ALGEBRA 1 LESSON 8-1 Simplify each expression. 1.2 3 2.3.4 2 2 2 4.(–3) 3 5.–3 3 6.6 2 12 1 4...

ALGEBRA 1 LESSON 8-1ALGEBRA 1 LESSON 8-1

Simplify each expression.

1. 23 2. 3. 42 22

4. (–3)3 5. –33 6. 62 12

1 42

Evaluate each expression for a = 2, b = –1, c = 0.5.

7. 8. 9. a 2a

bcc

abbc

(For help, go to Lessons 1-2 and 1-6.)

Zero and Negative ExponentsZero and Negative Exponents

8-1


1. 23 = 2 • 2 • 2 = 8

2. = =

3. 42 22 = = = = 4

4. (–3)3 = (–3)(–3)(–3) = 9(–3) = –27

5. –33 = –(3 • 3 • 3) = –(9 • 3) = –27

6. 62 12 = 36 12 = 3

7. for a = 2: =

8. for b = –1, c = 0.5: = –1

9. for a = 2, b = –1, c = 0.5: = = 4

1 42

1 4 • 4

1 16

42

22

4 • 42 • 2

164

a 2a

2 2 • 2

12

bcc

–1 • 0.50.5

abbc

2 • (–1) (–1) • 0.5

2 0.5


Solutions

8-1

Simplify.

a. 3–2

Simplify.19=


(–22.4)0b. Use the definition of zero as an exponent.

= 1


8-1

=Use the definition of negative exponent.

1 32

Simplify

a. b.

Simplify.3ab 2=

= 1 1x 3

Use the definition of negative exponent.

= x 3

Identity Property of Multiplication


= 1 • x 3

Multiply by the reciprocal of , which is x 3.

1 x3

Rewrite using a division symbol.

= 1 x –3


3ab –2 1 b2

Use the definition of negative exponent.

= 3a

8-1

1 x –3

Evaluate 4x 2y –3 for x = 3 and y = –2.

Method 1: Write with positive exponents first.

Substitute 3 for x and –2 for y.4(3)2

(–2)3=

36–8 –4

12= = Simplify.



8-1

4x 2y –3 = Use the definition of negative exponent.4x 2

y 3

(continued)

Method 2: Substitute first.

4x 2y –3 = 4(3)2(–2)–3 Substitute 3 for x and –2 for y.

4(3)2

(–2)3= Use the definition of negative exponent.

36–8 –4

12= = Simplify.



8-1

In the lab, the population of a certain bacteria doubles every

month. The expression 3000 • 2m models a population of 3000

bacteria after m months of growth. Evaluate the expression for m = 0

and m = –2. Describe what the value of the expression represents in

each situation.

3000 • 2m = 3000 • 20 Substitute 0 for m.

= 3000 • 1 Simplify.

= 3000

When m = 0, the value of the expression is 3000. This represents the initial population of the bacteria. This makes sense because when m = 0, no time has passed.



8-1

a. Evaluate the expression for m = 0.

(continued)

b. Evaluate the expression for m = –2.

3000 • 2m = 3000 • 2–2 Substitute –2 for m.

= 3000 • Simplify.

= 750

14

When m = –2, the value of the expression is 750. This represents the 750 bacteria in the population 2 months before the present population of 3000 bacteria.



8-1



1. 3–4 2. (–6)0

3. –2a0b–2 4.

5. 8000 • 40 6. 4500 • 3–2

k m–3

1

2 b2

– km3

8000 500


8-1

1 81


(For help, go to Lesson 1-6.)

Rewrite each expression using exponents.

1. t • t • t • t • t • t • t 2. (6 – m)(6 – m)(6 – m)

3. (r + 5)(r + 5)(r + 5)(r + 5)(r + 5) 4. 5 • 5 • 5 • s • s • s

Simplify.

5. –54 6. (–5)4

7. (–5)0 8. (–5)–4

Multiplication Properties of ExponentsMultiplication Properties of Exponents

8-3


1. t • t • t • t • t • t • t = t7

2. (6 – m)(6 – m)(6 – m) = (6 – m)3

3. (r + 5)(r + 5)(r + 5)(r + 5)(r + 5) = (r + 5)5

4. 5 • 5 • 5 • s • s • s = 53 • s3 = 53s3

5. –54 = –(5 • 5 • 5 • 5) = –(25 • 25) = –625

6. (–5)4 = (–5)(–5)(–5)(–5) = (25)(25) = 625

7. (–5)0 = 1

8. (–5)–4 = (– )4

= (– )(– )(– )(– )

= ( )( )

=

15

15

15

15

15

1 25

1 25

1 625


Solutions

8-3

Rewrite each expression using each base only once.


73 • 72

= 75 Simplify the sum of the exponents.

44 • 41 • 4–2



Use the definition of zero as an exponent.= 1

a.

b.

68 • 6–8c.


8-3

Add exponents of powers with thesame base.

73 + 2=

Think of 4 + 1 – 2 as 4 + 1 + (–2) to add the exponents.

44 + 1 – 2=

Add exponents of powers with thesame base.68 + (–8)=



p2 • p • p5a.

= p 8 Simplify.

Add exponents of powers with the same base.

= 20(x 6+(–4))

Simplify.= 20x 2

4x6 • 5x–4 b.


8-3

Commutative Property of Multiplication(4 • 5)(x 6 • x –4)=

Add exponents of powers with the same base.p 2 + 1 + 5=



a 2 • b –4 • a 5

= a 2 + 5 • b –4 Add exponents of powers with the same base.

Simplify.a 7

b 4=

2q • 3p3 • 4q4b.

= 24(p 3)(q 1 • q 4) Multiply the coefficients. Write q as q 1.

= 24(p 3)(q 1 + 4)Add exponents of powers with the same base.

= 24p 3q 5 Simplify.

a.


8-3

Commutative and Associative Properties of Multiplication

(2 • 3 • 4)(p 3)(q • q 4)=

Commutative Property of Multiplicationa 2 • a 5 • b –4=

Simplify (3 10–3)(7 10–5). Write the answer in scientific

notation.

= 21 10–8 Simplify.

= 2.1 101 • 10–8 Write 21 in scientific notation.

= 2.1 101 + (– 8)Add exponents of powers with the same base.

= 2.1 10–7 Simplify.


(3 10–3)(7 10–5) =Commutative and Associative Properties of Multiplication

(3 • 7)(10–3 • 10–5)


8-3

The speed of light is 3 108 m/s. If there are1 10–3 km in

1 m, and 3.6 103 s in 1 h, find the speed of light in km/h.

Speed of light = meters seconds

kilometersmeters

secondshour

• • Use dimensional analysis.

= (3 108) • (1 10–3) • (3.6 103) ms

kmm

sh

Substitute.

= (3 • 1 • 3.6) (108 • 10–3 • 103)Commutative and Associative Properties of Multiplication

= 10.8 (108 + (– 3) + 3) Simplify.



8-3

= 10.8 108 Add exponents.

= 1.08 101 • 108 Write 10.8 in scientific notation.

= 1.08 109 Add the exponents.

The speed of light is about 1.08 109 km/h.

(continued)



8-3



1. 34 • 35 2. 4x5 • 3x–2

3. (3 104)(5 102) 4. (7 10–4)(1.5 105)

5. (–2w –2)(–3w2b–2)(–5b–3)

6. What is 2 trillion times 3 billion written in scientific notation?

39 12x3

1.5 107 1.05 102

30b5–

6 1021


8-3



Rewrite each expression using each base only once.

1. 32 • 32 • 32 2. 23 • 23 • 23 • 23

3. 57 • 57 • 57 • 57 4. 7 • 7 • 7

Simplify.

5. x3 • x3 6. a2 • a2 • a2

7. y–2 • y–2 • y–2 8. n–3 • n–3

More Multiplication Properties of ExponentsMore Multiplication Properties of Exponents

8-4


1. 32 • 32 • 32 = 3(2 + 2 + 2) = 36

2. 23 • 23 • 23 • 23 = 2(3 + 3 + 3 + 3) = 212

3. 57 • 57 • 57 • 57 = 5(7 + 7 + 7 + 7) = 528

4. 7 • 7 • 7 = 73

5. x3 • x3 = x(3 + 3) = x6

6. a2 • a2 • a2 = a(2 + 2 + 2) = a6

7. y–2 • y–2 • y–2 = y(–2 + (–2) + (–2)) = y–6 =

8. n–3 • n–3 = n(–3 + (–3)) = n–6 =

1 y 6

1 n 6


Solutions

8-4

Simplify (a3)4.

Multiply exponents when raising a power to a power.

(a3)4 = a3 • 4

Simplify.= a12



8-4

Simplify b2(b3)–2.

b2(b3)–2 = b2 • b3 • (–2) Multiply exponents in (b3)–2.

= b2 + (–6) Add exponents when multiplying powers of the same base.

Simplify. = b–4

= b2 • b–6 Simplify.

1 b4= Write using only positive exponents.



8-4

Simplify (4x3)2.

(4x3)2 = 42(x3)2 Raise each factor to the second power.

= 42x6 Multiply exponents of a power raised to a power.

= 16x6 Simplify.



8-4

Simplify (4xy3)2(x3)–3.

(4xy3)2(x3)–3 = 42x2(y3)2 • (x3)–3 Raise the three factors to the second power.

= 42 • x2 • y6 • x–9 Multiply exponents of a power raised to a power.

= 42 • x2 • x–9 • y6 Use the Commutative Property of Multiplication.

= 42 • x–7 • y6 Add exponents of powers with the same base.

16y6

x7= Simplify.



8-4

An object has a mass of 102 kg. The expression

102 • (3 108)2 describes the amount of resting energy in joules the

object contains. Simplify the expression.

102 • (3 108)2 = 102 • 32 • (108)2Raise each factor within parentheses to the second power.

= 102 • 32 • 1016 Simplify (108)2.

= 32 • 102 • 1016 Use the Commutative Property of Multiplication.

= 32 • 102 + 16 Add exponents of powers with the same base.

= 9 1018Simplify.Write in scientific notation.



8-4



1. (x4)5 2. x(x5y–2)3

3. (5x4)3 4. (1.5 105)2

5. (2w–2)4(3w2b–2)3 6. (3 10–5)(4 104)2

x20 x16

y6

125x12 2.25 1010

432b6w2 4.8 103


8-4


(For help, go to Skills Handbook page 724.)

Division Properties of ExponentsDivision Properties of Exponents

8-5

Write each fraction in simplest form.

1. 2. 3. 4.

5. 6. 7. 8.

9. 10. 11. 12.

5 20

12525

60 100

1244

6 15

8 30

1035

1863

5xy15x

6y2

3x3ac12a

24m6mn2



8-5

1. 2.

3. 4.

5. 6.

7. 8.

9. 10.

11. 12.

= = 5 20

5 • 15 • 4

14

12525

25 • 525 • 1= = 5

60 100

20 • 320 • 5

35= = 124

4 = = 314 • 314 • 1

6 15

25

3 • 23 • 5= = = =

8 30

2 • 4 2 • 15

4 15

1035

27

5 • 25 • 7= =

1863

27

9 • 29 • 7= =

5xy15x

y3

5 • x • y5 • 3 • x= =

6y2

3x2y2

x3 • 2 • y2

3 • x= =

3ac12a

3 • a • c3 • 4 • a

c4= =

24m 6mn2

6 • 4 • m6 • m • n2

4 n2= =

Solutions



x4

x9a.

Simplify the exponents.= x–5

Rewrite using positive exponents. 1 x5=

p3 j –4

p–3 j 6

= p6 j –10 Simplify.

Rewrite using positive exponents.p6

j10=

b.

=Subtract exponents when dividing powers with the same base.

x4 – 9

=Subtract exponents when dividing powers with the same base.

p3 – (–3)j –4 – 6


8-5

A small dog’s heart beats about 64 million beats in a year. If

there are about 530 thousand minutes in a year, what is its average

heart rate in beats per minute?

64 million beats 530 thousand min

6.4 107 beats

5.3 105 min= Write in scientific notation.

6.45.3 107–5=

Subtract exponents when dividing powers with the same base.

6.45.3 102= Simplify the exponent.

1.21 102 Divide. Round to the nearest hundredth.

= 121 Write in standard notation.

The dog’s average heart rate is about 121 beats per minute.



8-5

Simplify . 3 y 3

4

34

y 12= Multiply the exponent in the denominator.

81y 12= Simplify.



3 y 3

4 34

(y 3)4=

Raise the numerator and the denominator to the fourth power.

8-5

a. Simplify .23

–3

33

23=Raise the numerator and the denominator to the third power.

Simplify.278 3

38or=



23

–3 32

3= Rewrite using the reciprocal of .

23

8-5

(continued)

b. Simplify .

Raise the numerator and denominator to the second power.

(–c)2

(4b)2=

Simplify.c2

16b2=



4bc

–

4bc

––2 c

4b–

2= Rewrite using the reciprocal of .4b

c–

Write the fraction with a negative numerator.

c 4b

– 2=

8-5



1. 2. 3.

4. 5. 6.

a8

a–2

w3

w7

(3a)4(2a–2)6a2

1.6 103

4 10–2

24

54x 3

3x 2

a10 1 w4 27

4 10425625 or 10

6 25

27 64x3


2 –3

8-5



Find the common difference of each sequence.

1. 1, 3, 5, 7, ... 2. 19, 17, 15, 13, ...

3. 1.3, 0.1, –1.1, –2.3, ... 4. 18, 21.5, 25, 28.5, ...

Use inductive reasoning to find the next two numbers in each pattern.

5. 2, 4, 8, 16, ... 6. 4, 12, 36, ...

7. 0.2, 0.4, 0.8, 1.6, ... 8. 200, 100, 50, 25, ...

Geometric SequencesGeometric Sequences

8-6


1. 1, 3, 5, 7, ... 2. 19, 17, 15, 13, ...7 – 5 = 2, 5 – 3 = 2, 3 – 1 = 2 13 – 15 = –2, 15 – 17 = –2,Common difference: 2 17 – 19 = –2

Common difference: –2

3. 1.3, 0.1, –1.1, –2.3, ... 4. 18, 21.5, 25, 28.5, ...–2.3 – (–1.1) = –1.2, –1.1 28.5 – 25 = 3.5, 25 – 21.5 = 3.5, – 0.1 = –1.2, 0.1 – 1.3 = –1.2 21.5 – 18 = 3.5Common difference: –1.2 Common difference: 3.5


Solutions

8-6



Solutions (continued)

8-6

5. 2, 4, 8, 16, ... 6. 4, 12, 36, ... 2(2) = 4, 4(2) = 8, 8(2) = 16, 4(3) = 12, 12(3) = 36, 16(2) = 32, 32(2) = 64 36(3) = 108, 108(3) = 324Next two numbers: 32, 64 Next two numbers: 108, 324

7. 0.2, 0.4, 0.8, 1.6, ... 8. 200, 100, 50, 25, ...(0.2)2 = 0.4, 0.4(2) = 0.8, 0.8(2) = 200 2 = 100, 100 2 = 50,1.6, 1.6(2) = 3.2, 3.2(2) = 6.4 50 2 = 25, 25 2 = 12.5,Next two numbers: 3.2, 6.4 12.5 2 = 6.25

Find the common ratio of each sequence.

a. 3, –15, 75, –375, . . .

3 –15 75 –375

(–5) (–5) (–5)

The common ratio is –5.

b. 3,32

34

38, , , ...

332

34

38

12 1

2 12

The common ratio is .12



8-6

Find the next three terms of the sequence 5, –10, 20, –40, . . .

5 –10 20 –40

(–2) (–2) (–2)

The common ratio is –2.

The next three terms are –40(–2) = 80, 80(–2) = –160, and –160(–2) = 320.



8-6

Determine whether each sequence is arithmeticor geometric.

a. 162, 54, 18, 6, . . .

62 54 18 6

13 1

3 13

The sequence has a common ratio.


The sequence is geometric.


8-6

(continued)

b. 98, 101, 104, 107, . . .

The sequence has a common difference.

98 101 104 107

+ 3 + 3 + 3

The sequence is arithmetic.



8-6

Find the first, fifth, and tenth terms of the sequence that has

the rule A(n) = –3(2)n – 1.

first term: A(1) = –3(2)1 – 1 = –3(2)0 = –3(1) = –3

fifth term: A(5) = –3(2)5 – 1 = –3(2)4 = –3(16) = –48

tenth term: A(10) = –3(2)10 – 1 = –3(2)9 = –3(512) = –1536



8-6

Suppose you drop a tennis ball from a height of 2 meters.

On each bounce, the ball reaches a height that is 75% of its previous

height. Write a rule for the height the ball reaches on each bounce. In

centimeters, what height will the ball reach on its third bounce?

The first term is 2 meters, which is 200 cm.

Draw a diagram to help understand the problem.



8-6

(continued)

The ball drops from an initial height, for which there is no bounce. The initial height is 200 cm, when n = 1. The third bounce is n = 4. The common ratio is 75%, or 0.75.

A rule for the sequence is A(n) = 200 • 0.75n – 1.

A(n) = 200 • 0.75n – 1 Use the sequence to find the height of the third bounce.

A(4) = 200 • 0.754 – 1 Substitute 4 for n to find the height of the third bounce.

= 200 • 0.753 Simplify exponents.

= 200 • 0.421875 Evaluate powers.

= 84.375 Simplify.

The height of the third bounce is 84.375 cm.



8-6


1. Find the common ratio of the geometric sequence –3, 6, –12, 24, . . .

2. Find the next three terms of the sequence 243, 81, 27, 9, . . .

3. Determine whether each sequence is arithmetic or geometric.a. 37, 34, 31, 28, . . .

b. 8, –4, 2, –1, . . .

4. Find the first, fifth, and ninth terms of the sequence that has the rule A(n) = 4(5)n–1.

5. Suppose you enlarge a photograph that is 4 in. wide and 6 in. long so that its dimensions are 20% larger than its original size. Write a rule for the length of the copies. What will be the length if you enlarge the photograph five times? (Hint: The common ratio is not just 0.2. You must add 20% to 100%.)

–2

3, 1, 13

arithmetic

geometric

4, 2500, 1,562,500

A(n) = 6(1.2)n-1; about 14.9 in.


8-6

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ALGEBRA 1 LESSON 8-1 Simplify each expression. 1.2 3 2.3.4 2 2 2 4.(–3) 3 5.–3 3 6.6 2 12 1 4...

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