.

**NUMBER SENSE**

Two numbers written in scientific notation have different powers of 10. Do you have to rewrite the numbers so they have the same power of 10 before multiplying or dividing? Explain.

Answer:

No need to rewrite the numbers so that they have

the same power of 10 before multiplying or dividing.

Explanation:

The steps to multiply two numbers in scientific notation is

multiply the coefficients round to the number of

significant figures in the coefficient with the smallest number

of significant figures and we add the exponents.

The steps to divide two numbers in scientific notation is

divide the coefficients–round to the number of significant

figures in the coefficient with the smallest number of significant figures

and we subtract the exponents.

No need to rewrite the numbers so that they have

the same power of 10 before multiplying or dividing.

In multiplication we add exponents with same base and

in division we subtract exponents.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

**OPERATIONS IN SCIENTIFIC NOTATION** Evaluate the expression. Write your answer in scientific notation.

Question 13.

(7.26 × 10^{4}) + (3.4 × 10^{4})

Answer:

(7.26 × 10^{4}) + (3.4 × 10^{4}) = 10.66 X 10^{4}

Explanation:

Given expressions as (7.26 × 10^{4}) + (3.4 × 10^{4}) as it is

(a × 10^{n}) + (b × 10^{n}) can be completed using the distributive property

of multiplication over addition, i.e., factor out the common factor 10^{n
}as (a × 10^{n}) + (b × 10^{n}) = (a + b ) X 10^{n} so (7.26 × 10^{4}) + (3.4 × 10^{4}) =

(7.26 + 3.4 ) X 10^{4} =10.66 X 10^{4} .

Question 14.

(2.8 × 10^{-5}) – (1.6 × 10^{-6})

Answer:

(2.8 × 10^{-5}) – (1.6 × 10^{-6}) = 26.4 X 10^{-6}

Explanation:

Given Expressions as (2.8 × 10^{-5}) – (1.6 × 10^{-6})

as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{-5} as 10^{-6} · So we take 10^{-5} and then grouping the 10^{-5 }with 2.8.

So 2.8 X 10^{-5} becomes 28 X 10^{-6} therefore (2.8 × 10^{-5}) – (1.6 × 10^{-6})=

(2.8 ×10 X 10^{-6}) – (1.6 × 10^{-6}) = (28 X 10^{-6}) – (1.6 × 10^{-6})

using the distributive property of multiplication over subtraction,

we get (28-1.6) X 10^{-6} = 26.4 X 10^{-6}.

Question 15.

(2.4 × 10^{4}) × (3.8 × 10^{-6})

Answer:

(2.4 × 10^{4}) × (3.8 × 10^{-6}) = 9.12 X 10^{-2}

Explanation:

Given expressions as (2.4 × 10^{4}) × (3.8 × 10^{-6})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (2.4 × 10^{4}) × (3.8 × 10^{-6}) = (2.4 X 3.8) X (10^{4} × 10^{-6}) =

9.12 X (10^{4-6}) = 9.12 X 10^{-2} .

Question 16.

(5.2 × 10^{-3}) ÷ (1.3 × 10^{-12})

Answer:

(5.2 × 10^{-3}) ÷ (1.3 × 10^{-12}) = 4 X 10^{9}

Explanation:

Given expressions as (5.2 × 10^{-3}) ÷ (1.3 × 10^{-12}) ,

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (5.2 × 10^{-3}) ÷ (1.3 × 10^{-12}) =

(5.2 ÷ 1.3 ) X ( 10^{-3+12}) = 4 X 10^{9}.

**Self-Assessment for Problem Solving**

Solve each exercise. Then rate your understanding of the success criteria in your journal.

Question 17.

It takes the Sun about 2.3 × 10^{8} years to orbit the center of the Milky Way. It takes Pluto about 2.5 × 10^{2} years to orbit the Sun. How ×many times does Pluto orbit the Sun while the Sun completes one orbit around the Milky Way?

Answer:

9.2 X 10^{5} times

Explanation:

Given It takes the Sun about 2.3 × 10^{8} years to orbit the

center of the Milky Way. It takes Pluto about 2.5 × 10^{2} years

to orbit the Sun. So number of times does Pluto orbit the Sun

while the Sun completes one orbit around the Milky Way is

dividing the number of years the sun takes by

the number of years Pluto takes. So we divide by dividing the factors and

dividing the powers of 10 then rewrite in scientific notation

as (2.3 × 10^{8} ) ÷ (2.5 × 10^{2}) = \(\frac{2.3}{2.5}\) X (10^{8-2}) =

0.92 X 10^{6} = 0.92 X 10 X 10^{5} = 9.2 X10^{5} times.

Question 18.

A person typically breathes about 8.64 × 10^{8} liters of air per day.

The life expectancy of a person in the United States at birth is about 29,200 days.

Estimate the total amount of air a person born in the United States breathes over a lifetime.

Answer:

The total amount of air a person born in the United States

breathes over a lifetime is 2.52288 X 10^{13 }liters

Explanation:

A person typically breathes about 8.64 × 10^{8} liters of air per day.

The life expectancy of a person in the United States at birth is

about 29,200 days. The total amount of air a person born in the

United States breathes over a lifetime is 29,200 X 8.64 × 10^{8} =

292 X 10^{2} X 8.64 X 10^{8} =(when bases are same powers are added)

2.92 X 8.64 X 10^{10} =2522.88 X10^{10} = 2.522 X 10^{13 }liters.

therefore the total amount of air a person born in the United States

breathes over a lifetime is 2.52288 X 10^{13 }liters.

Question 19.

In one week, about 4100 movie theaters each sold an average of 2200 tickets for Movie A. About 3.6 × 10^{7} total tickets were sold at the theaters during the week. An article claims that about 25% of all tickets sold during the week were for Movie A. Is this claim accurate? Justify your answer.

Answer:

Yes, the claim is accurate.

Explanation:

Given in one week, about 4100 movie theaters each sold

an average of 2200 tickets for Movie A. About 3.6 × 10^{7} total

tickets were sold at the theaters during the week.

Number of tickets for movie A =

4100 X 2200 = 9,020,000= 9.02 X 10^{6} =

is approximately equal to ≈ 9 X 10^{6 }tickets.

Now 25% of total movie tickets is \(\frac{25}{100}\) X 3.6 × 10^{7} =

25 X 3.6 X 10^{5} = 90 X 10^{5} = 9 X 10 X 10^{5} =9 X 10^{6} tickets.

Therefore, the claim of 25% of movie tickets sold for movie A

is approximately accurate.

### Operations in Scientific Notation Homework & Practice 8.7

**Review & Refresh**

**Write the number in scientific notation.**

Question 1.

0.0038

Answer:

0.0038 = 3.8 X 10^{-3}.

Explanation:

Given number is 0.0038 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.0038 = 3.8 X 0.001 = 3.8 X 10^{-3}.

Question 2.

74,000,000

Answer:

74,000,000 = 7.4 X 10^{7}

Explanation:

Given number is 74,000,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 74,000,000 = 7.4 X 10,000,000 = 7.4 X 10^{7} .

Question 3.

0.0000475

Answer:

0.0000475 = 4.75 X 10^{-5}

Explanation:

Given number is 0.0000475 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.0000475 = 4.75 X 0.00001 = 4.75 X 10^{-5}.

Find the values of the ratios (red to blue) of the perimeters and areas of the similar figures.

Question 4.

Answer:

Ratios of perimeter is \(\frac{3}{4}\),

Ratios of area is \(\frac{9}{16}\)

Explanation:

The ratios of corresponding sides are 9/12. These all reduce to 3/4.

It is then said that the scale factor of these two similar squares is 3 : 4.

The perimeter of red square is 4 X 9

the perimeter of blue square is 4 X 12

When we compare the ratios of the perimeters of these similar squares,

we get 4 X 9 : 4 X 12= 9 : 12 = 3 : 4

Now Area of square is side square

so the area of red square is 9^{2} = 81

the area of blue square is 12^{2}= 144

So the ratio of their areas is 81 : 144 = 9 :16.

Therefore Ratios of perimeter is \(\frac{3}{4}\),

Ratios of area is \(\frac{9}{16}\)

Question 5.

Answer:

Perimeter is \(\frac{3}{2}\),

Area is \(\frac{9}{4}\)

Explanation:

The ratios of corresponding sides are 6/4. These all reduce to 3/2.

It is then said that the scale factor of these two similar triangles is 3 : 2.

The perimeter of red triangle is 6 + 6 + 6 = 18 and

the perimeter of blue triangle is 4 + 4 + 4=12.

When we compare the ratios of the perimeters of these similar triangles,

we get 18 : 12 = 3:2.

We know if two similar triangles have a scale factor of a : b,

then the ratio of their areas is a^{2} : b^{2}.

We have scale factor as 3 : 2 so the ratio of their areas is 3^{2} : 2^{2}= 9 : 4.

Therefore the Perimeter is \(\frac{3}{2}\) and Area is \(\frac{9}{4}\).

**Concepts, Skills, & Problem Solving**

**OPERATIONS IN SCIENTIFIC NOTATION** Find the sum, difference, product, and quotient of Expression 1 and Expression 2. Write your answers in scientific notation. (See Explorations 1 and 2, p. 355.)

Question 6.

3 × 10^{3} Expression 1

2 × 10^{3} Expression 2

Answer:

Sum = (3 X 10^{3}) + (2 × 10^{3}) = 5 X 10^{3},

Difference = (3 X 10^{3}) – (2 × 10^{3}) = 1 X 10^{3 }= 10^{3},

Product = (3 X 10^{3}) X (2 × 10^{3}) = 6 X 10^{6},

Quotient = (3 X 10^{3}) ÷ (2 × 10^{3}) = \(\frac{3}{2}\) = 1.5

Explanation:

For sum:

Given expressions as (3 × 10^{3}) + (2 × 10^{3}) as it is

(a × 10^{n}) + (b × 10^{n}) can be completed using the distributive property

of multiplication over addition, i.e., factor out the common factor 10^{n
}as (a × 10^{n}) + (b × 10^{n}) = (a + b ) X 10^{n} so (3× 10^{3}) + (2 × 10^{3}) =

(3 + 2) X 10^{3} =5 X 10^{3}.

For difference:

Given expressions as (3 × 10^{3}) – (2 × 10^{3})

using the distributive property of multiplication over subtraction,

we get (3-2) X 10^{3} = 1 X 10^{3 }= 10^{3
}For product:

Given expressions as (3 × 10^{3}) × (2 × 10^{3})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (3 × 10^{3}) × (2 × 10^{3}) = (3 X 2) X (10^{3} × 10^{3}) = 6 X (10^{3+3}) = 6 X 10^{6}.

For quotient:

Given expressions as (3 × 10^{3}) ÷ (2 × 10^{3}) ,

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (3×10^{3}) ÷ (2 × 10^{3}) =

(3 ÷ 2 ) X ( 10^{3-3}) = (3 ÷ 2 ) X 10^{0 }= (3 ÷ 2 ) X 1 = (3 ÷ 2 ) = \(\frac{3}{2}\) =1.5.

Question 7.

6 × 10^{-4} Expression 1

1.5 × 10^{-4} Expression 2

Answer:

Sum = (6 X 10^{-4}) + (1.5 × 10^{-4}) = 7.5 X 10^{-4},

Difference = (6 X 10^{-4}) – (1.5 × 10^{-4}) = 4.5 X 10^{-4 },

Product = (6 X 10^{-4}) X (1.5 × 10^{-4}) = 9 X 10^{-8},

Quotient = (6 X 10^{-4}) ÷ (1.5 × 10^{-4}) = 4 X 10^{0},

Explanation:

For sum:

Given expressions as (6 × 10^{-4}) + (1.5 × 10^{-4}) as it is

(a × 10^{n}) + (b × 10^{n}) can be completed using the distributive property

of multiplication over addition, i.e., factor out the common factor 10^{n
}as (a × 10^{n}) + (b × 10^{n}) = (a + b ) X 10^{n} so (3× 10^{3}) + (2 × 10^{3}) =

(6 + 1.5) X 10^{-4} =7.5 X 10^{-4}.

For difference:

Given expressions as (6 × 10^{-4}) – (1.5 × 10^{-4})

using the distributive property of multiplication over subtraction,

we get (6-1.5) X 10^{-4} = 4.5 X 10^{-4}.^{
}For product:

Given expressions as (6 × 10^{-4}) × (1.5 × 10^{-4})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (6 × 10^{-4}) × (1.5 × 10^{-4}) = (6 X 1.5) X (10^{-4} × 10^{-4}) = 6 X (10^{-4-4}) = 9 X 10^{-8}.

For quotient:

Given expressions as (6 × 10^{-4}) ÷ (1.5 × 10^{-4}) ,

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (6×10^{-4}) ÷ (1.5 × 10^{-4}) =

(6 ÷ 1.5 ) X ( 10^{-4-(-4)}) = (6 ÷ 1.5 ) X 10^{0 }= (4) X 10^{0 }or 4.

**ADDING AND SUBTRACTING IN SCIENTIFIC NOTATION** Find the sum or difference. Write your answer in scientific notation.

Question 8.

(2 × 10^{5}) + (3.8 × 10^{5})

Answer:

(2 × 10^{5}) + (3.8 × 10^{5}) = 5.8 X 10^{5}.

Explanation:

Given expressions as (2 × 10^{5}) + (3.8 × 10^{5}) as it is

(a × 10^{n}) + (b × 10^{n}) can be completed using the distributive property

of multiplication over addition, i.e., factor out the common factor 10^{n
}as (a × 10^{n}) + (b × 10^{n}) = (a + b ) X 10^{n} so (2× 10^{5}) + (3.8 × 10^{5}) =

(2 + 3.8) X 10^{5} =5.8 X 10^{5}.

Question 9.

(6.33 × 10^{-9}) – (4.5 × 10^{-9})

Answer:

(6.33 × 10^{-9}) – (4.5 × 10^{-9}) = 1.83 X 10^{-9}.

Explanation:

Given expressions as (6.33 × 10^{-9}) – (4.5 × 10^{-9})

using the distributive property of multiplication over subtraction,

we get (6.33 – 4.5) X 10^{-9} = 1.83 X 10^{-9}.^{
}

Question 10.

(9.2 × 10^{8}) – (4 × 10^{8})

Answer:

(9.2 × 10^{8}) – (4 × 10^{8}) = 5.2 X 10^{8}.

Explanation:

Given expressions as (9.2 × 10^{8}) – (4 × 10^{8})

using the distributive property of multiplication over subtraction,

we get (9.2 – 4) X 10^{8} = 5.2 X 10^{8}.

Question 11.

(7.2 × 10^{-6}) + (5.44 × 10^{-6})

Answer:

(7.2 × 10^{-6}) + (5.44 × 10^{-6}) = 12.64 X 10^{-6} = 1.264 X 10^{-5}

Explanation:

Given expressions as (7.2 × 10^{-6}) + (5.44 × 10^{-6}) as it is

(a × 10^{n}) + (b × 10^{n}) can be completed using the distributive property

of multiplication over addition, i.e., factor out the common factor 10^{n
}as (a × 10^{n}) + (b × 10^{n}) = (a + b ) X 10^{n} so (2× 10^{5}) + (3.8 × 10^{5}) =

(7.2 + 5.44) X 10^{-6} =12.64 X 10^{-6 }= 1.264 X 10^{-5}.

Question 12.

(7.8 × 10^{7}) – (2.45 × 10^{6})

Answer:

(7.8 × 10^{7}) – (2.45 × 10^{6}) = 7.55 X 10^{7}

Explanation:

Given expressions as (7.8 × 10^{7}) – (2.45 × 10^{6})

as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{6} as 10^{7} · So we take 10^{7} and then grouping the 10^{7 }with 2.45.

So 2.45 X 10^{6} becomes 0.245 X 10^{7} therefore (7.8 × 10^{7}) – (0.245 × 10^{7})=

(7.8 X 10^{7}) – (2.45 X 10^{-1}X 10^{1} X 10^{6}) = (7.8 X 10^{7}) – (0.245 × 10^{7})

using the distributive property of multiplication over subtraction,

we get (7.8 – 0.245) X 10^{7} = 7.55 X 10^{7}.

Question 13.

(5 × 10^{-5}) + (2.46 × 10^{-3})

Answer:

(5 × 10^{-5}) + (2.46 × 10^{-3}) = 2.51 X 10^{-5}.

Explanation:

Given expressions as (5 × 10^{-5}) + (2.46 × 10^{-3})

as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{-5} as 10^{-3} · So we take 10^{-3} and then grouping the 10^{-3 }with 5.

So 5 X 10^{-5} becomes 0.05 X 10^{-3} therefore (5 × 10^{-5}) + (2.46 × 10^{-3})=

(5 X 10^{-2 }X 10^{-3}) + (2.46 X 10^{-3}) = (0.05 X 10^{-3}) + (2.46 × 10^{-3})

using the distributive property of multiplication over addition,

we get (0.05 + 2.46) X 10^{-5} = 2.51 X 10^{-5}.

Question 14.

(9.7 × 10^{6}) + (6.7 × 10^{5})

Answer:

(9.7 × 10^{6}) + (6.7 × 10^{5}) = 103.7 X 10^{5 }or 1.037 X 10^{7}

Explanation:

Given expressions as (9.7 × 10^{6}) + (6.7 × 10^{5})

as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{6} as 10^{5} · So we take 10^{5} and then grouping the 10^{5 }with 9.7.

So 9.7 X 10^{6} becomes 97 X 10^{5} therefore (9.7 × 10^{6}) + (6.7 × 10^{5})=

(9.7 X 10^{1 }X 10^{5}) + (6.7 X 10^{5}) = (97 X 10^{5}) + (6.7 × 10^{5})

using the distributive property of multiplication over addition,

we get (97 + 6.7) X 10^{5} = 103.7 X 10^{5 }= 1 .037 X 10^{2 }X 10^{5 }= 1.037 X 10^{7 }.

Question 15.

(2.4 × 10^{-1}) – (5.5 × 10^{-2})

Answer:

(2.4 × 10^{-1}) – (5.5 × 10^{-2}) = 1.85 X 10^{-1}

Explanation:

Given expressions as (2.4 × 10^{-1}) – (5.5 × 10^{-2})

as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{-2} as 10^{-1} · So we take 10^{-1} and then grouping the 10^{-1 }with 5.5.

So 5.5 X 10^{-2} becomes 0.55 X 10^{-1} therefore (2.4 × 10^{-1}) – (5.5 × 10^{-2}) =

(2.4 X 10^{-1}) – (5.5 X 10^{-1}X 10^{-1}) = (2.4 X 10^{-1}) – (0.55 × 10^{-1})

using the distributive property of multiplication over subtraction,

we get (2.4 – 0.55) X 10^{-1} = 1.85 X 10^{-1}.

Question 16.

**YOU BE THE TEACHER**

Your friend adds 2.5 × 10^{9} and 5.3 × 10^{8}. Is your friend correct? Explain your reasoning.

Answer:

Yes, Friend is correct

Explanation:

Given expressions as 2.5 × 10^{9} and 5.3 × 10^{8
}we do sum as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{8} as 10^{9} · So we take 10^{9} and then grouping the 10^{9 }with 5.3.

So 5.3 X 10^{9} becomes 0.53 X 10^{9} therefore (2.5 × 10^{9}) + (5.3 × 10^{8}) =

(2.5 X 10^{9}) + (5.3 X 10^{-1 }X 10 X 10^{8}) = (2.5 X 10^{9}) + (0.53 × 10^{9})

using the distributive property of multiplication over addition,

we get (2.5 + 0.53 ) X 10^{9} = 3.03 X 10^{9}.

As the results are same friend is correct.

**
MULTIPLYING AND DIVIDING IN SCIENTIFIC NOTATION** Find the product or quotient. Write your answer in scientific notation.

Question 17.

5 × (7 × 10

^{7})

Answer:

5 × (7 × 10

^{7}) = 3.5 X 10

^{8 }

Explanation:

As given expression is 5 × (7 × 10

^{7}) so we use

associative law of multiplication as a X (b X c) = (a X b ) X c,

So 5 × (7 × 10

^{7}) = (5 X 7) × 10

^{7}= 35 X 10

^{7}.= 3.7 X 10 X 10

^{7}= 3.7 X 10

^{8}.

Question 18.

(5.8 × 10^{-6}) ÷ (2 × 10^{-3})

Answer:

(5.8 × 10^{-6}) ÷ (2 × 10^{-3}) = 2.9 X 10^{-3}.

Explanation:

Given expressions as (5.8 × 10^{-6}) ÷ (2 × 10^{-3}) ,

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (5.8 × 10^{-6}) ÷ (2 × 10^{-3}) = (5.8 ÷ 2 ) X ( 10^{-6+3}) = 2.9 X 10^{-3}.

Question 19.

(1.2 × 10^{-5}) ÷ 4

Answer:

(1.2 × 10^{-5}) ÷ 4 = 3 X 10^{-6}

Explanation:

Given expression is (1.2 × 10^{-5}) ÷ 4 so (1.2 ÷ 4) X 10^{-5}= 0.3 X 10^{-5}=

3 X 10^{-1}X 10^{-5}= 3 X 10^{-6}.

Question 20.

(5 × 10^{-7}) × (3 × 10^{6})

Answer:

(5 × 10^{-7}) × (3 × 10^{6}) = 15 X 10^{-1} = 1.5

Explanation:

Given expressions as (5 × 10^{-7}) × (3 × 10^{6})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (5 × 10^{-7}) × (3 × 10^{6}) = (5 X 3) X (10^{-7} × 10^{6}) = 15 X (10^{-7+6}) = 15 X 10^{-1} = 1.5.

Question 21.

(3.6 × 10^{7}) ÷ (7.2 × 10^{7})

Answer:

(3.6 × 10^{7}) ÷ (7.2 × 10^{7}) = 5 X 10^{-1}

Explanation:

Given expressions as (5.8 × 10^{-6}) ÷ (2 × 10^{-3}) ,

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (3.6 × 10^{7}) ÷ (7.2 × 10^{7}) = (3.6 ÷ 7.2 ) X ( 10^{-7+7}) = 0.5 X 10^{0 }= 5 X 10^{-1}.

Question 22.

(7.2 × 10^{-1}) × (4 × 10^{-7})

Answer:

(7.2 × 10^{-1}) × (4 × 10^{-7}) = 2.88 X 10^{-7}

Explanation:

Given expressions as (7.2 × 10^{-1}) × (4 × 10^{-7})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (7.2 × 10^{-1}) × (4 × 10^{-7}) = (7.2 X 4) X (10^{-1} × 10^{-7}) =

28.8 X (10^{-1-7}) = 28.8 X 10^{-8 }= 2.88 X 10^{-7}.

Question 23.

(6.5 × 10^{8}) × (1.4 × 10^{-5})

Answer:

(6.5 × 10^{8}) × (1.4 × 10^{-5}) = 9.1 X 10^{3}

Explanation:

Given expressions as (6.5 × 10^{8}) × (1.4 × 10^{-5})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (6.5 × 10^{8}) × (1.4 × 10^{-5}) = (6.5 X 1.4) X (10^{8} × 10^{-5}) =

9.1 X (10^{8-5}) = 9.1 X 10^{3}.

Question 24.

(2.8 × 10^{4}) ÷ (2.5 × 10^{6})

Answer:

(2.8 × 10^{4}) ÷ (2.5 × 10^{6}) = 1.12 X 10^{-2}

Explanation:

Given expressions as (2.8 × 10^{4}) ÷ (2.5 × 10^{6}) ,

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (2.8 × 10^{4}) ÷ (2.5 × 10^{6}) = (2.8 ÷ 2.5 ) X ( 10^{4-6}) = 1.12 X 10^{-2}.

**MATCHING** You use technology to ﬁnd four sums.

Match the sum with its standard form.

Question 25.

4.3E8

Answer:

4.3E8 matches with C. 430,000,000

Explanation:

Given number 4.3E8 here nearly or approximately equals to ≈ 4.3 X 10^{8},

The exponent “8” says to use the 10 by eight times in a multiplication as

4.3 E 8 = 4.3 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 = 430,000,000,

So 4.3E8 we will match with C.

Question 26.

4.3E – 8

Answer:

4.3E – 8 matches with B. 0.000000043

Explanation:

Given number 4.3E – 8 nearly or approximately equals to ≈ 4.3 X 10^{-8},

The exponent “-8” says to use the 10^{-1} by eight times in a multiplication as

4.3 E 8 = 4.3 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 = 0.000000043,

so 4.3E -8 we will match with B.

Question 27.

4.3E10

Answer:

4.3E10 matches with D. 43,000,000,000

Explanation:

Given number 4.3E10 nearly or approximately equals to ≈ 4.3 X 10^{10},

The exponent “10” says to use the 10 by ten times in a multiplication as

4.3 E 10 = 4.3 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 X 10 =

43,000,000,000 so 4.3E10 we will match with D.

Question 28.

4.3E – 10

Answer:

4.3E – 10 matches with A.0.00000000043

Explanation:

Given number 4.3E – 10 nearly or approximately equals to ≈ 4.3 X 10^{-10},

The exponent “-10” says to use the 10^{-1} by eight times in a multiplication as

4.3 E 8 = 4.3 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1 X 0.1

= 0.00000000043 so 4.3E -10 we will match with A.

A. 0.00000000043

B. 0.000000043

C. 430,000,000

D. 43,000,000,000

Question 29.

**MODELING REAL LIFE**

How many times greater is the thickness of a dime than the thickness of a dollar bill?

Answer:

The dime is about 12 times thicker than the dollar bill.

Explanation:

Given the thickness of dime is 0.135 = 1.35 X 10^{-1} and

the thickness of a dollar is 1.0922 X 10^{-2 }divide the thickness of

dime with the thickness of dollar as (1.35 × 10^{-1}) ÷ (1.0922 × 10^{-2})

now we separately divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (1.35 × 10^{-1}) ÷ (1.0922 × 10^{-2}) = (1.35 ÷ 1.0922 ) X ( 10^{-1+2}) =

1.2360 X 10^{1}= 12.36 approximately equals to ≈ 12.

Therefore the dime is about 12 times thicker than the dollar bill.

Question 30.

**MULTIPLE CHOICE**

On a social media website, Celebrity A has about 8.6 × 10^{6} followers and Celebrity B has about 4.1 × 10^{6} followers. Determine which of the following is the best estimate for the number of followers for Celebrity A compared to the number of followers for Celebrity B.

A. more than 2 times greater

B. less than 2 times greater

C. more than 20 times greater

D. less than 20 times greater

Answer:

The best estimate for the number of followers for Celebrity A

compared to the number of followers for Celebrity B is

A. more than 2 times greater.

Explanation:

Given on a social media website, Celebrity A has about

8.6 × 10^{6} followers and Celebrity B has about 4.1 × 10^{6} followers.

To find the best estimate for the number of followers for Celebrity A

compared to the number of followers for Celebrity B we divide them as

the number of followers for Celebrity A to the number of

followers for Celebrity B as (8.6 × 10^{6}) ÷ (4.1× 10^{6})

now we separately divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (8.6× 10^{6}) ÷ (4.1 × 10^{6}) = (8.6 ÷ 4.1) X ( 10^{6-6}) =

2.097 X 10^{0}= 2.097 X 1 = 2.097 approximately equals to ≈ 2.

The best estimate for the number of followers for Celebrity A

compared to the number of followers for Celebrity B is

A. more than 2 times greater.

**REASONING** Evaluate the expression. Write your answer in scientific notation.

Question 31.

5,200,000 × (8.3 × 10^{2}) – (3.1 × 10^{8})

Answer:

5,200,000 × (8.3 × 10^{2}) – (3.1 × 10^{8}) = 4.006 X 10^{9}

Explanation:

Given expressions as 5,200,000 × (8.3 × 10^{2}) – (3.1 × 10^{8}),

So first we solve 5,200,000 × (8.3 × 10^{2}) as 5.2 X 10^{6 }X (8.3 × 10^{2}),

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

(5.2 X 8.3) X (10^{6} X 10^{2}) = (43.16) X (10^{6+2}) = 43.16 X 10^{8},

now we subtract expressions as (43.16 X 10^{8}) – (3.1 × 10^{8}) =

using the distributive property of multiplication over subtraction,

we get (43.16 – 3.1) X 10^{8} = 40.06 X 10^{8 }= 4.006 X 10 X 10^{8 }= 4.006 X 10^{9}.

Question 32.

(9 × 10^{-3}) + (2.4 × 10^{-5}) ÷ 0.0012

Answer:

(9 × 10^{-3}) + (2.4 × 10^{-5}) ÷ 0.0012 = 7.52

Explanation:

Given (9 × 10^{-3}) + (2.4 × 10^{-5}) ÷ 0.0012 first we solve

(9 × 10^{-3}) + (2.4 × 10^{-5}) given expressions as 9 × 10^{-3} and 2.4 × 10^{-5
}we do sum as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{-5} as 10^{3} · So we take 10^{-5} and then grouping the 10^{-5 }with 2.4.

So 2.4 X 10^{-5} becomes 0.024 X 10^{-3} therefore (9 × 10^{-3}) + (2.4 × 10^{-5})=

(9X 10^{-3}) + (2.4 X 10^{-2 }X 10^{-3}) = (9 X 10^{-3}) + (0.024 × 10^{-3})

using the distributive property of multiplication over addition,

we get (9 + 0.024 ) X 10^{-3} = 9.024 X 10^{-3}.

So (9 × 10^{-3}) + (2.4 × 10^{-5}) = 9.024 X 10^{-3 }now we write 0.0012 as

1.2 X 10^{-3 }therefore we divide (9.024 X 10^{-3}) ÷ (1.2 X 10^{-3}),

now we separately divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (9.024 ÷ 1.2) X (10^{-3 }÷ 10^{-3}) so 7.52 X 10^{-3+3 }= 7.52 X 10^{0 }=

7.52 X 1 = 7.52.

Question 33.

**GEOMETRY**

Find the perimeter of the rectangle at the right.

Answer:

The perimeter of the rectangle at the right is 1.962 X 10^{8 }cm.

Explanation:

Given area of rectangle as 5.612 X 10^{14}cm^{2} and breadth as

9.2 X 10^{7 }cm , We will find the length first we take length as l,

we know area of rectangle is length X breadth so

5.612 X 10^{14 }cm^{2} = l X 9.2 X 10^{7 }

so l= (5.612 X 10^{14 }) ÷ (9.2 X 10^{7}) now we separately divide

the coefficients and exponents. We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

(5.612 X 9.2) X (10^{14 }÷ 10^{7}) = 0.61 X (10^{14-7}) = 0.61X 10^{7}=0.61 X 10^{7 }cm,

therefore length of rectangle is 0.61 X 10^{7 }cm now we will calculate

perimeter of rectangle as we know perimeter of rectangle =

2 x ( length + breadth)= 2 X ((0.61 X 10^{7})+(9.2 X 10^{7}))cm,

now we calculate first (0.61 X 10^{7})+(9.2 X 10^{7}) using the

distributive property of multiplication over addition,

we get (0.61 + 9.2) X 10^{7} = 9.81 X 10^{7 } now 2 X (9.81X 10^{7})=

19.62 X 10^{7}= 1.962 X 10^{8 }cm. Therefore the perimeter of the

rectangle at the right is 1.962 X 10^{8 }cm.

Question 34.

**DIG DEEPER!**

A human heart pumps about 7 × 10^{-2} liter of blood per heartbeat.

The average human heart beats about 72 times per minute.

How many liters of blood does a heart pump in 1 year? 70 years?

Answer:

2.65 X 10^{6} liters of blood a heart pump in 1 year.

1.855 X 10^{8 }liters of blood a heart pump for 70 years.

Explanation:

Given a human heart pumps about 7 × 10^{-2} liter of blood per heartbeat.

The average human heart beats about 72 times per minute.

The the number of liters per minute multiply the number of

heartbeats per minute by the number of liters per heartbeat as

72 X 7 × 10^{-2} = 504 × 10^{-2} =5.04 liters per minute,

now we multiply by 60 to get liters per hour since

there are 60 minutes in 1 hour so 60 X 5.04 = 302.4 liters per hour,

now we multiply by 24 to get liters per day since

there are 24 hours in 1 day so 302.4 X 24 = 7257.6 liters per day,

now we multiply by 365 to get liters per year since

there are 365 days in 1 year and write in scientific notation

and round to two decimal places as 7257.6 X 365 = 2649024 =

2.649024 X 10^{6} approximately equal to ≈ 2.65 X 10^{6} liters per year

Now for 70 years it is 70 X 2.65 X 10^{6} = 185.5 X 10^{6} = 1.855 X 100 X 10^{6} =

1.855 X 10^{2} X 10^{6} =1.855 X 10^{8 }liters for 70 years.

Question 35.

**MODELING REAL LIFE**

Use the Internet or another reference to find the populations and areas (in square miles) of India, China, Argentina, the United States, and Egypt. Round each population to the nearest million and each area to the nearest thousand square miles.

a. Write each population and area in scientific notation.

b. Use your answers to part(a) to find and order the population densities (people per square mile) of each country from least to greatest.

Answer:

a. In India : Population = 1.311 X 10^{9 }, Area = 1.269 X 10^{6
}In China : Population = 1.371 X 10^{9 }, Area = 3.705 X 10^{6 }

In Argentina : Population = 4.3 X 10^{7 }, Area = 1.074 X 10^{6
}In United States : Population = 3.21 X 10^{8}, Area = 3.797 X 10^{6
}In Egypt: Population = 9.2 X 10^{7 }, Area = 3.9 X 10^{6}

b. The population densities are

India = 1.033 X 10^{3}

China = 3.7 X 10^{2}

Argentina =4.0 X 10^{1}

United States = 8.45 X 10^{1}

and in Egypt = 2.36 X 10^{1}

Each country from least to greatest is

Egypt, Argentina, United States, China, India.

Explanation:

The following were found by doing a google search :

we write each population and area in scientific notation,

We know 1 billion = 10^{9 }and 1 million = 10^{6 },

a. In India : Population = 1.311 billion= 1.311 X 10^{9 }

Area = 1.268 million = 1.269 X 10^{6
}In China : Population = 1.371 billion = 1.371 X 10^{9 }

Area =3.705 million= 3.705 X 10^{6 }

In Argentina : Population= 43 million = 4.3 X 10^{7 }

Area = 1.074 million = 1.074 X 10^{6
}In United States : Population=321 million = 3.21 X 10^{8 }

Area =3.797 million= 3.797 X 10^{6
}In Egypt: Population = 92 million= 9.2 X 10^{7 }

Area =390,000= 3.9 X 10^{6}

b. Divide each country’s population by its area to

get its population density, and we separately divide

the coefficients and exponents. We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

India = ( 1.311 X 10^{9 }) ÷ ( 1.269 X 10^{6}) = 1.033 X 10^{3}

(1.311 ÷ 1.269) X (10^{9-6}) = 1.033 X 10^{3} ,

China = ( 1.371 X 10^{9 }) ÷ ( 3.705 X 10^{6}) =

(1.371 ÷ 3.705) X (10^{9-6}) = 0.370 X 10^{3} = 3.7 X 10^{2},

Argentina = ( 4.3 X 10^{7 }) ÷ ( 1.074 X 10^{6}) =

(4.3 ÷ 1.074) X (10^{7-6}) = 4 X 10^{1},

United States = ( 3.21 X 10^{8 }) ÷ ( 3.797 X 10^{6}) =

(3.21 ÷ 3.797) X (10^{8-6}) = 0.845 X 10^{2} = 8.45X 10^{1},

Egypt = (9.2 X 10^{7 }) ÷ ( 3.9 X 10^{6}) =

(9.2 ÷ 3.9) X (10^{7-6}) = 2.36 X 10^{1},

Now we write each country from least to greatest density as

2.36 X 10^{1 }< 4 X 10^{1 }< 8.45X 10^{1}< 3.7 X 10^{2},1.033 X 10^{3} so

Egypt, Argentina, United States, China, India.

### Exponents and Scientific Notation Connecting Concepts

**Using the Problem-Solving Plan**

Question 1.

Atoms are made of protons, neutrons, and electrons. The table shows the numbers of protons and the masses of several atoms. Use a line of best fit to estimate the mass (in grams) of an atom that has 29 protons.

Understand the problem.

You know the numbers of protons and the masses of several atoms. You are asked to use the line of best fit to estimate the mass of an atom that has 29 protons.

Make a plan.

Use a graphing calculator to find an equation of the line of best fit.

Then evaluate the equation when x = 29.

Solve and check.

Use the plan to solve the problem. Then check your solution.

Answer:

Mass of an atom that has 29 protons is 4.843 X 10^{-23 }grams

Explanation:

Given 1 proton has 1.67 X 10^{-24 }grams of mass,

So 29 protons has 29 X 1.67 X 10^{-24 }grams,

48.43 X 10^{-24 }= 4.843 X 10^{-23 }grams.

Question 2.

Modoc Country, California, is 74.9 miles long and 56.2 miles wide. 263A map of the county is drawn using a scale factor of 2.11 × 10^{-6}. What is the perimeter of the county on the map? Express your answer using more appropriate units.

Answer:

The perimeter of the country on the map is 6.66631 X 10^{11 }inches

Explanation:

Given Modoc Country, California, is 74.9 miles long and 56.2 miles wide.

First we calculate perimeter it is 2 X ( 74.9 + 56.2) mile = 2 X 131.1 =

262.2 miles, So perimeter of Modoc Country, California is 262.2 miles.

Now on map we will use in inches , we know 1 mile = 5280 feet and

1 feet is equal to 12 inches so 262.2 we will convert into inches.

262.2 X 5280 X 12 = 1406592 inches. Now we are given with

scaling factor on the map as 2.11 × 10^{-6 }equals to 1 inch on the map,

So now perimeter of the country on the map is

(1406592 ) ÷ 2.11 × 10^{-6 }= 666631.27 X 10^{6 }= 6.66631 X 10^{11 }inches.

Question 3.

A research company estimates that in the United States, 7733 about 8.37 × 10^{7} adult males and 6.59 × 10^{7} adult females watch NFL football, while 3.13 × 10^{7} adult males and 5.41 × 10^{7} adult females do not watch NFL football. Organize the results in a two-way table. Include the marginal frequencies.

Answer:

Explanation:

Given A research company estimates that in the United States, 7733 about

8.37 × 10^{7} adult males and 6.59 × 10^{7} adult females watch NFL football,

while 3.13 × 10^{7} adult males and 5.41 × 10^{7} adult females do not watch NFL football.

Organized the results in two-way table above,

Entries in the “Total” row and “Total” column are called

marginal frequencies or the marginal distribution.

Entries in the body of the table are called joint frequencies.

So total adult watched NFL football match are

(8.37 X 10^{7}) + (6.59 X 10^{7})=14.96 X 10^{7} and do not

watch NFL football are(3.13 X 10^{7}) + (5.41 X 10^{7}) = 8.54 X 10^{7}.

**Performance Task**

Elements in the Universe

At the beginning of this chapter, you watched a STEAM Video called “Carbon Atoms.” You are now ready to complete the performance task related to this video, available at BigIdeasMath.com. Be sure to use the problem-solving plan as you work through the performance task.

Answer:

### Exponents and Scientific Notation Chapter Review

**Review Vocabulary**

Write the definition and give an example of each vocabulary term.

Let us take an example : 5^{2} ,

An expression that represents repeated multiplication

of the same factor is called a power.

Here the number 5 is called the base,

and the number 2 is called the exponent.

The exponent corresponds to the number of times

the base is used as a factor.

A method for expressing very large or very small numbers

as a product of decimal less than 10 and multiplied

by a power of 10.

Example: 8 X 10^{7 }or 5 X 10^{-2}.

**Graphic Organizers**

You can use a Definition and Example Chart to organize information about a concept.

Here is an example of a Definition and Example Chart for the vocabulary term power.

Choose and complete a graphic organizer to help you study the concept.

1. Product of Powers Property

2. Power of a Power Property

3. Power of a Product Property

4. Quotient of Powers Property

5. negative exponents

6. scientific notation

7. adding and subtracting numbers in scientific notation

8. multiplying and dividing numbers in scientific notation

Graphic Organizer:

1. Product of Powers Property

2. Power of a Power Property

3. Power of a Product Property

4. Quotient of Powers Property

5. negative exponents & power of a quotient

6. scientific notation

7. adding and subtracting numbers in scientific notation

8. multiplying and dividing numbers in scientific notation

**Chapter Self-Assessment**

As you complete the exercises, use the scale below to rate your understanding of the success criteria in your journal.

**8.1 Exponents (pp. 319–324)**

Learning Target: Use exponents to write and evaluate expressions. Write the product using exponents.

Question 1.

(- 9) • (- 9) • (- 9) • (- 9) • (- 9)

Answer:

(- 9) • (- 9) • (- 9) • (- 9) • (- 9) =(-9)^{5}

Explanation:

As -9 is multiplied by 5 times we write as

(-9)^{5 }here -9 is base and 5 is exponent.

Question 2.

2 • 2 • 2 • n • n

Answer:

2 • 2 • 2 • n • n = (2)^{3 }X (n)^{2 }

Explanation:

As 2 is multiplied 3 times we write as (2)^{3 }and n

is multiplied twice so (n)^{2
}so the expression is (2)^{3 }X (n)^{2 .}

**Evaluate the expression.**

Question 3.

11^{3}

Answer:

11^{3 }= 1331

Explanation:

Given 11^{3 }means 11 is multiplied thrice

as 11 X 11 X 11 we get 1331.

here base is 11 and exponent is 3.

Question 4.

-(\(\frac{1}{2}\))^{4}

Answer:

-(\(\frac{1}{2}\))^{4 }= – \(\frac{1}{16}\)

Explanation:

Given -(\(\frac{1}{2}\))^{4 }means \(\frac{1}{2}\) is multiplied four times

as- (\(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) X \(\frac{1}{2}\) )^{ }= – \(\frac{1}{16}\).

Question 5.

|\(\frac{1}{2}\left(16-6^{3}\right)\)|

Answer:

|\(\frac{1}{2}\left(16-6^{3}\right)\)| = 100

Explanation:

Given expression as |\(\frac{1}{2}\left(16-6^{3}\right)\)|=

first we solve 6^{3} =6 X 6 X 6 = 216 now (16-216) = -200

so |\(\frac{1}{2}\) X -200|= |-100|= 100.

Question 6.

The profit P(in dollars) earned by a local merchant selling x items is 3 represented by the equation P = 0.2x^{3} – 10. How much more profit does he earn selling 15 items than 5 items?

Answer:

More profits he earn by selling 15 items than 5 items is $650

Explanation:

Given P = 0.2x^{3} – 10

Profit by selling 5 items is

0.2 X (5)^{3} – 10 = 0.2 X 125 – 10 = 25 – 10 = $15.

Profit by selling 15 items is

0.2 X (15)^{3} – 10 = 0.2 X 3375 – 10 = 675 – 10 = $665.

More profits he earn by selling 15 items than 5 items is

$665 -15 = $650.

**8.2 Product of Powers Property (pp. 325–330)**

Learning Target: Generate equivalent expressions involving products of powers.

**Simplify the expression. Write your answer as a power.**

Question 7.

p^{5} • p^{2}

Answer:

p^{5} • p^{2 }= p^{7}

Explanation:

We write the given p^{5} • p^{7 }expression as a power,

we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added,

so p^{5} • p^{2 }as bases are same p powers are added p^{5+2} = p^{7}.

Question 8.

(n^{11})^{2}

Answer:

(n^{11})^{2 }= n^{22}

Explanation:

We write the given expression (n^{11})^{2}^{ }as a power,

General rule for finding (a^{m})^{n }power of a power,

If two powers have the same base then

we can multiply the powers as (a^{m})^{n }= a^{m x }^{n} .

so (n^{11})^{2 }has powers of powers therefore powers

are multiplied as (n)^{11 X 2}= (n)^{22}

Question 9.

\(\left(-\frac{2}{5}\right)^{3} \cdot\left(-\frac{2}{5}\right)^{2}\)

Answer:

\(\left(-\frac{2}{5}\right)^{3} \cdot\left(-\frac{2}{5}\right)^{2}\) = –

Explanation:

Given expression is \(\left(-\frac{2}{5}\right)^{3} \cdot\left(-\frac{2}{5}\right)^{2}\)

we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added, so we add powers as (-\(\frac{2}{5}\))^{3+2 }=

(-\(\frac{2}{5}\))^{5 }So \(\left(-\frac{2}{5}\right)^{3} \cdot\left(-\frac{2}{5}\right)^{2}\) =

Question 10.

Simplify (- 2k)^{4}.

Answer:

(- 2k)^{4 }= (-2)^{4 }X k^{4
}

Explanation:

Given to simplify the expression (- 2k)^{4 }we use general rule to

write (ab)^{m } power of a product as a^{m }X b^{m },So (- 2k)^{4 }= (-2)^{4 }X k^{4}

Question 11.

Write an expression that simplifies to x^{24} using the Power of a Power Property.

Answer:

x^{24} = (x^{12})^{2}

Explanation :

To simplify x^{24} using the Power of a Power Property,

we use general rule for finding (a^{m})^{n }power of a power,

If two powers have the same base then

we can multiply the powers as (a^{m})^{n }= a^{m x }^{n} ,here m X n = 24,

lets take m, n as 12,2 so that 12 X 2 = 24, therefore (x^{12})^{2} = x^{12 x 2 }= x^{24}.

therefore x^{24} = (x^{12})^{2} .

Question 12.

You send an email with a file size of 4 kilobytes. One kilobyte is 2^{10} bytes. What is the file size of your email in bytes?

Answer:

The file size of my email in bytes is

Explanation:

Given I send an email with a file size of 4 kilobytes. One kilobyte is 2^{10} bytes.

The file size of my email in bytes is 4 X 2^{10} bytes as 4 can be written as 2 X 2 =2^{2},

So 2^{2} X 2^{10} we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added, 2^{2} X 2^{10} = 2^{2+10} = 2^{12} bytes.

Question 13.

Explain how to use properties of powers 2 to simplify the expression 27 • 3^{2}.

Answer:

27 • 3^{2 }= 3^{3 }X 3^{2} = 3^{5
}

Explanation:

Given expression as 27 • 3^{2 }first we write 27 as power of 3,

so 27 = 3 X 3 X 3 = 3^{3 }now 27 • 3^{2 }= 3^{3 }X 3^{2 },

we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added, 3^{3 }X 3^{2 }= 3^{3+2}= 3^{5}.

8.3 Quotient of Powers Property (pp. 331–336)

Learning Target: Generate equivalent expressions involving quotients of powers.

**Simplify the expression. Write your answer as a power.**

Question 14.

Answer:

= (8)^{8-3 }= (8)^{5 }

Explanation:

Given \(\frac{8^{8}}{8^{3}}\) we use rule for finding \(\frac{a^{m}}{a^{n}}\),

a quotient of two powers with the same base as a^{m-n}

so \(\frac{8^{8}}{8^{3}}\) = (8)^{8-3 }= (8)^{5 }

Question 15.

Answer:

= 5^{10}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (5)^{2 }X (5)^{9 }^{
}we have same bases as 5 so we add powers as (5)^{2+9 }= (5)^{11
}we have denominator (5)^{1
}Now we have \(\frac{5^{11}}{5^{1}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{5^{11}}{5^{1}}\) = (5)^{11-1 }= 5^{10}.

Question 16.

Answer:

= w^{4}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (w)^{8 }X (w)^{5 }^{
}we have same bases as w so we add powers as (w)^{8+5 }= (w)^{13
}we have denominator (w)^{7}^{ }X (w)^{2
}we have same base as w so we add powers as (w)^{7+2 }= (w)^{9}

Now we have \(\frac{w^{13}}{w^{9}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{w^{13}}{w^{9}}\) = (w)^{13-9 }= w^{4}.

Question 17.

Answer:

= m^{5}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (m)^{8 }X (m)^{10 }X (m)^{2
}we have same bases as m so we add powers as (m)^{8+10+2 }= (m)^{20
}we have denominator (m)^{6}^{ }X (m)^{9
}we have same base as m so we add powers as (m)^{6+9 }= (m)^{15}

Now we have \(\frac{m^{20}}{m^{15}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{m^{20}}{m^{15}}\) = (m)^{20-15 }= m^{5}.

Question 18.

Write an expression that simplifies to x^{3} using the Quotient of Powers Property.

Answer:

x^{3} = \(\frac{x^{6}}{x^{3}}\)

Explanation:

To simplify x^{3} using the Quotient of Powers Property,

we use rule for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n }here we have m-n = 3,So lets take

m, n as 6,3 so that 6-3 = 3 therefore \(\frac{x^{6}}{x^{3}}\)= x^{6-3 }= x^{3}.

therefore x^{3} = \(\frac{x^{6}}{x^{3}}\).

Question 19.

At the end of a fiscal year, a company has made 1.62 × 7^{7} dollars in profit. The company employs 7^{3} people. How much will each person receive if the company divides the profit equally among its employees?

Answer:

Each person will receive $3889.62 if the company divides

the profit equally among its employees.

Explanation:

Given at the end of a fiscal year, a company has made 1.62 × 7^{7} dollars in profit,

The company employs 7^{3} people. To find how much each person

receive if the company divides the profit equally among its employees is

1.62 × 7^{7} ÷ 7^{3} we use the division rule of exponents,

where the exponents are subtracted. 1.62 X ( 7^{7} ÷ 7^{3} ) = 1.62 X ( 7^{7-3}) =

1.62 X 7^{4} = 1.62 X 7 X 7 X 7 X 7 = 3889.62, Therefore each person

will receive $3889.62 if the company divides the profit equally among its employees.

**8.4 Zero and Negative Exponents (pp. 337–342)**

Learning Target: Understand the concepts of zero and negative exponents.

Evaluate the expression.

Question 20.

2^{-4}

Answer:

4^{-2 }

Explanation:

Given expression as 2^{-4 }so we write as

Question 21.

95^{0}

Answer:

95^{0 }= 1

Explanation:

Given expression is 95^{0 }it is proven that any number or

expression raised to the power of zero is always equal to 1.

In other words, if the exponent is zero then the result is 1.

So 95^{0 }= 1.

Question 22.

\(\frac{8^{2}}{8^{4}}\)

Answer:

\(\frac{8^{2}}{8^{4}}\) = \(\frac{1}{8^{2}}\) = \(\frac{1}{64}\).

Explanation:

Given expression as \(\frac{8^{2}}{8^{4}}\) now we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n}

as 8^{2-4 }we get 8^{-2 }we write as \(\frac{1}{8^{2}}\) = \(\frac{1}{64}\).

Question 23.

(- 12)^{-7} • (- 12)^{7}

Answer:

(- 12)^{-7} • (- 12)^{7}= 1

Explanation:

Given expression is -12^{-7 }X -12^{7 },

Here we use Product of Powers Property for a^{m} • a^{n }= a^{m+n
}if we have product of two powers with the same base then

powers are added. So (12)^{-7 }X (12)^{7 }= (12)^{-7+7 =} (12)^{0}= 1.

Question 24.

\(\frac{1}{7^{9}} \cdot \frac{1}{7^{6}}\)

Answer:

\(\frac{1}{7^{9}} \cdot \frac{1}{7^{6}}\) = \(\frac{1}{7^{15}}\)

Explanation:

Given expression is \(\frac{1}{7^{9}} \cdot \frac{1}{7^{6}}\)

first we calculate separately values of denominators

then divide with numerator, we have denominator 7^{9 }X 7^{6}

we have same bases as 7 so we add powers as 7^{9+6 }= 7^{15
}as numerator is 1 we write as \(\frac{1}{7^{15}}\).

Question 25.

\(\frac{9^{4} \cdot 9^{2}}{9^{2}}\)

Answer:

\(\frac{9^{4} \cdot 9^{2}}{9^{2}}\) = 9^{4 }

Explanation:

Given Expression as \(\frac{9^{4} \cdot 9^{2}}{9^{2}}\)

first we calculate separately values of numerators

then divide with denominator, we have numerator (9)^{4 }X (9)^{2 }

we have same bases as 9 so we add powers as (9)^{4+2 }= (9)^{6
}Now we have \(\frac{9^{6}}{9^{2}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n},

So \(\frac{9^{6}}{9^{2}}\) = (9)^{6-2 }= (9)^{4}.

Simplify. Write the expression using only positive exponents.

Question 26.

x^{-2} • x^{0}

Answer:

x^{-2} • x^{0 }= \(\frac{1}{x^{2}}\)

Explanation:

Given expression is x^{-2} • x^{0 }= \(\frac{1}{x^{2}}\).

Question 27.

y^{-8}y^{3}

Answer:

y^{-8}y^{3 }= \(\frac{1}{y^{5}}\)

Explanation:

Given expression is y^{-8} • y^{3 }= \(\frac{1}{y^{5}}\).

Question 28.

\(\frac{3^{-1} \cdot z^{5}}{z^{-2}}\)

Answer:

\(\frac{3^{-1} \cdot z^{5}}{z^{-2}}\) =

Explanation:

Given expression is \(\frac{3^{-1} \cdot z^{5}}{z^{-2}}\)

we use rule for finding \(\frac{a^{m}}{a^{n}}\),

a quotient of two powers with the same base as a^{m-n}

for \(\frac{z^{5}}{z^{-2}}\) as z= z^{7
}now we have 3^{-1 }X z^{7 }

Question 29.

Write an expression that simplifies to x^{-4}.

Answer:

x^{-4 }=

Explanation:

To simplify x^{-4} we use rule for finding \(\frac{a^{m}}{a^{n}}\),

a quotient of two powers with the same base as a^{m-n},So m-n = -4 we take

m, n as 4,8 which makes (4-8) = -4 therefore x^{-4 }=

Question 30.

Water flows from a shower head at a rate of 24^{-1} gallon per second. How many gallons do you use when taking a 15-minute shower? a 20-minute shower?

Answer:

In 15 minutes we use in 20 minutes shower we use 50

Explanation:

Given water flows from a shower head at a rate of 24^{-1} gallon per second,

to know how many gallons we used in 15-minutes first

we convert minutes to seconds as 1 minute is equal to 60 seconds,

so 15 minutes is equal to 15 X 60 = 900 seconds now number of gallons we

used in 15 minutes are 900 X 24^{-1} =

Question 31.

Explain two different methods for simplifying w^{-2} • w^{5}.

Answer:

One is product of powers = w^{-2} • w^{5 }= w^{-2+5} = w^{3 }or

other is quotient of two powers with the same base = w^{5-2} = w^{3}

Explanation:

To simplify w^{-2} • w^{5 }first we use

other method is we write as .

8.5 Estimating Quantities (pp. 343–348)

Learning Target: Round numbers and write the results as the product of a single digit and a power of 10.

Round the number. Write the result as a product of a single digit and a power of 10.

Question 32.

29,197,543

Answer:

29,197,543 = 3 X 10^{7}

Explanation:

Given number is 29,197,543 is nearly or approximately

equal to ≈ 30,000,000 so we have 3 followed by 11 zeros,

so we write as 3 X 10^{7}.

Question 33.

0.000000647

Answer:

0.000000647 = 6 X 10^{-7}

Explanation:

Given number 0.000000647 is nearly or approximately

equal to ≈ 0.0000006, we have divided 6 by 10 followed

by 7 zeros so we write as 6 X 10^{-7}.

Question 34.

The speed of light is 299,792,458 meters per second. About how far can a light beam travel in 3 seconds? Write your answer as a product of a single digit and a power of 10.

Answer:

A light beam can travel 9 X 10^{8 }meters in 3 seconds.

Explanation:

Given the speed of light is 299,792,458 meters per second.

to know about how far can a light beam travel in 3 seconds is

3 X 299,792,458 now we write 299,792,458 is nearly or approximately

equal to ≈ 300,000,000 so 3 X 300,000,000 = 900,000,000

so we have 9 followed by 8 zeros so we write as 9 X 10^{8 }meters.

Therefore a light beam can travel 9 X 10^{8 }meters in 3 seconds.

Question 35.

The population of Albany, New York is about 98,989 and the population of Moscow, Russia is about 12,235,448. Approximately how many times greater is the population of Moscow than the population of Albany?

Answer:

100 times greater is the population of Moscow than the population of Albany.

Explanation:

Given the population of Albany, New York is about 98,989 and

the population of Moscow, Russia is about 12,235,448 if we see the

the population of Moscow is greater than the population of Albany.

So 12,235,448

**8.6 Scientific Notation (pp. 349–354)**

Learning Target: Understand the concept of scientific notation.

**Write the number in scientific notation.**

Question 36.

0.00036

Answer:

0.00036= 3.6 X 10^{-4}

Explanation:

Given number is 0.00036 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 0.00036 = 3.6 X 0.0001 = 3.6 X 10^{-4}.

Question 37.

800,000

Answer:

800,000 = 8 X 10^{5}

Explanation:

Given number is 800,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 800,000 = 8 X 100000 = 8 X 10^{5}.

Question 38.

79,200,000

Answer:

79,200,000 = 7.92 X 10^{7}

Explanation:

Given number is 79,200,000 in scientific notation

we write a number so that it has single digit to

the left of decimal sign and is multiplied by an integer power of 10.

So 79,200,000 = 7.29 X 10000000 = 7.29 X 10^{7}.

**Write the number in standard form.**

Question 39.

2 × 10^{7}

Answer:

2 X 10^{7} = 20,000,000

Explanation:

Given 2 × 10^{7 }the number in standard form is

2 X 10,000,000 = 20,000,000.

So 2 X 10^{7} = 20,000,000.

Question 40.

4.8 × 10^{-3}

Answer:

4.8 × 10^{-3 }= 0.0048

Explanation:

Given 4.8 × 10^{-3 }^{ }the number in standard form is

4.8 × 0.001^{ }= 0.0048.

Question 41.

6.25 × 10^{5}

Answer:

6.25 × 10^{5 }= 6,25,000

Explanation:

Given 6.25 10^{5 }^{ }the number in standard form is

6.25 X 100,000 = 6,25,000.

So 6.25 X 10^{5} = 6,25,000.

Question 42.

The mass of a single dust particle is 7.52 × 10^{-10} kilogram. What is the mass of a dust ball made of 100 dust particles? Express your answer using more-appropriate units.

Answer:

The mass of a dust ball made of 100 dust particles is 7.52 10^{-8} kilogram

Explanation:

Given the mass of a single dust particle is 7.52 × 10^{-10} kilogram.

So the mass of a dust ball made of 100 dust particles is

7.52 × 10^{-10} X 100 = 7.52 X 10^{-10} X 10^{2} = 7.52 × 10^{-10+2 }= 7.52 × 10^{-8}.

Therefore the mass of a dust ball made of 100 dust**
**particles is 7.52 10

^{-8}kilogram.

**8.7 Operations in Scientific Notation (pp. 355–360)**

Learning Target: Perform operations with numbers written in scientific notation.

**Evaluate the expression. Write your answer in scientific notation.**

Question 43.

(4.2 × 10^{8}) + (5.9 × 10^{9})

Answer:

(4.2 × 10^{8}) + (5.9 × 10^{9}) = 6.32 X 10^{9}

Explanation:

Given Expressions as (4.2 × 10^{8}) + (5.9 × 10^{9})

as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{8} as 10^{9} ·So we take 10^{9} and then grouping the 10^{9 }with 4.2.

So 4.2 X 10^{8} becomes 0.42 X 10^{9} therefore

(4.2 × 10^{8}) + (5.9 × 10^{9}) =(0.42 X 10^{1}X 10^{8}) + (5.9 X 10^{9}) =

(0.42 X 10^{9}) + (5.9 × 10^{9})

using the distributive property of multiplication over addition,

we get (0.42+5.9) X 10^{9} = 6.32 X 10^{9}.

Question 44.

(5.9 × 10^{-4}) – (1.8 × 10^{-4})

Answer:

(5.9 × 10^{-4}) – (1.8 × 10^{-4}) = 4.1 X 10^{-4}

Explanation:

Given expressions as (5.9 × 10^{-4}) – (1.8 × 10^{-4})

using the distributive property of multiplication over subtraction,

we get (5.9 – 1.8 ) X 10^{-4} = 4.1 X 10^{-4}.

Question 45.

(7.7 × 10^{8}) × (4.9 × 10^{-5})

Answer:

(7.7 × 10^{8}) × (4.9 × 10^{-5}) =3.773 X 10^{4}

Explanation:

Given expressions as (7.7 × 10^{8}) × (4.9 × 10^{-5})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (7.7 × 10^{8}) × (4.9 × 10^{-5}) = (7.7 X 4.9) X (10^{8} × 10^{-5}) =

37.73 X (10^{8-5}) = 3.773 X 10 X 10^{3} = 3.773 X 10^{4}.

Question 46.

(3.6 × 10^{5}) ÷ (1.8 × 10^{9})

Answer:

(3.6 × 10^{5}) ÷ (1.8 × 10^{9}) = 2 X 10^{-4}

Explanation:

Given expressions as (3.6 × 10^{5}) ÷ (1.8 × 10^{9}) ,

Separately we divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (3.6 × 10^{5}) ÷ (1.8 × 10^{9}) = (3.6 ÷ 1.8 ) X ( 10^{5-9}) = 2 X 10^{-4}.

Question 47.

A white blood cell has a diameter of about 0.000012 meter. How many times greater is the diameter of a white blood cell than the diameter of a red blood cell?

Answer:

1.5 times greater is the diameter of a white blood cell

than the diameter of a red blood cell

Explanation:

Given a white blood cell has a diameter of about 0.000012 meter and

red blood cell has a diameter of about 8 X 10^{-6 }meter,

the number of times greater is the diameter of a white blood cell

than the diameter of a red blood cell is 0.000012 ÷ 8 X 10^{-6 }we write

0.000012 approximately equal to ≈ 1.2 X 10^{-5 }now separately we

divide the coefficients and exponents.

We use the division rule of exponents,

where the exponents are subtracted.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

so (1.2 X 10^{-5}) ÷ 8 X 10^{-6 }= (1.2 ÷ 8) X (10^{-5 }÷ 10^{-6}) =

0.15 X (10^{-5+6}) = 0.15 X 10 = 1.5 therefore 1.5 times greater is

the diameter of a white blood cell than the diameter of a red blood cell.

**Exponents and Scientific Notation Practice Test**

**Write the product using exponents.**

Question 1.

(- 15) • (- 15) • (- 15)

Answer:

(- 15) • (- 15) • (- 15) = (-15)^{3}

Explanation:

As -15 is multiplied by 3 times we write as

(-15)^{3 }here -15 is base and 3 is exponent.

Question 2.

4 • 4 • x • x • x

Answer:

4 • 4 • x • x • x = (4)^{2 }X (x)^{3 }

Explanation:

As 4 is multiplied 2 times we write as (4)^{2 }and x

is multiplied thrice so (x)^{3
}so the expression is 4 • 4 • x • x • x = (4)^{2 }X (x)^{3 }

**Evaluate the expression.**

Question 3.

10 + 3^{3} ÷ 9

Answer:

10 + 3^{3} ÷ 9 = 13

Explanation:

Given expression 10 + 3^{3} ÷ 9 first we write 9 as

3 X 3 = 3^{2} ÷now we divide 3^{3} ÷ 3^{2} so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n}, 3^{3} ÷ 3^{2} = ( 3^{3-2}) = 3^{1}= 3,

now we add 10 so 10 + 3 = 13.

Question 4.

Answer:

= (-2)^{-1}=

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (-2)^{1 }X (-2)^{-4 }^{
}we have same bases as -2 so we add powers as (-2)^{1-4 }= (-2)^{-3
}we have denominator (-2)^{-2
}Now we have \(\frac{-2^{-3}}{-2^{-2}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
}\(\frac{-2^{-3}}{-2^{-2}}\) = (-2)^{-3+2 }= (-2)^{-1}n

**Simplify the expression. Write your answer as a power.**

Question 5.

9^{10} • 9

Answer:

9^{10} • 9 = 9^{11}

Explanation:

We write the given 9^{10} • 9^{1 }expression as a power,

we have general rule for a^{m} • a^{n }= a^{m+n
}If product of two powers with the same base then

powers are added, so 9^{10} • 9^{1 }as bases are same 9

powers are added 9^{10+1} = 9^{11}.

Question 6.

(6^{6})^{5}

Answer:

(6^{6})^{5 }= 6^{30}

Explanation:

We write the given expression (6^{6})^{5}^{ }as a power,

General rule for finding (a^{m})^{n }power of a power,

If two powers have the same base then

we can multiply the powers as (a^{m})^{n }= a^{m x }^{n} .

so (6^{6})^{5 }has powers of powers therefore powers

are multiplied as (6)^{6 X 5}= (6)^{30}

Question 7.

Answer:

= (-3.5)^{6}

Explanation:

First we calculate separately values of numerators

and denominators then divide, we have numerator (-3.5)^{13 }X (-3.5)^{2 }^{
}we have same bases as -3.5 so we add powers as (-3.5)^{13+2 }= (-3.5)^{15
}we have denominator (-3.5)^{9
}Now we have \(\frac{-3.5^{15}}{-3.5^{9}}\) so we use rule

for finding \(\frac{a^{m}}{a^{n}}\) a quotient of

two powers with the same base as a^{m-n
} \(\frac{-3.5^{15}}{-3.5^{9}}\) = (-3.5)^{15-9 }= (-3.5)^{6}.

Question 8.

Simplify (2y)^{7}.

Answer:

(2y)^{7 }=(2)^{7 }X y^{7
}

Explanation:

Given to simplify the expression (2y)^{7 }we use general rule to

write (ab)^{m } power of a product as a^{m }X b^{m },So (2y)^{7 }= (2)^{7 }X y^{7}

**Round the number. Write the result as a product of a single digit and a power of 10.**

Question 9.

4,610,428,970

Answer:

4,610,428,970 = 5 X 10^{9}

Explanation:

Given number is 4,610,428,970 is nearly or approximately

equal to ≈ 5,000,000,000 so we have 5 followed by 9 zeros,

so we write as 4,610,428,970 = 5 X 10^{9 }.

Question 10.

0.00000572

Answer:

0.00000572 = 5.72 X 10^{-6}

Explanation:

Given number 0.00000572 is nearly or approximately

equal to ≈ 0.000006, we have divided 6 by 10 followed

by 6 zeros so we write as 6 X 10^{-6}.

**Write the number in standard form.**

Question 11.

3 × 10^{7}

Answer:

3 X 10^{7 }= 30,000,000

Explanation:

Given 3 × 10^{7 }the number in standard form is

3 X 10,000,000 = 30,000,000.

So 3 X 10^{7} = 30,000,000.

Question 12.

9.05 × 10^{-3}

Answer:

9.05 × 10^{-3 }= 0.00905

Explanation:

Given 9.05 × 10^{-3 }^{ }the number in standard form is

9.05 × 0.001^{ }= 0.00905.

**Evaluate the expression. Write your answer in scientific notation.**

Question 13.

(7.8 × 10^{7}) + (9.9 × 10^{7})

Answer:

(7.8 × 10^{7}) + (9.9 × 10^{7}) = 1.77 X 10^{8}

Explanation:

Given expressions as (7.8 × 10^{7}) + (9.9 × 10^{7})

using the distributive property of multiplication over addition,

we get (7.8 × 10^{7}) + (9.9 × 10^{7}) = (7.8 + 9.9 ) X 10^{7 }= 17.7 X 10^{7 }=

1.77 X 10 X 10^{7 }=1.77 X 10^{8}.

Question 14.

(6.4 × 10^{5}) – (5.4 × 10^{4})

Answer:

(6.4 × 10^{5}) – (5.4 × 10^{4}) = 5.86 X 10^{5}

Explanation:

Given expressions as (6.4 × 10^{5}) – (5.4 × 10^{4})

as the powers of 10 differ we need to modify before we factor.

We work around this by using our exponent property b^{m} · b^{n} = b ^{(m+n)}

to rewrite the 10^{4} as 10^{5} · So we take 10^{4} and then grouping the 10^{4 }with 5.4.

So 5.4 X 10^{4} becomes 0.54 X 10^{5} therefore = (6.4 × 10^{5}) – (5.4 × 10^{4}) =

(6.4 X 10^{5}) – (5.4 X 10^{-1}X 10 X 10^{4}) = (6.4 X 10^{5}) – (0.54 × 10^{5})

using the distributive property of multiplication over subtraction,

we get (6.4 – 0.54) X 10^{5} = 5.86 X 10^{5}.

Question 15.

(3.1 × 10^{6}) × (2.7 × 10^{-2})

Ans;

(3.1 × 10^{6}) × (2.7 × 10^{-2}) = 8.37 X 10^{4}.

Explanation:

Given expressions as (3.1 × 10^{6}) × (2.7 × 10^{-2})

we work out the coefficients and the exponents separately.

and use the product rule; b^{m }x b ^{n} = b^{(m + n)} to multiply the bases.

So (3.1 × 10^{6}) × (2.7 × 10^{-2}) = (3.1 X 2.7) X (10^{6} × 10^{-2}) =

8.37 X (10^{6-2}) = 8.37 X 10^{4}.

Question 16.

(9.6 × 10^{7}) ÷ (1.2 × 10^{-4})

Answer:

(9.6 × 10^{7}) ÷ (1.2 × 10^{-4}) = 8 X 10^{11}

Explanation:

Given expressions as (9.6 × 10^{7}) ÷ (1.2 × 10^{-4}),

Separately we divide the coefficients and exponents.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator.

So (9.6 × 10^{7}) ÷ (1.2 × 10^{-4}) = (9.6 ÷ 1.2 ) X ( 10^{7-(-4)}) = 8 X 10^{11}.

Question 17.

Is (xy^{2})^{3} the same as (xy^{3})^{2}? Explain.

Answer:

(xy^{2})^{3} is not same as (xy^{3})^{2}

Explanation:

Given expressions (xy^{2})^{3} the same as (xy^{3})^{2 }

first we solve the expressions with general rule for

finding (a^{m})^{n }power of a power,

If two powers have the same base then

we can multiply the powers as (a^{m})^{n }= a^{m x }^{n} as

(xy^{2})^{3 }= (x)^{3} X (y^{2})^{3} = (x)^{3} X (y^{2x}^{3}) = x^{3} X y^{6}.

Now (xy^{3})^{2 }= (x)^{2} X (y^{3})^{2} = (x)^{2} X (y^{3×2}) = x^{2} X y^{6}.

Now comparing x^{3} X y^{6} with x^{2} X y^{6 }as x powers are not same so

(xy^{2})^{3} is not same as (xy^{3})^{2}.

Question 18.

One scoop of rice weighs about 3^{9} milligrams.

a. Write a linear function that relates the weight of rice to the number of scoops. What is the weight of 5 scoops of rice?

b. A grain of rice weighs about 3^{3} milligrams. About how many grains of rice are in 1 scoop?

Answer:

a. The linear function that relates the weight of rice to the

number of scoops is y= 3^{9}x ,The weight of 5 scoops of rice

is 98415 grams,

b. There are 729 grains in 1 scoop of rice

Explanation:

Given One scoop of rice weighs about 3^{9} milligrams.

Let us take x as number of scoops and y represent

the weight of rice,

a.The linear function that relates the

weight of rice to the number of scoops is y= 3^{9}x ,

so the weight of 5 scoops of rice = y= 3^{9} X 5 =

19683 X 5 = 98415 grams,

The weight of 5 scoops of rice is 98415 grams,

b. A grain of rice weighs about 3^{3} milligrams the number of

grains of rice are in 1 scoop is 3^{9} ÷ 3^{3},We use the division rule

of exponents, where the exponents are subtracted.

So (3^{9} ÷ 3^{3}) = 3^{9-3} = 3^{6 }= 3 X 3 X 3 X 3 X 3 X 3 = 729,

therefore there are 729 grains in 1 scoop of rice.

Question 19.

There are about 10,000 taste buds on a human tongue.

Write this number in scientific notation.

Answer:

10,000 = 1 x 10^{4} taste buds are there on a human tongue.

Question 20.

From 1978 to 2008, the amount of lead allowed in the air in the

United States was 1.5 × 10^{-6} gram per cubic meter.

In 2008, the amount allowed was reduced by 90%.

What is the new amount of lead allowed in the air?

Answer:

1.5 X 10^{-7} gram per cubic meter is the new amount of lead allowed in the air.

Explanation:

Given from 1978 to 2008, the amount of lead allowed in the air in the

United States was 1.5 × 10^{-6} gram per cubic meter.

In 2008, the amount allowed was reduced by 90%.

to know the new amount of lead allowed in the air, we see

as it was reduced by 90% then 10% is still allowed, to find 10% of

original amount multiply by 0.1 so 0.1 X 1.5 × 10^{-6} = 0.15 X 10^{-6} =

= 1.5 X 10^{-7} gram per cubic meter, Therefore 1.5 X 10^{-7} gram per cubic meter

is the new amount of lead allowed in the air.

### Exponents and Scientific Notation Cumulative Practice

Answer :

Bit B, 3072 years.

Explanation:

Cats were tamed 3 X 2^{10} years ago in Egypt, So it was

3 X 2^{10} = 3 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 X 2 = 3072 Years,

So bit b is matched.

Question 1.

Mercury’s distance from the Sun is approximately 5.79 × 10^{7} kilometers. What is this distance in standard form?

A. 5,790,000 km

B. 57,900,000 km

C. 579,000,000 km

D. 5,790,000,000 km

Answer:

The distance of 5.79 × 10^{7} kilometers in standard form is approximately

equal to ≈ 57,900,000 km so matches with B bit.

Explanation:

Given Mercury’s distance from the Sun is approximately 5.79 × 10^{7} kilometers,

So the distance in standard form is the number in standard form is

5.79 X 10,000,000 = 57,900,000 km. So 5.79 X 10^{7} km matches with B bit

57,900,000 km.

Question 2.

Your friend solves the problem. What should your friend change to correctly answer the question?

F. The left side of the equation should equal 360° instead of 180°.

G. The sum of the acute angles should equal 90.°

H. Evaluate the smallest angle when x = 15.

I. Evaluate the largest angle when x = 15.

Answer:

I. Evaluate the largest angle when x = 15.

Explanation:

Friend already solved x = 15°,So the largest angle in the triangle

is 8x = 8 X 15°= 120°,So friend needs to

Evaluate the largest angle when x = 15. So bit I is the correct answer.

Question 3.

Which expression is equivalent to the expression 2^{4}2^{3}?

A. 2^{12}

B. 4^{7}

C. 48

D. 128

Answer:

2^{4}2^{3 }= 128, D bit

Explanation:

Given expression is 2^{4}2^{3 }we know when bases are same

product of powers property says add the powers.

So 2^{4}2^{3 }= 2^{4 }X 2^{3 }= 2^{4+}^{3}= 2^{7 }= 2 X 2 X 2 X 2 X 2 X 2 X 2 = 128,

therefore 2^{4}2^{3 }=128, so it matches with bit D.

Question 4.

You randomly survey students in your school about whether they have a pet. You display your results in the two-way table. How many female students took the survey?

Answer:

Female 46 students took the survey.

Explanation:

As per two-way table it displays number of students in school

about whether they have a pet or not , So yes are 35 and no are 11,

Therefore total number of female students are 35 + 11 = 46,

So 46 female students took the survey.

Question 5.

A bank account pays interest so that the amount in the account doubles every 10 years. The account started with $5,000 in 1940. Which expression represents the amount (in dollars) in the account n decades later?

F. 2^{n} • 5000

G. 5000 (n + 1)

H. 5000^{n}

I. 2^{n} + 5000

Answer:

F. 2^{n} • 5000

Explanation:

Given a bank account pays interest so that the amount in the account

doubles every 10 years. The account started with $5,000 in 1940,

The expression that represents the amount (in dollars) in

the account n decades later is since the amount is getting doubled

every decade, So it will be increasing in multiples of 2,

so the correct answer is 2^{n} X 5000 so bit F.

Question 6.

The formula for the volume V of a pyramid is V = \(\frac{1}{3}\)Bh.

Which equation represents a formula for the height h of the pyramid?

A. h= \(\frac{1}{3}\) VB

B. h = \(\frac{3 V}{B}\)

C. h = \(\frac{V}{3B}\)

D. h = V – \(\frac{1}{3}\)B

Answer:

B. h = \(\frac{3 V}{B}\)

Explanation:

Given the formula for the volume V of a pyramid is V = \(\frac{1}{3}\)Bh,

If we cross multiply 3 with V divide by B we get height,

So the formula for the height h of the pyramid is h = 3 V X \(\frac{1}{B}\).

Therefore the correct answer is bit B, h = \(\frac{3 V}{B}\).

Question 7.

The gross domestic product (GDP) is a way to measure how much a country produces economically in a year. The table below shows the approximate population and GDP for the United States.

Part A Write the population and the GDP using scientific notation.

Part B Find the GDP per person for the United States using your answers from PartA. Write your answer in scientific notation. Show your work and explain your reasoning.

Answer:

Part A : Population is 3.24 X 10^{8},

GDP is $1.86 X 10^{13}

Part B :

The GDP per person for the United States is 5.74 X 10^{5}

Explanation:

Part A:

Given the population of United States, 2016 is 324,000,000 the scientific

notation is 3.24 X 100,000,000 = 3.24 X 10^{8 }and GDP is

$18,600,000,000,000 in scientific notation is $1.86 X 10,000,000,000,000 =

$1.86 X 10^{13}.

Part B :

The GDP per person is dividing the GDP of a country by its population,

So ($1.86 X 10^{13}) ÷ (3.24 X 10^{8}) ,Separately we divide the coefficients and exponents.

Noted that when we are dividing exponential terms,

always subtract the denominator from the numerator,

Therefore (1.86 ÷ 3.24) X (10^{13} ÷ 10^{8}) = 0.574 X (10^{13-8}) =

0.574 X 10^{5} = 0.574 X 10 X 10^{4} = 5.74 X 10^{5}.

Question 8.

What is the equation of the line shown in the graph?

Answer:

F. y = –

Explanation:

The equation of line is y = mx + b, Where m is slope and

b is y – intercept So m= (y2-y1) ÷ (x2 -x1)

from graph we have (x1,y1)= (3,2) and (x2,y2) = (-3,4)

m= (4-2) ÷ (-3-3) = 2 ÷ (-6) = (-1 ÷ 3) = now y-intercept is y value of the point where the line intersects the y- axis.

So here it is 3, Now the equation of the line is y =-y = -(1 ÷

Question 9.

Which graph represents a linear function?

Answer:

Graph In B bit represents a linear function

Explanation:

Linear function is one where if there is a constant rate of change,

If we see all the graphs only in graph B there is constant rate of change,

in graph A it is increasing, decreasing and again increasing,

in graph C it is higher and at a time decreased

in graph D it decreased and again increased,

Therefore only in graph B , therefore only in graph B

it represents a linear function.

Question 10.

Find (- 2.5)^{-2}.

Answer:

(- 2.5)^{-2 }=

Explanation:

Given expression as (- 2.5)^{-2 }we write the expression as

positive exponents by using So (- 2.5)^{-2 }is

Question 11.

Two lines have the same y-intercept. The slope of one line is 1, and

the slope of the other line is – 1. What can you conclude?

F. The lines are parallel.

G. The lines meet at exactly one point.

H. The lines meet at more than one point.

I. The situation described is impossible.

Answer:

G. The lines meet at exactly one point.

Explanation:

If two lines have different slopes, they cannot be the same line.

However, if they share a y-intercept, that means they cross the

y-axis at the same y value. Since the x value is constant on the

y-axis (0), they also share an x-value here. If they share an x value

and corresponding y value, they intersect at that point,

in this case their y-intercept and two lines can’t intersect

at more than one point, So we conclude the lines meet at exactly one point.

Question 12.

Which list of ordered pairs represents the mapping diagram?

A. (1, 2), (2, 0), (3, – 2)

B. (1, 0), (2, 2), (3, – 2)

C. (1, 0), (2, 2), (2, – 2), (3, – 2)

D. (0, 1), (2, 2), (- 2, 2), (- 2, 3)

Answer:

C. (1, 0), (2, 2), (2, – 2), (3, – 2)

Explanation:

A mapping shows how the elements are paired.

Its like a flow chart for a function, showing the input and output values.

A mapping diagram consists of two parallel columns..

Lines or arrows are drawn from domain to range,

to represent the relation between any two elements.

So by seeing as 1 is related 0, 2 is related to 2, -2 and 3 is related to -2,

therefore the ordered pairs are (1,0),(2,2),(2,-2),(3,-2) matches with

bit C.