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NEW YORK
Regents Test Prep Workbook
Teacher’s Guide
HMH_NY_AGA_Answer_Key.indd iHMH_NY_AGA_Answer_Key.indd i 5/19/10 8:06:34 PM5/19/10 8:06:34 PM
Copyright © by Houghton Mifflin Harcourt Publishing Company
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Printed in the U.S.A.
ISBN 978-0-547-48609-3
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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
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
Copyright © by Holt McDougal. 2 NY Regents Test Prep Workbook for Integrated Algebra
All rights reserved.
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.
HMH_NY_AGA_Answer_Key.indd 2HMH_NY_AGA_Answer_Key.indd 2 5/19/10 8:06:45 PM5/19/10 8:06:45 PM
Copyright © by Holt McDougal. 3 NY Regents Test Prep Workbook for Integrated Algebra
All rights reserved.
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
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Copyright © by Holt McDougal. 4 NY Regents Test Prep Workbook for Integrated Algebra
All rights reserved.
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}
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Copyright © by Holt McDougal. 5 NY Regents Test Prep Workbook for Integrated Algebra
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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.
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Copyright © by Holt McDougal. 6 NY Regents Test Prep Workbook for Integrated Algebra
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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
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Copyright © by Holt McDougal. 7 NY Regents Test Prep Workbook for Integrated Algebra
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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)
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Copyright © by Holt McDougal. 8 NY Regents Test Prep Workbook for Integrated Algebra
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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
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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
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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)
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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.
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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
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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}
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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
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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
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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
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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
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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
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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.
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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
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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
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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.
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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
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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.
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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.
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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
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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.
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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
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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.
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Practic Test pp. 191–195 e
Copyright © by Holt McDougal. 30 NY Regents Test Prep Workbook for Integrated Algebra
All rights reserved.
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.
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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.
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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
6. The diagram indicates
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
perpendicular to plane
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
perpendicular to plane
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.
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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.
5. The diagram indicates
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
perpendicular to the
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
5. The diagram indicates
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
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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
perpendicular to the
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
perpendicular to the
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
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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.
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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
perpendicular to the
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).
The measure of angle 1
equals the measure of
angle 2, the measure
of angle 3 equals the
measure of angle 4
(Defi nition of angle
bisector).
The measure of angle 2
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).
The measure of angle 2
equals the measure of
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
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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.
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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
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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:
a. Statements Reasons
1. _
AB � _
AF , _
BD � _
FD
2. _
AD � _
AD
3. ADF � ADB
1. Given
2. Refl exive
Prop. of
Congruence
3. SSS
Congruence
Postulate
b. Statements Reasons
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.
a. Statements Reasons
1. ∠1 � ∠3,
∠2 � ∠4
2. _
AG � _
AG
3. AGC � AGE
1. Given
2. Refl exive
Property of
Congruence
3. AAS
Congruence
Theorem
b. Statements Reasons
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
equals the measure of
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
Copyright © by Holt McDougal. 39 NY Regents Test Prep Workbook for Geometry
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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°
8. The diagram indicates
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.
Copyright © by Holt McDougal. 40 NY Regents Test Prep Workbook for Geometry
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HMH_NY_AGA_Answer_Key.indd 40HMH_NY_AGA_Answer_Key.indd 40 5/19/10 8:08:41 PM5/19/10 8:08:41 PM
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.
Interior and Exterior Angles
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.
Interior and Exterior Angles
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.
Copyright © by Holt McDougal. 41 NY Regents Test Prep Workbook for Geometry
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HMH_NY_AGA_Answer_Key.indd 41HMH_NY_AGA_Answer_Key.indd 41 5/19/10 8:08:43 PM5/19/10 8:08:43 PM
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.
Copyright © by Holt McDougal. 42 NY Regents Test Prep Workbook for Geometry
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HMH_NY_AGA_Answer_Key.indd 42HMH_NY_AGA_Answer_Key.indd 42 5/19/10 8:08:44 PM5/19/10 8:08:44 PM
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
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Copyright © by Holt McDougal. 44 NY Regents Test Prep Workbook for Geometry
All rights reserved.
8. a. FH, GF, EF. Each
segment has a vertex
as an endpoint and is
perpendicular to the
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°.
HMH_NY_AGA_Answer_Key.indd 44HMH_NY_AGA_Answer_Key.indd 44 5/19/10 8:08:45 PM5/19/10 8:08:45 PM
Copyright © by Holt McDougal. 45 NY Regents Test Prep Workbook for Geometry
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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.
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Copyright © by Holt McDougal. 46 NY Regents Test Prep Workbook for Geometry
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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)
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Copyright © by Holt McDougal. 47 NY Regents Test Prep Workbook for Geometry
All rights reserved.
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
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Copyright © by Holt McDougal. 48 NY Regents Test Prep Workbook for Geometry
All rights reserved.
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 .
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Copyright © by Holt McDougal. 49 NY Regents Test Prep Workbook for Geometry
All rights reserved.
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
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Copyright © by Holt McDougal. 50 NY Regents Test Prep Workbook for Geometry
All rights reserved.
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.
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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
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Copyright © by Holt McDougal. 52 NY Regents Test Prep Workbook
All rights reserved. for Algebra 2 and Trigonometry
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
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Copyright © by Holt McDougal. 53 NY Regents Test Prep Workbook
All rights reserved. for Algebra 2 and Trigonometry
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
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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
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Copyright © by Holt McDougal. 55 NY Regents Test Prep Workbook
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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
the second equation.
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
the second equation.
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
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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
Expressions pp. 41–42
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.
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Copyright © by Holt McDougal. 57 NY Regents Test Prep Workbook
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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
and the power is even,
the expression is defi ned
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
and the power is even,
the expression is defi ned
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
Expressions pp. 45–46
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
Expressions pp. 47–48
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
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Copyright © by Holt McDougal. 58 NY Regents Test Prep Workbook
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��
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
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Copyright © by Holt McDougal. 59 NY Regents Test Prep Workbook
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-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
-3 is not a solution.
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
-5 is not a solution.
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)
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-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
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= -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
Equations pp. 65–66
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
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Solve Logarithmic
Equations pp. 67–68
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
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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
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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
Expressions pp. 79–80
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
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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
Expressions pp. 87–88
1. 4
2. 1
3. 3
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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
Expressions pp. 93–94
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
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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
Expressions pp. 95–96
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
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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
Functions pp. 103–104
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.
7. Answers may vary.
Check students’ work
for two points with
different x-coordinates
and the same
y-coordinates.
8. Answers may vary.
Check students’ work
for a relation that fails
the vertical line test.
Sample:
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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
Functions pp. 107–108
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
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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 .
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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.
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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
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�
� 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
Functions pp. 131–132
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.
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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
Equations pp. 135–136
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
Functions pp. 139–140
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
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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.
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∠ 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.
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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
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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
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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
deviations above the
mean. So,
x = 63.2 + 2(2.9) = 69.
9. a. About 69% of the data
in a normal distribution
is below .5 standard
deviations above the
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.
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Copyright © by Holt McDougal. 80 NY Regents Test Prep Workbook
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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
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Copyright © by Holt McDougal. 81 NY Regents Test Prep Workbook
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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
.
The probability that
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
.
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The probability that
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
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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
deviations above the
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
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Copyright © by Holt McDougal. 84 NY Regents Test Prep Workbook
All rights reserved. for Algebra 2 and Trigonometry
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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