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NEW YORK

Regents Test Prep Workbook

Teacher’s Guide

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0982

iii

To the TeacherIncluded in this booklet are answer keys for the following books:

• New York Regents Test Prep Workbook for Integrated Algebra

• New York Regents Test Prep Workbook for Geometry

• New York Regents Test Prep Workbook for Algebra 2 and Trigonometry

Integrated Algebra answers 1

Geometry answers 31

Algebra 2 and Trigonometry answers 51

HMH_NY_AGA_Answer_Key.indd iiiHMH_NY_AGA_Answer_Key.indd iii 5/19/10 8:06:35 PM5/19/10 8:06:35 PM

HMH_NY_AGA_Answer_Key.indd ivHMH_NY_AGA_Answer_Key.indd iv 5/19/10 8:06:35 PM5/19/10 8:06:35 PM

Copyright © by Holt McDougal. 1 NY Regents Test Prep Workbook for Integrated Algebra

All rights reserved.

NY Regents Test Prep

Workbook for Integrated Algebra

Diagnostic Test pp. 1–6 1. 3 2. 3 3. 4 4. 1 5. 4 6. 2 7. 3 8. 1 9. 1 10. 4 11. 4 12. 2 13. 2 14. 2 15. 3 16. 2 17. 1 18. 3 19. 2 20. 4 21. 1 22. 4 23. 1 24. 1 25. 2 26. 3 27. 1 28. 3 29. 4 30. 1 31. 92.4 m

2

32. c > 2

33. 6400 34. $2917 35. Possible

answer: y =2

3x + 8 ;

36. relative error of length =

|18 20|

20100 = 10%;

relative error of area = 2 relative error of length = 2 10 = 20%; relative error of volume = 3 relative error of length = 3 10 = 30%

37. The score 68 is the 19

th

percentile; Possible justification: There are 6 scores lower

and6

3219% . First

quartile = 71; Possible justification: There are 32 scores, and the 8

th and

9th

scores are 70 and 72. The median or second quartile = 78; Possible justification: There are 32 scores, and the 16

th and

17th

scores are 76 and 80. Third quartile = 91; Possible justification: There are 32 scores, and the 24

th and 25

th scores

are 90 and 92.

38. 2,1

2

39.

(0, –3) and (4, 5)

Properties of Real Numbers pp. 7–8 1. 2 2. 3 3. 2 4. 1 5. 3

6.

Commutative Property

1

328( ) 9( ) = 28

1

39( )

Associative Property

= 281

39

= 28 3( )

Distributive Property

= 20 + 8( ) 3( ) = 60 + 24

= 84

7. 4 9 5 = 4 5 9 = 20 9 = 180

8. 4 88( )1

2= 4 44( )

= 4 40 + 4( ) = 4 40( ) + 4 4( ) = 160 +16

= 176

9. Eliza should use Order of Operations and multiply 3(9x) to get 27x. She can then combine like terms using the Distributive Property to get 2x + 27x = 29x.

2x + 3(5x + 7x 3x)

= 2x +15x + 21x 9x

= 38x 9x

= 29x

Both methods give the correct answer since they both used Order of Operations to simplify the expression. Simplify Radical Terms pp. 9–10 1. 1 2. 3

3. 2 6

4. 8

5. 7

10

6. 4 5

5

7. 6

6

··

·

·

HMH_NY_AGA_Answer_Key.indd 1HMH_NY_AGA_Answer_Key.indd 1 5/19/10 8:06:35 PM5/19/10 8:06:35 PM



8.

3

2

9. Answers may vary. Possible answers: Method 1:

800 = 400 2

= 20 2

Method 2:

800 = 100 4 2

= 10 2 2

= 20 2

10. Answers may vary. Possible answers: Method 1:

48

45

=16 3

9 5

=4 3

3 5

5

5

=4 15

15

Method 2:

48

45

=48

45

=16

15

=4

15

=4 15

15 15

=4 15

15

Adding and Subtracting Radical Expressions pp. 11–12 1. 2 2. 1 3. 2

4. 6 13

5. 5 2

6. 11 5

7. 16 2 + 2

16 2 + 2

4 2 + 2

5 2

8. 9 3 3

9 3 3

3 3 3

2 3

9. 25 5 + 5

25 5 + 5

5 5 + 5

6 5

10. 12 + 300

4 3 + 100 3

4 3 + 100 3

2 3 +10 3

12 3

11. 48 27

16 3 9 3

16 3 9 3

4 3 3 3

3

Multiplying and Dividing Radical Expressions pp. 13–14 1. 2 2. 4

3. 3 12 = 3 12 = 36 = 6

4. 5 10 = 5 10

= 50 = 25 2

= 25 2 = 5 2

5. 8 11 = 8 11

= 88 = 4 22

= 4 22 = 2 22

6. 7

2

2

2

=14

2

7. 8

3

3

3

=24

3=

4 6

3

=2 6

3

8.

12

5

5

5

=60

5=

4 15

5=

2 15

5

9. 2( 2 + 14)

= 2 2 + 2 14

= 4 + 28 = 2 + 4 7

= 2 + 2 7

10. (5 + 10)(8 + 10)

= 5(8) + 5 10 + 8 10 + 10 10

= 40 +13 10 +10

= 50 +13 10

Scientific Notation pp. 15–16 1. 3 2. 3

3. 93,000,000 = 9.3 107

4. 365(1.3 10–2

)

= 474.5 10–2

= 4.745 102 10

–2

= 4.745 100

= 4.745 in.

5 (4.5 10-6 )(5.6 108)

= (4.5 5.6)(10-6 108)

= 25.2 102

= 2.52 10 102

= 2.52 103

6. 1.275 10

4

1.7 103

=1.275

1.7

104

103

= 0.75 107

= 7.5 101

107

= 7.5 108

7. 1 101

millimeters (mm) =

1centimeter (cm)

1 102

centimeters (cm) =

1 meter (m)

1 103 meters (m) = 1 kilometer

(km) 1,000,000 dollars = 100,000,000 pennies. 100,000,000 pennies is

1 108 pennies. A stack of

1million dollars in pennies would

be 1.55 108 mm tall.




1.55 108mm

1 101cm

= 1.55 107cm or

15,500,000 cm tall.

1.55 107cm

1 102m

= 1.55 105mor

155,000 m tall

1.55 105m

1 103km

= 1.55 102km or

155 km tall

Solving Algebraic Problemspp. 17–181. 32. 3

3.

42

100=

x

300

100x = 42 300

100x = 126,000

x = 126

4. 15

100=

x

680

100x = 15 680

100x = 10,200

x = 102

5. 3

4=

a + 5

21

3 21= 4(a + 5)

63 = 4a + 20

43 = 4a

43

4= a

6. 3

y 3=

1

9

3 9 = (y 3)1

27 = y 3

30 = y

7. The original selling price ofthe camera

x x20

100= 144

x 0.2x = 144

0.8x = 144

x =144

0.8

x = 180

Price dealer originally paid

x + x50

100= 180

x + 0.5x( ) = 180

1.5x = 180

x =180

1.5

x = 120

Evaluating Expressionspp. 19–201. 12. 2

3. 22+ 6(8 5) ÷ 2

22+ 6 3 ÷ 2

4 + 6 3 ÷ 2

4 +18 ÷ 2

4 + 9

13

4. 5!+ 3(8 2) ÷ ( 3)

5! + 3 6 ÷ ( 3)

120 + 3 6 ÷ ( 3)

120 +18 ÷ ( 3)

120 6

114

5.

(3 + 2)(4 + 3) + 52

6 22

5 7 + 52

6 22

5 7 + 25

6 4

35 + 25

6 4

60

2

30

6. 4(3 | 2 6 | +5)

4(3 | 4 | +5)

4(3 4 + 5)

4 4

167. Substitute for the Variables

(5 4 23)

3 2

Evaluate exponents inside parentheses

(5 4 8)

3 2

Multiply inside parentheses

20 8( )3 2

Subtract inside parentheses

12

3 2

Multiply from left to right

12

6

Divide from left to right

2

Permutations pp. 21–221. 42. 43.

6!

3!=

6 5 4 3 2 1

3 2 1= 6 5 4 = 120

4.

9!

4!= 9 8 7 6 5 = 15,120

5.

15!

14!= 15




6. Using the FundamentalCounting Principle, Johncan make 4 _ 3 _ 2 = 24different outfits.

7. 6720 Yes.; 8; 8P5 =

8!

(8 5)!=

8!

3!= 8 7 6 5 4 = 6,720

8. 6P4 =

6!

(6 4)!=

6!

2!= 6 5 4 3 = 360

Solve and Graph Multi-StepInequalities pp. 23–241. 42. 33. 1

4. 6 2y > 0

2y > 6

y < 3

5. 3x + 5 17

3x 12

x 4

6. No. The solution set isx > 2, so 2 is not a solution.

7. 0.25x +11.75 + 42.50 75Cole can spend amaximum of $75, so thetotal cost of the balloons,the tank rental, and thetable cloths and streamersmust be less than or equalto $75.0.25x + 54.25 75

0.25x + 54.25 54.25 75

54.25

0.25x 20.75x 20.75 ÷ 0.25x 83

The maximum number ofballoons Cole canpurchase is 83.

0.25(83) + 54.25 7520.75 + 54.25 7575 75

Linear Inequalitiespp. 25–261. 12. 33. 44. 2

5. 4y + 3(y 2) < 5y

4y + 3y 6 < 5y

7y 6 < 5y

7y 5y 6 < 5y 5y

2y 6 + 6 < 0 + 62y < 62y ÷ 2 < 6 ÷ 2y < 3

6. 3p 2(2p 1)

3p 4p 2

3p 4p 4p 4p 2

p 2

p ÷ 1 2 ÷ 1p 2

7. Write the inequality

2(x 4) 6x + 9

Distribute the 2 to both terms

2x 8 6x + 9

Subtract 6x to both sides

4x 8 9

Add 8 to both sides

4x 17

Divide both sides by 4 and

reverse the inequality symbol

x17

48. I can group the r terms on

the right side of theinequality. This will give 5 9r, and I can divide bothsides of the inequality by 9to find the solution set for r.

9. 5(x 2) > 3(x + 4)

5x 10 > 3x + 12

5x 3x 10 > 3x 3x + 12

2x 10 + 10 > 12 + 10

2x > 222x ÷ 2 > 22 ÷ 2x > 11

Check x = 5

5(5 2) >?

? 3(5 + 4)

5(3) >?

3(9)15 > 27This is false. x=5 is not inthe solution set

Check x = 11

5(11 2) >?

3(11 + 4)

5(9) >?

3(15)45 > 45This is false but it doesshow that x = 11 is theendpoint on the graph.

Check x = 20

5(20 2) >?

3(20 + 4)

5(18) >?

3(24)90 > 72This is true. x = 20 is in thesolution set.

Domain and Rangepp. 27–281. 32. 13. Domain: from –2 to 2

D: –2 x 2 Range: from 1 to 4 R: 1 y 4

4. D: { 3, 0, 1, 4}

R: { 5, 2, 6}




5. Answers may vary.

6. D: {0, 50, 125, 250, 350} R: {2, 6, 11, 14, 16} The domain represents the

number of miles traveled. The range represents the number of gallons of gas remaining in her tank. The tank holds 16 gallons and all of the values in the range are 16 or less.

At 350 miles traveled, she has 2 gallons remaining in her tank. This means she used 16 – 2 = 14 gallons to travel 350 miles. Since 350 14 25 , she averaged 25 miles per gallon. This confirms her belief.

Multiple Representations of Functions pp. 29–30 1. 3 2. 2 3. x 4 2 0 1 5 10

f(x) 12 8 4 2 6 16

4.

5.x 2x 4 g(x)0 4 4 1 2 2 2 0 0 3 2 2 4 4 4

6. You can use an equation of a function to make a table of values by substituting different values for x and determine the corresponding values of y. Then you can plot the ordered pairs on a coordinate plane. To create a table of values from an equation, you can substitute different values for x into the equation to calculate the values of y. To create a table of values from a graph, you can read the graph at different points and record the values of x and y.

Writing Algebraic Expressions pp. 31–32 1. 3 2. 43. 1 4. 35m 5. h 0.5 6. c ÷ 25 7. 152 + b 8. r 50 9. 0.05i

I can substitute 23 for i in my expression and simplify. 0.05(23) = 1.15. The sales tax on a $23 shirt is $1.15.

10. 12

d 5

To write Gina’s age in terms of d, I first wrote Gina’s age in terms of Fiona age f. Then I substituted my expression for Fiona’s age in terms of d and simplified it. 2f 3

2( 12

d 5 ) 3

d + 7 If Danielle is 14 then Fiona is 12 years old,

because 12

(14) 5 = 7 +

5 = 12 If Danielle is 14 then Gina is 21 years old, because (14) + 7 = 21. I know this is correct because Gina’s age, 21, is also 3 less than twice Fiona’s age, 12.

Writing Verbal Expressions pp. 33–34 1. 3 2. 2 3. 1 4. 2 5. for example, 7 more than a

number 6. for example, 8 less than

two times a number 7. for example, a number

divided by 2 8. for example, 4 more than a

number divided by 3 9. The expression means $75

more than $85 times the number of hours work, or $85 per hour plus $75.

10. Laura’s age is 7 years less than twice Aaron’s age.




11. For example, 20 less than 5 times a number; the product of 5 and a number, decreased by 20. For example, the amount of money Garret will earn mowing lawns if he charges $5 per lawn and has expenses of $20; the number of people at a recital this year was 20 fewer than 5 times the number at the recital last year

Writing Mathematical Equations and Inequalities pp. 35–36 1. 3 2. 4 3. 3 4. 2 5. 3r + 2 and 6t 1

6. x2

+ 4 < 12

7. 9 + x = 15 8. 10 > 3x 9. 5x < x + 16 10. t 65

m > t j = m + 5

The first sentence states that Tanya is “no more than” 65 inches tall. No more than is the same as less than or equal to, so I used the “less than or equal to” symbol .

The second sentence states that Mila is taller than Tanya. That means Mila’s height is greater than Tanya’s height, so I used the greater than symbol > The third sentence states that Jamal is 5 inches taller than Mila. Is means equals, so I used the “equals” symbol =.

Writing Equations and Inequalities to Represent Situations pp. 37–38 1. 4 2. 2 3. 2 4. 3 5. 6.75h 25 6. 2c + 9 = 45 7. 1.80 + 0.2n 20 8. 3s = 54 9. 5.25l 500 4d + 100 > 500 5.25h = 4h + 100

Answers may vary. Possible answer: If Lara earned $600, how many hours did she work?

5.25l = 600 The number in the

sentence is an exact amount, so I used an equal sign. Her wages are $5.25 per hour, so I multiplied the number of hours, l, by her hourly wage.

Solving Verbal Problems in One Variable pp. 39–40 1. 4 2. 4 3. 512 + x = 2,115 x = 1,603

The store had 1,603 printers before the delivery.

4. y = 2.50 + 0.1(5) y = 3.00 The call cost $3.00. 5. 4.25s 34 s 8 Portia could buy at most 8 sandwiches. 6. 9 2 + k k 7 Kevin makes at most $7 per hour. 7. 50 + 25h = 100 + 20h 25h 20h = 100 50 5h = 50 h = 10 The two amounts will be

equal at hour 10. The second electrician would be cheaper for any hours after that.

8. 242 + 4.5x 578 4.5x 336 x

74 2

3

They will have to wash at least 75 cars if they charge $4.50 per car. 242 + 6x 578 6x 336 x 56 They will have to wash at least 56 cars if they charge $6 per car. The chorus will have to wash at least 19 more cars if they charge $4.50 instead of $6.

Solving Problems with Systems of Equations pp. 41–42 1. 2 2. 1 3. x y 1 375 10x 5y 10,875

Rewrite first equation x 1375 y 10(1375 y ) 5y 1087513750 10y 5y 1087513750 5y 10875

5y 10875 137505y 2875

y 575

x 1375 575x 800




4. x = the number of standard printsy = the number of enlarged prints

x y 280.20x 1.10y 8.30

x 28 y0.20(28 y ) 1.10y 8.305.6 0.2y 1.10y 8.305.6 0.9y 8.300.9y 8.30 5.60.9y 2.7y 3

x y 28x 3 28x 28 3x 25

5. x = the number of footballs y = the number of hockey sticks 5x + 3y = 69 3x + 6y = 96 x = 6 y = 13 The cost of one football and one hockey stick is $19.

6.

x y 81.89x 1.53y 13.32

Answers may vary. Example: Substitution method is a convenient method to solve this system. I can solve the first equation for y, since it has a coefficient of 1, and then substitute for y into the second equation.

x y 81.89x 1.53y 13.32

y 8 x1.89x 1.53(8 x) 13.321.89x 12.24 1.53x 13.32.36x 12.24 13.32.36x 1.08x 3y 8 xy 8 (3)y 5

3 packs of AA batteries and 5 packs of AAA batteries were bought. Quadratic and Exponential Equations pp. 43–44 1. 2 2. 4 3. 3 4. v = 4.9t2 = 4.9(10)2 = 4.9(100) = 490 m/s

How fast is the fly traveling at t=4s v = t2 4.9t + 27 = 42 4.9(4) + 27 = 16 19.6 + 27 = 23.4 m/s

5. (6 + 2x)(8 + 2x) = 80

4x2 + 28x + 48 = 80 4x2 28x 32 = 0 x2 7x 8 = 0

6. The function y 31,000(0.98x ) shows a current population of 31,000 for Lavanda and a decrease in population y of 2% each year for x years, since 100% – 2% = 98% or 0.98. The population of Verndon can be modeled by the function y 31,000(1.03x ) . This model represents exponential growth in population for the city of Verndon. The model for Lavanda’s population shows exponential decay.

Solving Systems of Equations Algebraically pp. 45–46 1. 3 2. 4 3. 4 4. (y 1) + 2y = 8 3y 1 = 8 3y = 9 y = 3 x = y 1 x = 3 1 x = 2 The solution to the system is (2, 3). 5. x + 2 = 2x 5 x = 7 x = 7 y = x + 2 y = 7 + 2 y = 9 The solution to the system is (7, 9). 6. 2y = 6 y = 3 2x + y = 1 2x + ( 3) = 1 2x = 4 x = 2 Ordered pair solution: (2, 3)

7. 8x = 8 x = 1 3x + 4y = 13 3( 1) + 4y = 13 3 + 4y = 13 4y = 16 y = 4 Ordered pair solution: ( 1, 4)




8. Kari’s next step is to substitute 2x + 1 for y in the second equation.

Olivia’s next step is to solve the resulting equation for y.

y = 2x + 1 3( 2x + 1) 2x = 19 6x + 3 2x = 19 8x = 16 x = 2 y = 2( 2) + 1 y = 4 + 1 y = 5

OR

Ordered pair solution: ( 2, 5) y + 2x = 1 3y –2x =19 4y = 20 y = 5 y + 2x = 1 5 + 2x = 1 2x = –4 x = –2 Ordered pair solution: ( 2, 5) Solving Systems of Quadratic and Linear Equations pp. 47–48 1. 4 2. 2 3. 1 4. 1 5. x2 + x 3 = 2x + 3

x2 x 6 = 0 (x + 2)(x 3) = 0 x = 2 or x = 3 y = 2x + 3 y = 2( 2) + 3 or y = 2(3) + 3 y = 1 or y = 9 The solutions are ( 2, 1) and (3, 9).

6. x y 8 0

y x 8y x 8

x2 + 10x 2 = x + 8 x2 + 9x 10 = 0 (x + 10)(x 1) = 0 x = 10 or x = 1 y = x+ 8 y = 10 + 8 or y = 1 + 8 y = 2 or y = 9 The solutions are ( 10, 2) and (1, 9).

7. 14x + 2y + 14 = 0 7x + y + 7 = 0 7x 7 7x 7

y = 7x 7 7x 7 = x2 + 5

x2 + 7x + 12 = 0 (x + 4)(x + 3) = 0 x = 4 or x = 3 y = x2 + 5 y = ( 4)2 + 5 or y = ( 3)2 + 5 y = 21 or y = 14 The solutions are ( 4, 21) and ( 3, 14)

8. 3x + 3y 3 = 0 x + y 1 = 0

x + 1 x + 1 y = x + 1 x2 5x + 5 = x + 1 x2 4x + 4 = 0 (x 2)(x 2) = 0 x = 2 y = x + 1 y= 2 + 1 y = 1 The solution to the system is (2, 1). The lines intersect at one point, since there is only one solution. Willa might have solved the linear equation for x and then substituted that value for x into the quadratic equation. Or, she might have substituted x2 5x + 5 for y in the linear equation.

Multiplying and Dividing Monomials pp. 49–50 1. 3 2. 2 3. 1 4. 2 5. 1 6. 10x3y4 7. 8x3y2z2 8. 3x3y3

9. 3xy2

10. 4x2y3z11. 4x2z 12. Nathan’s steps:

6x2(x2yz2)2xy 2z

6x4yz2

2xy 2z

3x3zy

Dean’s steps: 6x2(x2yz2)

2xy 2z3x(x2yz2)

y 2z

3x3yz2

y 2z3x3z

y

Both expressions

simplified to

3x3zy

so the

order in which they simplified the expression did not matter.

Simplifying Monomials and Polynomials pp. 51–52 1. 1 2. 2 3. 4 4. 2x3 + 5x2 + 5x + 1 5. 6x3 + 4x2 5 6. x3 + 6x2 + 2x 2 7. 5x3 3x2 + 6 8. (x 2)(x 3)

= x 2 2x 3x + 6 = x2 5x + 6

9. (x 7)(x + 7) = x2 7x + 7x 49 = x2 49

10. (x + 2)(x + 1) = x2 + 2x + x + 2 = x2 + 3x + 2




11. (P)(S) R Q

P x3 3x2 7x

PS 2x3 6x2 14x

R 2x 2y 2 4y 4x2

Q y 2 2y 8PS R Q

2x3 2x2 3y 2 16x 6y 8 Dividing Polynomials pp. 53–54 1. 3 2. 1 3. 4 4. 4x 1 5. 2x + 1 6. 3x + 10 7. 2x + 1 8. 4x + 3 9. 2x + 4 10.

x 2 5x2 0x 20

5x2 10x 10x 20 10x 20

5x 10

0

It is necessary to rewrite the polynomial because the divisor has an x term while the dividend does not. The 0x is a placeholder for the middle term.

Polynomial division and arithmetic division both use the same steps of divide, multiply, subtract, and bring down. Polynomial division uses variables and constants, while arithmetic division uses only numbers. Polynomial division needs placeholders if any of the terms are missing in the dividend.

Algebraic Fractions pp. 55–56 1. 2 2. 3 3. 4 4. 3 5. 0 6. 5 7. 7 8. 0 9. 3x2 + 15x = 0

x2 + 5x = 0 x(x + 5) = 0

5 xx (x 5)

0

Excluded value is 5. 10. x2 5x 14 = 0

(x + 2)(x 7) Excluded values are 2 and 7.

11. The excluded value is 6. I factored the numerator and the denominator to get:

2xx(x 6)

When x = –6 the function is undefined since 6 – 6 = 0. 0 is not an excluded value because an x is factored out of the numerator and denominator and divided out.

Simplifying Polynomial Fractions pp. 57–58 1. 3 2. 1 3. 2 4. 4

5. x 3x2 9

x 3(x 3) (x 3)

1x 3

6. x2 4xx2 x 12

x (x 4)

(x 3) (x 4)

xx 3

7. x2 2x 15x2 5x 6

(x 3) (x 5)

(x 3) (x 2)

(x 5)(x 2)

8. x 9x2 81

x 9(x 9) (x 9)

1(x 9)

9. x2 7x 6x 1

(x 1) (x 6)

x 1x 6

10. x 3x2 7x 12

x 3(x 3) (x 4)

1x 4

11. The excluded value of the function is x = 4 since (4)2 2(4) 8 16 8 8 0 I can factor the numerator and the denominator of the function.

y x2 8x 12

x2 2x 8

y(x 6) (x 2)

(x 4) (x 2)

The excluded value of the function is x = 4




The value –2 is notexcluded since an (x +2)can be factored from thenumerator and thedenominator and divideout.

Add and SubtractFractional Expressionspp. 59–601. 32. 43. 24. 3

5.

6 + 5

5x=

11

5x

6. 2x

25x

x + 5

7. 3x +10

3x + 7

x7

3

8. 3

x 3

x 3

9.

2x 5

x2

1

x 1,x 1

10. Before simplifying, Theexpression is undefined at x

= –1

2, because the

denominator would be

61

2+ 3 = 0 at x = –

1

2.

6(x + 2)

6x + 3+

2(x + 5)

6x + 3

=6x 12 + 2x +10

6x + 3

=4x 2

6x + 3

=2(2x +1)

3 (2x +1)

=2

3

The expression is defined forall values of x aftersimplifying.

Multiplying and DividingAlgebraic Fractionspp. 61–621. 22. 13. 34. 3

5.

6x4

y4

6.

2(x 1)

x + 4i

x (x + 4)

(x 2) (x 1)

=2x

x 2

7.

8 (x + 2)

(x +1) (x 1)i

x +1

4 (x + 2)

=2

x 1

8.

3x3y

5xy2i

15y

9xy3

=x

y3

9.

4x 8

x2

4

ix + 2

3x

=4 (x 2)

(x + 2) (x 2)i

x + 2

3x

=4

3x

10.

x2+ 2x 3

x2

9

ix

22x 3

x2+ 3x 4

=(x 1) (x + 3)

(x + 3) (x 3)i

(x +1) (x 3)

(x + 4) (x 1)

=x +1

x + 4

11. X1 = 4 r

2

4

3r

3

=3

r

X2 =

6(2r )2

(2r )3=

24r2

8r3

=3

r

X1

X2

=

3

r

3

r

=3

r

•r

3

=1

1= 1

Difference of Squarespp. 63–641. 32. 23. 4

4. (x + 10)(x 10)

5. (x + y)(x y)

6. (3x2 + 8)(3x

2 8)

7. Yes, it’s a difference ofsquares.

25x2 = 5x

81 = 9

(5x + 9)(5x 9)

8. No, it is not a difference ofsquares because 30x

2 is not

a perfect square.9. Yes, it is a difference ofsquares.

4x2 = 2x

121 = 11

(2x + 11)(2x 11)




10. The dimensions of the new park are 100 ft 100 ft, 10,000 100 .

The dimensions of the new park are (100 + x) ft (100 – x) ft, since 10,000 100 and

x2 x .

The perimeters of both parks remain the same since the same number of feet was subtracted from the width and added to the length of the new park.

The area of the new park will be less than the area of the current park since x2 will be subtracted from the current area.

Factoring Polynomials pp. 65–66 1. 2 2. 4 3. 4 4. 20x2 15x = 5x(4x) 5x(3) = 5x(4x 3) 5. 44a2 + 11a = 11a(4a) + 11a(1) = 11a(4a + 1) 6. (x + 1)(x + 12) 7. 3(x – 4)(x + 2) 8. 2(x 9)(x 4) 9. The factors of 6 are 6,

3, 2, 1, 1, 2, 3, and 6. The factors of 6 that add to 5 are 3 and 2, and 6 and 1

x2 + 5x 6 = (x + 6)(x 1)

The factors represent the dimensions of the field. The length is (x 1) m and the width is (x + 6) m.

(x 6)(x 1) x2 5x 6

(20 6)(20 1) 202 5(20) 6(26)(19) 400 100 6

494 494

Determining Whether a Given Value is a Solution pp. 67–68 1. 3 2. 4 3. 4 4. 1 5. no; the equation is not true

when x = 2 6. yes; if x = 3 then

4 6x = 14, which greater than 15

7. Stephen substituted 9 for x in the equation and evaluated each side. Since the two sides did not have the same value, 9 is not a solution to the equation.

8. x + 6.95 + 0.08x 70, or 1.08x + 6.95 70 Yes, she can buy the necklace. I substituted 55 for x in the inequality. 1.08x 6.95 701.08(55) 6.95 7059.4 6.95 7066.35 70

$66.35 $70 , so 55 is a solution. The tax on $30 earrings is 30(.08) = $2.40 $66.35 + $2.40 = $68.75, which is still less than $70, so she will be able to include the earrings with her purchase.

Solving Equations pp. 69–70 1. 4 2. 1 3. 1 4. 1 5. y = 5 6. m = 12 7. y = 4 8. m = 2 9. x = 48 10. y = 2

11. The slope of the equation for Matt’s health club is 18 and the y-intercept is 80. These represent the $18 per month dues and the $80 initiation fee. The slope for Kayla’s health club is $23 for monthly dues, and the y-intercept is the $45 initiation fee. To find when both health clubs cost the same, I set 18m 80 23m 45 and solved for m. 18m 80 23m 45

80 5m 4535 5m7 m

After 7 months, they will both cost $206. 18(7) 80 23(7) 45

126 80 161 45206 206

Solving Literal Equations for a Variable pp. 71–72 1. 4 2. 3 3. 2 4. 3

5. s P4

6. b = 180 a c

7. K PVT

8. a c3b

9. z 3(y x) 10. m pn 3

11. l 3Vwh

The original formula divides l by 3, so used the inverse operation of multiplication. The original formula multiplies l by wh, so I used the inverse operation of division.




l =3V

wh

=3i560

10i14

=1,680

140

= 12

Its base length is 10 in.

Solving Equations withFractional Expressionspp. 73–741. 42. 23. 14. 45. 26.

4

x+ 6 =

16

x

4

x(x) + 6(x) =

16

x(x)

4 + 6x = 16

6x = 12

x = 27.

3x 7( ) 2x +18

3x 7

= 12 3x 7( )

2x +18 = 36x 84

102 = 34x

3 = x

8.

5

3+

2

x + 4= 2

LCD = 3(x + 4)

5

3i3(x + 4) +

2

x + 4i3(x + 4)

= 2[3(x + 4)]

5(x + 4) + 6 = 6x + 24

5x + 20 + 6 = 6x + 24

5x + 26 = 6x + 24

x = 2

x = 2

9. The LCD is 2(x 2). They

are the same for bothfractional expressions

3

x 2i2(x 2) + 2i2(x 2) =

5

x 2i2(x 2) +

3

2i2(x 2)

6 + 4(x 2) = 10 + 3(x 2)

6 + 4x 8 = 10 + 3x 6

4x 2 = 3x + 4

x = 6

I checked my answer bysubstituting 6 for x in theequation.

3

6 2+ 2 =

5

6 2+

3

2

3

4+ 2 =

5

4+

3

2

3

4+

8

4=

5

4+

6

4

11

4=

11

4

Yes, just as for other types ofequations, I could group liketerms and then solve theequation.

3

x 2+ 2 =

5

x 2+

3

2

2i2

2=

5 3

x 2+

3

2

4

2

3

2=

2

x 2

2

x 2=

1

2

Solving Proportionspp. 75–761. 12. 33. 24. 4

5. 8x = 20

x =5

2

6. 5k = 15k = 3

7. 4a + 20 = 634a = 43a = 10.75

8. x(x + 2) = 15x

2 + 2x = 15

x2 + 2x 15 = 0

(x + 5)(x 3) = 0

x = 5 or x = 3

9. t(2t + 4) = 702t

2 + 4t = 70

2t2 + 4t 70 = 0

t2 + 2t 35 = 0

(t + 7)(t 5) = 0

t = 7 or t = 5

10.

x 2

x=

x

x + 3

x2= x

2+ x 6

0 = x 6

6 = x

x = 6

11.

Jake's height

Jake's shadow=

tree's height

tree's shadow

6

15=

h

50

I placed like measurementsin the numerator and in thedenominator. I placed Jake’sheight and the tree’s height inthe numerators and Jake’sshadow length and the tree’sshadow length in thedenominators. I placed bothof Jake’s measurements inone fraction and both of thetree’s measurements in theother fraction.

6

15=

h

50

15h = 50 6

15h = 300

15h

15=

300

15

h = 20

The tree is 20 feet tall.

Solving QuadraticEquations pp. 77–781. 22. 33. 24. 45. 1




6. (x 6) = 0 or (x 3) = 0 x = 6 or x = 3

7. (x + 8) = 0 or (x 5) = 0 x = 8 or x = 5

8. 3x = 0 or (x 7) = 0 x = 0 or x= 7

9. (2x 3) = 0 or (x + 9) = 0 x = 3/2 or x = 9

10. (x + 4)(x 3) = 0 x + 4 = 0 or x 3 = 0 x= 4 or x = 3

11. (x + 5)(x + 5) = 0 x + 5 = 0 x = 5

12. (x + 8)(x 1) = 0 x + 8 = 0 or x 1 = 0 x = 8 or x = 1

13. Yes, the equations share a solution. I factored each equation to find its solutions. x2 3x 10 = 0 (x 5)(x + 2) = 0 x 5 = 0 or x + 2 = 0 x = 5 or x = 2

x2 8x + 15 = 0 (x 5)(x 3) = 0 x 5 = 0 or x 3 = 0 x = 5 or x = 3 Both equations have x = 5 as a solution. The graphs will intersect at x = 5 because both graphs have x = 5 as a solution, so both graphs pass through (5, 0).

Quadratic Roots and Factors pp. 79–80 1. 3 2. 2 3. 1 4. 4 5. 4x(x 6) = 0 4x = 0 or x 6 = 0 x = 0 or x = 6 6. (x + 3)(x + 1) = 0 x= 3 or x = 1

7. x2 5x + 4 = (x 1)(x 4) (x 1)(x 4) = 0 x = 1 or x = 4 The roots are 1 and 4. 8. 3x2 + 12x = 3x(x + 4) 3x(x + 4) = 0 3x = 0 or x + 4 = 0 x = 0 or x= 4 The roots are 0 and 4. 9. The equation is y = x2 + 4x 5. If an equation has roots of 1 and 5, it has factors of (x 1) and (x + 5).

(x 1)(x + 5) = x2 + 4x 5 I can set each factor equal to zero to verify the roots of the equation.

(x 1) 0 (x 5) 0x 1 x 5

Set-Builder and Interval Notation pp. 81–82 1. 4 2. 3 3. 3 4. 2 5. [ 5, 1] 6. ( 3, 2) 7. ( 3, 6] 8. {x | x 15} 9. {2x | x is a real number} 10. {2x+1| x >0} 11. Yes, (i) and (ii) can be

represented using set-builder notation.

(i) {x | –5 x < 20} (ii) {3x| x is an integer} No, each set cannot be represented using interval notation. (i) [–5, 20) Complements of Sets pp. 83–84 1. 1 2. 2 3. 3 4. 1 5. 4 6. Jc = {1, 3, 5, 7, 9}

7. Kc is the set of irrational numbers. 8. Ac = {1, 4, 6, 8, 9, 10, 12,

14, 15, 16, 18} 9. Mc = {5x| x is an integer} 10. Qc = {2} 11. The complement of the

null set is the universe. The null set is empty and contains no elements. That means that the complement of the null set contains every element in the universe, so the complement of the null set is the universe itself.

12. Ac = . If A contains all of the elements in the universe, then there are no elements in the complement of A, and Ac is the null set.

13. The Venn diagram shows the universe within the rectangle and a set within the circle. The area outside the circle and inside the rectangle represents the complement of the set within the circle. Together, the set and its complement make up the universe.

The circle represents set A.

The area outside of the circle represents the elements that are not within set A, or the complement of A.

Intersections and Unions of Sets pp. 85–86 1. 2 2. 1 3. 4 4. 2 5. 2 6. 1 7. A C = {x | x is a real number}




8. A B = {12x | x 0, x is an integer}

9. A B C = { 48, 36, 24, 12}

10. You can use a Venn diagram by drawing overlapping circles. The individual circles represent each set and then writing the elements of each set within the appropriate circle. If there are elements in more than one set, they are written within the overlap of the circles. To find the elements in a union set, write all of the elements that are within any of the circles. To find elements in an intersection set, write all of the elements that are within the overlapping sections.

11. Yes, (A B) C is the same as A (B C) . If you are finding all of the elements that are in at least one of the sets, then the order in which you examine the sets does not matter. Yes, (A B) C is the same as A (B C) . If you are finding the elements that are in each of three different sets, then the order in which you examine the sets does not matter.

12. A (B C) = {2, 3, 4, 6, 8, 10}

A (B C) = {2, 4, 6} The sets are not the same. The first set is the union of A with the intersection of B and C, that are in both B and C combined with all the elements of A. The second set is the intersection of A with the union of B and C, so it is all of the elements that are in both B and C, then the intersection of these elements with those in A.

Rates of Change pp. 87–88 1. 3 2. 2 3. Slope is the rate of change

of the dependent variable relative to the x value. A horizontal line has the same y value for every different x value, so the rate of change is 0 and the slope is 0.

4. y = 0.15x+ 75 Slope = 0.15 The slope represents the charge per mile, or the rate of change of the cost relative to the number of miles.

5. The slope of the line represents the change in distance over time.

The slope of the line would be steeper if the number of miles traveled over the same time period doubled. This is because the y value would change twice as much for each change in the x value.

6. Cellular phone plan 1 has a slope that increases as the amount of minutes increase. The rate of change of the charge is relative to the minutes used. Cellular phone plan 2 has a slope of 0 since charges remain unchanged as the minutes increase.

Finding Equations of Lines pp. 89–90 1. 2 2. 3

3. 3 4. 4 5. y = mx + b

6 = 3(4) + b 6 = 12 + b b = 6 y = 3x 6

6. y = mx + b

8 12

( 2) b

8 1 bb 9

y 12

x 9

7. y = mx+ b 2 = 1(5) + b 2 = 5 + b

b = 3 y = x + 3

8. m = 2 y = mx+ b 0 = 2(4) + b b = 8 y = 2x 8

9. y = 25h + b 200 = 25(5) + b 200 = 125 + b 75 = b y = 25h + 75 The y-intercept is 75. It represents the flat fee the electrician charges.

y 25h 75y 25(3) 75y 75 75y 150

The electrician would charge $150 for a 3 hour service call.

Find the Slope of a Line Given Two Points on the Line pp. 91–92 1. 1 2. 2 3. 2 4. 4 5. 4 6. y = 3x + 5

7. y 14

x 1




8. y = 4y 9. y = 5x 3

10. y 1

2x 1

11. y = x + 5 12. I could use two points in

the table to write an equation for the relationship and then substitute value of x to find the unknown value of y.

m 7 30 ( 2)

42

2

y = 2x + b 3 = 2( 2) + b 3 = 4 + b b = 7 y = 2x + 7 The slope of the line is 2 and the y-intercept is 7.

Horizontal and Vertical Lines pp. 93–94 1. 2 2. 4 3.

4.

5.

Possible answer: (0, 0) and (0, 1). 6.

Possible answer: (0, 0) and (0, 1). 7. A horizontal line is of the form y = b so for any value of x, y = b. Using the slope formula:

m b b

x2 x1

0x2 x1

0 .

The slope of any horizontal line is 0.

8. A vertical line is of the form x = a so for any value of y, x = a. Using the slope formula:

my2 y1a a

y2 y10

which

is undefined. The slope of any vertical line is undefined. Finding the Slope of a Line pp. 95–96 1. 4 2. 1 3. m = 4 4. 4x 2y 3

2y 4x 3

y 2x 32

m = –2 5. 5x y 30

y 5x 30 slope = –5; y-intercept = 30

6. x y 7

y x 7y x 7

slope = 1; y-intercept = –7 7. 4x 3y 12

3y 4x 12

y 43

x 4

slope = 43

; y-intercept = 4

8. 2x y 3

y 2x 3y 2x 3

slope = 2; y-intercept = –3




Parallel Lines pp. 97–98 1. 2 2. 4 3. 1 4. Slope = 6 5. Slope = –2 6. y = 2x + 1 and y = 2x – 3

are parallel.

7. y = –x + 11

Testing Solutions to Linear

Equations pp. 99–100 1. 2 2. 2 3. Yes

y 4x4 4(1)4 4

4. No

y 2x 31 2(3) 31 6 31 31 3

5. Yes

y 2(2) 48 4 48 8

6. Sample: (2, 0) 7. Sample: (0, 9) 8. Sample: (4, 16)

9.x 2x y Ordered pair

(x, y) 1 2(1) 2 (1, 2) 2 2(2) 4 (2, 4) 3 2(3) 6 (3, 6) 4 2(4) 8 (4, 8) The ordered pair (5, 10)

means that Cynthia can buy 5 markers for $10.

Cynthia can buy 12 markers

for $24.

y 2x24 2x

x 12

Systems of Linear Inequalities pp. 101–102 1. 4 2. 1 3. Region A is the solution to

the system. I shaded above the line y = –x – 1 and above the line y = 2x + 1. Then I tested a point (–1, 1) in region A.

y x 11 ( 1) 11 0

True

y 2x 11 2( 1) 11 1

True

4. Sample:y x 1y 2x 2

( 1) (3) 1( 1) 2(3) 2

True

5.

Sample: Test (2, 0): y x 10 2 10 1

True

y 20 2

True

Quadratic Functions in Standard Form pp. 103–104 1. 4 2. 3 3. (0, 0); x = 0 4. (–2, 1); x = –2 5. a = 1; b = –4; c = 3

The graph opens upward because a > 0. Axis of symmetry x = 2 f(2) = –1 Vertex: (2, –1) y-intercept: 3




6. y = –x2 + 1; The vertex of curve 2 is at (0, 1) and the y-intercept is 1, so c = 1. The axis of symmetry for curve 2 is that same as the axis of symmetry for curve 1. Curve 2 opens downward so a is negative.

Trigonometric Functions pp. 105–106 1. 3 2. 4 3. cos A

4. sin = 35

; cos = 45

; tan

= 34

5. sin = 1213

; cos = 5

13;

tan = 125

6. sin A a

c;

cos A b

c

ac

2bc

2

1

a2

c2b2

c21

a2 b2

c21

c2

c21

1 1

Solving Right Triangles pp. 107–108 1. 3 2. 60

cos( 1) 36

sin( 1) 5.26

tan( 1) 5.23

3. 30

cos( 2) 5.26

sin( 2) 36

tan( 2) 35.2

4. tan A 4023

A tan 1 4023

A 60.1

5. tan F 48.2

F 26

tan H 8.24

H 64FH2 FG2 HG2

FH2 (8.2)2 42

FH2 67.24 16

FH2 83.24FH 9.12

6. sin X 3136

X 59

cos Y 3136

Y 31

XY 2 WY 2 WX 2

362 312 WX 2

1296 961 WX 2

335 WX 2

WX 18.3

Using Trigonometric Functions to Find Side Lengths of Right Triangles pp. 109–110 1. 2 2. 4

3. Sample:

sin27 18x

x 18sin27

x 39.65 m

4. Sample:

tan31 x10

10 tan31 xx 6.01 in.

5. Sample:

sin40 x10

10sin40 xx 6.4 cm

6. Sample:

sin35 x33

33sin35 xx 18.9 in.

The Pythagorean Theorem pp. 111–112 1. 1 2. 3 3. x2 162 202

x2 256 400

x2 144x 12

4. 22 52 x2

4 25 x2

29 x2

29 x

5. 52 32 625 9 6

16 64 610 cm

6. 252 72

57624 cm




7. East to west

72 42 49 16

65 8.06 miles

North to south

102 62 100 36

64 8 miles

Since 65 64 , the distance across the lake from north to south is shorter than east to west.

Geometric Constructions pp. 113–114 1. 2 2. 2 3. 3 4. I can place the point of my

compass on X and use my compass to measure the distance from X to Y. I used my straight edge to draw a line segment and I drew a point on it. Then I used the preserved compass span to measure the same distance as segment XY. I made a mark on the line. I can label my new endpoints A and B. AB XY

5. I can place the point of my compass on point C and spread the compass more than halfway across the line segment. I can draw an arc that goes above and below the line. Then I can my compass with the same distance on point D and draw an arc that goes above and below the line. I can then use my straight edge to join the two points of intersection.

6. I can place the point of my compass on vertex S and draw an arc that passes through segments RS and ST. Then I can place the point of my compass at the intersection of the arc and segment RS and draw an arc. I can repeat this with segment ST. I can use my straight edge to join the intersection of the arcs to vertex S.

7. I can use my compass and straight edge to construct a perpendicular bisector of segment AB that is halfway between A and B. I label the intersection C. This will divide AB into two equal segments, AC and BC . Then I use my compass and straight edge to construct a perpendicular bisector of AC . This divides AC into two congruent segments. Finally, I can construct a perpendicular bisector of BC . This will divide BC into two congruent segments. Since each new segment is equal to one-half of one-half of the original segment, the original segment was divided into four equal sections.

Determining Slopepp. 115–116 1. 2 2. 2 3. 3

4

4. 12

or 0.5 5. The slope of a horizontal

line is 0 because the y value does not change, so the numerator of the slope is 0.

6. The slope of the line represents the change in distance traveled over time, or the speed.

The slope would double because the change in distance would double while the change in time would remain the same.

Finding the y-intercept pp. 117–118 1. 4 2. 2 3. 3 4. 2 5. 4 6.

7. The y-intercept is 4.9. The y-intercept represents

the original value of the copier at t = 0, or when it was first purchased. I know because the y-intercept is the point where the line crosses the y-axis, which occurs when x = 0.

Between 6 and 7 years; I know this because when the value of x is between 6 and 7 years on the graph, the value of y is 0.

Using a Table of Values to Graph a Line pp. 119–120 1. 2 2. 1




3.

x y = –2x + 2 (x, y) 4 y = (–2)(–4) + 2 (–4, 10) 3 y = (–2)(–3) + 2 (–3, 8)

–2 y = (–2)(___) + 2 (–2, 6) –1 y = (–1, 4) 0 y = (0, 2)

4. Sample:

x y = 12

x 3 (x, y)

4 5 ( 4, 5) 2 4 ( 2, 4)

0 3 (0, 3) 2 2 (2, 2) 4 1 (4, 1)

5.C° F ° 0 32 5 41 10 50 15 59 20 68 25 77 30 86 35 95

Using multiples of 5 returns whole number temperatures in degrees Fahrenheit.

62° C is about 143°F.

Slope-Intercept Form pp. 121–122 1. 2 2. 1 3. 3 4. 2

5. y 14

x 3

6. y = 5x 7. y = 7x 2 8. 1 9. (0, 5) 10. y = 4 The slope is 0.

Since the equation is in the form y = mx + b, and 4 represents the value of b, the value of mx must be 0. Since x is a variable, m must equal zero, so the slope of the line is zero.

11. In the equation y = mx + b, m represents the value of the slope and b represents the y value of the y-intercept. The slope of a line written in slope-intercept form is given by the coefficient of x.

12. y 3.99x 1.99 ; The

slope of the equation is 3.99. It represents the increase in the total cost for every DVD that Juan rents. The y-intercept is 1.99. It represents the initial cost of the popcorn. If Juan rents 3 movies, his total cost will be $13.96.

y 3.99x 1.99y 3.99(3) 1.99y 11.97 1.99y 13.96

Graphing from Slope-Intercept Form pp. 123–124 1. 3 2. 4 3.

4.




5. The y-intercept of a line tells you the y value of the point where the line crosses the y axis. To graph the line, I can draw a point at the y-intercept. The slope of the line tells you how much the y value changes for each change in the x value. From the y-intercept, I can count the rise, up for a positive slope or down for a negative slope, and then the run, to the right, and make another point. Then I can join the two points to draw the line.

6. The variable x represents the number of school lunches Danielle can purchase. The variable y represents the amount of money that remains in Danielle’s account. The slope is –5 and it represents the cost of a school lunch that is deducted from her account. 60 is the y-intercept and it represents the initial amount of money in Danielle’s account.

According to the graph,

Danielle can purchase 12 lunches.

Solving Systems of Equations pp. 125–126 1. 4 2. 4 3. ( 1, 3) 4. (2, 3)

5. (5, 7)

6. (2, 4)

7.

Based on the graph, the

solution to the first system is (2, 3).

Based on the graph, the solution to the second system is (3, 2).

Since the two systems

have different solutions, Mario is correct. Justin may have reversed the order of the coordinates since each solution had a value of 3 and 2.

Graphing Inequalities in One Variable pp. 127–128 1. 2 2. 3 3. 2 4. x 2 5. x < 4 6. x > 3 7.

8.

9. The graph has a closed circle on 4 with an arrow to the left. 10. x 8 7.5 or x 60 No, 50 is not a solution to

the inequality because 50 60.

The graph has a closed circle on 60 with an arrow to the right.

Graphing Inequalities in One Variable pp. 129–130 1. 2 2. 4




3. Yes, because the graph is a straight line and each x value is paired with only one y value.

4. No, because the graph is not a straight line.

5. No, because the equation is for a linear function, and the function in the graph is non linear and curved.

6. The graph of a quadratic function is curved because the rate at which the y-value increases is not constant. For example, when the x value increases by 1 from 4 to 5 to 6, the y-value increases by 9, from 17 to 26, then increases by 11 from 26 to 37.

Quadratic Equations pp. 131–132 1. 4 2. 4 3. 4 4. Yes. The second-order differences are constantly increasing by 2. A quadratic equation can model this situation.

Figure Marbles 1 1 2 4

3 9 4 16

5. Only one representation is

needed. The equation y x2 x 3

contains an x2 term, making it a quadratic equation.

In the equation y x2 x 3 , the second-order differences, of the y-values, from the table of values are all +2.

The graph of y x2 x 3 is in the shape of a parabola.

6. The equation y x2 1

contains an x2 term, making it a quadratic equation.

The second-order

differences for the y-values constantly increase by 2.

x x2 + 1 (x, y) –2 5 (–2, 5) –1 2 (–1, 2) 0 1 (0, 1) 1 2 (1, 2) 2 5 (2, 5)

The graph of the plotted points is in the shape of a parabola.

Area of Composite Figures pp. 133–134 1. 3 2. 3 m 3. 254.34 4. 18.84 mm 5. Area of the first figure is

approximately 356.5 ft2. Area of the second figure is

approximately 35.2 cm2. To find both areas, you find

the areas of a rectangle and a semicircle. In the first figure you add both areas. In the second figure, you subtract both areas.

Surface Area of Prisms and Cylinders pp. 135–136 1. 4 2. 4 3. 150 ft2 4. 345.6 5. Each of the ten bases has

an area of 4 in.2; each of the twelve lateral faces has an area of 6 in.2; (10 4) (12 6) 112 in.2

6. The height of the stacked cylinders is 30 in; the surface area =

2 (4)(30) 2 (4)2; SA = 854.1 cm2 7. 26 + 42 = 68 cm2 8. 24 + 16 + 12 – 3.1 = 48.9;

12.6 + 3.1 = 15.7; 48.9 + 15.7 = 64.6

Volumes of Prisms and Cylinders pp. 137–138 1. 4 2. 3 3. 576 cm3 4. 2,010.6 mm3 5. 4(3) × 4w × 4h =

192wh6. 150 + 50.3 200.3 m3 7. 25.1 + 84.8 109.9 ft3 Functions and Relations pp. 139–140 1. 3 2. 2 3. 4 4. The relation is not a

function. The x-value of –3 is paired with two different values for y.

5. This is a function. All values of x are paired with exactly one value for y.

6. This is not a function. I can draw a vertical line at x = 3 and it will intersect with the graph at two different places.

in.2

in.2

in.2

in.3

π π




7. The following relations are functions: {(1, 2), (2, 3), (3, 4), (4, 5), (5, 6)}

{(3, 2), (5, 2), (8, 6), (10, 6)}

Each value of x is paired with only one value of y. The following relations are not functions:

{(1, 2), (1, 3), (2, 3), (3, 4), (4, 5), (5, 6)}

{(3, 2), (5, 6), (8, 6), (8, 2), (10, 6)}

Each relation has one value of x that is paired to two different values for y.

Parent Functions pp. 141–142 1. 1 2. g(x) = x2. The

transformation is a refection across the x-axis.

3. g(x) = x . The transformation is a translation 2 units left.

4. The graph of the function f(x) = x2 + 1 is translation of the parent function f(x) = x2, 1 unit up on the coordinate plane.

5. The graph shows the function f(x) = x2 – 3, and is a translation of the parent function f(x) = x2, 3 units down on the coordinate plane.

Transforming Functions pp. 143–144 1. 1 2. 3 3. g(x) 2(2x4 6x2 4)

4x4 12x2 8

g(x) is a vertical stretch of f(x).

4. g(x) 2(2x)4 6(2x)2 4

32x4 24x2 4

g(x) is a horizontal compression of f(x).

5. Since g(x) = 13

f(x), g(x) is

a vertical compression of f(x).

6.

y = x2 is the parent

quadratic function with vertex at (0, 0) and x-intercept = 0.

y = –x2 is the reflection of y

= x2 across the x-axis. The vertex is at (0, 0) and the x-intercept = 0.

y = 3x2 is a vertical compression of y = x2. The vertex is at (0, 0) and the x-intercept = 0.

y = x2– 4 is a horizontal

translation of y = x2. The vertex is at (0, –4) and the x-intercepts are 2 and –2.

Graph Linear Inequalities pp. 145–146 1. 2 2. 4 3. 1 4.




5.

6. y < –5x + 10 7. y = 4x – 1 y 4x 1

The point (2, 7) lies on the

boundary line. Since it is a solution to the inequality, the boundary line is solid.

Solving Systems by Graphing pp. 147–148 1. 2 2. 2 3. Answers may vary.

Possible answers: Solutions: (1, –2) and (1, –3)

Not solutions: (0, 0) and (5, 5)

4. Answers may vary. Possible answers:

Solutions: (0, 0) and (–1, 0) Not solutions: (5, 0) and

(5, –2)

5.

6.

7.

The boundary lines are

parallel. A system of equations whose lines are parallel has no solution. A system of inequalities whose lines are parallel may have solutions.

A system of inequalities may also have no solution. For example, the system

y x 3y x 3

has no solution.

Solving Quadratic Equations by Graphing pp. 149–150 1. 2 2. x = 3 and x = –3 3. 2x2 6x 0 x = 0 and x = 3

2x2 6x 0

2(0)2 6(0) 00 0 0

2x2 6x 0

2(3)2 6(3) 018 18 0

4.

The x-intercepts of the

function are at x = –1 and at x = 5.

x2 4x 5 0(x 5)(x 1) 0

The roots of the equation x2 – 4x – 5 = 0 are at x = –1 and 5.

5. A quadratic function can have, at most, two roots because the graph of a parabola can intersect the x-axis at zero points, one point, or two points.

Solving Systems by Graphing pp. 151–152 1. 3 2.

(1, –2) and (5, –2) 3. no solution 4.

4




4

2

6

24

26

2426 4 6

y

x222(0, 21)

(23, 2)

y 52x21

O

y 5 x22113

1

2

22

21

24

22 2124 1 2

y

xO

y 52x11

y 52(x12)211

Finding the Vertex and Axis of Symmetry of a Parabola pp. 153–154 1. 3 2. x = –2 3. x = 2 4. Vertex (–3, – 10) Axis of

symmetry x = –3 5. Vertex (0, – 3) Axis of

symmetry x = 0 6.

The parabolas have the same vertex and axis of symmetry because they are reflections of each other across the line y = 3. Rates pp. 155–156 1. 2 2. 4 3. 28,000 mph 4. 8 cranes per hour 5. c = 28.5m Cost of fabric per meter: $28.50 Total cost: $330.10 6. Gallons in tank: 13 gallons

Distance traveled on a liter of gas: 10.3 km and 24.5 miles Gas price per liter: 69 cents

Using Proportional Relationships pp. 157–158 1. 4 2. 2

3. 1 in.16 in.

x in.540 in.

16x 540 in.x 33.75 in.

1 in.

16 in.x in.

336 in.16x 336 in.x 21 in.

33.75 in. 21 in.

4. 1 in.3 ft

x in.45 ft

3x 45x 15

1 in.3 ft

x in.28 ft

3x 28

x 9 13

15 in. 9 1

3in.

5. 1 in.80in.

x in.540 in.

80x 540 in.x 6.75 in.

1 in.80in.

x in.336 in.

80x 336 in.x 4.2 in.

6.75 in. 4.2 in.

6. 1300

x30(128)

1300

x3840

300x 3840x 12.8 ozor 13 oz

5.

6.




1300

x200(33.8)

1300

x6760

300x 6760x 22.5 ozor 23 oz

She needs to send her about 23 ozs.

7. 1 lb 12 oz = 16 oz + 12 oz = 28 oz.

28 oz$3.79

18 ozx

28x 68.22x 2.44

The cost of the peanut butter Randi used was $2.44. 28 oz 28.35 g = 793.8 grams

793.8g$3.79

209.4gdollar

750g$3.20

234.4gdollar

The 750 gram jar of peanut butter is the better deal.

Absolute and Relative Error pp. 159–160 1. 3 2. 2 3. 42% 4. relative error of area = 6%;

relative error of volume = 9%

5. relative error of length = 13%; relative error of volume = 39%

6. relative error of length = 1.5%; relative error of area = 1.5(2) = 3%

7. Jerome’s calculated

volume: V (4)2(10)V 160

Actual volume of the can: V (3.5)2(10)V 122.5

Relative Error: 160 122.5

122.537.5

122.50.306

or about 30.6% relative error.

8. Margo’s surface area: 1006 in.2 Actual surface area: 900 in.2 Margo’s volume: 2145 in.3

Actual volume: 1800 in.3

Relative error of surface area: 1006 900

900106900

0.118

or about 11.8% relative error. Relative error of volume: 2145 1800

1800345

18000.192

or about 19.2% relative error.

Categorizing Data pp. 161–162 1. 2 2. 3 3. The survey is quantitative

because the data is represented by counts or measurements.

4. Assign a number to each category:

strongly agree 4 somewhat agree 3 somewhat disagree 2 strongly disagree 1

5. There are five different responses in the table: poor, fair, average, good, and excellent. Assign the numbers 1–5 to each response:

poor 1 fair 2 average 3 good 4 excellent 5 The quantitative data is: {1,3,3,5,3,5,4,2,4,2} The mean user rating is

1 3 3 5 3 5 4 2 4 210

3.2

Univariate and Bivariate Data pp. 163–164 1. 2 2. univariate 3. univariate 4. univariate 5. bivariate 6. Because the shoe size is

the only variable, this is univariate data. The data can be displayed in a frequency table.

The frequency table makes

it easy to see that the most common shoe size on the team is size 9.




7. The data set is bivariate.

As the humidity increases,

the heat index increases. Sampling Methods and Bias pp. 165–166 1. 2. population: all restaurant

patrons; sample: those who take the Web site survey; voluntary response sample

3. population: all members of Serafina’s class; sample: every third person in Serafina’s class; convenience sample

4. population: all residents of the city; sample: those who fill out cards; voluntary response sample

5. Population: all people at the shopping center Sample: people who leave from one door at the shopping center, The sample is a convenience sample because the people leaving through the chosen door are easily accessed. The sample is biased because people leaving through another door do not have an opportunity to be chosen.

6. Yes; the graph is biased because the scale used on the y-axis is very large. A scale this large will make points appear to be closer together. This gives the impression that the expenses have remained fairly constant during the time period. A better graph would be to use a y-axis scale from 0 to 1700. This would present the expenses in an unbiased way.

Measures of Central Tendency pp. 167–168 1. 2 2. Mean =

75 63 89 914

79.5

No Mode Median = 82 3. Mean =

1 2 2 2 3 3 3 48

2.5

Modes = 2 and 3 Median = 2.5 4. Mean =

19 25 31 19 34 22 31 34

826.88

Modes = 19, 31, and 34 Median = 28 5. Mean =

58 58 60 60 60 61 65

760.29

Mode = 60 Median = 60 6. The median and the mode

best describe the distance that occurs most often.

None of the values describe the data well. There is an outlier of 15, so the range should be looked at along with the measures of central tendency when describing the data.

7. Mean = 153 145 148 158

4151

There is no mode Median = 150.5 The mean describes

Lamont’s average score. The mean and the median

are very close in value and both describe Lamont’s scores well.

Histograms pp. 169–170 1. 1 2.

2




3. Spelling Test Grades

Grade Cumulative Frequency

70 3 75 3 80 8 85 11 90 19 95 21

100 25

4.

Total Sales Day Amount

(dollars) 1 958 2 1,983 3 3,173 4 4,309 5 5,219 6 6,139 7 7,187 8 8,365 9 9,341 10 10,446 11 11,480 12 12,587

Box-and-Whisker Plots pp. 171–172 1. 2 2. 1 3.

4.

11 is an outlier. The interquartile range is

49.5 – 31.5 = 18 5. The least value is 2 years.

The greatest value is 22 years. The lower quartile is 5. The upper quartile is 13. The median is 7.; The interquartile range is 13 – 5 = 8; The data point 29 is an outlier.; 7 is the median, so 50% of the teachers have at least 7 years of experience.33 × 0.25 = 8.25 8; Eight teachers have taught for 5 years or less.

Create a Scatter Plot pp. 173–174 1. 3 2. $80,000; 60 3.

4.

Using point-slope form:

m = 32

and point (0, 2)

y 2 32

(x 0)

y 2 32

x

y 32

x 2

If x is 16, then the value of y would be

y 32

(16) 2 24 2 26

Evaluating Published Reports pp. 175–176 1. 4 2. Ms. Wiggins gave no

information regarding how many students were polled. Ms. Wiggins’ sample size may have been too small to accurately reflect the student population. Therefore, it may be unreasonable to assume 35% of the high school’s enrollment drives their own cars to school.

3.The report gives no information on the number of people who were surveyed. The sample size many be too small for the study.

4.The graph appears to substantiate the conclusions in the report. The graph is misleading because the percents do not total to 100%. The percentages do not match the areas of each cereal category. The graph should include additional cereal brands and an “other” category to accurately reflect 100% of the people surveyed.




5. There is no indication of why the Tierra is 50% safer than its competitors. There is no indication of what number “Nearly all” represents. The report may reflect bias because customers who test drive the Tierra already like the car;

What does 50% safer mean?

How was “safer” defined? How many people were

surveyed? Were those who test drove

the Tierra chosen randomly?

Include the number of people surveyed.

Indicate how 50% safer was measured.

Include the trunk’s dimensions.

How many customers test drove the vehicle and how many of those bought the car. Indicate whether or not the customers were chosen randomly.

Scatter Plots and Trend Lines pp. 177–178 1. 4 2. negative 3. no correlation 4. The correlation is negative

because the data shows that the number of mishandled baggage increases and the percent of on-time arrivals decrease.

5. The correlation is positive because the data shows that the length of a person’s femur increases as their height increases.

6. The greater the number of pages the greater the cost of the novel. The correlation is positive.

The price of a 300-page novel on the trend line is $27. The actual price $30. The difference is $3.

Correlation and Causation pp. 179–180 1. 3 2.(1) correlation; (2)

causation; (3) causation; (4) correlation

3. There is a moderate causation and strong correlation between April showers and May flowers because the water is necessary for the flowers to grow and bloom. However there are other factors such as temperature which affect the growth of May flowers. Both students are correct.

4. The correlation is negative. As the temperature

decreases, the number of cups of coffee sold increases. The increase in sales is caused by colder temperatures. This is a causal relationship.

Linear Transformations pp. 181–182 1. 3 2. The mean, median, and mode all increase by 50. 3. Mean = 9 pounds Median = 8.75 pounds Mode = 8 pounds Mean = 4.5 pounds Median = 4.375 pounds Mode = 4 pounds Original range = 7.5 New range = 3.75 If the weights are multiplied by 0.5, the range changes by a factor of 0.5.

4. Mean = $445; Mode = $395; Median = $425 $370 + $25 = $395; $395 + $25 = $420; $395 + $25 = $420; $455 + $25 = $480; $495 + 25 = $520; $560 + $25 = $585 Mean = $470; Mode = $420; Median = $450 The mean, mode, and median have all increased by $25. The range of the original salaries is 560 – 370 = 190. The range of the new salaries is 585 – 395 = = 190. The range stayed the same. Using Trend Lines to Make Predictions pp. 183–184 1. 4 2. $0; At 9 years the line of

best fit indicates that he would receive close to $0. At 10 years, he should probably expect the same amount.

3. $30,000; $2,500; $15,000 4. Graphs will vary. Sample:

Answers will very

depending on the position of the trend line.

Sample: 19 mi/gal; 5L




Independent and Dependent Events pp. 185–186 1. 3 2. 0.5

3. 16

; 16

;

136

4. 16

; 12

; 1

12; The events are

independent because rolling a 3 on the first roll does not affect the probability of rolling an even number on the second roll.

5. 4052

1013

; 1251

;

1013

1251

120663

; The events

are dependent because not replacing the first card drawn affects the probability of the second event by reducing the total outcomes from 52 to 51.

Theoretical Probability pp. 187–188 1. 1 2. 3 3. 1

4. P(odd) = 59

P(even) = 49

;

Spinning an odd number is more likely to occur.

5. P(yellow) = 1

12 P(blue)

= 1

12; Selecting a blue or a

yellow candy is equally likely.

6. (2, 1) (1, 2) , 2

361

18;

(3, 1) (1, 3) (2,

2), 336

112

;

(4, 5) (5, 4) (3, 6) (6, 3),

4

3619

7. (HHH) (HHT) (HTH) (THH) (TTT) (TTH) (THT) (HTT); outcomes with 2 heads (HHT) (HTH) (THH)

(TTT) (TTH) (THT) (HTT)

P(exactly 3 heads) = 18

P(at least 1 tail) = 78

P(two or more tails) = 12

P(no heads) = 18

Finding the Probability of an Event and its Complement pp. 189–190 1. 3

2. 1 10100

1 110

910

3. 1 62365

303365

4. 926

; 1 726

1926

5. 14

; 34

; 19100

; 1625

1. 12. 33. 3 4. 45. 1 6. 3 7. 1 8. 4 9. 2 10. 3 11. 3 12. 1 13. 4 14. 1 15. 2 16. 4 17. 2 18. 4 19. 2 20. 3 21. 4 22. 3 23. 2 24. 1

25. 3 26. 2 27. 2 28. 329. 2 30. 3 31. 15 2

32. 45

33. x = –1 and 3

34. 72 miles; 60 miles per

hour; 2.67 hours 35.

36.


Practic Test pp. 191–195 e



37. probability of either a

spade or a king = 413

;

probability of both hearts

= 351

; The most likely

result is a spade with a number less than 7. Justification: The probability of a red card with a number less than 4

is 452

, the probability of a

queen is 452

, and the

probability of a spade

is 552

.

38. x 0.25x 50 , x = 40, The original cost is $40. Final sale price =

$50 0.25 $50 $37.50. Possible explanation: The original cost and final sale price are different because the 25% increase is based on $40 and the 25% decrease is based on $50. The same percent applied to different amounts results in different amounts.

39. The x-intercepts are 1 and 3. The roots of the equation are 1 and 3.


New York Regents

Geometry Test Prep Review

and Practice Answer Key

Diagnostic Test pp. D1–D10

1. 2

2. 4

3. 2

4. 4

5. 3

6. 4

7. 4

8. 4

9. 2

10. 3

11. 4

12. 3

13. 2

14. 3

15. 1

16. 2

17. 2

18. 1

19. 2

20. 1

21. 2

22. 3

23. 3

24. 4

25. 2

26. 1

27. 1

28. 1

29. Plane R is perpendicular

to line n.

30.

31. If a dilation was

performed, the new

image would be similar

to the original. To

determine similarity,

I would calculate the

ratio of the shortest

side of one triangle to

the shortest side of the

other. Similarly, I would

calculate the ratio of

the largest sides and

fi nally of the third sides.

If these ratios are equal,

then a dilation was

performed.

32. Their slopes must be

equal.

33. y = 3x + 16

34. If a polygon is a

pentagon, then it has

fi ve interior angles. The

converse: If a polygon

has fi ve interior angles,

then it is a pentagon. The

converse is true because

part of the defi nition of

a pentagon is that it has

fi ve interior angles.

35. a. (x + 2) 2 + (y - 4) 2

= 169

b. y = - 12x ____ 5 -

4 __

5 ; the

slope of CP is - 12 ___

5 .

Use the slope and one

of the points to fi nd

the equation of the

line.

c. y = 5x ___ 12

- 37

___ 4 ; the

slope of the tangent

line is the negative

reciprocal of the slope

of CP. Using a slope

of 5 ___

12 and point P fi nd

the equation of the

line.

36.

(x - 3) 2 + (y - 4) 2 = 16

37. a. Yes, by defi nition, a

prism is a polyhedron

with two congruent

faces, called bases,

that lie in parallel

planes. By defi nition,

a polyhedron is a

solid that is bounded

by polygons, called

faces, that enclose a

single region of space.

The fi gure is bounded

by polygons and the

bases are congruent

and parallel.

b. No, the fi gure does not

have two congruent,

parallel bases.

38. (2, 1), (0, 11), (6, 3);

Lines Perpendicular to a

Plane pp. 1–2

1. 3

2. 1

3. 2

4. 4

5. 2

6. It is given that line AB is

perpendicular to plane S;

therefore, line AB is

perpendicular to every

line in the plane that

intersects it at point B.

Copyright © by Holt McDougal. 31 NY Regents Test Prep Workbook for Geometry



7. FALSE; RP cannot be

perpendicular to plane

W, because plane W

contains RP.

8. Plane R is perpendicular

to line n. If a line (n) is

perpendicular to each of

two intersecting lines

(l and m) at their point of

intersection (A), then the

line is perpendicular to

plane R.

9. Sample answer: Abby

can create 2 lines with

string leading away from

the birdbath that intersect

at the birdbath. Using the

level, she must determine

that the birdbath is

plumb, or perpendicular,

with each of the lines

where the line meets the

birdbath. If the birdbath is

plumb, or perpendicular,

with each of the lines,

then it is perpendicular

with the plane that

contains the two lines. If it

is perpendicular with the

plane, it is perpendicular

with every line in the

plane and would,

therefore, appear plumb

from any direction.

Plane Perpendicular to a

Given Line pp. 3–4

1. 2

2. 2

3. 1

4. 1

5. The diagram indicates

that HC is perpendicular

to GE. It is given that BC

is perpendicular to HC.

Plane P is the only plane

that exists through points

B, G, and E. Therefore,

plane P is the only plane

that exists through C

perpendicular to HC.

6. We cannot assume that

plane M is perpendicular

to line FG, since there

is nothing to indicate

a right angle at the

intersection of lines AB

and FG.

7. Statements (Reasons)

1. HC is perpendicular

to GE at C and HC is

perpendicular to BC at

C. (Given.)

2. Points B, G, and E lie

in plane P. (Given.)

3. GE and BC lie in plane

P. (Postulate 10)

4. Plane P is

perpendicular to HC at

C. (Defi nition)

5. Plane P is the only

plane perpendicular to

HC at C. (Postulate 8)

8. a. Two lines in the same

plane are either

parallel or intersect in

a point. Since AM and

BO do not intersect,

they must be parallel.

b. If lines AM and BO

are parallel, they

have the same slope.

Therefore, if AM is

perpendicular to line

D, then BO must be

perpen dicular to

line D.

Line Perpendicular to a

Given Plane pp. 5–6

1. 3

2. 4

3. 4

4. 1

5. 1


that GE is perpendicular

to HC. If there is a line

and a point not on the

line, then there is exactly

one line through the

point perpendicular to

the given line. Since

GE is the only line

that is perpendicular

to HC at C, GE is the

only line that can be


M at C.

7. First, MP is contained in

plane X and, therefore,

cannot be perpendicular

to plane X. Second,

there is nothing on the

diagram to indicate that

MP forms a right angle

with either of the lines it

intersects.

8. Statements (Reasons)

1. HC is perpendicular to

GE at C. (Given.)

2. Plane M contains HC.

(Given.)

3. HC is the only line

perpendicular to GE

at C. (Postulate 14)

4. GE is the only line


M at C. (2, 3)

9. Sample answers:

a. point B; b. AB; c. BX;

d. BX can be the only

line perpendicular

to plane M at point

B because there

is exactly one line

(BX) through point B

perpendicular to AB.

Coplanar Lines pp. 7–8

1. 4

2. 2

3. 3

4. 2

5. Yes. Through any three

noncollinear points, there

exists exactly one plane.

The plane containing the

parallel lines could be

parallel to the plane or

intersect

the plane.




6. Yes. Three non-linear

points determine a

unique plane.

7. No. No. It is possible

for two lines to lie in the

same plane and

not be perpendicular

to the same plane nor

intersect.

8. a. CD and AD, AD and

AB, AH and GH.

b. AB and CD, BC and

AD, AH and BG, AB

and HG, AD and HE,

AH and DE, DE and

CF, EF and DC, EF

and HG, FG and EH,

BC and GF, BG

and CF.

Perpendicular Planes

pp. 9–10

1. 1

2. 3

3. 2

4. 4

5. 1

6. AHG, DEF, CDA, and

FEH7. BAD and BFG

8. a. line HC

b. Yes. The diagram

indicates that AB

is perpendicular to

plane S. Since plane T

contains AB, plane S

is also perpendicular

to plane T.

9. Joshua must make sure

that a vertical edge of

the wall is perpendicular

to the fl oor. If we think

of a wall and the fl oor

as planes and the edge

of the wall as a line

within the wall plane,

and Joshua follows this

procedure with each

wall, the walls will be

perpendicular to the

fl oor since two planes

are perpendicular to

each other if and only

if one plane contains a

line perpendicular to the

second plane.

Perpendicular Lines within

Planes pp. 11–12

1. 4

2. 4

3. 1

4. Since AB is perpen-

dicular to plane S at B,

and HC is perpendicular

to AB at B, HC must be

in plane S.


that AB is perpendicular

to plane S at B.

Therefore, if AB is also

perpendicular to BC at B,

BC must be in plane S.

6. a. ML, NO, HK, and IJ

b. JK, KO, OM, and

MJ. If a line is

perpendicular to a

plane, then any line


given line at its point

of intersection with the

given plane is in the

given plane.

7. By defi nition, a prism is

a polyhedron with two

congruent faces, called

bases, that lie in parallel

planes. The other faces,

called lateral faces,

are parallelograms

formed by connecting

the corresponding

vertices of the bases. If

one edge of the prism

is perpendicular to its

base, then that edge

must be perpendicular

to the other base, since

the bases are parallel.

Since the lateral faces

are parallelograms, each

angle of intersection

between an edge and a

base must therefore be

a right angle. In a right

prism, each lateral edge

is perpendicular to both

bases, therefore, this

pentagonal prism is a

right pentagonal prism.

Lines Common to Planes

pp. 13–14

1. 2

2. 3

3. 4

4. 2


that LN is perpendicular

to plane KON. Since

planes MON and IHN

contain LN, they are also

perpendicular to KON.

6. No. Line k is not

contained in either plane

T or plane U.

7. a. HIJb. Plane HIJ does

not contain a line

perpendicular to

plane LMO.

8. a. the line AD

b. DEF and ABG

c. because both planes

ABC and AHE

contain AD which is

perpendicular to both

DEF and ABG

Parallel Lines pp. 15–16

1. 3

2. 2

3. 4

4. 1

5. AB ‖ HG and AH ‖ BG6. BC and AE




7. a. If the intersection is

two parallel lines, then

a plane intersects two

parallel planes.

b. No, as in the case

of a regular pyramid

with a square base.

The base (plane)

intersects two opposite

lateral faces (planes)

in two parallel lines.

However, the two

lateral faces converge

to a point and are not

parallel.

8. a. Yes, by defi nition, a

prism is a polyhedron

with two congruent

faces, called bases,

that lie in parallel

planes.

b. ADF

c. Yes, if a plane

intersects two parallel

planes, then the

intersection is two

parallel lines.

Parallel Planes pp. 17–18

1. 2

2. 4

3. 3

4. The bases of a right

cylinder are parallel

because the segment

joining the centers of the

bases is perpendicular

to the bases. Therefore,

the bases are perpen-

dicular to the same line

and thus, parallel.

5. ABC and EFG6. Parallel. Plane M

and plane X are

perpendicular to both

AB and CE and if two

planes are perpendicular

to the same line, they

are parallel.

7. a. If two planes are

parallel, they are


same line.

b. By defi nition, the

bases of a prism are

parallel. The defi nition

of an oblique prism

is a prism with lateral

edges that are not


bases. The bases are

parallel, but the lines

in the illustration are

not perpendicular to

the bases.

Lateral Edges of a Prism

pp. 19–20

1. 2

2. 1

3. 1

4. 1

5. 1

6. 27513.6 sq. ft

7. 52.79 cm 2

8. a. a right pentagonal

prism

b. S = 286.02 ft 2

9. At least two sides would

be longer or wider

than the corresponding

sides of the folded box.

Several sides have to

fold over to enclose

the box and make its

structure more rigid.

Prisms of Equal Volume

pp. 21–22

1. 3

2. 4

3. 2

4. 2

5. 4

6. The two stacks have

the same height and the

same cross-sectional

area at every level.

Therefore, by Cavalieri’s

Principle, the two stacks

have the same volume.

45 in. 3

7. 4. The altitudes are

equal, so the base areas

must be equal. The base

area of the prism on the

left is 1 __

2 (6 in + 12 in) ×

3 in = 27 in 2 . The value

of x is calculated

1 __

2 (x in + 14 in) × 3 in =

27 in 2 .

8. 35 ft 3

9. a. The volume of each

prism is 180 ��

3 cm 3 .

b. The volumes are

equal.

c. Prisms have equal

volumes if their bases

have equal areas

and their altitudes

are equal.

Volume of a Prism

pp. 23–24

1. 2

2. 4

3. 2

4. 1

5. 1

6. about 644.83 mm 3

7. 27.53 yd 3

8. 15 ft by 36 ft

9. a. 4500 in 3

b. 75 in 3

c. 20 rocks

Properties of a Regular

Pyramid pp. 25–26

1. 2

2. 1

3. 1

4. 4

5. 4

6. 16.37 ft 3

7. 12 in. 2

8. 288 ft 3




9. a. The volume doubles.

b. The volume is

multiplied by 4.

c. If you replace the

height h by 2h in the

volume formula, it will

multiply the volume

by 2. If you replace

the side length s by 2s

in the volume formula,

it will multiply the

volume by 4 because

(2s) 2 = 4s 2 .

Properties of a Cylinder

pp. 27–28

1. 4

2. 1

3. 4

4. 1

5. 1

6. 16.69 in

7. 13.27 cm

8. a. about 351.86 ft 2

b. doubling the radius

c. If the radius is 8 and

height is 10, then

S ≈ 904.78; if the

radius is 4 and

height is 20, then

S ≈ 603.19.

9. a. 22.6 ft 3 . The formula

for the volume of a

cylinder is V = πr 2 h =

π � 2.4 ___

2 � 2 · 5 ≈ 22.6.

b. about 169 gal.;

7.48 · 22.6 ≈ 169.

c. 6 times. Sample

answer: Since

approximately

169 gallons fi ll the

tank, it would take

6 times to amount to

1000 gallons.

Properties of a Right

Circular Cone pp. 29–30

1. 2

2. 1

3. 1

4. 2

5. 3

6. 1005.31 cm 3

7. about 1178 yd 2

8. about 28.44 mi 2

9. a. 5 in

b. 15π

c. 12π

Properties of a Sphere

pp. 31–32

1. 4

2. 3

3. 2

4. 4

5. 2

6. 137,942,558.20 mi 2

7. 0.75

8. a. 4.8 in

b. 9.6 in

c. about 144.76 in 2

9. a. about 10,306 cm 3

b. You would expect the

volume to be 8 times

as much because (2r) 3 = 8r 3

c. about 82,448; yes; 8:1

Construct a Bisector of an

Angle pp. 33–34

1. 4

2. AC and AB3. Yes, the radius should

be noticeably greater

than half the length of

segment CB. Otherwise,

you cannot see clearly

where the arcs intersect.

The arcs will not even

intersect if the radius is

less than half of CB.

4. 1. _

AB � _

AC , _

BG � _

CG (Given)

2. _

AG � _

AG (Refl exive

Property of Segment

Congruence)

3. ΔACG � ΔABG (SSS)

4. ∠CAG � ∠BAG

(Corr. parts of � Δs

are �.)

5. ___

› AG bisects ∠A.

(Defi nition of angle

bisector)

5.

A

EC

F

BD

By the construction,

AD = AE (radii of

same circle) and

DF = EF (arcs of equal

length). AF = AF. All

of these sets of equal

segments are also

congruent. We have

congruent triangles

by SSS. Since the

triangles are congruent,

any of their leftover

corresponding parts

are congruent which

makes angle BAF

equal (or congruent)

to angle CAF.

Construct the

Perpendicular Bisector

of a Segment pp. 35–36

1. 3

2. 2

3. 3

4. Yes. Triangle STU is

isosceles.




5.

A B

Step 1

A B

Step 2

A M B

Step 3

Sample answer: In

the construction of the

segment bisector four

congruent triangles are

created. In the process,

four pairs of congruent

angles are formed

which are right angles

making the bisector


segment.

6.

Label the bushes A, B,and C, as shown. Drawsegments AB and BC.

Step 1

A

B C

Step 2

A

B C

Draw the perpendicularbisectors of AB and BC.By Theorem 10.4, theseare diameters of the circle containing A, B,and C.

Step 3

sprinklerA

B C

Find the point wherethese bisectors intersect.This is the center of the circle through A, B, andC, and so it is equidistantfrom each point.

Construct Parallel or

Perpendicular Lines

pp. 37–38

1. 1

2. When two lines are cut

by a transversal and the

corresponding angles

are congruent, the lines

are parallel.

3. Theorem 6.6 guarantees

that parallel lines divide

transversals propor-

tionally. Since AD/DE =

DE/EF = EF/FG = 1

implies AJ/JK = JK/KL =

KL/LB = 1 means AJ =

JK = KL = LB.

4 a.

Q

P

2

3

4

1

B

D

n

�

C

b. Yes. If two parallel lines

are cut by a transversal,

the angle bisectors of

alternate interior angle

pairs are parallel.

Proof:

Statements (Reasons)

� is parallel to n (Given).

Angle AQP is congruent

to angle BPQ (Alternate

Interior Angles Theorem).

The measure of angle 1

plus the measure of

angle 2 equals the

measure of angle AQP,

the measure of angle 4

plus the measure of

angle 3 equals the

measure of angle

BPQ (Angle Addition

Postulate).


equals the measure of

angle 2, the measure

of angle 3 equals the

measure of angle 4

(Defi nition of angle

bisector).


plus the measure of

angle 2 equals the

measure of angle AQP,

the measure of angle 3

plus the measure of

angle 3 equals the

measure of angle BPQ

(Substitution).

Two times the measure

of angle 2 equals two

times the measure

of angle 3 (Transitive

Property of Equality).



angle 3 (Division

Property of Equality).

Angle 2 is congruent to

angle 3 (Defi nition of

congruent angles).

QC is parallel to PD

(Alternate Interior Angles

Converse Theorem).

Construct an Equilateral

Triangle pp. 39–40

1. 3

2. The construction

indicates that all sides

of the triangle are

congruent, or equal.

3.

O

C




4. Triangular pyramid. Four

equilateral triangles in

different planes with their

vertices connected.

Construct an Equilateral

Triangle pp. 41–42

1. 2

2. 3

3. 4

4. 2

5. Theorem 5.4 shows

you that you can fi nd a

point equidistant from

three points by using the

perpendicular bisectors

of the sides of the

triangle formed by the

three points.

6. Right. The orthocenter is

on the right angle.

7. a. Find the intersection

of the perpendicular

bisectors of the

triangle formed by the

three points.

b. approximately (7, 6.5)

8. Sample answer: the

midpoint of FG is L(3, 7),

so the equation of the

median from H(6, 1) to

L(3, 7) is y = -2x + 13.

P(4, 5) lies on this

median. The midpoint

of GH is J(5, 5), so the

equation of the median

from F(2, 5) to J(5, 5)

is y = 5. P(4, 5) lies on

this median, so all three

medians intersect at the

centroid.

Compound Loci pp. 43–44

1. 2

2. 4

3. 3

4. 3

5. 1

6. Let d be the distance

from R to k. The locus

of points is 4 points if

d < 1, 3 points if d = 1,

2 points if 1 < d < 3,

1 point if d = 3, and

0 points if d > 3.

7. Let d be the distance

from R to the perpen-

dicular bisector of PQ.

The locus of points is

2 points if d < 4, 1 point

if d = 4, and 0 points if

d > 4.

8.

angle bisector of ∠ABC

A

B C

9. The epicenter is at

about (–6, 1). Each

seismograph gives you a

locus that is a circle.

Circle A has center

(-5, 5) and radius 4.

Circle B has center

(-4, -3.5) and radius 5.

Circle C has center

(1, 1.5) and radius 7.

Draw the three circles in

a coordinate plane. The

point of intersection of

the three circles is the

epicenter.

x2

2

y

A

B

C

Graph Compound Loci

pp. 45–46

1. 1

2. 4

3. 3

4. 4

5. the points on the x-axis

with x-coordinate greater

than -4 and less than 4

6. 2 points, (2, 2) and (4, 4),

the intersections of y = x

with y = 2 and y = 4

7. (-1, 1)(3, 1)

8.

x

1

1

(3, 1)(–1, 1)

y

a. x = 3

b. (x - 2) 2 + (y - 2) 2 = 4

c. y = x and y = -x

Negation of a Statement

and Truth Value pp. 47–48

1. 1

2. 2

3. 2

4. 3

5. 3

6. No. It is false when the

hypothesis is true while

the conclusion is false.

7. a. Polygon ABCDE is not

both equiangular and

equilateral.

b. false

8. “Right triangles do not

have two acute angles”,

the negation is false.

Consider a 30°-60°-90°

triangle.

9. a. If you get caught

exceeding the speed

limit, then you will get

a speeding ticket.




b. If you get a speeding

ticket, then you will pay

higher insurance rates.

c. If you do not get

caught exceeding the

speed limit, then you

will not get a speeding

ticket, ~p → ~q.

d. If you do not get a

speeding ticket, then

you will not pay higher

insurance rates,~q → ~r.

e. true, false

Compound Statements

pp. 49–50

1. 3

2. 4

3. 2

4. 1

5. 1

6. False; observe the

drawing below:

C

AB

D

∠ACD and ∠BCD share _ CD in common, but they

are not adjacent angles.

7. True.

8. “If the fi gure is a

pentagon, then it has

six interior angles.”;

Hypothesis—“the

fi gure is a pentagon”;

Conclusion—“it has six

interior angles”; negate

the conclusion.

9. a. true

b. false; d = 1 mm (ash)

Inverse, Converse, and

Contrapositive of

Conditional Statements

pp. 51–52

1. 1

2. 2

3. 4

4. 4

5. 3

6. If a polygon is not a

quadrilateral, then it

does not have four sides.

7. If a polygon is a

quadrilateral, then it has

four sides.

8. If two angles are

complementary, then

they add to 90°. If two

angles add to 90°, then

they are complementary.

If two angles are not

complementary, then

they do not add to 90°.

If two angles do not add

to 90°, then they are not

complementary.

9. a. If two angles form a

linear pair, then they

are supplementary.

If two angles are

vertical, then they

are congruent.

If two adjacent

angles form a right

angle, then they are

complimentary.

b. If two angles are

supplementary, then

they form a linear pair.

This is true because

the defi nition of

supplementary angles

is that they form a linear

pair. If two angles are

congruent, then they

are vertical angles. This

is false because there

can be angles that are

congruent without being

vertical angles. If two

adjacent angles are

complimentary, then

they form a right angle.

This is true because

the defi nition of

complimentary angles

is that they form a right

angle.

c. Sample answer: If

an angle measures

less than 90°, then

it is an acute angle.

The converse is true

because the defi nition

of an acute angle is

that is measures less

than 90°.

Writing a Proof pp. 53–54

1. 4

2. 3

3. 2

4. 4

5. Symmetric Property of

Congruence

6. Transitive Property of

Congruence

7. Defi nition of midpoint,

Substitution Property

of Equality, Transitive

Property of Congruence

or Substitution Property

of Congruence

8. 1. D; 2. A; 3. F; 4. C; 5. G;

6. B; 7. E

Compound Statements,

pp. 55–56

1. 3

2. 2

3. 4

4. 3

5. 4

6. yes; AAS

7. angle F, angle L8. Sample answers:

a. Statements Reasons

1. _

AB ‖ _

CD , _

CB ‖ DE,

_

AB � _

CD

2. ∠ABC � ∠BCD,

∠BCD � ∠CDE

3. ∠ABC � ∠CDE

4. ∠BAC � ∠DCE

5. ABC � CDE

1. Given

2. Alt. Int.

Angles

Theorem

3. Transitive

prop. of

congruence

4. Corresp.

Postulate

5. ASA

Congruence

Postulate




b. Statements Reasons

1. _

AC � _

CE

2. AC = CE

3. C is midpoint

of A.

1. Def. of

congruent

triangles

2. Def. of �

3. Def. of

midpoint

9. Sample answers:


1. _

AB � _

AF , _

BD � _

FD

2. _

AD � _

AD

3. ADF � ADB

1. Given

2. Refl exive

Prop. of

Congruence

3. SSS

Congruence

Postulate


1. _

AD � _

AD

2. ADF � ADB

3. ∠DAF � ∠DAB

4. _

AB � _

AF , _

BC � _

FE

5. AB = AF,

BC = FE

6. AF + FE = AE,

AB + BC = AC

7. AF + BC = AC

8. AF + FE = AC

9. AE = AC

10. _

AE � _

AC

11. ACD � AED

1. Refl exive

Prop. of

Congruence

2. Proven in

part (a)

3. Def. of

congruent

triangles

4. Given

5. Def. of

congruent

segments

6. Segment

Addition

Postulate

7. Substitution

prop. of

equality

8. Substitution

prop. of

equality

9. Transitive

prop. of

equality

10. Def. of

congruent

segments

11. SAS

Congruence

Postulate

c. Statements Reasons

1. _

BC � _

FE , _

BD � _

FD

2. ACD � AED

3. _

CD � _

ED

4. BCD � FED

1. Given

2. Proven in

part (b)

3. Def. of

congruent

triangles

4. SSS

Congruence

Postulate

Corresponding Parts of

Congruent Triangles

pp. 57–58

1. 3

2. 4

3. 2

4. 1

5. 4

6. AB & CD, BG & DE,

GH & EF, HA & FC;

angle A & angle C,

angle B & angle D,

angle G & angle E,

angle H & angle F.

7. 25, 105°

8. D, A, B, C, A, E, E, B, C

9.


1. ∠1 � ∠3,

∠2 � ∠4

2. _

AG � _

AG

3. AGC � AGE

1. Given

2. Refl exive

Property of

Congruence

3. AAS

Congruence

Theorem


1. ∠2 � ∠4

2. ∠BCG � ∠FGE

3. _

GC � _

GE

4. BGC � FEG

1. Given

2. Vert. Angles

Thm.

3. Corresp.

parts of �

triangles

are �.

4. ASA

Congruence

Postulate

c. Statements Reasons

1. _

GD � _

GD

2. ∠AGC � ∠AGE,

_

GC � _

GE

3. m∠AGC = m∠AGE

4. m∠AGC + m∠CGD = 180°

m∠AGE + m∠EGD = 180°

5. m∠AGC + m∠EGD = 180°

6. ∠EGD � ∠CGD

7. CDG � ∠EDG

1. Refl exive

Prop. of

Congruence

2. Corresp.

parts of �

triangles

are �.

3. Def. of

congruence

4. Def. of linear

pair

5. Substitution

Prop. of

equality

6. Congruent

Supplements

Thm.

7. SAS

Congruence

Postulate

Sum of the Measures of

the Angles of a Triangle pp.

59–60

1. 4

2. 1

3. 1

4. 3

5. 4

6. Set 3x + 2x = 90 and

solve for x. Then fi nd the

values of 3x and 2x.

7. 65°

8. a. 2 ��

2x + 5 ��

2x +

2 ��

2x = 180

b. 40°, 100°, 40°

c. obtuse

9. Sample answer: They

both reasoned correctly

but their initial plan was

incorrect. The measure

of the exterior angle

should be 150°.

The Isosceles Triangle

Theorem and Its Converse

pp. 61–62

1. 4

2. 2

3. 4

4. 4

5. 2

6. The measure of angle B


angle C equals 26°.

7. 18

8. a. A, D; Base Angles

Theorem

b. A, BEA; Base Angles

Theorem

c. CD, CE; Converse of

Base Angles Theorem

d. EB, EC; Converse of

Base Angles Theorem




9. a. Statements (Reasons)

1. AB is congruent to

CD, AE is congruent

to DE, angle BAE is

congruent to CDB

(Given)

2. Triangle ABE is

congruent to triangle

DCE (SAS)

b. Triangle AED, triangle

BEC c. Angle EDA, angle

EBC, angle ECB d. No, triangle AED and

triangle BEC remain

isosceles triangles

with angle BEC equal

to angle AED.

Geometric Inequalities

pp. 63–64

1. 3

2. 2

3. 1

4. 3

5. 4

6. x > y; x > z7. 114°


that exterior angle of the

triangle has the same

measure as one of the

nonadjacent interior

angles, which cannot be.

9. 50°, 50°, 80°; 65°, 65°,

50°. There are two

distinct exterior angles.

If the angle is supple-

mentary to the base

angle, the base angles

measure 50°. If the

angle is supplementary

to the vertex angle,

then the base angles

measure 65°.

The Triangle Inequality

Theorem pp. 65–66

1. 2

2. 4

3. 2

4. 4

5. 2

6. Yes, because the sum of

any two sides is greater

than the third side.

7. No, because 3 + 6 is not

greater than 9.

8. 4 < P < 24; 2 < FG < 8,

1 < GH < 7, and

1 < HF < 9

9. a. The sum of the other

two side lengths is

less than 1080.

b. No. The sum of the

distance from Granite

Peak to Fort Peck

lake and Granite Peak

to Glacier National

Park must be more

than 565.

c. d > 76 km,

d < 1054 km.

d. The distance is less

than 489 kilometers.

Longest Side or Largest

Angle of a Triangle

pp. 67–68

1. 1

2. 3

3. 3

4. 1

5. 2

6. angle A, angle C,

angle B7. AB, BC, CA; angle C,

angle A, angle B8. a. AC, BC, AB b. BC, AB, AC

9.

20 ft

27 ft longest side

largest angle

24 ft

46° 59°

75°

The peak angle is

opposite the longest side

so, by Theorem 5.10,

the peak angle is the

largest angle. The angle

measures sum to 180°,

so the third angle

measure is 180° -

(46° + 59°) = 75°.

Lines Cut by a Transversal

pp. 69–70

1. 4

2. 3

3. 2

4. 2

5. 3

6. 38°

7. 142°

8. angles 1 & 2, 1 & 3,

1 & 4, 2 & 3, 2 & 4, and

3 & 4

9. Corresponding Angles

Postulate; 180°; measure

of angle 3 plus measure

of angle 2; Consecutive

Interior Angles Converse

Theorem.

Interior and Exterior Angles

of Polygons pp. 71–72

1. 3

2. 2

3. 3

4. 4

5. 3

6. 68

7. 77°

8. 2n. No. Only n angles

are considered.




9. 3 sides. Solve the

equation (n + x - 2) · 180 = 540 + (n - 2) · 180 for x where n is the

number of original sides

and x is the number of

sides added.


of Regular Polygons

pp. 73–74

1. 1

2. 1

3. 2

4. 4

5. 1

6. Interior angle: 120°;

exterior angle: 60°.

7. 1620°

8. 40°. The sum of the

measures of the exterior

angles is always 360°,

and there are nine

congruent external

angles in a nonagon.

9. AH

G B

CDE

PF

90°. The measure of each

interior angle is 135°

making the measure of

each exterior angle 45°.

Since the interior angles

of triangle BPC contain

two exterior angles and

angle BPC, the measure

of angle BPC equals 90°.

Parallelograms pp. 75–76

1. 3

2. 3

3. 4

4. 2

5. 2

6. x = 15, y = 8

7. (5, 1)

8. a. 3 in

b. 70°

c. It decreases. It gets

shorter. The sum

of the measures of

the interior angles

always is 360°. As the

measure of angle Q

increases, so does

the measure of angle

S. Therefore, the

measure of angle

P must decrease to

maintain the sum of

360°. As the measure

of angle Q increases,

the measure of angle

P decreases, moving

Q closer to S.

9. Sample answer: Given

that PQRS is a parallelo-

gram, you know that QR

is parallel to PS with

QP being a transversal.

By defi nition and the

fact that they are

consecutive interior

angles, angle Q and

angle P are supplemen-

tary by the Consecutive

Interior Angles Theorem.

So x° + y° = 180° by

the defi nition of supple-

mentary angles.

Special Parallelograms

pp. 77–78

1. 1

2. 3

3. 2

4. 2

5. 4

6. 4

7. 6

8. x = 50, y = 5

9. a. Angle ABD is

congruent to angle

CDB using the

Alternate Interior

Angles Congruence

Theorem: angle

ADB is congruent

to angle ABD; AB is

congruent to AD; and

ADB is congruent to

angle ABD using the

Transitive Property of

Angle Congruence

and AB is congruent

to AD using the

Converse of Base

Angles Theorem.

b. Rhombus; if AD is

parallel to BC, then

the quadrilateral is

a parallelogram by

defi nition. Using

the fact that opposite

sides of a parallelo-

gram are congruent

along with the fact that

AB is congruent to AD,

as shown in part (a),

means all four sides of

the parallelogram are

congruent. Therefore,

ABCD is a rhombus by

defi nition.


of Regular Polygons

pp. 79–80

1. 4

2. 3

3. 2

4. 2

5. 3

6. 8

7. 12, 36

8. DC = 2MN - AB since

MN = (AB + DC)

_________ 2

;

DC = 2(8) - 14, DC = 2.

9. 6, 8; 50; solve the

equation ( x 2 + 36)

________ 2

=

7x - 6, the solution

x = 6 must be rejected

because the midsegment

will equal 36 and that is

not possible.




Identify Special

Quadrilaterals pp. 81–82

1. 1

2. 4

3. 3

4. 3

5. 2

6. Parallelogram

7. Rectangle

8. In the fi rst case, the

diagonals bisect each

other. Therefore, quadri-

lateral WXYZ could be a

parallelogram, rhombus,

rectangle, or square. In

the second case, the

diagonals are congruent,

so WXYZ could be a

rectangle or a square.

9. a. 3 __

2

b. 1

c. DG is congruent to

EF because opposite

sides of a rectangle

are congruent. The

measure of angle

DGF equals the

measure of angle

EFG equals 90°

by defi nition of

a rectangle. The

measure of angle

EFC + the measure

of angle EFG = 180°

and the measure of

angle DGA + the

measure of angle

DGF = 180° because

linear angles are

supplementary. The

measure of angle

DGA + 90° = 180°

and the measure of

angle EFC + 90°

by the Substitution

Property of Equality.

The measure of angle

DGA = the measure

of angle EFC = 90°

by the Subtraction

Property of Equality.

The measure of angle

A = the measure

of angle B = the

measure of angle

C = 60° because

regular triangles have

3 congruent angles.

Triangle ADG is

congruent to triangle

CEF by AAS.

Midsegment Theorem

pp. 83–84

1. 4

2. 1

3. 4

4. 4

5. 3

6. 120

7. 22

8. Solve for x by 2x + 5 =

5x - 1. Substitute x = 2

into BC = 2x + 5 and

BC = 9. Similarly, AB = 9

and CA = 3. Since each

side of the exterior

triangle is twice the

length of the parallel side

of the interior triangle,

GJ = GH = 18

and HJ = 6, so, HJ is

the shortest side of

triangle GHJ.

9. Sample answer: A

midsegment of EN; the

length of the quarter-

segment will be 3 __

4 the

length of LN and the

length of the eighth-

segment will be 7 __

8 the

length of LN; triangle

LMN, midsegment XY,

quarter-segment DE, and

eighth-segment FG.

L(0, 0), M(a, b), and

N(c, 0) leading to

X( a __ 2 ,

b __ 2 ),

Y( (a + c)

______ 2 ,

b __ 2 ),

D( a __ 4 ,

b __ 4 ),

E( (a + 3c)

_______ 4 ,

b __ 4 ),

F( a __ 8 ,

b __ 8 ), and

G( (a + 7c)

_______ 8 ,

b __ 8 ),

LM = c, XY = c __ 2 ,

DE = 3c ___ 4 , and FG =

7c ___ 8

.

Centroid of a Triangle

pp. 85–86

1. 4

2. 4

3. 4

4. 4

5. 4

6. BP = 10, FP = 5

7. ( 10

___ 3 ,

10 ___

3 )

8. (4, 5) or two-thirds of the

distance from F to the

midpoint of GH.

9. a. (4, 2) Sample answer:

The equation of the

line passing through

midpoint (1, 5) of JK

and L with slope -1

is y = -x + 6. The

equation of the line

passing through the

midpoint (4, -1) of JL

and K with undefi ned

slope is x = 4. The

intersection is the

centroid. 4, 2. They

are the same as the

x- and y-coordinates

of the centroid.




b. mean of the

x-coordinate found

by [(-2) + 4 + 10]

_____________ 3 =

12

___ 3 = 4; mean of the

y-coordinate found by

[2 + 8 + (-4)]

____________ 3 =

6 __

3 = 2

c. Yes. The centroid of

triangle JKP is (1, 3).

The mean of the

x-coordinates is 1

and the mean of the

y-coordinates is 3.

Establish Similarity of

Triangles pp. 87-88

1. 1

2. 3

3. 3

4. 2

5. 1

6. Similar. Triangle ABC is

similar to triangle EDC.

7. Angle RSQ is congruent

to angle UST by the

Vertical Angles Theorem

and it was given that

angle Q is congruent to

angle T making triangle

SQR similar to triangle

STU using the AA

Similarity Theorem.

8. Similar. Triangle DEF is

similar to triangle SRT.

3 __

5

9. The two right triangles

formed by the altitudes

and the two sides

measuring a and b

are similar by the AA

Similarity Postulate.

Since the ratio of

the hypotenuses is b __ a ,

then the ratio of

corresponding sides,

which are the altitudes

of the original triangles

is the same ratio by

Corresponding Lengths

in similar Polygons.

Similar Triangles pp. 89–90

1. 4

2. 4

3. 4

4. 2

5. 4

6. No. The ratio of

corresponding sides

would be the same

but they would

not necessarily be

congruent.

7. You would need to

know that one pair of

corresponding sides is

congruent.

8.

x

y

Since the two triangles

are similar, the ratio of

corresponding sides are

the same. Therefore,

compare the vertical rise

to the horizontal run.

9. Sample answer:You can

use the AA Similarity

Postulate if you know

that two pairs of

corresponding angles are

congruent. If you know

that the three pairs of

corresponding sides are

proportional, you can

use the SSS Similarity

Theorem. If you know

that one pair of angles

are congruent and that

the sides including these

angles are proportional,

you can use the SAS

Similarity Theorem.

Use Proportionality

Theorems pp. 91–92

1. 4

2. 4

3. 4

4. 4

5. 2

6. 2

7. 40

8. Parallel; 8

__ 5

= 12

___ 7.5

,

so the Converse of the

Triangle Proportionality

Theorem applies.

9. a. about 4.7 cm

b. Sample answer: The

line connecting the top

left to the bottom left of

Car 1 is parallel to the

line connecting the top

left to the bottom left

of Car 2.

The triangle with

vertices consisting of

the vanishing point,

the top left of Car 1,

and the bottom left of

Car 1 is similar to the

triangle with vertices

consisting of the

vanishing point, the

top left of Car 2, and

the bottom left of

Car 2.

c. about 4.3 cm

Mean Proportionality

Theorems pp. 93–94

1. 4

2. 1

3. 3

4. 4

5. 1

6. 1

7. 2






8. a. FH, GF, EF. Each

segment has a vertex

as an endpoint and is


opposite side.

b. ��

35

c. about 35.5

9. a. 9.8 ft

b. 17.8 ft

c. 12.3 ft

d. triangle HTG is similar

to triangle AHG is

similar to triangle ATH.

Pythagorean Theorem and

Its Converse pp. 95–96

1. 4

2. 4

3. 2

4. 3

5. 4

6. 615 m 2

7. No. (2y) 2 + (4y) 2 ≠

(4 ��

5 y) 2

8. 36 and 60 have a

common factor of 12.

The Pythagorean triple

3, 4, 5, can be multiplied

by 12 to get 36, 48, 60.

So, the third side could

be 48 ft.

9. 0.72 mi. Since the block

is rectangular, halfway

through the bike ride,

I am at the opposite

corner. The path that I

rode forms two legs of

a right triangle and the

distance from my house

forms the hypotenuse.

The fi rst leg is 0.4 mile.

The second leg is found

by solving.

2(0.4 mi) + 2(d2nd leg

) =

2 mi. The second leg is

0.6 mile. The distance

from my house is

then solved using the

Pythagorean Theorem.

Chords of Circles pp. 97–98

1. 4

2. 3

3. 2

4. 4

5. 3

6. Diameter. AB bisects

and is perpendicular to

chord CD.

7. 5.83

8. a. x = 1.75

b. 80°

9. a. 100°

b. 245°

c. x = 3

d. 8

Tangent Lines to a Circle

pp. 99–100

1. 4

2. 3

3. 4

4. 1

5. 2

6. r = 4.5

7. They are perpendicular.

8. Yes.

9.

a. 4 __

3

b. y = 4x ___ 3 +

25 ___

3

c. 5

d. 10

___ 3

Angle Relationships in

Circles pp. 101–102

1. 2

2. 3

3. 4

4. 1

5. 4

6. 95°

7. 32°

8. The measure of angle

LPJ is less than or

equal to 90°. If PL is

perpendicular to KJ at K,

then the measure of LPJ

equals 90°. Otherwise,

it would measure less

than 90°.

9. 76°, 104°. The handle

and ground form an

angle outside the circle.

Let one arc measre x°

and the other arc

measure (180 - x)°. Use

Theorem 10.13 to fi nd

each arc length.

Arcs of a Circle Cut by Two

Parallel Lines pp. 103-104

1. 2

2. 3

3. 4

4. 4

5. 2

6. 95°

7. 50°

8. about 16.3°

9. 100°. AB is parallel to

CD, so the measure

of arc AC equals the

measure of arc BD

equals 25°. Since

CE = DE, the measure

of arc CE = the measure

of arc DE = 105°. If we

total the known number

of degrees, we have

105 + 105 + 25 + 25 =

260. 360 - 260 = 100.

The measure of arc AB

is then 100°.




Segment Lengths in

Circles pp. 105–106

1. 1

2. 2

3. 4

4. 4

5. 4

6. 46 ft

7. 8

8. 4 cm/sec. By the

Segments of Chords

Theorem, 6 × 8 = 4 ×

CN, so CN = 12 cm. It

takes 3 sec. for sparkles

to move the 6 cm from

C to D, so the sparkles

need to travel 12 cm in

3 sec. or 4 cm/sec.

9. a. 60°

b. Using the Vertical

Angles Theorem,

angle ACB is

congruent to angle

FCE. Since the

measure of angle

CAB = 60° and the

measure of angle

EFD = 60°, then angle

CAB is congruent

to angle EFD. Using

the AA Similarity

Postulate, triangle

ABC is similar to

triangle FEC.

c. y __

3 =

(x + 10) _______

6 ;

y = (x + 10)

_______ 2 .

Isometries in the Plane

pp. 107–108

1. 1

2. 3

3. 1

4. 3

5. 4

6. (6, -3)

7. J'(2, -2), K'(1, -5), and

L'(5, -3)

8. a. 5 squares to the right

followed by 4 squares

down

b. about 12.8 mm

c. about 0.522 mm/sec.

9. a. (12, 4)

b. (4, -2)

c. (2, 3)

Properties that Remain

Invariant under Isometries

pp. 109–110

1. 3

2. 4

3. 2

4. 3

5. 4

6. (x, y)→(x + 3), (y - 6)

7. (x, y)→(x - 3), (y + 6)

8. The distance from X to

line m is equal to the

distance from X ' to

line m.

9. a. (x, y)→(x + 6), (y - 2)

b. y = (x - 7) 2 + 1

c. y = (x + 3) 2 + 11

d. (x, y)→(x - 6), (y - 6)

The graph is moved

left 6 units and down

6 units.

Orientation, Numbers

of Invariant Points, and

Parallelism in Isometries

pp. 111–112

1. 2

2. 4

3. 3

4. 4

5. 3

6. (-4, 3)

7. r = 5, s = 8, t = 5,

w = 54

8. A' (1, 9), B' (-4, 7),

C' (-1, 10)

9. a. A'(1, -4);

b. A' B'B'(5, -6), C'(3, -1)

x

1

A´

B´

C´1

y

c. A"(-1, 4), B"(-5, 6),

C"(-3, 1)

x

1

A˝

B˝

C˝1

y

d. A'"(3, 4), B'"(-1, 6),

C'"(1, 1)

A´˝

B´˝

C´˝x1

1

y

e. (1, 0)

Geometric Relationships

and Transformations

pp. 113–114

1. 3

2. 4

3. 3

4. 4

5. 2

6. It preserves length and

angle measure.

7. 3.2 cm. They are

perpendicular.

8. Refl ect the object across

two parallel lines, and

then refl ect it across a

third line perpendicular

to the fi rst two lines.




9. a. triangle A"B" C"

b. line k and line m

c. sample answer: AA',

AA"

d. 5.2 in.

e. Yes. Defi nition of

refl ection of a point

over a line.

Dilations pp. 115–116

1. 2

2. 3

3. 2

4. 1

5. 4

6. The scale factor of a

reduction is between

0 and 1.

7. The scale factor of an

enlargement is greater

than 1.

8. a. 6 __

1

b. 8.75 in

9. C' (-2, -2), D' (-2, 2)

Invariant Properties of

Similarities pp. 117–118

1. 2

2. 4

3. 4

4. 3

5. 3

6. No. The lengths are not

proportional.

7. 11.4

8. If a dilation was

performed, the new

image would be similar

to the original. To

determine similarity,

I would calculate the

ratio of the shortest

side of one triangle to

the shortest side of the

other. Similarly, I would

calculate the ratio of the

largest sides and fi nally

of the third sides. If these

ratios are equal, then a

dilation was performed.

9. 9

8

7

6

5

4

A

B

CD

PN

M

L

1

1 2 3 5 7 8 9

y

x

Similarity Transformations

pp. 119–120

1. 3

2. 3

3. 2

4. 2

5. 1

6. No. Dilation does not

preserve length.

7. 25

___ 2 . Yes. The center

point is (0, 0) with scale

factor 25

___ 2 .

8. O

y

x

Perspective drawings

use converging lines

to give the illusion that

an object is three-

dimensional. Since the

back of the drawing is

similar to the front, a

dilation can be used to

create this illusion with

the vanishing point as

the center of dilation.

9. a. enlargement

b. 3 __

2 , fi nd A' B'/AB =

3 �

� 5 ____

2 ��

5 =

3 __

2

Because the dilated

triangle is larger than the

original, the dilation is an

enlargement. Each

coordinate of the original

triangle is multiplied by

3 __

2 to get the coordinates

of the dilated triangle.

Thus, the scale factor

is 3 __

2 .

Analytical Representations

of Transformations

pp. 121–122

1. 1

2. 4

3. 1

4. 3

5. [ -4 -3 -6

0 5 6]

6. [ 3 6 - 9 __ 2

- 3 __ 2

3 __

4 3

] 7. [4 6 3 4

1 3 2 8]

8. [-1 0

0 1 ]

[3 4 6

1 3 2] [1 0

0 -1

] [ 3 4 6

-1 -3 -2

] Slope of Perpendicular

Lines pp. 123–124

1. 4

2. 2

3. 1

4. 1

5. 2

6. -2

7. 1 __

3

8. Slope = - 2 __ 3

y

x

1

1

h

(2, 5)

–2

3(5, 3)

(3, 0)

(7, 6)




9. a. 1 __

2

b. y = � 1 __ 2 � x + 2

yk

j

x

1

2

y = 1–2

x + 2

y = –2x + 2

(2, 3)

Use a protractor to

measure one of the

angles formed by the

lines.

Equations of Parallel and

Perpendicular Lines

pp. 125–126

1. 2

2. 3

3. 4

4. 2

5. 1

6. Their slopes must be

equal.

7. Perpendicular. The

product of their slopes

is -1.

8. a. y = ( 2 __

3 ) x - 1

b. y = -x + 3

c. y = 3x + 2

d. Sample answer: x = 4

is a vertical line and

y = 2 is a horizontal

line.

9. a.

x

y

1

1

B(5, 2)

A(2, 1)

b. y = 1 __

3 x +

1 __

3

c. y = -3x + 7

d. y = -3x + 17

e. m = 1 __

3

Finding Equations, Given a

Point and a Perpendicular

Line pp. 127–128

1. 2

2. 4

3. 1

4. 1

5. 2

6. y = � 1 __ 4 � x + 1

7. y = � - 1 __ 3 � x + 6

8. a. y = � - 1 __ 6 � x -

1 __

2

b. y = � 7 __ 4 � x + 3

1 __

2

9. a.

x

y

–2

–2

–4

–6

–8

–4–6–8

2

2 4 6 8

4

6

8

b. No.

c. y = 3x + 13

Finding Equations, Given

a Point and a Parallel Line

pp. 129–130

1. 3

2. 3

3. 3

4. 3

5. 1

6. y = � - 1 __ 3 � x + 9

7. y = 2x - 1

8. a. y = 6x - 19

b. y = � - 4 __ 7 � x + 1

1 __

7

9. a.

x

y

–2

–2

–4

–6

–8

–4–6–8

2

2 4 6 8

4

6

8

y = — 1–3

x + 2

b. No.

c. y = � - 1 __ 3 � x + 3

Use the Midpoint Formula

pp. 131–132

1. 4

2. 2

3. 3

4. 4

5. 2

6. � 11 ___

2 , 5 �

7. 5

8. when x 2 and y

2 are

replaced by zero in the

Midpoint Formula and

x 1 and y

1 are replaced

by m and n, the result is

� m __ 2 ,

n __ 2 �

9. (4, 7, 3)

Use the Distance Formula

pp. 133–134

1. 4

2. 2

3. 3

4. 2

5. 4

6. 6.7 units

7. ��

29

8. Yes; they have the same

length of 4.5 units.

9. a.

x

y

–2

–2

–4

–6

–8

–4–6–8

2

2 4 6 8

4

6

8

A

B

C D

b. Yes

c. ��

10 . The difference

in the coordinates

of the endpoints are

the same for both

segments.

Perpendicular Bisectors

pp. 135–136

1. 1

2. 3

3. 2

4. 3

5. y = - 2 __ 3 x +

19 ___

3




6. y = - 3 __ 2

x + 13

___ 4

7. No. Sample answer: The

midpoint of the line

is (5, 3 __

2 ). When we plug

the x-value into the

equation, the result is 10

and not 5. Therefore, the

line is not a bisector of

the line segment.

8. a. (1, 1)

b. 5 __

7

c. - 7 __ 5

d. y - 1 = (- 7 __ 5 )(x - 1)

e. y = -7x ____

5 +

12 ___

5

Triangles and Quadrilaterals

in the Coordinate Plane

pp. 137–138

1. 1

2. 1

3. 3

4. 3

5. 2

6. (-3, 2). Since DA

must be parallel and

congruent to BC, use

the slope and length of

BC to fi nd point D by

starting at point A.

7. (-5, -3). Since AB

must be parallel and

congruent to CD, use

the slope and length of

AB to fi nd point D by

starting at point C.

8. PQ and QR are not

opposite sides. PQ

and RS are opposite

sides, so they should be

parallel and congruent.

The slope of PQ is

(4 - 2)

}}}} (3 - 2)

= 2. The slope of

RS is (5 - 3)

}}}} (6 - 5)

= 2.

They are parallel, so

PQRS is a parallelogram.

9. AB = ��

(p2 + q2) , q

__ p ,

� p __ 2 ,

q __

2 � .

BC = ��

(p2 + q2) , -q

___ p ,

� 3p ___

2 ,

q __

2 � .

CA = 2p, 0, (p, 0). No.

Yes. It’s not a right

triangle because none

of the slopes are nega-

tive reciprocals, and it

is isosceles because

two of the sides are the

same length.

Solving Systems of

Equations Graphically

pp. 139–140

1. 3

2. 4

3. 3

4. 1

5. 1

6. 2

7. 1

8. (-2, 1), (1, 4)

9. No solution;

4

6

2

0 1 2 3 4

–2

–1–2–3

–4

–6

Writing Equations of

Circles with Center and

Radius pp. 141–142

1. 4

2. 1

3. 1

4. 4

5. 1

6. (x - 2) 2 + (y - 3) 2 = 49

7. (x - 2) 2 + (y - 2) 2 = 9

8. If (h, k) is the center of

a circle with a radius r, the equation of the circle

should be (x - h)2 +

(y - k) 2 = r 2 . (x + 3) 2 +

(y + 5) 2 = 9.

9. Tower A = (x - 0) 2 +

(y - 0) 2 = 9. Tower B =

(x - 5) + (y - 3) 2 = 6.25.

Tower C = (x - 2) 2 +

(y - 5) 2 = 4

Writing Equations of

Circles with the Graph

pp. 143–144

1. 4

2. 3

3. 4

4. (x + 3) 2 + (y - 2) 2 = 4

5. x 2 + (y - 1) 2 = 4

6. a. (x + 3) 2 + y 2 = 1

b. (x - 3) 2 + y 2 = 49

7. a. The distance from the

center to the known

point is the radius of

the circle.

b. (x - 2) 2 + (y - 3) 2 = 4.

Finding the Center and

Radius of a Circle

pp. 145–146

1. 3

2. 4

3. 1

4. 2

5. 3

6. (4, -2), 5

7. (5, 3), 4

8. a. False, The center is

(-4, 1).

b. False. The radius is

3 ��

3 .




9. a.

2

2

4

4

6

6

8

8

x

y

b. ( x - 3) 2 + ( y - 4) 2 = 16

c. (3, 4), 4

Graphing Circles

pp. 147–148

1. 4

2. 1

3. 3

4.

2 x

y

–2

(1, –3)

5.

1

1 x

y

(2, 7)

6.

2

2

x

y Pigpen

Silo

House

Barn

(6, 7)

7. a. - 1

b. (x + 1) 2 + (y - 2) 2 = 9

c.

x

y

1

2




Practice Test pp. 149–156

1. 4

2. 4

3. 3

4. 2

5. 2

6. 3

7. 4

8. 1

9. 2

10. 2

11. 1

12. 4

13. 3

14. 4

15. 3

16. 3

17. 1

18. 2

19. 4

20. 3

21. 4

22. 4

23. 4

24. 4

25. 2

26. 1

27. 2

28. 1

29.

30. 2 ��

262

31. y = � 1 __ 2 � x + 2

32. 1 __

2 . Yes, use the SSS

Similarity Theorem.

33.

34. CD and EF, CF and DE 35. 180 - x, 180 - x,

2x - 180; x __

2 ,

x __

2 , 180 - x;

0 < x < 180

36. 160°

37. BC =

[AD · (AD + DE)] -AB2

}}}}}}}}}}} AB

38. Sample answer: Draw

AB with length 1 in.

Open compass to 1 inch

and draw a circle with

that radius, continue with

this setting and mark off

equal parts on the circle.

Connect 2 consecutive

points with the center of

the circle.


Copyright © by Holt McDougal. 51 NY Regents Test Prep Workbook

All rights reserved. for Algebra 2 and Trigonometry

New York Regents Test Prep

Workbook for Algebra 2/

Trigonometry

Diagnostic Test pp. 1–5

1. 2

2. 2

3. 3

4. 2

5. 3

6. 2

7. 4

8. 3

9. 1

10. 1

11. 1

12. 1

13. 1

14. 4

15. 4

16. 2

17. 2

18. 3

19. 3

20. 4

21. 4

22. 4

23. 2

24. 4

25. 3

26. 3

27. 2

28. two complex roots

29. Using the Law of Sines,

a ____

sin A =

b ____ sin B

, or

12 _____

sin 96 =

9 ____

sin B , so

sin B ≈ .746, and

m∠B ≈ 48°. The sine

function is also positive

in Quadrant 2. Using

reference angle 48°, m∠B = 132° in Quadrant

2. Since this would

cause the angle sum of

the triangle to be greater

than 180°, this is not a

solution.

30. The probability of

selecting a red marble

followed by a purple

marble is 5 ___

16 · 3 ___

15 =

1 ___

16 .

The probability of

selecting a green marble

followed by a black

marble is 6 ___

16 · 2 ___

15 =

3 ___

40 .

Since 3 ___

40 >

1 ___

16 ,

selecting a green marble

followed by a black

marble is more likely.

31. Solve the equation for y.

3x 2 - y = 1

-y = - 3x 2 + 1

y = 3x 2 - 1

The equation in function

notation is f(x) = 3x 2 – 1.

32. sin θ = y __ r =

3 ____

��

13 =

3 ��

13 _____

13

cos θ = x __ r =

-2 ____

��

13

= - 2 �

� 13 _____

13

tan θ = y __ x = -

3 __

2

33. log12

4 = log 4

_____ log 12

≈ 0.602

_____ 1.079

≈ 0.558

34. x 2 + y 2 - 4x + 6y

+ 4 = 0

( x 2 - 4x) + ( y 2 + 6y)

= -4

( x 2 - 4x + 4) + ( y 2 + 6y

+ 9) = -4 + 4 + 9

(x - 2) 2 + (y + 3) 2 = 9

35. log5125 = 7x

125 = 5 7x 5 3 = 5 7x 3 = 7x

x = 3 __

7 ≈ 0.43

36. csc 0° is undefi ned.

Using reciprocal

relationships,

csc 0° = 1 _____

sin 0° =

1 __

0 ,

which is undefi ned.

37. Since the editor is

placing the articles

in specifi c spots in

the magazine, the

order of the articles is

important. The number

of permutations of 6

objects chosen 4 at a

time is

6P

4 =

6! _______

(6 - 4)!

= 6!

__ 2!

= 360

38. P = 500(1.03)t; 633 elk

39. a. Use a graphing

calculator to fi nd the

power regression

model to fi nd a

regression equation of

y = 1.13x 2.30

b. If x = 21.5,

y = 1.13(21.5) 2.30

≈ 1311.25

c. If x = 142.4,

y = 1.13(142.4) 2.3

= 101426.5.

This value is not reliable.

It is extrapolated, and

the x-value is far from the x-values in the table.

Negative and Fractional

Exponents pp. 7–8

1 . 3

2. 4

3. 1

4. 3

5. 3

6. � 1 __ 8 � -3

= 8 3 = 8 · 8 · 8

= 512

7 . � 1 __ 8 � -2/3

= 8 2/3 = ( 8 1/3 ) 2

= 22 = 4

8. ( 6 -2 )( 8 1/3 ) � 1 __ 3

� -3

= � 1 __ 6 � 2 3 ��

8 (3) 3

= 1 ___

36 (2)(27) =

3 __

2




9. a. � 52

__ 34

� -3/2

= � 25 ___

81 � -3/2

= � 81 ___ 25

� 3/2

= � ��

81 ___ 25

� 3 = � 9 __

5 � 3

= 729

____ 125

b. � 52

__ 34

� -3/2

= (52)-3/2

_______ (34)-3/2

= 5-3

___ 3-6

= 36

__ 53

= 729 ____

125

Operations with Radicals

pp. 9–10

1. 2

2. 1

3. 4

4. 3

5. 1

6. 3 ��

40 - 3 ��

81

= 3

��

23 · 5 - 3

��

33 · 3

= 3

��

23 · 3 ��

5 - 3

��

33 · 3 ��

3

= 2 3 ��

5 - 3 3 ��

3

7. 3 ��

25 3 ��

20 = 3 ��

25 · 20

= 3 ��

500

= 3

��

53 · 4

= 3

��

53 · 3 ��

4

= 5 3 ��

4

8. ( ��

125 - ��

200 + ��

72 )

( ��

5 )

= ��

125 · ��

5

- ��

200 ��

5 + ��

72 ��

5

= ��

125 · 5 - ��

200 · 5

+ ��

72 · 5

= ��

625 - ��

1000

+ ��

360

= ��

(25)2 - ��

102 · 10

+ ��

62 · 10

= 25 - ��

10 2 ��

10

+ ��

62 ��

10

= 25 - 10 ��

10 + 6 ��

10

= 25 - 4 ��

10

9. a. a + b = ��

3 + 2 ��

5

+ ��

3 - 2 ��

5

= 2 ��

3

b. a - b = ��

3 + 2 ��

5 -

( ��

3 - 2 ��

5 )

= ��

3 + 2 ��

5

- ��

3 + 2 ��

5

= 4 ��

5

c. ab = ( ��

3 + 2 ��

5 )

( ��

3 - 2 ��

5 )

= ( ��

3 ) 2 - 2 �

� 3 �

� 5

+ 2 ��

3 ��

5

- (2 ��

5 ) 2

= 3 - 2 ��

15

+ 2 ��

15 - 20

= -17

d. a2 = ( ��

3 + 2 ��

5 ) 2

= ( ��

3 ) 2 + 2( �

� 3 )

(2 ��

5 ) + (2 ��

5 ) 2

= 3 + 4 ��

15 + 20

= 23 + 4 ��

15

Operations with

Polynomial

Expressions pp. 11–12

1. 2

2. 3

3. 2

4. 1

5. 1

6. (5y - 3)(5y + 3)

= (5y) 2 - 3 2

= 25y 2 - 9

7. (3x 2 - 5x + 2)

- (x 2 - 3x + 4)

= 3x 2 - 5x + 2 - x 2 + 3x - 4

= 3x 2 - x 2 - 5x + 3x

+ 2 - 4

= 2x 2 - 2x - 2

8. (x 2 + 3)(x 2 - 3)(x + 4)

= ((x2) 2 - 3 2 )(x + 4)

= (x 4 - 9)(x + 4)

= x 4 (x) + 4x 4 - 9x - 9(4)

= x 5 + 4x 4 - 9x - 36

9. a. c = b - a

= (x 2 + 3x)

- (2x 2 - x + 5)

= x 2 + 3x - 2x 2 + x - 5

= x 2 - 2x 2 + 3x

+ x - 5

= - x 2 + 4x - 5

b. d = ab

= (2x 2 - x + 5)

(x 2 + 3x)

= 2x 2 (x 2 + 3x)

- x(x 2 + 3x)

+ 5(x 2 + 3x)

= 2x 4 + 6x 3 - x 3 - 3x 2 + 5x 2 + 15x

= 2x 4 + 5x 3 + 2x 2 + 15x

Irrational Expressions

pp. 13–14

1. 3

2. 1

3. 4

4. 2

5. 1

6. (2e - 1)(2e + 1)

= (2e)(2e) + (-1)(2e)

+ (2e)(1) + (-1)(1)

= 4e 2 - 1

7. (π 2 - 3π + 3e + 4)

- (-2π 2 + 3e)

= π 2 - 3π + 3e + 4

+ 2π 2 - 3e

= π 2 + 2π 2 - 3π + 3e -

3e + 4 = 3π 2 - 3π + 4

8. (π + 1)(π + 2)(π + 3)

= (π 2 + 2π + π + 2)

(π + 3)

= (π 2 + 3π + 2)(π + 3)

= π 3 + 3π

2 + 3π 2

+ 9π + 2π + 6

= π 3 + 6π 2 + 11π + 6




9. a. f(π)g(π) = (4π + 1)

(4π - 1) = 16π 2 - 1

The result is irrational.

b. f( ��

3 )g( ��

3 )

= (4 ��

3 + 1)(4 ��

3 - 1)

= 16( ��

3 ) 2 -1 = 16(3) -1

= 47

The result is rational.

Rationalizing

Denominators pp. 15–16

1. 3

2. 4

3. 2

4. 3

5. 1

6. 12

___ 3 ��

3 =

12 3 ��

9 ______

3 ��

3 3 ��

9 =

12 3 ��

9 _______

3 ��

27

= 12

3 ��

9 ______

3 = 4

3 ��

9

7. 2 - �

� 5 _______

4 - ��

5

= (2 - �

� 5 )(4 - �

� 5 ) ______________

(4 - ��

5 )(4 + ��

5 )

= 8 + 2 �

� 5 - 4 �

� 5 - 5 __________________

16-5

= 3 - 2 ��

5 ________ 11

= 3 ___ 11

- 2 ��

5 _____ 11

8. 4 _______________ (2 + �

� 5 )(3 - �

� 2 )

= 4(3 + �

� 2 ) ______________________

(2 + ��

5 )(3 - ��

2 )(3 + ��

2 )

= 4(3 + �

� 2 ) _____________

(2 + ��

5 )(9 - 2)

= 4(3 + �

� 2 ) _________

7(2 + ��

5 )

= 4(3 + �

� 2 )(2 - �

� 5 ) ________________

7(2 + ��

5 )(2 - ��

5 )

= 4(6 - 3 �

� 5 + 2 �

� 2 - �

� 10 ) ______________________

7(4 - 5)

= 4(6 - 3 �

� 5 + 2 �

� 2 - �

� 10 ) ______________________

-7

= - 24 ___ 7 +

12 ��

5 _____

7 - 8 �

� 2 ____

7

+ 4 �

� 10 _____

7

9. a. 1 ______ 1 +

3 ��

3

= 1(1 -

3 ��

3 + 3 ��

9 ) ___________________

(1 + 3 ��

3 )(1 - 3 ��

3 + 3 ��

9 )

= 1 -

3 ��

3 + 3 ��

9 ___________

1 + 3

= 1 - 3 ��

3 + 3 ��

9 __________

4

= 1 __ 4 -

3 ��

3 ___

4 +

3 ��

9 ___

4

b. 1 ______ 1-

3 ��

3

= 1(1 +

3 ��

3 + 3 ��

9 ) ____________________

(1 - 3 ��

3 ) (1 + 3 ��

3 + 3 ��

9 )

= 1 +

3 ��

3 + 3 ��

9 ____________

1 - 3

= 1 +

3 ��

3 + 3 ��

9 ____________

-2

= - 1 __ 2 -

3 ��

3 ___

2 -

3 ��

9 ___

2

Square Roots of Negative

Numbers pp. 17–18

1. 4

2. 3

3. 1

4. 2

5. ��

-525

=

��

-1 · 52 · 21

= ��

-1

��

52 ��

21

= 5i ��

21

6. ��

-72 = ��

-1 · 62 · 2

= ��

-1 ��

62 ��

2

= 6i �� 2

7. -3 ��

-8 + 5 ��

-45

= -3 ��

-1 · 22 · 2

+ 5 ��

-1 · 32 · 5

= -3 ��

-1 ��

22 ��

2

+5 ��

-1 ��

32 ��

5

= - 6i ��

2 + 15i ��

5

8. x 2 + 500 = 0

x 2 = -500

x = ± ��

-500

x = ± ��

-1 · 102 · 5

x = ± ��

-1 ��

102 ��

5

x = ±10i ��

5

Powers of i pp. 19–20

1. 1

2. 4

3. 2

4. 4

5. 3

6. i 232 = i 4·58 = (i 4 ) 58 = 1 58 = 1

7. -i 214 = -i 4 · 53 + 2

= -(i4)53i 2 = -1 53 · -1

= 1

8. i 231 · i 65 = i 231+ 65 = i 296

= i 4 · 74 = (i 4) 74 = 1 74 = 1

9. i -1 = 1 __ i = i3 ____

i · i3 = i

3

__ i4

= - i __ 1 = -i;

i -2 = 1 __ i2 = 1 ____

-1 = -1;

i -3 = 1 __ i3 = i ____

i3 · i = i __

i4

= i __ 1 = i;

i -4 = 1 __ i4 = 1 __

1 = 1

Operations on Complex

Numbers pp. 21–22

1. 3

2. 2

3. 4

4. 4

5. 2

6. -5 - (3 + i) - (8 + 2i) = -5 - 3 - i - 8 - 2i = -16 -3i

7. (5 - 2i)(1 + 3i) = 5 +15i - 2i + 6

= 11 + 13i8. (2 + 3i)(5 - i)(5 - 3i)

= (10 + 15i - 2i + 3)

(5 - 3i) = (13 + 13i)(5 - 3i) = 65 - 39i + 65i + 39

= 104 + 26i




9. (11 - 2i) - (4i + 3)

________________ (-2 + 3i)(2 - i)

= 11 - 2i - 4i - 3 ______________ -4 + 2i + 6i + 3

= 8 - 6i _______ -1 + 8i

= (8 - 6i)(-1 - 8i)

________________ (-1 + 8i)/(-1 - 8i)

= -8 - 64i + 6i - 48

________________ 1 + 64

= -56 - 58i __________

65

= - 56 ___ 65

- 58 ___ 65

i

Sigma Notation pp. 23–24

1. 2

2. 1

3. 2

4. 4

5. 3

6. Observe that

an = 3n + 3 and there are

11 terms: � n=1

11

(3n+3)

7. � n=1

5

n2 = 1 + 4 + 9

+ 16 + 25

= 45

8. � n=1

5

(3n2- 4n)

= (3 - 4) + (12 - 8)

+ (27 - 12) + (48 - 16)

+ (75 - 20) = -1 + 4

+ 15 + 32 + 55 = 95

9. a. � n=1

5

an = (-2) 1 + (-2) 3

+ (-2) 4 + (-2) 5

= -2 + 4 - 8 + 16 - 32

= -22

b. � n=1

5

bn = 2 + 2 + 2

+ 2 + 2

= 10

c. � n=1

5

(an-bn) = -4 + 2

+ (-10) + 14 + (-34)

= -32

d. Yes

Solving Absolute Value

Equations and Inequalities

pp. 25–26

1. 3

2. 2

3. 4

4. 4

5. 2

6. � 2 - 3y � + 4 = 7

� 2 - 3y � = 3

2 - 3y = 3 or 2 - 3y = 3

-3y = 1 or -3y = -5

y = - 1 __ 3 or y = 5 __

3

7. 1 __ 3 � x - 5 � - 4 __

3 ≥ 1

1 __ 3 � x - 5 � ≥ 7 __

3

� x - 5 � ≥ 7

x - 5 ≥ 7 or x - 5 ≤ -7

x ≥ 12 or x ≤ -2

8. � x + 5 � = 7 - 2x x + 5 = 7 - 2x or

x + 5 = -7 + 2x 3x = 2 or -x = -12

x = 2 __ 3 or x = 12

|x + 5| = 7 - 2x

� 2 __ 3 + 5 � = 7 - 2 � 2 __

3 �

Check x = 2 __ 3 :

� 17 ___ 3 � = 7 - 4 __

3

17 ___ 3 = 17 ___

3 �

� x + 5 � = 7 - 2x � 12 + 5 � = 7 - 2(12)

Check x = 12:

� 17 � = 7 - 24

17 ≠ -17

The solution is x = 2 __ 3 .

9. � 2x + 3 � = � x - 2 � 2x + 3 = x - 2 or

2x + 3 = -x + 2

x = -5 or 3x = -1

x = -5 or x = - 1 __ 3

Check x = -5:

� 2x + 3 � = � x - 2 �

� 2(-5) + 3 � = � -5 - 2 �

� -7 � = � -7 � 7 = 7 �

Check x = - 1 __ 3 :

� 2x + 3 � = � x - 2 �

� 2 (- 1 __ 3 ) + 3 � = � - 1 __

3 - 2 �

� 7 __ 3 � = � - 7 __

3 �

7 __ 3 = 7 __

3 �

The solution is x = -5 or

x = - 1 __ 3 .

The Discriminant pp. 27–28

1. 4

2. 1

3. 2

4. 3

5. 3

6. a = 1, b = -4, c = 5;

b 2 - 4ac = (-4) 2 - 4(1)(5)

= 16 - 20 = -4

7. a = 1, b = 5, c = -1 ;

b 2 - 4ac = (5) 2 - 4(1)(-1)

= 25 + 4

= 29 > 0

There are two real roots.

8. 3x 2 + 1 = 2x - 5

3x 2 - 2x + 6 = 0

a = 3, b = -2, c = 6;

b 2 - 4ac = (-2) 2 - 4(3)(6)

= 4 - 72 = -68 < 0

There are two imaginary roots.

9. The discriminant is

b 2 - 12.

a. If the equation has no

real roots, then

b 2 - 12 < 0.

b 2 - 12 < 0

b 2 < 12

- ��

12 < b < ��

12

-2 ��

3 < b < 2 ��

3

b. If the equation has

two real roots, then

b 2 - 12 > 0.

b 2 - 12 > 0

b 2 > 12

b < - ��

12 or b > ��

12

b < - 2 ��

3 or b > 2 ��

3




Systems of Equations

pp. 29–30

1. 3

2. 3

3. 4

4. 2

5. 3

6. Substitute x + 2 for y in

the second equation.

x - 1 = x 2 + 2x + 1

0 = x 2 - x + 2

x = 1 ± �

� -7 ________

2

There is no solution.

7. Substitute x + 2 for y in


x + 2 = x 2 + 2x + 1

0 = x 2 + x - 1

x = 1 ± �

� 5 _______

2

y = 1 + �

� 5 _______

2 + 2 = 5 + �

� 5 _______

2

y = 1 - �

� 5 _______

2 + 2 = 5 - �

� 5 _______

2

( 1 + �

� 5 _______

2 ,

5 + ��

5 _______

2 ) and

( 1 - �

� 5 _______

2 ,

5 - ��

5 _______

2 )

8. y = x2 + x - 6

_________ x - 2

= (x + 3)(x - 2)

____________ x - 2

, so it is

equivalent to y = x + 3

for x ≠ 2. Substitute x + 3

for y in the second

equation.

x + 3 = x 2 + 2x - 3

0 = x 2 + x - 6

0 = (x + 3)(x - 2)

x = 2 or x = -3

x = 2 is extraneous.

When x = -3, y = 0.

The solution is (–3, 0).

9. Substitute x - 1 for y in


x - 1 = x 2 + bx + 1

0 = x 2 + bx + 1

x = -b ± �

� b2 - 4 _____________

2

There is no solution if

b 2 - 4 < 0

b 2 < 4

- 2 < b < 2.

Quadratic Inequalities

pp. 31–32

1. 1

2. 4

3. 2

4. 4

5. 4

6. Graph a dashed

boundary y = 2x 2 + 5x- 1 because the points

on the boundary are

not solutions. Shade

the area above the

boundary.

x8642

y

2826

242826

8642

O

7. Graph a dashed

boundary

y = -x 2 + 3x - 2

because the points on

the boundary are not

solutions. Shade the

area below the boundary.

x8642

y

28

22242826

8642

O

8. First solve the related

equation.

x 2 - 3x - 5 = 3(x - 2)

x 2 - 3x - 5 = 3x - 6

x 2 - 6x + 1 = 0

x = 6 ± �

� 32 ________

2

= 3 ± 2 ��

2

Divide the number line

into three intervals:

x < 3 - 2 ��

2 , 3 - 2 ��

2

< x < 3 + 2 ��

2 , and

x > 3 + 2 ��

2 . Test the

points -1, 3, and 7.

Test x = -1:

(-1) 2 -3(-1) -5 ≤ 3(-1 -2)

-1 ≤ -9

Test x = 3:

(3) 2 - 3(3) - 5 ≤ 3(3 - 2)

-5 ≤ 3 �

Test x = 7:

(7) 2 - 3(7) - 5 ≤ 3(7 - 2)

23 ≤ 15

The solution is the interval

3 - 2 ��

2 < x < 3 + 2 ��

2 .

9. Solve the related

equation.

x 2 - 4x - c = 3

x 2 - 4x - c - 3 = 0

x = 4 ± ��

16 - 4(1)(-c - 3) ____________________

2

= 4 ± �

� 28 + 4c ____________

2

= 2 ± ��

7 + c

a. Observe that if c = -7,

then the quadratic

equation has only one

solution. Hence, the

inequality has only

one solution.

b. If c > -7, then there

are two solutions to

the quadratic equation

and the solution to

the inequality is the

interval 2 - ��

7 + c ≤ x ≤ 2 + �

� 7 + c ,

but if c < -7, there

are no solutions to the

inequality.

Direct and Inverse Variation

pp. 33–34

1. 1

2. 2

3. 1

4. 1

5. 4




6. Substitute in the known

values.

y = kx 3 = k(9)

1 __ 3 = k

7. Substitute in the known

values to fi nd the

constant of variation.

y = k __ x

-6 = k ____ -12

72 = k8. Substitute in the known

values to fi nd the

constant of variation.

y = k __ x

-3 = k __ 7

-21 = k The inverse variation

equation is y = - 21 ___ x . Substitute 6 for y and

solve for x.

6 = - 21 ___ x

x = - 7 __ 2

9. Observe that the product

xy = 20. So y = 20 ___ x and

the variables are related

by inverse variation.

Substitute 8 for x.

y = 20 ___ 8 = 5 __

2

Applications of

Exponential Functions

pp. 35–36

1. 2

2. 2

3. 2

4. 20 = 40(0.75)t

0.5 = (0.75)t

ln 0.5 = t ln 0.75

t = ln 0.5 ______ ln 0.75

≈ 2.4

The population will be 20

after 2.4 hours.

5. This is exponential

decay, so the model is

y = a(1-r) t .

Substitute 200 for a

and 0.08 for r and

simplify.

y = 200(0.92) t When t = 5, y

= 200(0.92) 5 ≈ 132.

6. 2289 _____ 2180

= 1.05, so r =

0.05. The exponential

growth model is

y = 2180(1.05) t . When t = 10,

y = 2180(1.05) 10 ≈ 3551.

Solve for t when

y = 3000.

3000 = 2180(1.05) t

3000 _____ 2180

= 1.05 t

ln � 3000 _____ 2180

� = t ln1.05

t = ln � 3000 _____ 2180

� ______

ln1.05 ≈ 6.5

The population will

reach 3000 after

6.5 years.

Factor Polynomials pp. 37–38

1. 1

2. 3

3. 2

4. 3

5. 4

6. 20x 4 - 125y 2 = 5(4x 4 - 25 y 2 ) = 5(2x - 5y)(2x + 5y)

7. z 5 - 4z

= z( z 4 - 4)

= z( z 2 + 2)( z 2 - 2)

8. x 4 y 4 - 18x 2 y 2 + 81

= (x 2 y 2 - 9)(x 2 y 2 - 9)

9. x 4 - 25x 2 y 2 - 4x 2 + 100y 2 = x 2 (x 2 - 25y 2 ) -

4(x 2 - 25 y 2 ) = (x 2 - 4)(x 2 - 25y 2 ) = (x - 2)(x + 2)

(x - 5y)(x + 5y)

Negative and Fractional

Exponents pp. 39–40

1. 3

2. 2

3. 1

4. 3

5. 3

6. 5a-2/3b-1/2

__________ a-5/3b-3/2

= 5a -2/3 + 5/3 b -1/2 + 3/2

= 5ab

7. � x-2/3

____ y5/3

� -6

= x(-2/3) (-6)

________ y(5/3)(-6)

= x4

____ y-10

= x 4 y 10

8. (2pq-3)-5(-2p1/3q-1)6

___________________ (-3p5/2q)-2

= � 1 ___

32 p-5q15 � � 64p2q-6 �

___________________ 1 __

9 p-5q-2

= 2p-3q9

________ 1 __

9 p-5q-2

=18p 2 q 11

9. a. a = f 4 g 3 h -3 = f4g3

____ h3

b. a -2 x = 1

a 2 a -2 x = a 2 · 1 x = a 2 = � f4g3h-5 � 2 = f 8 g 6 h -10

= f8g6

____ h10

Rewrite Algebraic

Expressions as Radical


1. 1

2. 3

3. 3

4. 2

5. 1

6. Since the root is even

and the power is even,

the expression is defi ned

over the real numbers for

all values of w.




7. � 25x 6 y 8 � 1

__ 8

� 5 2 � 1

__ 8

· � x 6 �

1 __ 8 · � y 8 �

1 __

8 =

5 1

__ 4

· x

3 __ 4 · |y| = |y|

4

��

5x3

8. � 48x 6 � 1

__ 4

=

4

��

48x6

= 4

��

24 · 3 · x2 · x4

= 2|x| 4

��

3x2

Since the root is even



over the real numbers for

all values of x.

9. a. ��

121x9 y4 z2

b. For x, the root is even

and the power is odd,

so the expression is

defi ned when x ≥ 0.

For y and z, the root is

even and the power is

even, so the expression

is defi ned for all values

of y and z.

c. ��

121x9 y4 z2

= ��

11 2 · x 8 · x · � y 2 � 2 z 2

= ��

112 · (x4)2 · x · (y2)2 z2

= 11x4y2 |z| �� x

Rewrite Radical

Expressions as

Expressions with Fractional

Exponents pp. 43–44

1. 3

2. 2

3. 4

4. 4

5. 1

6. 3

��

40z6 = � 40 6 � 1

__ 3

= � 5 · 8 · z 2 · z 2 · z 2 � 1 __

3

= � 5 · 2 3 · � z 2 � 3 � 1

__ 3

= 2z 2 5 1

__ 3

7. 6x

3 ��

y4 ______ y =

6xy 4

__ 3 ____ y

= 6xy 4 __ 3

y -1 = 6xy

4 __ 3

y

- 3 __ 3

= 6xy 4 __

3 - 3 __

3 = 6xy

1 __ 3

8. 4

��

162x6 = � 162x 6 � 1

__ 4

=

162 1

__ 4

· � x 6 �

1 __

4 = � 2 · 81 �

1

__ 4

· x 6 __

4 = � 2 · 3 4 �

1 __

4 · x

3

__ 2

= 2 1 __

4 · 3

4 __

4 · |x| · x

1

__ 2

= 2 1 __

4 · 3 · |x| · x

2 __

4

= 3 |x| � 2x 2 � 1

__ 4

Since the root is even



for all values of x.

9. a. The expression is

defi ned for all b and

c, b ≠ 0 and c ≠ 0

(the roots of b and c are

odd; neither can equal

0 or the denominator

will be 0) and for all a

> 0 (the root is even

and the power is odd;

a cannot equal 0 or the

denominator will be 0).

b. �

� a3 ·

3 ��

(bc)2 ___________

abc

= a 3 __ 2 · b

2 __ 3

· c

2 __ 3 __________

abc

= a 3 __

2 · b

2

__ 3 · c

2

__ 3

· a-1

· b-1 · c-1 = a 3

__ 2

- 1

· b 2

__ 3

- 1 · c

2

__ 3

- 1 = a

1

__ 2

· b - 1 __

3 · c

- 1 __ 3

Evaluate Exponential


1. 2

2. initial amount a = 46,000,

percent decrease

r = 0.18, y = a(1 - r)t

= 46,000(1 - 0.18)t

y = 46,000(0.82)t

3. y = a (1 + r)t

= 7500(1 + 0.04)t

= 7500(1.04)t

For t = 8,

y = 7500(1.04)t

= 7500(1.04)8

≈ $10,264

4. a. y = a(1 + r)t

= 4000(1 + 0.0505)t

y = 4000(1.0505)t

y = 4000(1.0505)8

≈ $5932

b. A = Pert

A = 4000 · e0.05t

A = 4000e0.05 · 8

= 4000e0.4

A ≈ $5967

c. The second account

is the better choice

because it earns more

interest, even though the

interest rate is higher for

the fi rst account.

Operations with Radical


1. 3

2. 1

3. 3

4. 1

5. 4

6. ��

72x2 y4 z3 =

��

2 · 62 · x2 · y2 · y2 · z · z2

= 6 · x · y · y · z �� z

= 6xy2 z �� z

7. 3

��

24x8 + 3

��

81x8

= 3

��

3 · 23 · x2 · x6

+ 3

��

3 · 33 · x2 · x6

= 2x2 3

��

3x2 + 3x2 3

��

3x2

= 5x2 3

��

3x2

8. ��

x5 ____

��

y4 ·

��

y ___

��

x3

= ��

x5y ____

x3y4 = �

�

x2

__ y3

= ��

x2

__ y2

· 1 __ y = x __ y · 1 ___ �

� y

= x __ y · 1 ___ �

� y ·

��

y ___

��

y

= x __ y · �

� y ___ y =

x ��

y ____

y2

9. a. ��

a2x2 - 2abx + b2

= ��

(ax - b)2

= |ax - b|

b. |ax - b| cannot be

negative, so




��

a2x2 - 2abx + b2

< 0 has no solution.

c.

��

a2x2 - 2abx + b2 = 0

|ax - b| = 0

ax - b = 0

x = b __ a d.

��

a2x2 - 2abx + b2 > 0

|ax - b| > 0

ax - b > 0 -(ax - b) > 0

x > b __ a -ax + b > 0

-ax > -b x < b __ a

x can be any value but b __ a .

Operations with

Rational Expressions

pp. 49–50

1. 3

2. 3

3. 3

4. 4

5. 1

6. 5x - 2 ______ x + 4

- 2x + 1

______ x

= 5x - 2 ______ x + 4

· x __ x

- 2x + 1

______ x · x + 4 _____

x + 4

= 5x2 - 2x ________ x(x + 4)

- 2x2 + 9x + 4

___________ x(x + 4)

= 3x2 - 11x - 4 ____________ x(x + 4)

7.

3 __ x + 1 ___ 2y

______ x ÷ xy

=

3 __ x + 1 ___ 2y

______ x · xy ·

2xy ___

2xy

= 6y + x

______ 2x3y2

8. 2 _____ x + 3

- 1 ___________ x2 + 7x + 12

= 2 _____ x + 3

- 1 ___________ (x + 3)(x + 4)

= 2 _____ x + 3

· x + 4 _____

x + 4

- 1 ___________ (x + 3)(x + 4)

= 2x + 8 ___________

(x + 3)(x + 4)

- 1 ___________ (x + 3)(x + 4)

= 2x + 8 - 1

___________ (x + 3)(x + 4)

= 2x + 7 ___________

(x + 3)(x + 4)

= 2x + 7 ___________

x2 + 7x + 12

9. a. First simplify a and b.

a = 2 _____

3 __ x + 1 · x __ x = 2x _____

3 + x

= 2x _____ x + 3

b = 4 _____

1 __ x + 2 · x __ x

= 4x ______ 1 + 2x

= 4x ______ 2x + 1

a + b = 2x _____ x + 3

+ 4x ______ 2x + 1

= 2x _____ x + 3

· 2x + 1 ______

2x + 1

+ 4x ______ 2x + 1

· x + 3 _____

x + 3

= 4x2 + 2x _____________

(x + 3)(2x + 1)

+ 4x2 + 12x ______________

(x + 3)(2x + 1 )

= 8x2 + 14x _____________

(x + 3)(2x + 1)

b. a · b = 2x _____ x + 3

· 4x ______ 2x + 1

= 8x2

_____________ (x + 3)(2x + 1)

c. a ÷ b = 2x _____ x + 3

÷ 4x ______ 2x + 1

= 2x _____ x + 3

· 2x + 1 ______

4x

= 12 x _____

x + 3 · 2x + 1

______ 24 x

= 2x + 1

______ 2x + 6

Evaluate and Rewrite

Logarithmic Expressions

pp. 51–52

1. 4

2. 4

3. 1

4. 3

5. 1

6. about 1.594

7. 2 log x + log y

- 1 __

2 log z

8. logb189 = logb (7 · 27)

= logb (7 · 33)

= logb7 +

3 logb 3

≈ 0.845 +

3(0.477)

≈ 2.276

9. a. In Step 1, the student

did not apply the

power property of

logarithms. The correct

step is

3 log 5 4 -

2 __

3 log

5 64

= log 5 � 4 3

_____ 64 (2/3)

� b. 3 log

5 4 -

2 __

3 log

5 64

= log 5 � 64

___ 16

� = log

5 (4)

Use the Sum and Product

of the Roots of a Quadratic

Equation pp. 53–54

1. 3

2. 2

3. 1

4. 3

5. 1

6. a = 3, b = -5, c = -9

r1 + r

2 = - b __ a = - -5 ___

3 = 5 __

3

r1r

2 = c __ a = -9 ___

3 = -3

7. x2 - (r1 + r

2) x + r

1r

2 = 0

x2 - � - 5 ___ 12

� x + � - 1 __ 4 � = 0

x2 + 5 ___ 12

x - 1 __ 4

= 0

12x2 + 5x - 3 = 0

8. x2 - 10x - k = 0,

a = 1, b = -10, c = -k,

r1 + r

2 = - b __ a

-2 + r2 =

-(-10) _______

1




-2 + r2 = 10

r2 = 12

The other root is 12.

9. 2x2 + 5x + 6 = 0,

a = 2, b = 5, c = 6,

r1 + r

2 = -

b __ a = -

5 __

2

r1r

2 =

c __ a =

6 __

2 = 3

The new equation has

roots 1 __

r 1 and

1 __

r 2 :

x2 - � 1 __ r 1

+ 1 __

r 2 � x

+ 1 __ r 1

· 1 __ r 2

= 0

x2 - � r 1 + r 2 ______

r 1 r 2 � + 1

____ r 1 r 2

= 0

Substitute - 5 __

2 for r 1 + r 2

and 3 for r 1 r 2 :

x2 - � r 1 + r 2 ______

r 1 r 2 � + 1

____ r 1 r 2

= 0

x2 - � - 5 __

2 ____

3 � +

1 __

3 = 0

x2 + 5 __

6 x +

1 __

3 = 0

6x2 + 5x + 2 = 0

Solve Radical Equations

pp. 55–56

1. 4

2. 1

3. 4

4. 2

5. 2

x + 2 = ��

-3x - 8

(x + 2)2 = � �� -3x - 8 � 2

x2 + 4x + 4 = -3x - 8

6. x2 + 7x + 12 = 0

(x + 4)(x + 3) = 0

x + 4 = 0 or x + 3 = 0

x = -4 or x = -3

Check x = -4:

x + 2 = ��

-3x - 8

-4 + 2 � ��

-3(-4) - 8

-2 � ��

4

-2 ≠ 2

-4 is not a solution.

Check x = -3:

x + 2 = ��

-3x - 8

-3 + 2 �

��

-3(-3) - 8

-1 � ��

1

-1 ≠ 1


The equation has no real

solutions.

��

x2 - 8 = ��

-2x + 7

� ��

x2 - 8 � 2

= � �� -2x + 7 � 2

x2 - 8 = -2x + 7

7. x2 + 2x - 15 = 0

(x + 5)(x - 3) = 0

x + 5 = 0 or x - 3 = 0

x = -5 or x = 3

Check x = −5:

��

x2 - 8 = ��

-2x + 7

��

(-5)2 - 8 �

��

-2(-5) + 7

��

17 = ��

17

−5 is a solution.

Check x = 3:

��

x2 - 8 = ��

-2x + 7

��

(3)2 - 8 � ��

-2(3) + 7

��

1 = ��

1

3 is a solution.

The solution set is

{−5, 3}.

8. ��

24 - 5x = x - 2

� �� 24 - 5x � 2 = (x - 2)2

24 - 5x = x2 - 4x + 4

x2 + x - 20 = 0

(x + 5)(x - 4) = 0

x + 5 = 0 or x - 4 = 0

x = - 5 or x = 4

Check x = −5:

��

24 - 5x = x - 2

��

24 - 5(-5) � -5 - 2

��

49 ≠ -7


Check x = 4:

��

24 - 5x = x - 2

��

24 - 5(4) � 4 - 2

��

4 � 2

2 = 2

4 is a solution.

The solution is x = 4.

9. ��

5x - 8 = ��

x2 - x

� �� 5x - 8 � 2 = � �� x 2 - x � 2 5x - 8 = x2 - x

x2 - 6x + 8 = 0

(x - 4)(x - 2) = 0

x - 4 = 0 or x - 2 = 0

x = 4 or x = 2

Check x = 2:

��

5x - 8 = ��

x2 - x ��

5(2) - 8 � ��

(2)2 - 2

��

2 = ��

2

2 is a solution.

Check x = 4:

��

5x - 8 = ��

x2 - x ��

5(4) - 8 � ��

(4)2 - 4

��

12 = ��

12

4 is a solution.

Check by graphing:

IntersectionX=2 Y=1.4142135

IntersectionX=4 Y=3.4641016

The solution set is {2, 4}.

Solve Rational Equations

and Inequalities pp. 57–58

1. 3

2. 3

3. 2

4. 4

5. 1

6. -6 _______

x2 + 3x =

2 _____

x + 3 -

1 __ x

-6 ________

x (x + 3) =

2 _____

x + 3 -

1 __ x

-6 ________

x (x + 3) =

2x ________ x (x + 3)

- 1(x + 3)

________ x (x + 3)




-6 = 2x - x - 3

-3 = xBut x = -3 makes two

of the denominators in

the original equation 0,

so it is not a solution.

There is no real solution.

7. 17

___ x = 4 + 1 __ x

17

___ x = 4x ___ x +

1 __ x

17 = 4x + 1

16 = 4x 4 = x

8. 3x2

_____ x + 2

- 3x2

____ x-2

= x

3x2 (x - 2)

_____________ (x + 2 )(x - 2)

- 3x2 (x + 2)

____________ (x + 2) (x - 2)

= x (x + 2) (x - 2)

______________ (x + 2) (x - 2)

3x3 - 6x2 - 3x3 - 6x2

= x (x2 - 4)

-12x2 = x3 - 4xx3 + 12x2 - 4x = 0

x(x2 + 12x - 4) = 0

x = 0 or

x = -12 ± �

� 122 - 4(1)(-4) ___________________

2(1)

= -12 ± �

� 160 ___________

2

= -6 ± 2 ��

10

x = 0, x = -6 ± 2 ��

10

9. x - 5 ≤ 14

___ x

x - 5 = 14

___ x

x(x - 5) = 14

x2 - 5x - 14 = 0

(x - 7)(x + 2) = 0

x = 7 or x = -2

x = 0 cannot be a

solution, so plot -2, 0,

and 7 on a number line.

Use an open circle for

0, since it cannot be

a solution. Use closed

circles for -2 and 7,

since the inequality is

less than or equal to.

Check a value in each

interval:

−14:

-14 - 5 ≤ 14 ____ -14

-19 ≤ -1 True

−1:

- 1 - 5 ≤ 14 ___ -1

-6 ≤ -14 False

2:

2 - 5 ≤ 14 ___ 2

-3 ≤ 7 True

14:

14 - 5 ≤ 14 ___ 14

9 ≤ 1 False

Shade the intervals that

tested true.

02224 2 4 6 8

x ≤ 2 or 0 < x ≤ 7

Complete the Square

pp. 59–60

1. 4

2. 3

3. 2

4. 2

5. 1

6. b = 1 __ 3

� b __ 2 � 2 = � 1 __

3 __

2 �

2

= � 1 __ 6 � 2 = 1 ___

36

7. � b __ 2 � 2 = � - 3 __

2 � 2 = 9 __

4

x2 - 3x = 28

x2 - 3x + 9 __ 4 = 28 + 9 __

4

� x - 3 __

2 � 2 = 121 ____

4

x - 3 __ 2 = ± 11 __

2

x = 3 __ 2 ± 11 __

2

x = -4 or x = 7

8. � b __ 2 � 2 = � - 6 __

2 � 2 = 9

x2 - 6x + 45 = 0

x2 - 6x = -45

x2 - 6x + 9 = -45 + 9

(x - 3)2 = -36

x - 3 = ±6ix = 3 ± 6i

9. a. 2x2 + ax - a2 = 0

x2 + a __ 2 x =

a2

__ 2

� a __ 2 __

2 �

2

= � a __ 4 � 2 =

a2

___ 16

x2 + a ___ 2 x+

a2

___ 16

= a2

__ 2 +

a2

___ 16

� x + a __ 4 � 2 =

9a2

___ 16

x + a __ 4 = ±

3a ___ 4

x = - a __ 4 ±

3a ___ 4

x = - a __ 4 +

3a ___ 4 =

a __ 2 or

x = - a __ 4 -

3a ___ 4 = -a

x = - a __ 2 x = -a

b. 2x = a x + a = 0

2x - a = 0

(2x - a)(x + a) = 0

2x2 + 2xa - ax - a2 = 0

2x2 + ax - a2 = 0

Use the Quadratic Formula

pp. 61–62

1. 2

2. 1

3. 4

4. 3

5. 4

6. x2 + 9 = -x x2 + x + 9 = 0

x = -1 ± �

� (1) 2 - 4(1)(9) _________________

2(1)

= -1 ± �

� -35 __________

2

= -1 ± i �

� 35 _________

2

7. 5x 2 + 10x = 3

5x 2 + 10x - 3 = 0

x = -b ± �

� b 2 - 4ac ________________

2a

= - 10 ± �

� 10 2 - 4(5)(-3) _______________________

2(5)

= -10 ± �

� 160 ___________

10

= -10 ± �

� 410 ___________

10




= -5 ± 2 �

� 10 __________

5

h = -16t 2 + v0t + h0

3 = -16t 2 1 40t 1 5

0 = -16t 2 1 40t 1 2

3 = -16t 2 1 40t 1 5

0 = -16t 2 1 40t 1 2

8.

t = -40 ± �

� (40) 2 - 4(-16)(2) _____________________

2(-16)

t = -40 ± �

� 1728 ____________

-32

t = -0. 049 or t ≈ 2.55

Reject the solution -0.049

because the ball’s time in

the air cannot be negative.

The ball is in the air for about

2.55 seconds.

Use the formula for area,

A 5 lw, to write an equation

that represents the situation.

new area 5 new length · new width

↓ ↓ ↓ 2(100)(150) 5 (150 + x) ·

(100 + x)

The equation that represents

the situation is

0 5 x 2 1 250x - 15,000.

9. Solve for x to fi nd the

new dimensions of the

practice fi eld.

0 5 x2 1 250x - 15,000

x 5

-250 ± �

�� (250) 2 - 4(1)(-15,000) ___________________________

2(1)

x 5 -250 ± �

� 122,500 _______________

2

x 5 50 or x 5 -300.

Reject the solution -300.

The practice fi eld’s length and

width should be increased by

50 feet. The new dimensions

are 150 feet by 200 feet.

Solve Polynomial

Equations pp. 63–64

1. 4

2. 3

3. 1

4. 3

5. 4

6. x4 - 17x2 + 16 = 0

� x2 � 2 - 17 � x2 � + 16 = 0

t = x2

t 2 - 17t + 16 = 0

(t - 1)(t - 16) = 0

t = 1 or t = 16

x2 = 1 or x2 = 16

x = ±1 or x = ±4

7. x3 - 4x2 + 2x = 0

x(x2 - 4x + 2) = 0

x = 0 or

x = -b ± �

� b 2 - 4ac ________________

2a

= -(-4) ± �

� (-4) 2 - 4(1)(2) ______________________

2(1)

= 4 ± �

� 8 _______

2

= 2 ± ��

2

8. x 6 - x 4 - 6x 2 = 0

x 2 ( x 4 - x 2 - 6) = 0

x 2 = 0 or x 4 - x 2 - 6 = 0

x 2 = 0 or ( x 2 ) 2 - x 2 - 6 = 0

x 2 = 0 or ( x 2 - 3)

( x 2 + 2) = 0

x 2 = 0, x 2 - 3 = 0,

or x 2 + 2 = 0

x 2 = 0 or x 2 = 3 or x 2 = -2

x = 0 or x = ± ��

3 or

x = ±i ��

2

9. a. The number of zeros

is equal to the degree

of the polynomial,

which is 4.

b. Since the equation

has 4 roots, one of

the roots must be

repeated.

c. The possible factors of

P(x) are x, (x − 1), and

(x − 2). Since one root

is repeated, one of

these factors must be

squared. The possible

equations are:

P(x) = x2(x - 1)(x - 2) = 0

P(x) = x(x - 1)2(x - 2) = 0

P(x) = x(x - 1)(x - 2)2 = 0

Solve Exponential


1. 1

2. 1

3. 3

4. 3

5. 4

e2x = 10

ln e2x = ln 106.

2x = ln10

x = ln10 ____ 2

≈ 1.151

7. 43x - 1 = � 1 __ 2

� 3x + 1

(22)3x - 1 = (2-1)3x + 1

22(3x - 1) = 2(-1)(3x + 1)

26x - 2 = 2-3x -1

6x-2 = -3x -1

9x = 1

x = 1 __

9

8. 102x - 13 · 10x + 40 = 0

(10x)2 - 13(10x) + 40 = 0

Let t = 10x.

t 2 - 13t + 40 = 0

(t - 8)(t - 5) = 0

t = 8 or t = 5

10x = 8 or 10x = 5

log 10x = log 8 or log 10x

= log 5

x = log 8 or x = log 5

x ≈ 0.903 or x ≈ 0.699

9. 2n = ��

3n + 2

2n = � 3 n + 2 � 1

__ 2

2n = 3 n + 2

_____ 2

log 2n = log 3 n+2

____ 2

n log 2 = � n + 2 _____

2 � log 3

2n log 2 = (n + 2) log 3

2 log 2 = � n + 2 _____ n � log 3

2 log 2 = � 1 + 2

__ n � log 3

n(2 log 2 - log 3) = 2 log 3

n = 2 log 3

____________ 2 log 2 - log 3

≈ 7.638




Solve Logarithmic


1. 3

2. 2

3. 4

4. 4

5. 3

6. 7 = 4 log 16

(x + 3)

7 __

4 = log

16 (x + 3)

16 7 __

4 = x + 3

128 = x + 3

x = 125

7. log3 2 = log

3 (7x) -

log3 (x2 - 2)

log3 2 = log

3 � 7x _____

x2 - 2 �

3log32 = 3

log3 � 7x _____ x2 - 2

�

2 = 7x _____

x2 - 2

2x 2 - 4 = 7x

2x 2 - 7x - 4 = 0

(2x + 1)(x - 4) = 0

2x + 1 = 0 or x - 4 = 0

2x = - 1 or x = 4

x = - 1 __ 2 or x = 4

8. log 4 (x 2 + 16) =

log 4 (2x + 1) + log

4

(x - 2)

log 4 (x 2 + 16) =

log 4 [(2x + 1)(x - 2)]

4ln(x2 + 16) = 4ln[(2x + 1)(x - 2)]

x2 + 16 = (2x + 1)

(x - 2)

x2 + 16 = 2x2 - 3x - 2

0 = x2 - 3x - 18

0 = (x - 6)(x + 3)

x - 6 = 0 or x + 3 = 0

x = 6 or x = -3

Discard the solution x = -3

because it makes ln (2x + 1)

and ln (x -2) negative.

9. a. log 3 (2x + 7) 2 = 6

2 log 3 (2x + 7) = 6

log 3 (2x + 7) = 3

3 log 3 (2x + 7) = 3 3

2x + 7 = 27

2x = 20

x = 10

b. Graph both sides of the

equation and use the

intersection feature on

the CALC menu.

c.

Intersection

Y=6X=-13.54277

Intersection

Y=6X=6.54277

The solution in part a. checks.

Identify a Sequence and

Find the Formula for Its nth

Term pp. 69–70

1. 4

2. 1

3. 2

4. 4

5. 3

6. The difference between

consecutive terms

is constant, so the

sequence is arithmetic.

a 1 = 112 and d = −17.

a n = a 1 + (n - 1)d

= 112 + (n - 1) · (-17)

= 112 - 17n + 17

= 129 - 17n7. The ratio of consecutive

terms is constant, so the

sequence is geometric.

a 1 = 5 and r = −9.

a n = a1 · r n-1

= 5 · (-9) n-1

8. Rewrite the formula:

a n = 3(2n - 7) = 6n - 21

= 6n - 6 - 15

= -15 + (n - 1)6

The fi rst term is −15 and the

common difference is 6.

9. a. Rewrite the formula:

a n = 3n _____

2n-1

= 3 · 3n-1

_______ 2n-1

= 3 · � 3 __ 2 �

n-1

Compare this formula

to the general formula

for the nth term of a

geometric sequence:

a n = a 1 · (r) n-1 . It is in

the correct form, so the

sequence is geometric.

b. a1 = 3 in the formula,

so the fi rst term is 3.

c. r = 3 __

2 in the formula,

so the common ratio

is 3 __

2 .

Determine the Common

Difference or Common

Ratio in a Sequence

pp. 71–72

1. 2

2. 3

3. 3

4. 1

5. 3

6. -2 - (-5) = 3,

1 - (-2) = 3, 4 - 1 = 3

The common difference is 3.

7. 0.03

____ 0.3

= 0.1, 0.003

_____ 0.03

= 0.1,

0.0003

______ 0.003

= 0.1

The common ratio is 0.1.

8. Find the common

difference using the last

two terms given:

d = 2 - (-3) = 5

Find k by adding d to the

fi rst term:

k = -13 + 5 = -8

an = -13 + (n - 1)5

= -13 + 5n - 5

= 5n - 18




9. Write two equations with

a 1 and r:

a n = a 1 · r n-1

a 3 : 3 = a

1 · r 3-1

3 = a 1 · r 2

a 7 : 1875 = a

1 · r 7-1

1875 = a 1 · r 6

Divide the second

equation by the fi rst

equation and solve for r:

1875

_____ 3 =

a1 · r6

______ a

1 · r 2

625 = r 6-2

625 = r 4

4 ��

625 = 4

��

r4

r = 5

The common ratio is 5.

Determine a Specifi ed Term

of a Sequence pp. 73–74

1. 1

2. 1

3. 4

4. 3

5. 2

a n = a 1 + (n - 1)d

a 4 = 6 = a

1 + (4-1)(-5)

6 = a 1 - 15

6.

a 1 = 21

a n = 21 + (n - 1)(-5)

a n = 26 - 5n

7. a 1 = 102, d = −18,

a n = 102 + (n - 1)(-18)

= 102 + 18 - 18n

= 120 - 18n-60 = 120 - 18n

-180 = -18n

n = 10

a n = a 1 (r) n-1

a 3 = a

1 (r) 3-1

8. 2304 = 3600(r) 2

2304

_____ 3600

= r 2

r = 4 __

5

a n = a 1 (r) n-1

a n = 3600 � 4 __ 5 � n-1

= 3600 · � 4 __ 5 � -1

· � 4 __ 5 � n

= 3600 · � 5 __ 4 � · � 4 __

5 � n

= 4500 � 4 __ 5 � n

9. a. The ratio between

consecutive terms is

constant:

48 ____

144 =

16 ___

48 =

16

___ 3 ___

16 =

1 __

3

The common ratio is 1 __

3 .

a n = a1(r) n - 1

= 144 � 1 __ 3 � n-1

b. = 16 · 3 2 · � 1 __

3 � n-1

= 16 · � 1 __ 3 � -2

· � 1 __ 3 � n-1

= 16 � 1 __ 3 � n - 3

c. As n gets larger, the

terms of the sequence

get very small.

Recursive Sequences

pp. 75–76

1. 3

2. 1

3. 3

4. 3

5. 1

6. a 1 = 2

a 2 = ( a

1 ) 2 = 2 2 = 4

a 3 = ( a

2 ) 2 = 4 2 = 16

a 4 = ( a

3 ) 2 = 16 2 = 256

The fi rst four terms of

the sequence are 2, 4,

16, 256.

7. a 1 = 1

a 2 = (2 + 1) a

1

= (2 + 1)1 = 3

a 3 = (3 + 1) a

2

= (3 + 1)3 = 12

a 4 = (4 + 1) a

3

= (4 + 1)12 = 60

a 5 = (5 + 1) a

4

= (5 + 1)60 = 360

The fi rst fi ve terms of the

sequence are 1, 3, 12,

60, 360.

8. Since the plant is 10 cm

tall when n = 1 week,

a1 = 10. Each week

the plant is 1.05 times

the plant’s height the

week before. The plant’s

growth generates a

geometric sequence.

An explicit formula is

a n = a 1 (1.05) n – 1 . A

recursive formula is

a 1 = 10, a n = 1.05an –1

(or a 1 = 10, a n +1

= 1.05 a n .) 9. a. a

1 = 1, a

2 = 1

a 3 = a

2 - (-1) 3 ( a

1 )

= 1- (-1)(1) = 2

a 4 = a

3 - (-1) 4 ( a

2 )

= 2 - (1)(1) = 1

a 5 = a

4 - (-1) 5 ( a

3 )

= 1 - (-1)(2) = 3

a 6 = a

5 - (-1) 6 ( a

4 )

= 3 - (1)(1) = 2

a 7 = a

6 - (-1) 7 ( a

5 )

= 2 - (-1)(3) = 5

a 8 = a

7 - (-1) 8 (a

6)

= 5 - (1)(2) = 3

a 9 = a

8 - (-1) 9 ( a

7 )

= 3 - (-1)(5) = 8

a 10

= a 9 - (-1) 10 ( a

8 )

= 8 - (1)(3) = 5

a 11

= a 10

- (-1) 11 ( a 9 )

= 5 - (-1)(8)

= 13

a 12

= a 11

- (-1) 12 ( a 10

)

= 13 - (1)(5) = 8

a 13

= a 12

- (-1) 13 ( a 11

)

= 8 - (-1)(13)

= 21

a14

= a13

- (-1) 14 (a12

)

= 21 - (1)(8) = 13

The fi rst fourteen

terms of the sequence

are 1, 1, 2, 1, 3, 2, 5,

3, 8, 5, 13, 8, 21, 13.

b. The even-numbered

terms are 1, 1, 2, 3,

5, 8, and 13. The odd-

numbered terms are

1, 1, 2, 3, 5, 8, and 13.

The even- and odd-

numbered terms form

sequences with the

same terms, and the




terms of the sequences

are terms of the

Fibonacci sequence.

The Sum of a Series

pp. 77–78

1. 2

2. 1

3. 1

4. 2

5. 3

6. The series is arithmetic

with a1 = 12 and

common difference -8.

an = 12 + (n - 1)(-8)

a7 = 12 + (7 - 1)(-8)

= -36

S7 = 7 [ 12 + (-36)

_________ 2 ] =

7(-12) = -84

7. a1 = - 2 __

3

S 12

= - 2 __

3 [ 1 - � - 2 __

3 � 12

__________

1 - � - 2 __ 3 � ]

≈ -0.4

8. The sequence is

arithmetic with common

difference -18.

an = 102 + (n - 1)(-18)

an = 120 - 18n

Use ai = 120 - 18i. The

nth partial sum of the

series 102 + 84 + 66 +

48 + … is

� i=1

n

(120 - 18i ) .

The 20th term of the

sequence is a20

= 120

- 18(20) = -240.

S20

= 20 [ 102 + (-240) ___________

2 ]

= -1380

9. a. The common ratios

a

2 _____

3600 and

2304 _____ a

2

are

equivalent. So,

a 2 _____

3600 =

2304 _____

a 2

(a2) 2 = 8,294,400

a2 = ±2880

The common ratio is

2880

_____ 3600

= 4 __

5 or

- 2880 _____

3600 = - 4 __

5 .

b. a1 = 3600, r = ± 4 __

5 :

Sn = 3600 [ 1 - � 4 __ 5 � n _______

1 - 4 __

5 ]

= 3600 [ 1 - � 4 __ 5 � n _______

1 __

5 ]

= 18,000 [ 1 - � 4 __ 5 � n ]

or

Sn = [ 3600 1- (- 4 __

5 )

n

________

1-(- 4 __ 5 ) ]

= 3600 [ 1 - � - 4 __ 5 �

n

________ 9 __

5 ]

= 2000 [ 1 - � - 4 __ 5 � n ]

c. As the value of n gets

larger, the sum gets

closer to 18,000 when

r = 4 __

5 , and the sum

gets closer to 2000

when r = - 4 __ 5 .

Evaluate Numerical


1. 1

2. 4

3. 1

4. 3

5. 2

6. Since x 6 y 7 = x13 – 7y 7 , r = 7.. The coefficient is

13C

7 = 1716.

7. The eighth term is

20

C18

-1

(3x) 13 - (18-1) (1) 18-1

=20

C17

(3x) 3 (1) 17 = 1140(27x 3 ) = 30,780x 3 .

8. By the Binomial Theorem,

the x2y term of the

expansion is

3C

1x2(2y) 1 = 3 · x 2 · 2y

5 6x 2 y. So the coefficient

of the x 2 y term is 6, not 3.

9. a. ( �� x +2) 4 =

4C

0( ��

x ) 4

(2) 0 + 4C

1 ( ��

x ) 3 (2) 1

+ 4C

2 ( ��

x )2 (2) 2

+ 4C

3 ( ��

x )1 (2) 3

+ 4C

4 ( ��

x )0 (2) 4

= 1x 2 (1) + 4 (x �� x )(2)

+ 6x(4) + 4( �� x )(8)

+ 1(1)(16)

= x 2 + 8x �� x + 24x

+ 32 �� x +16

= x 2 + 24x +

(8x + 32) �� x + 16.

b. (x1/3 - y1/3) 3 = 3C

0(x1/3) 3

(-y1/3) 0 + 3C

1(x1/3) 2

(-y1/3) 1 + 3C

2(x1/3) 1

(-y1/3) 2 + 3C

3(x1/3) 0

(-y1/3) 3 = (1)(x)(1)

+ 3x 2/3 (-y 1/3 ) + 3x 1/3

(-y 2/3 ) + 1(1) (y)

= x – 3x 2/3 y 1/3 + 3x 1/3

y 2/3 – y

Relations and Functions

pp. 81–82

1. 3

2. 1

3. 4

4. The domain is the set of

x-values: {–10, –7, –4, 4,

6, 9}

The range is the set of

y-values: {–4, 0, 1, 2, 4}

Since each element in

the domain is paired with

exactly one element in

the range, the relation is

a function.

5. y = (–1.5) 2 – 1

= 2.25 – 1 = 1.25




6. There are fewer

elements in the domain

than in the range. This

means that at least one

element in the domain

is paired with more

than one element in the

range. Therefore, the

relation is not a function.

7. a. y 2 = x

��

y 2 = �� x

y = ± �� x

b. x = 0: y = ± ��

0

x = 1: y = ± ��

1 = ±1

x = 2: y = ± ��

2

x = 3: y = ± ��

3

x = 4: y = ± ��

4 = ± 2

{(0, 0), (1, 1), (1, –1), (2,

��

2 ), (2, - ��

2 ), (3, ��

3 ),

(3, - ��

3 ), (4, 2), (4, –2)}

c. With the exception

of 0, each element

in the domain is

paired with two

values in the range.

The relation is not a

function.

Determine Domain and

Range from a Function’s

Equation pp. 83–84

1. 3

2. 4

3. 1

4. 1

5. 2

6. y = -5x 2 - 30x + 3 is a

quadratic function;

a = −5, and b = -30.

f � - (-30)

____ 2(-5)

� = f � - 30 ___

10 � = f (-3)

= -5 (-3)2 - 30 (-3) + 3

= -45 + 90 + 3 = 48

Since a is negative, the

range is y ≤ 48.

7. Domain: all real numbers.

a = 1 __

3 , which is greater

than 0. Range: y > -5

8. Find values of x that

make the denominator 0.

3x 3 - x 2 - 4x = 0

x(3x 2 - x - 4) = 0

x(3x - 4)(x + 1) = 0

x = 0 or 3x - 4 = 0 or

x + 1 = 0

x = 0 or x = 4 __

3 or x = -1

The domain is all real

numbers, x ≠ -1, x ≠ 0,

x ≠ 4 __ 3 .

9. The domain is restricted

by g(x) > 0.

x 2 - x - 6 = 0

(x - 3)(x + 2) = 0

x - 3 = 0 or x + 2 = 0

x = 3 or x = -2

Test values in the

intervals (-∞, -2),

(-2 , 3), and (3, ∞) in

the original equation:

(-3) 2 - (-3) - 6 > 0

9 + 3 - 6 > 0

6 > 0 True

(0) 2 - (0) - 6 > 0

- 6 > 0 False

(4) 2 - (4) - 6 > 0

16 - 4 - 6 > 0

6 > 0 True

Domain: (-∞, -2) or

(3, ∞)

Range: all real numbers

Write and Evaluate

Functions pp. 85–86

1. 2

2. 1

3. 3

4. 2

5. 3

6. (p + 2) 2 - q = 16 -q = -( p + 2) 2 + 16

q = ( p + 2) 2 - 16

f(p) = ( p + 2) 2 - 16

7. d( p) = p - 2

_____ p2 - 1

d(3s - 1)

= (3s - 1) - 2

___________ (3s - 1) 2 - 1

= 3s - 3 ______________

9s 2 - 6s + 1 - 1

= 3s - 3

_______ 9s2 - 6s

= 3(s - 1)

________ 3s(3s - 2)

= s - 1

_______ s(3s - 2)

8. g(x) = x ___________

x2 + 2x + 1

= x _________

( x + 1 )2

h(t) = g(t + 1)

= t + 1 ___________

((t + 1) + 1)2

= t + 1

______ (t + 2)2

h(t) = t + 1

______ (t + 2)2

h(4) = 4 + 1

_______ (4 + 2)2

= 5 ___

36

9. a. f(x) = (4x - 1) 2 f(2x) = (4(2x) - 1) 2

= (8x - 1) 2

= 64x 2 - 16x + 1

b. 2f(x) = 2(4x - 1) 2

= 2(16x 2 - 8x + 1)

= 32x 2 - 16x + 2

c. No; the two functions

are not equal.

Doubling a function

is not the same as

doubling the variable

of a function.

Evaluate Numerical


1. 4

2. 1

3. 3




4. 3

5. 2

6. g(–a) = 2(–a) = –2a

f( g(–a)) = f(–2a)

= (–2a)2 – 4(–2a) + 1

= 4a2 + 8a + 1

7. (g ° f)(x) = g(f(x))=

g(3 �� x ) = 9 - 2(3 ��

x ) 2

= 9 – 2(9x) = 9 – 18x

8. (f ° g ° h)(x) = f(g(h(x)))

g(h(x)) = g(x – 1)

= -(x - 1) 2 = –(x 2 – 2x

+ 1) = –x 2 + 2x – 1

f(g(h(x))) = f(–x 2 + 2x – 1)

= 2(–x 2 + 2x – 1) + 1

= –2x 2 + 4x – 2 + 1

= –2x 2 + 4x – 1

9. a: C(x(t)) = 3000t + 550

b: 24,550

c: C(x(8)) = $24,550 is

the cost of producing

400 shoes in 8 hours.

Determine if a Function is

One-to-One, Onto, or Both

pp. 89–90

1. 4

2. 2

3. 4

4. 2

5. The function is neither

one-to-one nor onto. It is

not one-to-one because

it has y-values paired

with more than one

x-value. It is not onto

because values less

than 0 on the y-axis are

not used.

22

23

3

5

1

0

6

1

2

P Q

6. y

x

The function has y-values

that are paired with more

than one x-value, so it

is not one-to-one. Every

possible y-value is used, so

it is onto.

7. a. Answers may vary.

Sample:

22

23

3

5

1

0

2

P Q

b. Answers may vary.

Sample:

22

23

3

5

1

0

6

1

2

P Q

c. Answers may vary.

Sample:

422

23

3

5

1

0

2

P Q

Inverses of Functions

pp. 91–92

1. 3

2. 2

3. 1

4. 2

5. 1

6. f –1(–2) = 3 means that

the point (–2, 3) is on

the graph of f –1. So, the

point (3, –2) is on the

graph of f. Therefore,

f(3) = –2.

7. The inverse is {(1.2, –2),

(–1.8, –1), (–2.8, 0),

(–1.8, 1), (1.2, 2)}. Since

the x-values 1.2 and –1.8

are paired with more

than one y-value, the

inverse is not a function.

8. The range of the inverse

is the domain of f. To

fi nd the range of f –1, fi nd

the domain of f, which is

x ≥ 1.

9. a. The inverse of f is a

function because the

graph of f passes the

horizontal line test.

b.

x2

2

y

Evaluate Numerical


1. 2

2. 3

3. 2

4. 4

5. 1

6. f(g(x)) = 4 � 1 __ 4 x -

7 __

4 � + 7

= x – 7 + 7 = x

g(f(x)) = 1 __

4 (4x + 7) -

7 __

4

= x + 7 __

4 -

7 __

4

Since f(g(x)) = x and

g(f(x)) = x, g is the

inverse of f.7. y = 0.25x – 10

x = 0.25y – 10




x + 10 = 0.25y

x + 10

______ 0.25

= y

x ____

0.25 +

10 ____

0.25 = y

4x + 40 = y

8. R is not a function

because there are two

outputs mapped to one

input; R–1 is a function

because each of its

inputs is mapped to one

output.

9. a. f(x) = 2x – 1

y = 2x - 1

x = 2y - 1

x + 1 = 2y

log2 (x + 1) = log

2 2y

log2 (x + 1) = ylog

2 2

log2 (x + 1) = y

f -1 (x) = log2 (x + 1)

b. f(f -1 (x)) = 2log2 (x + 1) - 1

= (x + 1) - 1 = x

f -1(f(x)) =

log2 [(2x - 1) + 1]

= log22x = x

c. f –1(3) = log2 (3 + 1)

= log24 = 2

f –1(–0.5) = log2 (–0.5 + 1)

= log20.5 = log

2 1 __ 2

= log22–1 = –1

Evaluate Numerical


1. 1

2. 3

3. 4

4. 2

5. y = - 1 ____

x - 5 + 8

6. The graph was shifted to

the left 4 units and down 3

units. An equation for the

graph is y = (x + 4) 2 – 3.

7. f(x – 2): (0, 0), (2, 3),

(4, 1), (6, 5)

2f(x – 2): (0, 0), (4, 3),

(8, 1), (12, 5)

2f(x – 2) + 3: (0, 3),

(4, 6), (8, 4), (12, 8)

2 xO

y

2

8. a. For f(x) = x 2 , f(–x)

= (–x) 2 = x 2 . Since

f(–x) = f(x), the

graphs are identical.

For f(x) = x 3 , f(–x)

= (–x) 3 = –x 3 . Since

f(–x) ≠ f(x), the graphs

are not identical.

b. For f(x) = x 2 , –f(x)

= –x 2 and f(–x) = x 2 . Since –f(x) ≠ f(–x),

the graphs are not

identical.

For f(x) = x 3 , –f(x)

= –x 3 and f(–x) =

–x 3 . Since –f(x) =

f(–x), the graphs are

identical.

c. The graph of

f(x) = x 2 is symmetric

about the y-axis. The

graph of f(x) = x3

has 180° rotational

symmetry about the

origin.

Circles pp. 97–98

1. 3

2. 3

3. 4

4. 1

5. 1

6. (x – 4) 2 + [y – (–1)] 2

= (2 ��

3 ) 2

(x – 4) 2 + (y + 1) 2 = 12

7. The center of the given

circle is (0, 0). The

location of this point

when shifted to the left 3

units and down 7 units is

(–3, –7).

8. The center of the circle is

(1, –4) and the radius is

4. The distance from the

center to the given point is

��

(3 - 1)2 + (-4 + 4)2

= ��

4 = 2.

Since the distance from

the center to the point is

less than the radius, the

point lies inside the circle.

9. a. The center is at the

midpoint of the diameter

with endpoints (1, 4) and

(–3, –4):

� 1 + -3 ______

2 ,

4 + (-4) _______

2 � ,

or (–1, 0).

b. The radius is the

distance from the

center to a point on

the circle:

��

(-1 - 1)2 + (0 - 4)2

= ��

4 + 16 = ��

20

= 2 ��

5

c. The center-radius form

of the equation of the

circle is (x + 1) 2 + y 2 = 20.

Expand and simplify the

center-radius form.

x 2 + 2x + 1 + y 2 = 20

x 2 + y 2 + 2x + 1 – 20 = 0

x 2 + y 2 + 2x – 19 = 0

Approximate Solutions to a

Polynomial Equation

pp. 99–100

1. 3

2. 1

3. 1




4. Graph the function and

fi nd the zeros:

ZeroX=-.6666667 Y=0

ZeroX=.33333333 Y=0

ZeroX=.75 Y=0

5. The x-intercepts are the

zeros of the function,

such that if b is an

x-intercept,

(x - b) is a factor.

f(x) = (x + 6)(x + 3)(x - 0)

(x - 2)(x - 3) = x(x + 6)

(x + 3)(x - 2)(x - 3)

6. Graph the function and

fi nd the zeros:

Y=0ZeroX=-1

Y=0ZeroX=1.5

Y=0ZeroX=2

7. a. The x-intercepts, or

roots, are −2, −1, and 1.

b. Since the polynomial

is 4th degree, it must

have four factors.

The point where the

graph touches the

x-axis but does not

cross it represents a

repeated factor. The

factorization is

f(x) = (x + 2)(x + 1)

(x - 1)(x - 1)

c. A new 4th-degree

function with the same

x-intercepts must also

have a repeated factor.

Possible functions:

g(x) = (x + 2)(x + 2)

(x + 1)(x - 1) or

h(x) = (x + 2)(x + 1)

(x + 1)(x - 1)

Determine Domain and

Range pp. 101–102

1. 4

2. 2

3. 1

4. The function has a

vertical asymptote at

x = -3 and extends

without limit along the

positive x-axis. The

domain is (-3, ∞).

5. The holes at (-1, 3) and

(1, -3) indicate that the

function is not defi ned at

these points. Exclude -1

and 1 from the domain.

Exclude -3 and 3 from

the range. The function

has a maximum at

(-1, 3) and a minimum

at (1, -3), so the range

falls between -3 and 3.

The domain is (-∞, ∞),

x ≠ -1 and x ≠ 1. The

range is (-3, 3).

6.

p2

p2

p2

3p2

3p2

3p2

3p2

y

O x2

2

p2p2

2

a. The function has

vertical asymptotes at

odd multiples of π __ 2 .

The domain is

(-∞, ∞), x ≠ ... - 3π ___

2 , π __

2 ,

π __ 2 , 3π

___ 2 , ...

b. The function never

gets closer to the

x-axis than a distance

of 2. The range is

(-∞, -2] and [2, ∞).

Identify Relations and


1. 1

2. 3

3. 4

4. 1

5. Yes; there is no vertical

line that intersects the

graph in more than one

point.

6. No, almost any vertical

line x = a, where a is in

the domain of the relation,

intersects the graph in

more than one point.


Check students’ work

for two points with

different x-coordinates

and the same

y-coordinates.


Check students’ work

for a relation that fails

the vertical line test.

Sample:




y = ± �� x

y

xO 4222

22

24

2

4

The relation is not a

function because you

can draw a vertical line

that intersects it in more

than one point.

Exponential Functions

pp. 105–106

1. 2

2. 3

3. 2

4. 1

5. 3

6.

y

x

8

7

6

5

1 2 3 4

3

4

21222324

2

1

O

7. y = -f(x) + 4

y = - � 1 __ 4 � x + 4

8. Graph the equations y

= 2 x and y = 4 x on the

same coordinate plane

and compare the graphs.

The solutions to the

inequality correspond

to the x-values in the

domain where the graph

of 4x is above the graph

of 2x.

The solutions to the

inequality are all x such

that x > 0.

9. a. The graph is

translated to the left 3

units and up 1 unit.

b. The domain is

(-∞, ∞) and the

range is (1, ∞).

Since g(0) = 9 0+3 +

1 = 729 + 1 = 730,

the y-intercept of the

graph is (0, 730).

The horizontal

asymptote is translated

up 1 unit. So the line

y = 1 is a horizontal

asymptote of the graph.

The graph increases

over the domain.

Graph Logarithmic


1. 1

2. 2

3. 3

4.

x 0 1 2

10x 1 10 100

x 1 10 100

log10

x 0 1 2

10 20 30 40 50 60 70 80 90100

21

1

2

(100, 2)

(10, 1)

(1, 0)

5. You cannot take the log

of a number less than or

equal to zero. So,

x + 2 > 0

x > -2.

The domain is all real

numbers greater than

-2. The range of any

log function is all real

numbers.

6. Two points on the graph

of g(x) = logbx are (1, 0)

and (b, 1). Use these

points and the refl ection

of the graph of f(x) = bx

over the line y = x to

sketch the graph of g(x).

212 3 4

22

2

3

4f(x)5bx

(b, 1)

(1, 0)g(x)5logb

x

y

x

7.

x 0 1 2

f-1(x) = 3x 1 3 9

x 1 3 9

f(x) = log3x 0 1 2

x 0 1 2

g-1(x) = 4x 1 4 16

x 1 4 16

g(x) = log4x 0 1 2

Graph f(x) and g(x).

O 84

1

2

12 16 18

21

2428

f (x)5log3x

g(x)5log4x

y

x

a. f(x) > g(x) when

x >1, or on the

interval (1, ∞).

b. f(x) < g(x) when x < 1,

or on the interval (0, 1).

c. f(x) = g(x) at x = 1.

Express and Apply the Six

Trigonometric Functions

pp. 109–110

1. 2

2. 4

3. 2

4. tan X = 0.7

= length of leg opposite ∠X

_________________________ length of leg adjacent to ∠X

= YZ ___ XZ




0.7 = YZ ___ 14

YZ = 9.8

5. cos 60°

= length adjacent to ∠60°

___________________ hypotenuse

= 30 ___ w

0.5 ≈ 30 ___ w

w ≈ 30 ___ .5

≈ 60

6. sin A = 2 __ 3

= length of leg opposite ∠ A

_____________________ length of hypotenuse

length of leg opposite ∠ A

= 2

length of hypotenuse = 3

C

23

bA

B

a 2 + b 2 = c 2 b 2 = c 2 - a 2

b =

��

c2 - a2

=

��

32 - 22

= ��

9 - 4

= ��

5

sec A =

length of hypotenuse

_______________________ length of leg adjacent to ∠ A

= 3 __ b

= 3 ___ �

� 5 = 3 ___

��

5 · �

� 5 ___

��

5

= 3 �

� 5 ____

5

7. a. sin S =

length of leg opposite ∠S

_____________________ length of hypotenuse

= TU ___ ST

= 5 ___ 13

cos S =

length of leg adjacent to ∠S

_______________________ length of hypotenuse

= US ___ ST

= 12 ___ 13

tan S =

length of leg opposite ∠S

_______________________ length of leg adjacent to ∠S

= TU ___ US

= 5 ___ 12

b.

sin S _____ cos S

= 5 ___ 13

___

12 ___ 13

= 5 ___

13 · 13 ___

12

= 5 ___ 12

= tan S

c. (sin S ) 2 + (cos S ) 2 =

� 5 ___ 13

�

2

+ ( 12 ___ 13

) 2

= 25 ____ 169

+ 144 ____ 169

= 169 ____ 169

= 1

Values of Sine, Cosine, and

Tangent Angles pp. 111–112

1. 1

2. 3

3. 1

4. 2

5. 2

6. sin2 45° + cos2 60° =

( ��

2 ___ 2 )

2

+ � 1 __ 2 � 2

= 2 __ 4 + 1 __

4

= 3 __ 4

7. tan 2 2π + 1 __ 2 cos 45° =

(0)2 + � 1 __ 2 � � ��

2 ___ 2 � = �

� 2 ___

4

8. cot 0° is undefi ned.

Using reciprocal

relationships,

cot 0° = 1 _____ tan 0°

= 1 __ 0 ,

which is undefi ned.

9. θ = 45°. Using

knowledge of special

angles, an angle θ for

which sin θ = cos θ is

θ = 45°.

Now verify tan 45°

= cot 45°:

cot 45° = 1 ______ tan 45°

= 1 __ 1 = 1 = tan 45°

Verify sec 45° = csc 45°:

sec 45° = 1 ______ sin 45°

=

1 ______ cos 45°

= csc 45°

Reference Angles in

Standard Position

pp. 113–114

1. 3

2. 2

3. 1

4. 4

5. 3

6.

x

y

2408

608

Referenceangle

7.

x

y

3158

458

Referenceangle

8. The angle θ = 924° is

coterminal with the angle

α = 204°. Therefore, θ′ = 204° - 180° = 24°.

9. a. The sign of the

trigonometric function

will be determined

by the quadrant in

which the original θ

terminates. Once the

value of the reference

angle θ′ is found, it is

important to apply the

sign of the quadrant

in which the original

angle was found.

b. sin = 5π ___

4 is in

quadrant III, where the

sign is negative. Its

reference angle is

5π ___

4 - 4π

___ 4 = π __

4 . The

sin π __ 4 = �

� 2 ___

2 , so

sin = 5π ___

4 = - �

� 2 ___

2 .




Co-Function and

Reciprocal Relationships

pp. 115–116

1. 2

2. 2

3. 1

4. 3

5. 4

6. cot x _____ csc x =

cos x _____ sin x

______

1 _____

sin x =

cos x _____ sin x

· sin x _____ 1 = cos x

7. csc 34° = sec (90° - 34°)

= sec 56°

8. To show that sine and

cosecant are reciprocal

functions, show (sin θ)

(csc θ) = 1. Using

the defi nitions of

trigonometric functions,

(sin θ)(csc θ) = opp

____ hyp

·

hyp

____ opp = 1. Thus,

sin θ = 1 ____ csc θ

.

9. tan (90° - θ)

= sin (90° - θ)

___________ cos (90° - θ)

= cos θ _____

sin θ = cot θ

Similarly,

cot (90° - θ)

= cos (90° - θ)

_____________ sin (90°- θ)

= sin θ _____

cos θ

= tan θ

Values of Secant,

Cosecant, and Cotangent

Angles pp. 117–118

1. 3

2. 2

3. 1

4. 4

5. 1

6. cot 45° - csc 30°

= 1 ______ tan 45°

- 1 ______ sin 30°

= 1 __ 1 - 1 __

1 __ 2 = 1 - 2

= - 1

7. cot 60° + csc 270°

= tan(90° - 60°)

+ 1 _______ sin 270°

= tan 30° + 1 ___

-1

= �

� 3 ___

3 - 3 __

3 =

��

3 - 3 ______

3

8. An equivalent expression

is cot 5°.

The angle 265° is in

quadrant III. Its reference

angle is 85°. Thus,

tan 265° = tan 85°.

To fi nd an angle of

less than 45°, use the

co-function relationship

tan 85° = cot(90° - 85°)

= cot 5°.

9. Show sec θ __ 2 ≠ 1 __

2 sec θ by

using a counter-example.

Let θ = 60°,

sec 60° ___ 2

= sec 30°

= 1 ______ cos 30°

= 1 ___

�

� 3 ___

2

= 2 __ 3

But,

1 __ 2 sec 60°

= 1 __ 2 · 1 ______

cos 60°

= 1 __ 2 · 1 __

1 __ 2

= 1 __ 2 · 2 = 1

Therefore, sec θ __ 2 = 1 __

2

sec θ is not an identity.

The Unit Circle pp. 119–120

1. 1

2. 3

3. 4

4. 2

5. 1

6.

y(0, 1)

(1, 0)

(0, 21)

(21, 0) x

7. θ = -π crosses the unit

circle at point (-1,0).

8.

x

y

(0, 1)

(1, 0)

(0, 21)

(21, 0)

α = 5π ___

2 crosses the unit

circle at point (0,1) and

shares a terminal side

with special angle π __ 2

.

9. a., b., c.

(cos(A), sin(A))

(cos(2A), sin(2A))

A2A

d. If A is an angle in

standard position

crossing the unit

circle at (x, y), then

x = cos A and y =

sin A. The x-values in

quadrant I and IV are

positive, thus cos(-A)

= cos A. y-values

in quadrant I are

positive and y-values

in quadrant IV are

negative. Therefore,

sin(-A) = -sin A.




Determine the Arc Length

of a Circle pp. 121–122

1. 1

2. 2

3. 3

4. 3

5. r = s __ θ

= 6 ___ 1.5

= 4

6. s = rθ

= 8 in. · 2 1 __

4

= 8 in.

7. The arc length s = 5 1 __ 2 .

The pie is divided into

8 equal pieces, so the

radian measure of the

central angle of one

piece is θ = 2π · 1 __ 8 = π __

4 .

Thus, the diameter

of the pizza is d = 2r

= s __ θ

= 5 1 __

2 ___

π __ 4 ≈ 14.01. To

the nearest inch, the

diameter of the pizza is

14 inches.

8. First use the arc length

formula to fi nd the

measure of the angle

in radians. The radius r of the track is given as

r = 150. The arc length

created by racing from

point A to point B is s =

247. Using the formula,

the angle measure is

θ = 247 ____ 150

radians. Now

convert radians into the

nearest degree:

θ = 247 ____ 150

· 180 ____ π ≈ 94°

Values of Trigonometric

Functions. pp. 123–124

1. 1

2. 3

3. 4

4. 1

5. 2

6. tan θ = y __ x = 0 __

5 = 0

7. cot θ = x __ y = 5 __ 0 , which is

undefi ned.

8. First fi nd r.

r = ��

(-3) 2 + (-2) 2

= ��

9 + 4

= ��

13

Thus, sin θ = -2 ____ �

� 13

= - 2 ____ �

� 13 ,

cos θ = -3 ____ �

� 13 = -3 ____

��

13 ,

and

tan θ = -2

___ -3

= 2 __ 3 .

9. The terminal side of

angle θ must lie in

quadrant IV; this is the

only quadrant for which

sin θ and tan θ are both

negative. Because

tan θ = 7 __ 5 =

y __ x , and the

terminal side of θ is in

quadrant IV, we know

that y must be negative

and x must be positive.

Thus, y = -7, x = 5, and

r = ��

5 2 - 7 2 = ��

74 .

Therefore, cos θ = x __ r

= 5 ____ �

� 74 and csc θ =

r __ y

= ��

74 ____ -7

= - ��

74 ____ 7 .

Inverse Functions

pp. 125–126

1. 3

2. 2

3. 4

4. 1

5. 3

6. The range for y = cos x

is the same as the

domain of y = cos-1 x

with x and y

interchanged. Thus, the

range is -1 ≤ y ≤ 1.

7. Yes; x = - π __ 2 in the

domain. The domain

for y = tan-1x is all real

numbers.

8. Over the interval - π __ 2

< x < π __ 2 , the range for

y = tan x is -∞ < y < ∞.

The domain for

y = tan-1x is the same

as the range for

y = tan x. Thus, the

domain for y = tan-1x

is -∞ < x < ∞.

9. Only one-to-one

functions have inverses.

If a function has the

same output for more

than one input, then it

is not one-to-one and,

therefore, cannot have

an inverse. For example,

sin π __ 4 = �

� 2 ___

2 and

sin 3π ___

4 = �

� 2 ___

2 . The

domain for which

y = sin x has a unique

output for each input is

- π __ 2 ≤ x ≤ π __

2 .

Using Inverse Functions

pp. 127–128

1. 3

2. 2

3. 3

4. 4

5. 1

6. cos-1 1 __ 2 = 60° = π __

3

because cos 60° = 1 __ 2

on

the interval [0, π].

7. tan-1 1 = 45° = π __ 4

because tan-1 45° = 1

on the interval [ - π __ 2 , π __

2 ] .

8. The angles by which

sin -1 x = cos -1 x is 45°

and 225°. At 45°, both

sin 45° and cos 45° is




�

� 2 ___

2 . At 225°, both

sin 225° and cos 225°

is - ��

2 ___ 2 .

9.

A C

B

5

u

2

Since sin-1 2 __ 5 is the angle

θ for which sin θ = 2 __ 5 and

sin θ = opp

____ hyp

, draw a

right triangle ABC with

angle θ and solve for

the adjacent side AC.

Using the Pythagorean

theorem,

AC = ��

25 - 4 = ��

21 .

Thus, tan � sin-1 � 2 __ 5 � �

= tan θ = opp

____ adj

= 2 ____ �

� 21

= 2 ��

21 _____ 21

.

Graphs of Inverses

pp. 129–130

1. 3

2. 2

3. 3

4. 4

5. 1

6. The graph of arctangent

has asymptotes. There

is no restriction on the

domain, but the values

of the range get closer

and closer to the values

y = - π __ 2 and y = π __

2

without ever reaching

these values.

7. No; by the defi nition of

inverse function,

y = tan-1x if and only if

x = tan y. Since tan π __ 2 is

undefi ned, there is no

value of x such that

tan-1x = π __ 2 .

8.

22p

2p

2p

p

121

y

x

A function must be one-

to-one in order to have

an inverse. Graphically

this means it must pass

the horizontal line test

and the vertical line test.

The graph does not

satisfy the vertical line

test. Therefore x = sin y

does not have an inverse

on the interval.

Determine Trigonometric


1. 1

2. 4

3. 3

4. 4

5. 1

6. sec 48° = 1 ______ cos 48°

= 1 ___________ 0.669130606

≈ 1.494

7. First convert to degrees:

π __ 8 · 180° ____ π = 22.5°. Then

evaluate

cot π __ 8 = 1 _____

tan π __ 8 = 1 _______

tan 22.5°

= 1 ___________ 0.414213562

= 2.414

8. Set up an equation to

solve for x using the

information given in the

triangle. By the defi nition

of trigonometric ratios,

sin 31.3° = opp

____ hyp

= x ____ 6.21

.

Now solve for x:

x = 6.21 · sin 31.3°

= 3.2262

9.

18 in.

17.5°

d

Using trigonometric

ratios, the height is:

sin 17.5° = opp

____ hyp

= 18 ___ d

(sin 17.5°)(d) = 18

d = 18 _______ sin 17.5°

≈ 60

To the nearest inch,

the length of the board

should be 60 in.

Pythagorean Identities

pp. 133–134

1. 2

2. 1

3. 3

4. 2

5. 4

6. sin 2 x + cos 2 x = 1

sin2 x _____ cos2 x

+ cos2 x _____ cos2 x

= 1 _____ cos2 x

tan 2 x + 1 = sec 2 x7. cos 2 x(1 + tan 2 x)

= cos 2 x(sec 2 x)

= cos 2 x ( 1 _____ cos2 x

) = 1

8. Assume the premise.

Use the identities

and defi nitions of trig

functions to rewrite both

sides of the equation to

prove the equation is true.

tan x + cot x = csc x sec x

sin x ____ cos x + cos x ____ sin x

= 1 ____ sin x

· 1 ____ cos x

cos2 x + sin2 x ___________

cos x · sin x = 1 _________

cos x · sin x

1 _________ cos x · sin x

= 1 _________ cos x · sin x

9. Assume the premise.

Use the identities

and defi nitions of trig

functions to rewrite both

sides of the equation to

prove the equation true.




sec x + csc x __________ tan x + cot x = sin x + cos x

1 ____ cos x + 1 ____

sin x ___________

sin x ____ cos x + cos x ____ sin x

= sin x + cos x

sin x + cos x __________

sin x cos x _____________

sin2 x + cos2 x _____________ sin x cos x

= sin x + cos x

sin x + cos x __________

sin x cos x · sin x cos x ________

1

= sin x + cos x sin x + cos x = sin x + cos x

Solve Trigonometric


1. 1

2. 1

3. 3

4. 2

5. 4

6. 2(2 + cos x) = 3 + cos x

4 + 2 cos x = 3 + cos x

cos x = -1

x = 180°

7. 3 tan x - 3 = 0

3 tan x = 3

tan x = 1

x = 45°, 225°

8. There are no like terms

to combine and there is

no common factor, so

substitute 1 ____ sin x

for csc x.

4 sin x - csc x = 0

4 sin x - 1 ____ sin x

x = 0

4 sin2 x - 1

_________ sin x

= 0

4 sin 2 x - 1 = 0

4 sin 2 x = 1

sin 2 x = 1 __ 4

sin 2 x = ± 1 __ 2

x = 30°, 150°, 210°, 330°

9. There are no like terms

to combine and there is

no common factor, so

substitute 1 ____ tan x for cot x.

tan x + 3 1 ____ tan x = 4

tan x + 3 ____ tan x - 4 = 0

tan2 x - 4 tan x + 3

_______________ tan x = 0

tan 2 x - 4 tan x + 3 = 0

(tan x - 3)(tan x - 1) = 0

tan x - 3 = 0

tan x - 1 = 0

tan x = 3 tan x = 1

The tangent of x is

positive in quadrants

I and III. Use inverse

functions.

For x = tan-1 3, use a

calculator:

x ≈ 72° and 72°

+ 180° = 252°

For x = tan-1 1:

x ≈ 45° and 45° + 180°

= 225°

Determine Amplitude,

Period, Frequency, and

Phase Shift pp. 137–138

1. 2

2. 2

3. 1

4. 3

6. |-3| = 3; 2π ___

2/3 = 3π

7. |-2| = 2; 2π ___

3/4 = 8π

___ 3

8. The graph of the cosine

function is just the graph

of the sine function but

shifted to the left

90° = π __ 2 radians.

9. a. 4°

b. t = 5°, 9°, 13°, …

c. There is no vertical shift

that will produce this

result. The function is

undefi ned at θ = 3°,

7°, 11°, …, and any

vertically shifted function

will also be undefi ned at

those values.

Sketch and Recognize


1. 1

2. 3

3. 2

4. 4

5. domain: all real numbers,

range: -32 ≤ x ≤ 32

6. domain: all real numbers,

range: - ��

2 ___ 2 ≤ x ≤ �

� 2 ___

2

7. a. The amplitude = 2

and the period = 3π

= 2π ___

b , A = 2 and

B = 2 __ 3 , so the

equation is

y = 2 sin 2 __ 3 x.

b.y

x

21.5

3 6 12

8. a., b.

1 2 3 4 5

1

3

4

5

2425

b

y

x21

23

24

25

21 O

c. y = -2 cos x

d. When x = π __ 6 ,

y = -2cos x

= -2cos π __ 6

= -2 ( �

� 3 ___

2 )

= - ��

3

The Graphs of y = sec(x),

y = csc(x), y = tan(x), and

y = cot(x) pp. 141–142

1. 2

2. 3

3. 1

4. 4




5.

2p2p2

y

x1

2

3

4

21

23

24

6.

2p2p2

y

x

2

3

1

4

22

23

21

24

7. a.

1 2 3 4 5

y

x

When f(x) 5 3, x <1.57

b. Using the graphing

calculator, trace the

graph f(x) = 3, when

x ≈ 1.57.

8. a. The period of the

function is

2π ___

C = 2π

___ π __ 2 = 4.

The phase shift is

D __ C

= π __ 2 __

π __ 2 = 1.

b. The asymptotes are:

π

__ 2 x + π __

2 = 0 π __

2 x + π __

2 = π

π __ 2 x = - π __

2 π __

2 x = π __

2

2x = -2 x __ 2 = 1 __

2

x = -1 x = 1

π __ 2 x + π __

2 = 2π and

π

__ 2 x = 3π

___ 2

x __ 2 = 3 __

2

x = 3

c.

3 7

y

x

22

21

1

3

4

23

24

y 5 csc( )p2

p21x1

2

Periodic Graphs pp. 143–144

1. 4

2. 3

3. 1

4. 2

5. 4

6. The period is 2π ; the

amplitude is 3.

7. Using the results from

problem 6, we fi nd that

A = 3 and B = 1.

In looking at the graph,

there is no vertical

shift. Thus, D = 0. The

equation is y = 3 sin x.

8. a. The vertical translation

D is the average of

the minimum and

maximum values of

the range. Thus,

D = 5 + 8

_____ 2 = 6.5.

b. The amplitude A is the

distance from D to the

minimum or maximum

of the range.

A = |8 - 6.5| = |5 - 6.5| = 1.5

c. The period of the

function is 4π and

period = 2π ___

B . Thus,

4π = 2π ___

B

B = 2π ___

4π = 1 __

2 .

9. The vertical translation

D is the average of the

minimum and maximum

values of the range.

Thus, D = 1 + 5

_____ 2

= 3.

The amplitude A is the

distance from D to the

maximum or minimum.

Thus, A = |5 - 3| = |1 - 3| = 2.

To compute B, use the

fact that the period is 4π

and that period = 2π ___

B .

Thus,

4π = 2π ___

b

B = 2π ___

4π = 1 __

2 .

The line creates a

wave with the equation

y = 2 sin 1 __ 2

x + 3.

Law of Sines and Law of

Cosines pp. 145–146

1. 4

2. 3

3. 3

4. 1

5. 4

6. Use the Law of Sines:

p ____

sin P = r ____

sin R

p ___

0.2 = 22 ___

0.4

0.4p = 0.2(22)

p = 0.2(22)

______ 0.4

= 11

7. Use the Law of Cosines:

c 2 = a 2 + b 2 - 2ab cosC

c 2 = 19 2 + 26 2

- 2 · 19 · 26 · cos 42°

c 2 = 302.7729

c = 17.400 ≈ 17

8. First use the measures

of angles B and C to fi nd

the measure of angle A.




∠ A = 180° - (∠ B + ∠ C)

= 180° - (80° + 34°)

= 66°

Now use the Law of

Sines:

a ____ sin A

= b ____ sin B

16 ______ sin 66°

= b ______ sin 80°

(sin 66°)b = 16(sin 80°)

b = 16(sin 80°)

_________ sin 66°

b = 16(0.948)

________ 0.9135

b = 17.248 = 17.2

9. a.

BA

CD

40

221168

b

b. We are given two

sides and one angle

in ABC. Use the

Law of Cosines to fi nd

side b.

b 2 = a 2 + c 2 - 2ac cos B

b 2 = 22 2 + 40 2 - 2 · 22 · 40 · cos 116°

b 2 = 484 + 1600

- 1760(-0.43837114)

(The cosine of an

obtuse angle is

negative.)

b 2 = 2855.5332

b ≈ 53.437 ≈ 53

The larger diagonal is

about 53 cm.

Areas of Triangles and

Parallelograms pp. 147–148

1. 2

2. 3

3. 1

4. 4

5. 3

6. Area = 1 __ 2 ef sin D

= 1 __ 2 · 10 · 8 · sin 45°

≈ 40 · (0.70710678)

≈ 28.3

7. Area = 1 __ 2 ab sin A

= 1 __ 2 · 24 · 24 · sin 30°

= 288 · (0.5)

= 144 m2

8. a. A

B

D

C

1308508

15

15

b. Use the area formula

for a triangle.

Area Rhombus = 2(Area

BCD)

= 2 ( 1 __ 2 ab sin C)

= 2 � 1 __ 2 15 · 15 sin 50° �

= 225 · (0.7660444431)

= 172.3599997

≈ 172 units 2

c. Using the area formula

for a parallelogram:

A = ab sin C A = 15 · 15 · sin 130°

A = 172.3599997

≈ 172 units 2

9. a.

CB

A

808

320

450

b. Area = 1 __ 2 ac sin B

= 1 __ 2 320 · 450 · sin 80°

= 72,000 · (0.984807753)

= 70,906 yards 2

c. 70,906 ÷ 100

≈ 709.06 bags

Since partial bags

cannot be purchased,

round up to 710.

710 · $4.50

= $3,195.00

Determine SSA Solutions

pp. 149–150

1. 1

2. 1

3. 3

4. 4

5. 1

6. 12 ____ sin A

= 7 _____ sin 46

, so

sin A ≈ 1.23. Since the

sine function has a

range of [-1, 1], this is

not a possible value, and

no such angle exists.

7. 23 ____ sin A

= 29 _____ sin 55

, so

sin A ≈ .65 and m∠ A

≈ 41°. Since sine is

positive in Quadrant 2,

180° - 41° = 139° also

has a sine of .65. But, if

m∠ A = 139°, the sum

of the angles of triangle

ABC will exceed 180°.

8. 9 ____ sin A

= 12 _____ sin 26

, so

sin A ≈ .33 and

m∠ A ≈ 19°. m∠ B

= 180° - 26° - 19°

= 135°, so

12 _____ sin 26

= b ______ sin 135

, and

b ≈ 19.

Using 19° as a reference

angle in Quadrant 2,

161° also has a sine

of .33, but if m∠ B =

161°, the angle sum

will exceed 180˚, so a

second triangle is not

possible.

9. 11 ____ sin C

= 7 _____ sin 29

,

so sin C ≈ .99 and m∠ C

≈ 82°. m∠ A = 180°

- 29° - 82° = 69°, so

7 _____ sin 29

= a _____ sin 69

, and

a ≈ 13.




Using 82° as a reference

angle in Quadrant 2,

98° also has a sine of

.99. A second triangle is

possible: m∠ C = 98°.

m∠ A = 180° - 29° -

98° = 53°, so

7 _____ sin 29

= a _____ sin 53

, and

a ≈ 12.

Apply Angle Sum and

Difference Formulas

pp. 151–152

1. 4

2. 4

3. 1

4. 2

5. 1

6. Proof:

cos 90° = cos(45° + 45°)

= cos 45° cos 45°

- sin 45° sin 45°

= ��

2 ___ 2 · �

� 2 ___

2

- ��

2 ___ 2 · �

� 2 ___

2

= 0

7. Proof:

tan 2π ___

3 = tan � π __

3 + π __

3 �

= tan π __

3 + tan π __

3 _____________

1 - tan π __ 3 · tan π __

3

= �

� 3 + �

� 3 __________

1 - ��

3 · ��

3

= 2 �

� 3 ____

1 - 3

= - 2 �

� 3 ____

2

= - ��

3

8. Use the formula

sin(A - B) = sin A cos B - cos A sin B

sin 98°cos 8° - cos 98° sin 8°

= sin (98° - 8°)

= sin 90°

= 1

9. To use the sum and

difference functions, we

will need the sine and

cosine of both A and B.

Using the Pythagorean

formula and knowledge

of the signs of

trigonometric functions

in the quadrants, angle

A is made by a triangle

with legs of length 8 and

15 and hypotenuse 17.

So, sin A = 8 ___ 17

and

cos A = 15 ___ 17

. Angle B is

made by a triangle with

legs of length 7 and

24 and hypotenuse 25.

Since it is in quadrant 2,

its cosine is negative.

sin B = 7 ___ 25

and

cos B = - 24 ___ 25

.

a. Use these values to

compute.

sin(A - B)

= sin A cos B - cos A sinB

= 8 ___ 17

� - 24 ___ 25

� - 15 ___ 17

� 7 ___ 25

� = -

192 - 105 _________

425

= - 397 ____ 425

b. Use the values

computed above to

compute cos(A + B).

cos(A + B)

= cos A cos B - sin A sin B

= 15 ___ 17

� - 24 ___ 25

� - 8 ___ 17

� 7 ___ 25

� = - 260 - 56

________ 425

= - 416 ____ 425

Apply Double-Angle and

Half-Angle Formulas

pp. 153–154

1. 2

2. 1

3. 4

4. 4

5. 1

6. Using the Pythagorean

theorem, if sin θ = 5 ___ 13

,

cos θ = - 12 ___ 13

since

cosine is negative in

Quadrant 2.

sin 2θ = 2 sin θ cos θ

= 2 � 5 ___ 13

� � - 12 ___ 13

� = - 120 ____

169

7. Use the half-angle

function. Choose the

positive root since

tangent is positive in

Quadrant 1.

tan 22.5° = tan � 1 __ 2

· 45° � = + �

�

1 - cos 45° __________ 1 + cos 45°

= + ��

1 - �

� 2 ___

2 _______

1 + ��

2 ___ 2

= + ��

2 - �

� 2 ______

2 _______

2 + �

� 2 _______

2

= + ��

2 - ��

2 _______ 2 + �

� 2

8. Use the double angle

function and simplify.

sin 2A _____ tan A

= 2 sin A cos A __________

sin A _____ cos A

= 2 sin A cos A · cos A _____ sin A

= 2 cos A cos A = 2 cos2 A9. Use the double angle

function and simplify.

sin 2B _________ cos 2B + 1

= 2 sin B cos B ____________ 2 cos2 B - 1 + 1

= 2 sin B cos B _____________ 2 cos B cos B

= sin B _____ cos B

= tan B




Defi ne and Convert

Radians pp. 155–156

1. 3

2. 1

3. 4

4. 2

5. 3

6. 73π ____

60 radians

7. 202.5°

8. m∠B = 11π ____

10 · 180 ____ π =

= 198°. Angle B is larger.

9. Area = πr 2 = 36π, so

the radius of circle T is

6 cm. A central angle

of 1 radian has an

intercepted arc of 6 cm.

The intercepted arc of

∠M is 5 cm, so

m∠M = 5 __ 6 radians.

Understand Kinds of

Studies and Survey

Factors pp. 157–158

1. 3

2. 2

3. 1

4. 1

5. 4

6. In a survey, data is

gathered by asking

questions. In an

observation, the data is

gathered by watching the

study subjects; they are

not asked any questions.

7. Shoppers at a sporting

goods store are not

representative of the

general population. They

may be more likely to

have a positive opinion

of a sport.

8. Students taking

advanced placement

classes may hold

different attitudes than

students in general. The

fact that the researcher

is videotaping responses

may make some student

uncomfortable in giving

an honest answer.

9. a. Send a written survey

to all addresses in the

city.

b. Survey people who

are leaving a library.

Calculate Measures

of Central Tendency

pp. 159–160

1. 4

2. 1

3. 2

4. 3

5. 1

6. There are an even

number of data elements

in this data set. The

middle two numbers are

15 and 22. The median

is the average of these

two numbers, so the

median is 18.5.

7. The missing data

element must be 18. If

it were 15, the mode

would be 15. If it were

any number other than

15 or 18, the data would

be bimodal with modes

of 15 and 18.

8. Call the missing data

element d.

24 + 13 + 30 + 21 + d ___________________

5

= 21.2

24 + 13 + 30 + 21 + d

= 106

d = 18

9. a. 62.8

b. 68

c. 72 and 89

Calculate Measures of

Dispersion pp. 161–162

1. 1

2. 4

3. 3

4. 4

5. 4

6. Quartile 1 is 20.5 and

Quartile 3 is 32.5, so the

interquartile range is 12.

7. The interquartile range

shows the range of the

middle half of the data.

8. If all of the data

values are doubled,

the standard deviation

is also doubled. This

happens because

doubling the numbers

causes the mean to

double, which causes

the difference between

the data values and

the mean to double.

Because this number is

squared, the variance

increases by a factor

of 4. Since the square

root of the variance is

taken to get the standard

deviation, the factor of

change for the standard

deviation is 2.

9. a. Quartile 1 = 9,

quartile 2 = 16.5,

quartile 3 = 22.5.

Interquartile range

= 22.5 - 9 = 13.5

Range = 23 - 4 = 19

b. Mean = 15.5

Variance = 50.1

(remember to divide

by 11, not 12, since

this is sample data)

c. Standard deviation

= ��

50.1 ≈ 7.1

Understand Normal

Distributions pp. 163–164

1. 4

2. 3

3. 4

4. 2

5. 2




6. The middle 86.6%

covers from 1.5 standard

deviations below the

mean to 1.5 standard

deviations above the

mean. So, 5.0 - 1.4

= 3.6 is 3 standard

deviations, and

1 standard deviation is

3.6 ÷ 3 = 1.2.

7. The mean is the center

of the curve, so mean

= 45. Since about 34%

of the data is 1 standard

deviation above or below

the mean, the standard

deviation is 9.

8. In a normal distribution,

about 2.3% of the data

is above 2 standard


mean. So,

x = 63.2 + 2(2.9) = 69.

9. a. About 69% of the data

in a normal distribution

is below .5 standard


mean. Maria’s score is

81 + .5(9.7) ≈ 86.

b. The lowest passing

score is

81 - 1.5(9.7) ≈ 66. In

a normal distribution,

about 93.3% are

above 1.5 standard

deviations below the

mean.

Determine and Interpret the

Function for a Regression

Model pp. 165–166

1. 3

2. 4

3. 3

4. 1

5. Data is interpolated

when a y-value is found

for an x-value that is

within the x-values that

are in the given data.

Data is extrapolated

when a y-value is found

for an x-value that is

larger or smaller than all

of the x-values that are

in the given data.

6. a. y = 4x + 85

b. $117

7. a. logarithmic

b. y = 5.1 + 2.2 ln x c. 10.6

Differentiate between

Combinations and

Permutations pp. 167–168

1. 2

2. 3

3. 2

4. 1

5. Combination; the order

of the shirts has no

meaning in this situation.

6. Permutation; the order

of the winners matters

because it determines

in which place they

fi nish.

7. a. The combination is

the group of eight

employees selected

for the special project,

because the order of

the eight employees

has no meaning.

b. The permutation is

selecting a project

manager and a time

recorder from the

group, because these

two positions are

different.

8. a. This scenario is a

permutation if the

order of the books

matters. For example,

if Joachim reads the

books in the order

they are selected, then

it is a permutation.

b. This scenario is a

combination if the

order of the books

does not matter. For

example, if Joachim

is simply selecting

the books now

and will decide the

order to read them

in later, then it is a

combination.

Calculate Number of

Permutations pp. 169–170

1. 4

2. 3

3. 4

4. 3

5. 2

6. Since there is no choice

for the fi rst fl oat and the

last fl oat, the remaining

nine fl oats must be

arranged in the middle

nine positions. There are

9P

9 = 9! __

0! = 362,880 ways

to do this.

7. All seven letters are

unique so there are 7P

4

ways of arranging the

seven letters into 4-letter

“words.” Therefore, there

are 7P

4 =

7! __

3! = 840

“words.”

8. For the two digits,

there are

10

P2 = 10! ___

8! = 90

possible choices.

For the four letters,

since uppercase and

lowercase letters are not

the same, there are

52

P4 = 52! ___

48! = 6,497, 400

possible choices. Since

any of the two digit

number permutations

can be combined with

any of the four letter

permutations, there are

90 · 6,497,400

= 584,766,000 possible

passwords.




9. If the quilter uses n

fabrics and each quilt

block uses 3 of these

fabrics as described,

there are nP3 possible

blocks she can make.

We need to fi nd n so

that nP3 ≥ 200. Solve by

guess and check:

5P

3 = 5! __

2! = 60

6P

3 = 6! __

3! = 120

7P

3 = 7! __

4! = 210

Therefore, she must use

7 different fabrics.

Calculate Number of

Combinations pp. 171–172

1. 3

2. 2

3. 1

4. 3

5. 1

6. One student (the student

who volunteered) has

already been selected,

so four more students

must be selected

from the remaining

18 students. There are

18

C4 = 18! _____

4!14! = 3060

ways to select this group.

7. The number of games

that need to be played is

the same as the number

of ways to select two

teams from 21 teams.

21

C2 = 21! _____

2!19! = 210

8. There are 4C

1 = 4! ____

1! 3!

= 4 ways to select the

center,

8C

2 = 8! ____

2!6! = 28 ways

to select the guards, and

9C

2 = 9! ____

2!7! = 36 ways

to choose the forwards.

Since each selection

of center can play with

each pair of guards and

each pair of forwards,

there are 4 · 28 · 36

= 4032 possible ways

to select players to play

together.

9. First the principal selects

the 5 students for the

trip. There are

15

C5 = 15! _____

5!10! = 3003

ways to do this. Then

the principal will select

4 students from the

remaining 10 students

to work on the service

project. There are

10

C4 = 10! ____

4!6! = 210

ways to select these

students. Finally, the

principal will select

3 students to plan

graduation from the

remaining 6 students.

There are

6C

3 = 6! ____

3!3! = 20 ways

to do this. Altogether,

there are 3003 · 210 · 20 = 12,612,600 ways to

select the three groups.

Number of Elements in a

Sample Space pp. 173–174

1. 4

2. 2

3. 1

4. 2

5. There are 7C

2 = 21 ways

she can choose the

animals and 9C

3 = 84

ways she can choose

the countries. Since

she can choose any

combination of animals

with any combination of

countries, there are 21

· 84 = 1764 ways she

can choose animals and

countries.

6. There are 17

P3 = 4080

ways to award the fi rst,

second, and third place

prizes. After that, 14

students will remain.

There are 14

C4 = 1001

ways to select the four

students to receive the

honorable mention prize.

Therefore, there are

4080 · 1001 = 4,084,

080 ways to award the

prizes.

7. There are 10

P2 = 90

permutations of digits

and 26

P4 = 358,800

permutations of letters,

so there are 90 · 358,800 = 32,292,000

possible passwords.

If the password contains

a “5”, it could be the

fi rst or second digit.

In either case, there

are 9 possible choices

for the other digit, so

there are 9 + 9 = 18

possible permutations of

digits that contain a “5”.

Likewise, if the password

contains an “e”, there

are 25

P3 = 13,800 ways

to select the other three

letters. Since the “e”

could be in any of the

four-letter positions, there

are 4 · 13,800 = 55,

200 letter permutations

that contain the letter

“e”. Therefore, there

are 18 · 55,200 = 993,

600 passwords that

contain “5” and “e”. The

probability of randomly

selecting one of them

is 993,600

__________ 32,292,000

= 2 ___

65 .

8. a. There are 14

C4

= 1001 ways to select




the four students

if there are no

restrictions.

b. If there must be one

student representing

each language, there

are 4 · 3 · 5 · 2 = 120

ways to select the

group.

c. Out of the 1001 ways

to select students

with no restrictions

calculated in part a.,

determine how many

of these combinations

refl ect only one

language. There are

not enough Chinese

or Italian speakers to

make a group of four

students from only one

of those languages.

Since there are four

French speakers, it

is possible to make

one group of only the

four French speakers.

There are fi ve Spanish

speakers, and there

are 5C

4 = 5 ways to

select a group of four

students from them.

Therefore, there are

5 + 1 = 6 possible

groups that contain

only students who

speak one language.

So there are 1001

- 6 = 995 ways to

select students so that

at least two languages

are represented.

Calculate Theoretical

Probability pp. 175–176

1. 1

2. 1

3. 1

4. 3

5. Since the fi rst number

cube does not show a 3,

it could show a 1, 2, 4,

5, or 6, so the probability

that it shows a 5 is 1 __

5 .

The probability that the

second cube shows a 5

is 1 __

6 . The probability that

they both show a

5 is 1 __

5 · 1 __

6 =

1 ___

30 .

6. Since three of the

possible numbers the

cube can show are odd

and the other three are

even, each cube has a

1 __

2 probability of showing

an even number and 1 __

2

probability of showing an

odd number. If at least

one cube shows an even

number, this means that

all three cubes cannot

show an odd number.

The probability of all

cubes showing an odd

number is 1 __ 2 · 1 __

2 · 1 __

2 = 1 __

8 .

Since “at least one

cube showing an even

number” contains every

event other than all

cubes being odd, the

answer is 1 - 1 __ 8 = 7 __

8 .

7. a. The probability that

the fi rst marble is

red is 4 ___ 26

= 2 ___ 13

.

The probability that

the second marble is

yellow is 5 ___

26 . Since

these events are

independent, the

probability of one

or the other of them

occurring is

2 ___ 13

+ 5 ___ 26

= 9 ___ 26

.

b. Since the fi rst marble

is not purple, the

probability that it is

red is 4 ___ 23

. Since the

second marble is not

black, the probability

that it is purple is 3 ___ 20

.


both of these events

occurs is

4 ___ 23

· 3 ___ 20

= 3 ___ 115

.

8. a. Let the side length of

the square be 4r; then

the area of the square

is 16 r 2 , the area of

each circle is π r 2 , and

the area of each of

the smaller squares

is 4 r 2 . The probability

that a point is in a

shaded region is

shaded area ___________ entire area

=

2(π r 2 ) + 2(4 r 2 )

____________ 16 r 2

= π + 4

_____ 8

.

b. The probability that

the fi rst point is in the

upper right square is

4r2

____ 16r2

= 1 __ 4

. The

probability that the

second point will not

be in the upper left

circle is

16 r 2 - π r 2 ________ 16 r 2

= 16 - π _____

16 .

c. The probability that

one point is in the

shaded region is

π + 4

_____ 8 , so the

probability that fi ve

points are in the

shaded region is

( π + 4

_____ 8

) 5

.





not all points will be in

the shaded region is

1 - [ π + 4 _____

8 ]

5

≈ .433.

Calculate Empirical

Probabilities pp. 177–178

1. 3

2. 3

3. 3

4. 2

5. 15 out of 125 students

drive. To compute how

many students out of

650 students drive,

set up and solve a

proportion.

15 ____ 125

= x ____ 650

x = 78

6. The empirical probability

of failure is

259 ________ 1000000

= .000259.

Since this is greater than

the allowed probability of

0.00001, the circuit does

not meet the reliability

requirement.

7. a. Adding the numbers in

all the cells, the total

number of students in

the school is found to

be 783. Since there

are 92 twelfth grade

girls, the probability

that a random student

is a twelfth grade girl is

92 ____ 783

.

b. There are 207 tenth

graders, and 101 of

them are boys. The

probability that a tenth

grader is a boy is

101 ____ 207

.

c. If a student is not

a girl and is not in

9th grade, then the

student must be a 10th,

11th, or 12th grade boy.

There are 101

+ 103 + 92 = 296

such students. Of

these, 103 are 11th

graders, so the

probability of a student

who is not a girl and is

not a boy in 11th grade

is 103 ____ 296

.

The Binomial Theorem

pp. 179–180

1. 3

2. 4

3. 2

4. 2

5. � i=0

g-2

gCi.5 i .5 g-i

6. The numbers 1, 3, and

5 are odd. Since the

sections on the spinner

are the same size, the

probability of landing on

an odd number is 3 __ 5 ,

and the probability of

landing on an event

number is 2 __ 5 . The

probability of landing on

an odd number exactly

15 out of 30 times is

30

C15

� 3 __ 5 � 15

� 2 __ 5 �

15

≈ .078.

7. a. Since 10% of the

codes give a free

bottle of juice and 1%

of the codes give $10,

11% of the codes give

a prize. The probability

of winning at least 2

prizes in 20 bottles is

� i=2

20

20Ci .11i.89 20-i

≈ .662.

b. The probability of

winning exactly 1

bottle of juice in 20

bottles is 20

C1.10 1 .90 19

≈ .270.

8. a. Of the 88 people

represented in the

table, 18 of them are

30 years old. So the

probability of being

30 years old is

18 ___ 88

= 9 ___ 44

, and the

probability of not

being 30 years old is

35 ___ 44

. The probability

that exactly 2 out of 8

people are 30 is

8C

2 ( 9 ___

44 )

2

� 35 ___ 44

� 6 ≈ .297.

b. The probability of

being less than

30 years old

is 38 ___ 88

= 19 ___ 44

, and the

probability of not

being less than 30

(that is, being 30

or older) is 25

___ 44

. The

probability that at least

15 of 20 people are

less than 30 is

� i=15

20

20Ci � 19

___ 44

� i � 25

___ 44

� 20 - i

≈ .004.

The Normal Distribution

pp. 181–182

1. 3

2. 4

3. 3

4. 2

5. 3

6. Since the probability

of success in each

experiment is close




to .5, the normal

distribution is a good

approximation. 125 is

18 above 107, and 18 is

1½ standard deviations.

The probability of being

more than 1½ standard


mean is 6.7%.

7. Since the number of

experiments is large,

the normal distribution

is a good approximation.

There is an 84%

probability that the

variable will be above

one standard deviation

below the mean. In this

situation, 188 is one

standard deviation below

the mean, so there is an

84% probability that the

variable will be above

188.

8. a. p = 22

___ 43

, so the

expected value for

the number of blue

marbles selected in 84

trials is

84 � 22 ___

43 � ≈ 43.

b. The binomial standard

deviation is

��

84 � 22 ___

43 � � 21

___ 43

� ≈ 46.

9. a. p = .513, so the

expected value for the

number of voters who

supported candidate

A is 5000(.513)

= 2565.

b. The binomial

standard deviation is

��

5000(.513)(1 - .513)

≈ 35.3.

c. If the number of

trials in increased to

8000, the standard

deviation will increase.

The standard

deviation would be

��

8000(.513)(1 - .513)

≈ 44.7.

Practice Test pp. 183–187

1. 2

2. 1

3. 2

4. 3

5. 1

6. 2

7. 3

8. 4

9. 4

10. 2

11. 4

12. 3

13. 2

14. 4

15. 1

16. 2

17. 1

18. 1

19. 3

20. 4

21. 2

22. 1

23. 2

24. 4

25. 2

26. 3

27. 4

28. {(0, 5), (3, 4)}

29. To calculate the number

of combinations of 20

items taken 5 at a time,

compute

20

C5 =

20! ________

5!(20 - 5)!

= 20! _____

5!15!

= 15,504

30. First, fi nd the mean:

40 + 28 + 51 + 6 + 19

___________________ 5 = 28.8.

Next, compute the variance:

(40 - 28.8)2 + (18 - 28.8)2 + (51 - 28.8)2 + (6 - 28.8)2 + (19 - 28.8)2

______________________________________________________ 5 = 246.

Finally, compute standard deviation as the square root of the variance. ��

246.96 ≈ 15.7.

31. - 1 __

2 +

1 __

2 i

32. 6 + 2 3x = 17

2 3x = 11

log 2 3x = log 11

3x log 2 = log 11

x = log 11

_____ 3 log 2

≈ 1.15

33. 80° · π radians ________

180°

= 4π

___ 9 radians

34.

21 0 1 2 32223

35. 3 ��

27x 5 y 4 ________ xy =

3 ��

3 3 x 5 y 4 _______ xy

= 3x

5 __ 3

y

4 __ 3

______ xy = 3x

2 __ 3 y

1 __ 3




37. The radius of the circle

is the distance between

(–1, 4) and (2, –1).

r = ��

(x - h)2 + (y - k)2

= ��

(-1 - 2)2 + [4 - (-1)]2

= ��

(-3)2 + 52 = ��

34

An equation of the circle

with center

(2, –1) and radius ��

34 is

(x – h) 2 + (y – k) 2 = r 2

(x – 2) 2 + [y – (–1)] 2

= � �� 34 � 2

(x – 2) 2 + (y + 1) 2 = 34.

38. Since sec x = 1 _____

cos x ,

fi rst graph y = cos x.

Then use the y-values

to graph y = sec x. The

asymptotes are at the

values where y = sec x

is not defi ned.

y = cos x

y

xO

1

21

p2

p 2p 3p 4p

3p2

5p2

7p2

y = sec x

p 3p 4p2p

p221

1

y

x3p2

5p2

7p2

39. a. In a normal

distribution, 86.6% of

the values are from 1.5

standard deviations

below the mean to 1.5

standard deviations

above the mean. So,

90 - 54 = 36 is 3

standard deviations,

and 1 standard

deviation is 12.

b. 6.7% of the values

are more than 1.5

standard deviations

above the mean.

c. The high score cannot

be determined. The

normal distribution

only shows the

distribution of values,

not specifi c values.

36. � 3 �

�� x2y4 +

3 ��

xy2 � � 3 ��

x2y - 3 ��

x4y2 � __________________________ xy

= 3 ��

x4y5 - 3 ��

x6y6 + 3 ��

x3y3 - 3 ��

x5y4 _________________________ xy

= xy

3 ��

xy2 - x2y2 + xy - xy 3 ��

x2y _______________________ xy

= xy � 3 �

�� xy2 - xy + 1 -

3 ��

x2y � ______________________ xy

= 1 - xy + 3 ��

xy2 - 3 ��

x2y


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