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Holt McDougalAlgebra 1
Steps to Success:Algebra 1 Teacher’s Guide
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Special thanks to:
Jacqueline Bush-Campbell, Mathematics Content Specialist, School District of Philadelphia
Marie Klump, Mathematics Content Specialist, School District of Philadelphia
Copyright © by Houghton Miffl in Harcourt Publishing Company
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0i_A1_MPAAETG548630_00i-0iii.indd ii0i_A1_MPAAETG548630_00i-0iii.indd ii 5/14/10 2:02:37 AM5/14/10 2:02:37 AM
© Houghton Mifflin Harcourt Publishing Company iii Holt McDougal Algebra 1
Chapter 1 Answers ......................................................1
Chapter 2 Answers ....................................................10
Chapter 3 Answers ....................................................17
Chapter 4 Answers ....................................................24
Chapter 5 Answers ....................................................32
Chapter 6 Answers ....................................................44
Chapter 7 Answers ....................................................53
Chapter 8 Answers ....................................................61
Chapter 9 Answers ....................................................67
Chapter 10 Answers ..................................................77
Chapter 11 Answers ..................................................85
Chapter 12 Answers ..................................................96
Contents
0i_A1_MPAAETG548630_00i-0iii.indd iii0i_A1_MPAAETG548630_00i-0iii.indd iii 5/14/10 2:02:37 AM5/14/10 2:02:37 AM
Chapter 1 Answers
LESSON 1–1
Reading Strategies 1. Possible answers: 10y, 10 • y, and 10(y) 2. Possible answers: k divided by 6, the
quotient of k and 6 3. b − 4 4. 4 − b 5. 20m 6. 58 + t
Review for Mastery 1. 35m 2. h − 0.5 3. c ÷ 25 4. 152 + b 5. r − 50 6. 8 7. 10 8. 29 9. 19 10. 18 11. 12 12. 18 13. 16 14. 2 15. 9 16. 36 17. 7
Practice 1. the sum of a and 3 2. 2 times x 3. y less than 5 4. the quotient of n and 4 5. 10 increased by t 6. the product of 3 and s 7. c + 2 8. 5m 9. 8 10. 4 11. 12 12. 3 13. 16 14. 8 15. a. j − 4 b.11 years old; 16 years old; 54 years old
Challenge 1. 70 + 0.35(m)
Expression for Plan A Cost of Plan A
40 + 0.45(200) $130 40 + 0.45(300) $175 40 + 0.45(400) $220 40 + 0.45(500) $265
Expression for Plan B Cost of
Plan B 70 + 0.35(100) $105 70 + 0.35(200) $140 70 + 0.35(300) $175 70 + 0.35(400) $210 70 + 0.35(500) $245
2. Plan A 3. Plan B 4. Possible answer: It depends on the
number of minutes used. Plan A is cheaper if you use less than 300 minutes each month. Plan B is cheaper if you use more than 300 minutes each month. If you use exactly 300 minutes each month, the plans cost the same.
LESSON 1–2
Reading Strategies 1. negative 2. because subtraction problems become
addition problems 3. −13 4. 30 5. 3 6. 11 7. −9 8. −19
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 11
Review for Mastery 1. −3
2. 5
3. 4 4. −1 5. −8 6. 6 7. −8 8. 11 9. −2 10. 0 11. −5 12. −4
13. −9.7 14. − 49
15. 3 16. −10.3
17. 2 18 4
− = −
Practice 1. −6 2. 6 3. 0 4. −3 5. −7 6. −18 7. −2 8. 4.2 9. 5 10. −29 11. −7 12. 22
13. −5 14. −7 12
15. −15 16. −11 17. 16 feet 18. −3 points
Challenge
Math Expression Balance $450 + (−$25) $425 $425 + (−$200) $225
$225 + $50 $275 $275 + $50 $325
$325 + (−$350) −$25 −$25 + $75 $50
$50 + (−$25) $25 $25 + $100 $125
$125 + (−$200) −$75 −$75 + $225 $150
1. no 2. yes; 3/23 and 5/17 3. $150 4. yes
LESSON 1–3
Reading Strategies 1. −5 + −5 + −5 + −5 2. negative 3. positive 4. 14 5. −15 6. −6 7. 36 8. B, C, E 9. A, D 10. A: 4, B: −16, C: −4, D: 4, E: −8 11. B
Review for Mastery 1. + 2. − 3. − 4. − 5. − 6. + 7. −21 8. 5 9. −24 10. −5 11. 48 12. −2.5 13. −5.1 14. 41 15. −2.1 16. −10.8 17. 3 18. −5.1
19. 2 9•5 4
20. 3 3•8 2
21. 618 •1
or 18 • 6 22. 1564
−
23. 14
24. 718
−
25. 916
26. −4
27. −21 28. 85
29. 127
− 30. 114
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 12
Practice 1. −24 2. 20
3. 0 4. − 38
5. 8 6. 6 7. 14.31 8. −63 9. 2 10. −5 11. 3 12. −3
13. − 23
14. − 116
15. 3 16. undefined 17. 1 18. 0 19. −10 points 20. 72
21. 12
22. 2
23. 0 24. 0 25. undefined
Challenge 1. Goal: Product of −8
−1 −1 2
2 Start
2 −1
−1 −2 −1 Finish
2. Goal: Product of 120
1 3 −2
−2 4 −1
−5 Start
−3 2 Finish
3. Goal: Quotient of 1
−4 3 −2
−24
Start 3 2
1 2 −2 Finish
4. Goal: Quotient of −3
−36
Start 2 −3
3 −2 −3
4 2 −1 Finish
LESSON 1–4
Reading Strategies 1. eighth 2. 56 3. base; exponent 4. 4 times 5. is not 6. 2 • 5 7. 13 8. 81 9. 16 10. 1
Review for Mastery 1. 7 • 7 • 7 • 7 • 7 • 7 2. (−3) • (−3)
3. 1 1 1 1 1 1 1• • • • • •2 2 2 2 2 2 2
4. −(6 • 6 • 6 • 6 • 6) 5. 243
6. −16 7. 481
8. −8 9. 1 10. 0
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 13
11. Power Multiplication Value
21 2 2 22 2 • 2 4 23 2 • 2 • 2 8 24 2 • 2 • 2 • 2 16 25 2 • 2 • 2 • 2 • 2 32
12. Power Multiplication Value
31 3 3 32 3 • 3 9 33 3 • 3 • 3 27 34 3 • 3 • 3 • 3 81 35 3 • 3 • 3 • 3 • 3 243
13. Power Multiplication Value
101 10 10 102 10 • 10 100 103 10 • 10 • 10 1000 104 10 • 10 • 10 • 10 10,000 105 10 • 10 • 10 • 10 • 10 100,000
14. 24 15. 33 16. 106 17. 28 18. (−3)7 19. 1002 20. 54
21. (−4)6 22. 31
2⎛ ⎞⎜ ⎟⎝ ⎠
Practice 1. 62 2. 102 3. 43 4. 9
5. 1000 6. 14
7. 1 8. 16 9. −16 10. 43 11. 102 12. 54 13. 25 14. (−3)3
15. 23
4⎛ ⎞⎜ ⎟⎝ ⎠
16. 8 ft3
Challenge
Year abx New price 2 50(1 + 0.04)2
50(1.04)2 50(1.0816)
$54.08
3 50(1 + 0.04)3 50(1.04)3 50(1.124864)
$56.24
4 50(1 + 0.04)4 50(1.04)4 50(1.16985856)
$58.49
5 50(1 + 0.04)5 50(1.04)5 50(1.216652902)
$60.83
6 50(1 + 0.04)6 50(1.04)6 50(1.265319019)
$63.27
7 50(1 + 0.04)7 50(1.04)7 50(1.315931779)
$65.80
8 50(1 + 0.04)8 50(1.04)8 50(1.36856905)
$68.43
9 50(1 + 0.04)9 50(1.04)9 50(1.423311812)
$71.17
1. year 7 2. $2.00 3. $2.74 4. 5 years 5. 15%
LESSON 1–5
Reading Strategies 1. true 2. false 3. false 4. true 5. false 6. false 7. true 8. true
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 14
Review for Mastery 1. 52; 62; 92; 102;
1; 4; 9; 16; 49; 64; 121; 144; 169
2. 9 ; 16 ; 64 ; 81 ; 121; 144 ; 169
1; 2; 5; 6; 7; 10; 11; 12 3. 11 4. −8 5. 16 6. −20
7. 113
8. − 512
9. real number, rational number, terminating decimal, integer, whole number, natural number
10. real number, rational number, repeating decimal
11. real number, irrational number
Practice 1. 5 2. −2
3. 16
4. 4
5. −10 6. − 23
7. 8.7 feet 8. 7.1 feet 9. integer, rational, terminating 10. irrational 11. rational, repeating 12. natural, whole, integer, rational,
terminating
Challenge 1. 730 2. 730 3. 1430 4. 532,900 5. 2,044,900 6. 2,577,800 7. 1606 8. 5226 ft 9. 5226 ft
Section Quiz: Lessons 1-1 to 1-5
1. C 2. F 3. C 4. G 5. A 6. H 7. D 8. J
9. D 10. G 11. A 12. F 13. B 14. F
LESSON 1–6
Reading Strategies 1. 19 2. −1 3. −40 4. 6 5. 1600 6. 2
Review for Mastery 1. 20; 23; 21 2. 16; 4; 7 3. 3; 9; 1 4. 9 5. −1 6. 14 7. 7 8. 97 9. 30 10. 13 11. 30 12. 16 13. 17.8 cm
Practice 1. 13 2. 5 3. 37 4. 1 5. 88 6. −72 7. −2 8. 69 9. −5 10. −25 11. −7 12. 32 13. 6 14. 8 + |−13| 15. 18 ÷ (−1 • 2) 16. −7(8 − n) 17. 592
Challenge
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 15
LESSON 1–7
Reading Strategies 1. Possible answer: 18 + 9 + 2 = 18 + 2 + 9 2. Possible answer: Mr. Wiley will distribute
the test to all the students. 3. Distributive Property 4. Commutative Property 5. Associative Property 6. combine like terms 7. combine like terms 8. 120 9. 27 + 7x 10. 52x + 8
Review for Mastery 1. 46 2. 35 3. 180 4. 192 5. 2300 6. 352 7. yes 8. no 9. no 10. 4st 11. 6y3 + 5y 12. 12x3 + 6x4 13. 3x + 16 14. 10y − 10
Practice 1. 30 2. 60 3. 38 4. 1600 5. 20 6. −560 7. 10; 275 8. 15; 300 9. 9; 27; 327 10. 30x 11. 5n2 12. 2y 13. 4t3 − 2t2 14. −3d 15. 7r3
16. Commutative; Associative 17. Distributive; Commutative 18. 4x + 5x + 2x; 11x 19. 2x + 1 + 3x + 4 + 2x + 1 + 3x + 4; 10x + 10
Challenge 1. (50 + 1)(70 + 4);
(50 + 1)(70) + (50 + 1)(4); 50(70) + 1(70) + 50(4) + 1(4); 3500 + 70 + 200 + 4; 3774
2. [40 + (−1)](90 + 2); [40 + (−1)](90) + [40 + (−1)](2); 40(90) + (−1)(90) + 40(2) + (−1)(2); 3600 − 90 + 80 − 2; 3588
3. [100 + (−4)][100 + (−2)]; [100 + (−4)](100) + [100 + (−4)](−2); 100(100) + (−4)(100) + 100(−2) + (−4)(−2); 10,000 − 400 − 200 + 8; 9408
4. (x + 3)(x) + (x + 3)(8) x(x) + 3(x) + x(8) + 3(8) x2 + 3x + 8x + 24 x2 + 11x + 24
5. [x + (−2)](x + 7) [x + (−2)](x) + [x + (−2)](7) x (x) + (−2)(x) + x(7) + (−2)(7) x2 − 2x + 7x − 14 x2 + 5x − 14
6. (a + b)(c) + (a + b)(d) ac + bc + ad + bd
LESSON 1–8
Reading Strategies 1.
x y (x, y) −2 −(−2) + 5 = 7 (−2, 7) −1 −(−1) + 5 = 6 (−1, 6) 0 −0 + 5 = 5 (0, 5) 1 −1 + 5 = 4 (1, 4) 2 −2 + 5 = 3 (2, 3)
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 16
2. x y (x, y)
−4 (−4)2 − 6 = 10 (−4, 10) −2 (−2)2 − 6 = −2 (−2, −2) 0 02 − 6 = −6 (0, −6) 2 22 − 6 = −2 (2, −2) 4 42 − 6 = 10 (4, 10)
Review for Mastery 1. move 5 right and 8 up 2. move 2 left and 9 up 3. move 6 left and 1 down 4–9.
10.
Input Output Ordered Pair
x y (x, y)
−2 (−2)2 − 1
4 − 1 3
(−2, 3)
−1 (−1)2 − 1
1 − 1 0
(−1, 0)
0 (0)2 − 1 0 − 1
−1 (0, −1)
1 (1)2 − 1 1 − 1
0 (1, 0)
2 (2)2 − 1 4 − 1
3 (2, 3)
The points form a U-shape.
Practice 1−6.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 17
7. Q II 8. Q IV 9. Q I 10. Q III 11. none 12. Q IV
13. Input
Output
Ordered
Pair x y (x, y)
−2 y = (−2)2 − 1 = 3 (−2, 3) −1 y = (−1)2 − 1 = 0 (−1, 0) 0 y = (0)2 − 1 = −1 (0, −1) 1 y = (1)2 − 1 = 0 (1, 0) 2 y = (2)2 − 1 = 3 (2, 3)
The points form a U-shape.
14. y = 50 + 15x; (3, 95), (4, 110), (5, 125),
(6, 140)
Challenge
1. y = 150 + 5x 2. (0, 150), (10, 200), (20, 250), (50, 400);
the points form a straight line. 3. y = 10x
4. sample ordered pairs: (0, 0), (10, 100), (20, 200), (50, 500); the points form a straight line.
5. when the line for income is above the line for cost
6. when the line for income is below the line for cost
7. a loss of $100 8. a profit of $50 9. 30 dolls
Section Quiz: Lessons 1-6 to 1-8
1. D 2. G 3. B 4. G 5. D 6. H 7. A 8. G 9. D 10. F 11. C
Chapter 1 Enrichment: All Kinds Of Numbers
T2
W
E
N
T
Y
H I
H
T1
W O
R
D
R T3
H
O
U N D R E D T H SH7
T
E
E V E N
E
E
F
IF6
F T H S
8
E
V
E
N
T
H
SHT14
H
R
E17
I G H T
Y
X
I
S15
E
NE
N
T
HTRUOF16
L
A
H12
T11
T13
F
S9
N
U
H
T4 5
S
A
N
D
TROF10
I
F
T
Y
Y
H
Y
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 18
Chapter 1 Big Ideas
1. When adding numbers with the same sign, add their absolute values and use the sign of the numbers. When adding numbers with different signs, find the difference of their absolute values and use the sign of the number with the greater absolute value.
2. The operations of squaring and finding a square root are inverse operations.
3. An irrational number cannot be
expressed in the form ab . π is an
irrational number because it cannot be
expressed in the form ab .
4. Start at the origin. Move 5 units left and 7 units up.
5. Natural numbers and whole numbers both include counting numbers. Whole numbers also include zero.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 19
Chapter 2 Answers
LESSON 2–1
Reading Strategies 1. addition 2. Add 5 to the left side. 3. Add 4 to both sides. 4. m = 13 5. g = 5 6. k = −3 7. t = −15 8. y = 2 9. h = 11
Review for Mastery 1.
4
2.
5
3. 8 4. 19 5. 5 6. x + −7 = 12; 7 7. x + 1 = −5; −1 8. −4 = x + −2; 2 9. −16 10. −6 11. −7
Practice 1. m = 7 2. t = 23 3. p = 4 4. a = 8
5. c = 5 6. y = 35
7. b = 0 8. w = 9 9. p = 4 10. x = 4 11. x = −2 12. r = 14.4 13. x + 18 = 86; 68; The score on the second
test was higher than the first, so the score on the first test should be less than 86.
14. x − 4 = 29; 33˚F; The actual temperature was lower than predicted, so the predicted temperature should be greater than 29˚F.
15. x + 72 = 122; $50; The amount of money raised in one week should be less than the two-week total of $122.
Challenge 1. a. Let x represent the distance traveled.
37,538 + x = 37,781 b. 243 miles 2. s + d = e 3. a. e − d
b. e − s c. s + d
4. a. d = 1697 b. s = 17,152 c. e = 63,777 d. d = 1111
5. 7 23
feet
6. f = o + d, o = f − d, and d = f − o
7. 9 512
feet
LESSON 2–2
Reading Strategies 1. less 2. greater 3. greater 4. b = 4 5. c = 18 6. k = 20
Review for Mastery 1. divided ; multiply ; −14
2. multiplied ; divide ; −8
3. 10 4. −35
5. −7 6. 52
7. − 75
8. 17
9. 12 10. −10
11. − 910
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 110
Practice 1. d = 18 2. n = −21 3. t = −15 4. r = 20 5. b = 17.5 6. v = 36 7. y = −5 8. p = 10 9. m = 1 10. x = 2
11. y = − 17
12. k = 12
13. 4x = 63; 252 students
14. 5x = 1.15; $0.23 15. 4x = 64; 16 mm
Challenge 1. x = 3 2. x = −5
3. x = 3 4. t = 65
5. w = 12
6. d = 30
7. ax = b; 1a
• ax = 1a
• b; 1 • x = 1a
• b;
x = ba
8. axa
= ba
; 1x = ba
; x = ba
9. Using the Multiplication Property of Equality, multiply each side of the equation by the reciprocal of the coefficient of the x-term. Using the Division Property of Equality, divide each side of the equation by the coefficient of the x-term. By either method, the solution is the same.
LESSON 2–3
Reading Strategies 1. like terms 2. subtraction 3. Subtract 3 from both sides, then multiply
by 5. 4. n = 5 5. d = 7 6. j = −2
Review for Mastery 1. 4 2. 60
3. 53
4. 5
5. 54
6. −1
7. 252
Practice 1. 2; 10; 2 2. 3; 8; 4 3. 21; 9; 3 4. t = −2 5. x = 5.4 6. r = −23 7. y = 3 8. b = 24
9. m = 18
10. x = 6
11. y = −3 12. d = −1 13. −8 14. 7x + 6 + 5x = 90 15. x = 7
Challenge 1. 4 inches 2. 8 inches 3. 1 inch 4. 9 inches 5. 15 inches 6. 7 inches 7. 3 inches 8. 7 inches 9. 6 inches 10. 11 inches 11. 15 inches 12. 4 inches 13. 5 sides 14. 12 sides 15. 7 sides 16. 18 sides
LESSON 2–4
Reading Strategies 1. Look for an opportunity to use the
Distributive Property. 2. Combine any like terms. 3. Collect the variable terms on one side of
the equal sign. 4. p = −6 5. x = 1 6. t = −2
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 111
Review for Mastery 1. Possible answers: add −3x to each side,
add −7x to each side 2. Possible answers: add 4x to each side,
add 10x to each side 3. Possible answers: add 3x to each side,
add −15x to each side 4. −48 5. 2 6. 5 7. no solution 8. 1 9. all real numbers 10. no solution 11. all real numbers 12. −1
Practice 1. 2a; 3; 10; 5 2. 4r; 9; −4 3. −5b; 30; 5b; 3b; 10 4. c = −19 5. all real numbers 6. no solution 7. a. 3 hours 8. a. 2 hours b. 75° F b. $12
Challenge
1. x = − 73
2. x = −4
3. x = 7 12
4. x = −2
5. 2x + (−6) = 0 or −2x + 6 = 0 6. 9x + 18 = 0 or −9x + (−18) = 0 7. 5x + 27 = 0 or −5x + (−27) = 0 8. 3x + (−21) = 0 or −3x + 21 = 0
9. a. x = − ba
b. After writing the equation in the form ax + b = 0, substitute the values of a and b into the formula found in part a.
10. x = 3 11. x = −2
12. x = −5 25
13. x = 7
14. a. infinitely many solutions b. no solution
LESSON 2–5
Reading Strategies 1. Possible answer: 3x + 2y = 9 2. The equation contains only one
variable, n. 3. Yes, because it has two or more
variables. 4. Divide both sides by r.
5. t = 83
b −
6. a. h = VIw
b. 3 cm
Review for Mastery
1. s = 4P 2. b = 180 − a – c
3. K = VPT
4. w = 3Vlh
5. 12 in. 6. add –x to both sides 7. multiply both sides by 2 8. add 3r to both sides
9. a = 3cb
10. z = 3(y − x)
11. m = pn − 3
Practice
1. C = K − 273 2. f = 1T
3. y = 5x 4. s = r − 4t
5. m = 73
p n+ 6. j = 6 kh−
7. w = 9v 8. a = bc − 3
9. a. t = dr
10. a. F = 2 + E − V
b. 3 b. 12
Challenge 1. l = 375 2. P = 2000 3. t = 5 years
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 112
4. r = 0.032 or 3.2% 5. $351 6. 3 years 7. $2200 8. 0.03 or 3% 9. $247 10. 5 years 11. 0.027 or 2.7%
LESSON 2–6
Reading Strategies 1. one 2. two 3. none 4. one 5. none 6. two
Review for Mastery 1. −6, 10 2. −15, 1 3. 0, 10 4. −3, 3 5. 2 6. no solution 7. no solution 8. −10
Practice 1. −3; 2; −2; 2 2. −7; 7; 4; 4; 4; 4; −11; 3 3. 6; −6; 6; −5; 7 4. −8, 8 5. −14, 14 6. −9, 9 7. −17, 17 8. −11, 7 9. −1, 11 10. −5, 5 11. −7, 3 12. −11, 9
13. x − 24 = 2
14. 22 miles per gallon; 26 miles per gallon
Challenge 1.
(−3) − 2 = −5 = 5; 7 − 2 = 5 = 5
2. yes; possible answer: x − 2 + 1 = 6
3. x − 5 = 4
4. x − 2 = 6
5. x − 3 = 8
6. x + 5 = 4
7. x = 13
8. x − 5.5 = 0.5
9. x − 2.5 = 7
10. x − 1 = 32
Section Quiz: Lessons 2-1 to 2-6
1. A 2. G 3. D 4. H 5. D 6. J 7. B 8. F 9. B 10. H 11. C 12. G 13. C 14. G 15. D
LESSON 2–7
Reading Strategies
1. 16 oz 1 lbor1 lb 16 oz
2. no; the second quality is not 1. 3. The two quantities are equal. 4. ratio; rate; ratio 5. 28 pg/hr
6. Possible answer: 1 122 24
=
Review for Mastery
1. 31
2. 1100
3. 81
4. 1000 m1 km
; 3,391,10083.5
; 40,612
5. 320 c/min 6. x = 2.5 7. k = 3 8. a = 10.75 9. y = 30
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 113
Practice 1. 20 2. 58 ft/s 3. $1.05/lb 4. 2.5 pages/min 5. y = 4 6. x = 18 7. m = 2 8. t = 75 9. b = −4 10. x = 1 11. 150 in. 12. 160 mi
Challenge 1. bc: 7 • 15 = 105; ad: 5 • 21 = 105
2. a. Multiply each side by bdac
b. Cross-Product Property; Multiplication Property of Equality; Identity Property of Multiplication
3. ab
= cd
ad = bc
1ab
⎛ ⎞⎜ ⎟⎝ ⎠
ad = 1ab
⎛ ⎞⎜ ⎟⎝ ⎠
bc
db
= ca
Cross-Product Property; Multiplication Property of Equality; Identity Property of Multiplication
LESSON 2–8
Reading Strategies
1. ΔABC 2. The measure of angle B is 93 degrees. 3. Triangle XYZ is similar to triangle ABC. 5. 49° 6. 38°
7. Possible answer: AB XYBC YZ
=
8. 1810 15X = ; 12 cm
Review for Mastery 1. 15.6 ft 2. 6.125 cm 3. 5; 25; (5)3 = 125
4. 23
; 22
3⎛ ⎞⎜ ⎟⎝ ⎠
= 49
; 827
5. 107
; 107
; 107
; 310
7⎛ ⎞⎜ ⎟⎝ ⎠
= 1000343
6. The ratio of the areas is the square of the ratio of corresponding dimensions:
2120
⎛ ⎞⎜ ⎟⎝ ⎠
= 1400
.
7. The ratio of the circumferences is equal to the ratio of corresponding dimensions: 4.
Practice 1. 4 2. 28
3. 5x
= 412
; 15 feet
4. width = 4 cm; length = 12 cm
5. 12
6. 16 cm; 32 cm; 12 cm2; 48 cm2
7. 12
8. They are the same. 9. 14
10. The ratio of the areas is the square of the ratio in problem 5.
Challenge
1. ST ; YZ 2. 1 in.; 2 in.; 3.5 in.
3. DE ; XY 4. 2 in.; 4 in.
5. 24
= 12
= 0.5; 3.57
= 12
= 0.5
6. 12
or 0.5 7. 0.865
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 114
8. a. 0.865
b. 12
or 0.5
9. The sine of a 30° angle equals the cosine of a 60° angle. The sine of a 60° angle equals the cosine of 30° angle.
LESSON 2–9
Reading Strategies 1. 56 2. 12 3. 20 4. percent 5. whole 6. part 7. 40 8. 15
Review for Mastery 1. percent; whole; 3.2 2. part; percent; whole; 40 3. percent; whole; part; 15% 4. part; percent; whole; 22.5 5. 20%; what number; 7 6. 50%; 92; what number 7. What percent; 40; 14 8. 65%; 80; what number 9. 35 10. 5%
Practice 1. 49 2. 148; 37; 25 3. 24; 0.24 4. 6.5; 0.065 5. x; 80; 25; 20 6. 22.5; 45; x; 50 7. x; 65%; 80;
x = 65100
• 80, or x = 0.65 • 80, 52
8. 30; x; 75; 30 = 75x; 40%
9. 49 10. 65 11. 25% 12. 20% 13. 45 14. 600 15. 700
Challenge 1. $162.50 2. $97.50 3. 85% of $130 is $110.50, not $97.50 4. a 30% markup followed by a 15%
markdown 5. Multiply the original amount by one plus
the decimal or fractional form of the markup and then multiply that amount by one minus the decimal or fractional form of the markdown.
6. $97.50 7. $58.50 8. a 30% markdown followed by a 15%
markdown 9. Multiply the original amount by one minus
the decimal or fractional form of the markdown and then multiply that amount by one minus the decimal or fractional form of the next markdown.
LESSON 2–10
Reading Strategies 1. total pay = 52,600 + 0.02 × 842,000;
$69,440 2. simple interest = 6300 × 0.05 × 5;
$1575 3. tip = 0.18 × 74.50; $13.41 4. tax = 0.045 × 24,500; $1102.50
Review for Mastery
1. 112
2. 14
3. 32
4. $15
5. 4 years 6. $1200 7. $0.48 8. $1.20 9. $13 10. $19.50 11. $26
Practice 1. 3%; 795; $795 2. 0.025; 3; 112.5; $112.50 3. $20; $20; $2.00; $2.00; $1.00; $3.00
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 115
4. $425 5. 8% 6. about $1.60
Challenge 1. $18.00 2. $0.70; $2.10; $16.80 3. $0.65; $1.95; $15.60 4. $0.60; $1.80; $14.40 5. $0.55; $1.65; $13.20 6. $13 7. 16 = x + 0.05x + 0.15x
8. $13.33 9. 56
10. $20 • 56
≈ $16.67
LESSON 2–11
Reading Strategies 1. decrease 2. discount 3. greater than 4. markup 5. $122.50 6. $13.50
Review for Mastery 1. 20% increase 2. 40% decrease 3. 50% increase 4. 75% decrease 5. 54.5% decrease 6. 150% increase 7. 98.7 8. $36.76 9. $49 10. 40%
Practice 1. 110; 110; 44; 110; 40; 2. 16.88; 4.22; 4.22; 12.66; $4.22; $12.66 3. 20% decrease 4. 300% increase 5. 26 6. 49 7. 70¢ 8. 20%
Section Quiz: Lessons 2-7 to 2-11
1. C 2. F 3. D 4. J 5. B 6. H 7. D 8. H
9. B 10. H 11. D 12. H 13. C
Chapter 2 Enrichment: Order Up!
1. ( )− + ÷ =4 3 1 6 2 11
2. ( )+ 4 ÷ + − =21 5 6 3 8
3. ( )+ + 12 ÷ − =210 6 4 8 11
4. × ÷ + ÷ − =8 6 4 10 2 12 5
5. ( )− + ÷ − =5 1 4 6 3 12 10
6. ( )÷ − + = 35 6 1 4 2 50
Chapter 2 Big Ideas
1. When adding, subtracting, multiplying, or dividing the same nonzero number on both sides of an equation, the statement will still be true.
2. When solving an equation that is always true, the original equation is an identity, and it has infinitely many solutions. But when solving an equation that is false, the original equation is a contradiction, and it has no solutions.
3. Step 1 Locate the variable you are asked to solve for in the equation.
Step 2 Identify the operations on this variable and the order in which they are applied.
Step 3 Use inverse operations to undo operations and isolate the variable.
4. You can solve proportions for a missing value by using the Cross Products Property, which states that in a proportion, cross products are equal. Then solve the equation for the missing value.
5. A percent change can be found by dividing the amount of increase or decrease by the original amount. Then express the quotient as a percent.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 116
Chapter 3 Answers
LESSON 3–1
Reading Strategies 1. x is greater than or equal to 5; x is at
least 5; x is no less than 5 2. empty; because the value 8 is not a
solution 3.
4. t < 6.5; t is less than 6.5 5. g ≥ 85
Review for Mastery 1.
all real numbers less than or equal to 2 2.
all real numbers less than −3 3. all real numbers greater than 12 4. all real numbers less than or equal to 1 5.
6.
7. x ≥ −2 8. x < −4
Practice 1. c 2. a 3. b 4. d
5.
6.
7.
8.
9. d 10. b 11. a 12. c 13. h = height; h > 4;
14. s = lawful speed; s ≤ 65;
Challenge 1. (4, ∞) 2. x ≤ −2 3. (−∞, 3)
4. x ≥ −5; [−5, ∞) 5. x > 0
6. (−∞,−1]
7. [−3, 2] 8. 0 < x ≤ 3 9. −4 ≤ x < 5
10. (−1, 1)
LESSON 3–2
Reading Strategies 1. Step 1 2. the direction of the inequality symbol
x = 0 x = 1 x = −3 x = −4 x = 2 x = 3 x = 1.5
0 ≤ 10 T
5 ≤ 10 T
−15 ≤ 10 T
−20 ≤ 10 T
10 ≤ 10 T
15 ≤ 10F
7.5 ≤ 10T
x = 0 x = 3 x = −4 x = −3 x = −2 x = −2.5 x = −5
1 < −2 F
4 < −2 F
−3 < −2 T
−2 < −2 F
−1 < −2 F
−1.5 < −2F
−4 < −2T
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 117
3. Substitute 8 for m in 14 = m + 6. 4. Possible answer: 9 5. 5 9
14 5 99 9
t + = −− + = −
− = −;
5 913 5 9
8 9
t + < −− + < −
− < −; no
6. 6 28 6 2
2 2
b − =− =
=;
6 29 6 2
3 2
b − ≥− ≥
≥; yes
Review for Mastery 1. b > 7
2. x < 3
3. x < −4
4. f ≥ 4
5. c ≤ −5
6. x < 2
7. w > −3
8. n ≥ 4
Practice 1. t > 8
2. x ≥ −7
3. p ≥ 5
4. m > −9
5. w ≤ 3
6. g > 0
7. 32 + d ≥ 40; d ≥ 8
8. 11 + m ≤ 20; m ≤ 9 9. 17 + h > 30; h > 13
Challenge 1. 0 < P ≤ 200 2. No; because 2(70 + 40) = 220 feet is
greater than 200 feet 3. 80 feet; 80 feet; 1600 square feet 4. 60 feet; 60 feet; 2400 square feet 5. If the dimensions are reversed, the area
remains the same. 6. 50 feet by 50 feet; a square;
2500 square feet 7. 40 feet by 40 feet
LESSON 3–3
Reading Strategies 1. No; because you are dividing by a
positive number. 2. t ≤ −20; p < −1.5 3. Correct 4. Incorrect; ≤ was reversed to >,
instead of ≥. 5. Incorrect; The sign was not reversed.
Review for Mastery 1. x ≥ −6
2. g > 4
3. v < 5 4. m > 12 5. q ≤ 2
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 118
6. x < −4
7. x ≥ −5 8. d > −3
Practice 1. x ≥ 3
2. a < 5
3. b > 4
4. y ≤ −2
5. x > −4
6. k < −3
7. n ≤ 6
8. x ≥ 0
9. 2b ≤ 15; b ≤ 7.5; 0, 1, 2, 3, 4, 5, 6, or 7 bags 10. 18c ≤ 153; c ≤ 8.5; 0, 1, 2, 3, 4, 5, 6, 7, or
8 CDs
Challenge 1. 2 > −6 2−3.
4. The segments cross each other showing
that the value that was lesser (−1)
now corresponds to the value that is greater (2), and vice versa.
5. The ratio of these lengths is the absolute
value of −2: bottomtop
= 2.4 cm1.2 cm
= 2.
6. yes; the segments cross each other and the ratio of the lengths of the pieces of the dashed segment is still the absolute
value of the factor:
bottomtop
= 0.9 cm2.7 cm
= 13
.
7. The segments would intersect at the midpoint of the dashed line.
Section Quiz: Lessons 3-1 to 3-3
1. A 2. J 3. B 4. J 5. A 6. F 7. C 8. H 9. B 10. F
LESSON 3–4
Reading Strategies 1. to show the infinitely many solutions 2.
3. −1 Possible Answer: 4. −2, −3, −4, −5, −6 5. yes; −2
Review for Mastery 1. e ≥ −2
2. c > 6
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 119
3. s ≤ −3
4. j > 4
5. x > 3 6. b ≤ −13 7. g < 5 8. k ≥ 1
Practice 1. 5; 5; 12; 6 2. 3; 3; 3; 12; > −4
3. 6; 6; 4; 5; 5; −1; 3; 3; n > − 13
4. x ≥ −1
5. z ≤ 2
6. a > 12
7. x < 9
8. 85 + 60 + x
3≥ 70; x ≥ 65; at least 65
9. 30s + 120 ≤ 360; s ≤ 8; 0, 1, 2, 3, 4, 5, 6, 7, or 8 sitcoms
Challenge 1.
2. 3, 4, 5, 6 3. x ≥ 3 and x ≤ 6 where x is an integer. 4. a. x > 55;
b. x > 0 and x < 40 5. No, since there is a lower limit as well as
an upper limit 6. a. x ≥ 40 and x ≤ 55;
b. x > 0 and x < 40 or x > 55
7. maximum: $190,000 minimum: $180,000 x ≥ $180,000 and x ≤ $190,000
8. x ≥ $75,000 and x ≤ $255,000
LESSON 3–5
Reading Strategies 1. Left; so he divides by a positive number. 2. t > 8 3. n > 7; left; you do not have to switch the
inequality symbol.
Review for Mastery 1. add −6y to both sides 2. add −3p to both sides 3. add 3r to both sides 4. c > −4
5. x < 32
6. a > 15 7. no solutions 8. all real numbers 9. all real numbers 10. all real numbers 11. no solutions
Practice 1. 3x; −3x; 8; ≥ −8 2. −6y; 14; 6y; 6y; 14; 14; 14; 1
3. 15n; 18; 5n; 18; 18; 18; 5; 5; n < −4 25
4. x ≤ −2
5. b < 3
6. contradiction 7. identity 8. identity 9. p − 0.15p + 12 < p; p > 80; greater than
$80
10. 6x > 12
(4)(x + 6); x > 3
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 120
Challenge 1. x = 4; the lines intersect 2. x < 4; the line for y1 is above
(greater than) the line for y2 3. x + 3 > 2x − 1 4. x < 4; same solutions as problem 2 5. Possible answer: Find the x-values for
which the line for y2 is either above (greater than) or intersects (equal to) the line for y1. The solutions are x ≥ 4.
6. y1 = x − 3 y2 = −3x − 11 x y x y −4 −7 −4 1 −2 −5 −3 −2 0 −3 −2 −5 2 −1 −1 −8 4 1 0 −11
x ≤ −2
LESSON 3–6
Reading Strategies 1. OR 2. Possible answer: 5, 6, 7 3. Possible answer: 3, 10, 11 4. AND 5. OR statement; AND statement
Review for Mastery 1. −3 < x − 4 x − 4 ≤ 10 2. 8 ≤ m + 4 m + 4 ≤ 15 3.
4. −4 < k < 1
5. 2 < x ≤ 7
6. x < 0 OR x ≥ 4 7. x < −6 OR x > −3 8.
9. x ≥ 3 OR x < −4
10. b ≥ 7 OR b ≤ −1
Practice 1.
2.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 121
3. x ≤ −1 OR x ≥ 5 4. x > −4 AND x ≤ −1 5. 5; 5; 5; 5; −3; 4
6. 1; 1; 1; 1; −10; 2; 2; 2; 2; 2; −5; 1
7. 400 ≤ m ≤ 600
8. 6.40 ≤ r ≤ 9.80
Challenge 1 − 6.
7. Answers may vary. Sample answer:
closed intervals centered at each integer
with each interval being 12
unit long
8. n ≤ x ≤ n + 12
9. 2n ≤ x ≤ 2n + 1 10. 4n − 1≤ x ≤ 4n + 1 11. a. Answers may vary. Sample answers:
The center of the interval is 1, and its
length is 2n
.
b. As n gets larger, the center remains at 1, but the length of the interval gets smaller.
LESSON 3–7
Reading Strategies 1. 5 2. 1 and 9 3.
4. −1
5. −3 and 1
6.
7.
Review for Mastery 1. x > −4 AND x < 4
2. x ≥ −3 AND x ≤ 5
3. x ≥ −3 AND x ≤ 3
4. x > −5 AND x < −3
5. x ≤ −1 OR x ≥ 1
6. x < −5 OR x > 1
7. x < −4 OR x > 4
8. x ≤ 1 OR x ≥ 7
Practice 1. 7; 7; 2; −2; 2 2. −3; 3; 1; 1; 1; 1; −2; 4 3. x > −4 AND x < 4
4. x ≥ −4 AND x ≤ 0
5. x ≥ −5 AND x ≤ 5
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 122
6. x < −2 OR x > 2
7. x ≤ −2 OR x ≥ 4
8. x < −5 OR x > −1
9. x − 85 ≤ 4; 81 ≤ x ≤ 88
Challenge 1. 3| x | < 6; | x | < 2; x > −2 AND x < 2
2. 9| x | ≥ 27; | x | ≥ 3; x ≤ −3 OR x ≥ 3
3. x ≥ −1 AND x ≤ 1
4. x > − 32
AND x < 32
5. x < −3 OR x > 1
6. all real numbers
Section Quiz: Lessons 3-4 to 3-7
1. A 2. G 3. D 4. F 5. C 6. G 7. C 8. H 9. A 10. G 11. D 12. H 13. A
Chapter 3 Enrichment: Magic Square
8 –5 –6 5
–3 2 3 0
1 –2 –1 4
–4 7 6 –7
The magic sum is 2
Chapter 3 Big Ideas
1. To show that an endpoint is a solution, draw a solid circle at the number. To show that an endpoint is not a solution, draw an empty circle.
2. To solve an inequality or an equation, you need to isolate the variable and use inverse operations.
3. When you multiply or divide both sides of an inequality by the same negative number, you reverse the inequality symbol. When you multiply or divide both sides of an inequality by the same positive number, you do not change the inequality symbol.
4. Graph each part of the inequality separately noting when to use a solid circle and when to use an empty circle.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 123
Chapter 4 Answers
LESSON 4–1
Reading Strategies 1. Possible answers: rose slowly; increased
gradually 2. 3. C 4. A 5. B
Review for Mastery 1. 2.
3. Graph B
4. continuous 5. discrete
Practice 1. falling 2. staying the same 3. rising 4. Graph B 5. Graph C 6. Graph A 7.
continuous
8. Possible answer: A subway train has up to 6 cars. Each car can hold 40 passengers.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 124
Challenge 1. 2. 3. 4.
LESSON 4–2
Reading Strategies 1−3. Possible answers are given. 1.
x 1 2 3 4 y 1 2 3 4
2. 3. because the domain value 1 is paired
with more than one range value. 4. no 5. no 6. yes
Review for Mastery 1. D: {−2, −1, 0, 1}; R: {4, 1, 0}
2. D: {4, −2, −5}; R: {5, 6, 12}
3. D: {0, 1, 2, 3}; R: {5, 6, 7, 8}
4. D: {−3, −4, −5}; R: {10, 11, 12, 13} 5. D: −2 ≤ x ≤ 2; R: 1 ≤ y ≤ 4 6. D: −3 ≤ x ≤ 2; R: 1 ≤ y ≤ 3 7. No; −3 is paired with both 2 and 3. 8. Yes; each domain value is paired with
exactly one range value. 9. No; several domain values are paired
with more than one range value.
Practice 1.
x y −2 5 −1 1 3 1
−1 −2
2.
x y 5 3 4 3 3 3 2 3 1 3
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 125
3. {0 ≤ x ≤ 4}; {0 ≤ y ≤ 4}; yes; each domain
value is paired with exactly one range value.
4. {8, 9}; {−3, −4, −6, −9}; no; both domain values are paired with more than one range value.
5. {0, 1, 2}; {4, 5, 6, 7, 8}; no; two domain values are paired with two range values.
Challenge 1. multiples of 5: . . . , −15, −10, −5, 0, 5, 10,
15, . . . 2. multiples of k: kn 3. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 4. If P has coordinate a, Q has coordinate
b, assign 2
a b+ .
5. If P has coordinate a, assign |a|. 6. If P has coordinate a and Q has
coordinate b, assign |a − b|. 7. a. If the length of a side of square X is s,
then P(square X) = 4s. b. If the length of a side of square X is s,
then A(square X) = s2. 8. a. The function rotates square A 120°
counterclockwise about point O. b. The set of all squares. 9. a.
1, 1 1, 2 1, 3 1, 4 1, 5 1, 6 2, 1 2, 2 2, 3 2, 4 2, 5 2, 6 3, 1 3, 2 3, 3 3, 4 3, 5 3, 6 4, 1 4, 2 4, 3 4, 4 4, 5 4, 6 5, 1 5, 2 5, 3 5, 4 5, 5 5, 6 6, 1 6, 2 6, 3 6, 4 6, 5 6, 6
b.{2, 3, 4, 5, 6, 7, 8, 9, 10, 12}
LESSON 4–3
Reading Strategies 1. dependent; independent 2. f(x) = 4x 3. D: {1, 2, 3, 4, 5} 4. R: {4, 8, 12, 16, 20} 5. 12 ribbons 6. 20 ribbons
Review for Mastery 1. The amount of food eaten depends on
the size of the animal.; pounds of food; size of animal
2. The number of firefighters depends on the size of the fire.; number of firefighters; size of the fire
3. The temperature of the water depends on the time it has been on the stove.; temperature of the water; time on the stove
4. The amount of the restaurant bill depends on the number of meals ordered.; amount of the restaurant bill; number of meals ordered
5. total charge; number of hours; f(x) = 90x; $180; $675
6. total charge; number of hours; f(x) = 295 + 95(x − 1); $508.75; $960
Practice 1. y = 3x 2. y = x − 3 3. y = −3x 4. independent; dependent 5. dependent; independent 6. 10; 2 7. −8; 2 8. −7; 20 9. f(b) = 10b; D: {0, 1, 2, 3, 4, 5, 6, 7};
R: {0, 10, 20, 30, 40, 50, 60, 70} 10. f(h) = 45 + 5h; D: {1, 2, 3, 4};
R: {50, 55, 60, 65}
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 126
Challenge 1. 104°F 2. Answers may vary. Sample answer: the
112°F may have been a “heat index” reading.
3. For each value of C there is exactly 1 value for F.
4. 77; 14; 64.4; 30 5. C = 59
(F − 32)
6. 15; 25; −20; 26 23
7. Yes; for each value of F, there is exactly one value of C.
8. The result is the original 35. 9. There is no temperature equivalent in
both systems.
LESSON 4–4
Reading Strategies 1. 2.
Review for Mastery 1.
x y = (x + 2)2 (x, y) −4 y = (−4 + 2)2 = (−2)2 = 4 (−4, 4) −3 y = (−3 + 2)2 = (−1)2 = 1 (−3, 1) −2 y = (−2 + 2)2 = (0)2 = 0 (−2, 0) −1 y = (−1 + 2)2 = (1)2 = 1 (−1, 1) 0 y = (0 + 2)2 = (2)2 = 4 (0, 4)
2.
x y = 12 x − 3 (x, y)
−4 y = 12
(−4) − 3 = −2 − 3 = −5 (−4, −5)
−2 y = 12
(−2) − 3 = −1 − 3 = −4 (−2, −4)
0 y = 12
(0) − 3 = 0 − 3 = −3 (0, −3)
2 y = 12
(2) − 3 = 1 − 3 = −2 (2, −2)
4 y = 12
(4) − 3 = 2 −3 = −1 (4, −1)
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 127
3. −3 4. 1 5. 0 6. 3 7. 1 8. 5
Practice 1.
x y = x + 2 (x, y) −2 y = −2 + 2 (−2, 0) −1 y = −1 + 2 (−1, 1) 0 y = 0 + 2 (0, 2) 1 y = 1 + 2 (1, 3) 2 y = 2 + 2 (2, 4)
2.
x y = x2 ÷ 2 (x, y)
−4 y = (−4)2 ÷ 2 (−4, 8) −2 y = (−2)2 ÷ 2 (−2, 2) 0 y = (0)2 ÷ 2 (0, 0) 2 y = (2)2 ÷ 2 (2, 2) 4 y = (4)2 ÷ 2 (4, 8)
3.
x y = 8x (x, y) 0 y = 8(0) (0, 0) 1 y = 8(1) (1, 8) 2 y = 8(2) (2, 16) 3 y = 8(3) (3, 24) 4 y = 8(4) (4, 32)
about 28 miles
Challenge 1.
x g(x) = −2x g(x) −1 g(−1) = −2(−1) = 2 2 0 g(0) = −2(0) = 0 0 1 g(1) = −2(1) = −2 −2 2 g(2) = −2(2) = −4 −4
g(x) f(x) = x2 −4 f(g(x))
2 f(2) = 22 − 4 = 4 − 4 = 0 0 0 f(0) = 02 − 4 = 0 − 4 = −4 −4
−2 f(−2) = (−2)2 − 4 = 4 − 4 = 0 0 −4 f(−4) = (−4)2 −4 = 16 − 4 = 12 12
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 128
2. 3. y = −2x2 + 8 4.
x y = g(f(x)) = −2x2 + 8 (x, g (f(x)))
−2 g(f(−2)) = −2(−2)2 + 8 = −8 + 8 = 0 (−2, 0)
−1 g(f(−1)) = −2(−1)2 + 8 = −2 + 8 = 6 (−1, 6)
0 g(f(0)) = −2(0)2 + 8 = 0 + 8 = 8 (0, 8)
1 g(f(1)) = −2(1)2 + 8 = −2 + 8 = 6 (1, 6)
2 g(f(2)) = −2(2)2 + 8 = −8 + 8 = 0 (2, 0)
5. No
Section Quiz: Lessons 4-1 to 4-4
1. D 2. G 3. C 4. H 5. B 6. F 7. C 8. F
LESSON 4–5
Reading Strategies 1. number of children in a family; monthly
cost of food; increases; positive
2. population of cities in the U.S.; average February temperature; shows no pattern; none
3. number of practice runs; finish time; decreases; negative
Review for Mastery 1. Negative correlation; each knot
decreases the length of the rope 2. No correlation; there is no relationship
between height and algebra skill 3. negative correlation 4. positive correlation
5.
6. Possible answer: 8 mi/gal Possible answer: 20 mi/gal Possible answer: 4.5 L Possible answer: 1.4 L
Practice 1.
2. positive 3. negative
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 129
4. positive; as the temperature goes up, more people would go in the pool to cool off.
5. no correlation; the height of a person has nothing to do with how many phone calls they make
6. Possible answer: about 38 batteries
Challenge 1. Yes, all the data points are either on the
lines or between them.
2. 3 units; 2 units 4. The closer together the parallel bounding
lines are, the more tightly the data cluster around a line. The farther apart the parallel bounding lines are, the more loosely the data points cluster around a line.
LESSON 4–6
Reading Strategies 1. The terms do not all differ by the same
number. 2. The same number is being multiplied, not
added, to each term. 3. Possible answer: 1, 6, 11, 16, 21, …
4. − 12
; 5 12
, 5, 4 12
5. 130
6. 88
Review for Mastery 1. −2, −2, −2; yes 2. +0.3, +0.4, +0.5; no 3. −5, −8, −11 4. 15, 20, 25 5. no 6. yes; 2.5; 11.25, 13.75, 16.25 7. an = 10 + (n − 1)(4) 8. −5; 5; an = −5 + (n − 1)(5) 9. 9 10. 99 11. 190
Practice 1. no 2. yes 3. 3 4. −6 5. d = −10; −20, −30, −40 6. d = −2; 92, 90, 88 7. 256 8. −19 9. 30 10. 20 11. a12 = 30 + 11(20) 12. $250
Challenge 1. 0, 0, 0, 0, . . . 2. 0, 1, 0, 1, 0, . . . 3. 0, 1, 2, 0, 1, 2, . . .
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 130
4. 0, 1, 2, 3, 0, 1, 2, 3, . . . 5. 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, . . . 6. 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, . . . 7. a. The sequence will range from 0 to 11
and then repeat the pattern from 0 to 11. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, . . .
b. The remainder will always be a whole number ranging from 0 to 1 less than the divisor. The remainders repeat because every whole number is either a multiple of the divisor of a multiple of the divisor plus a whole number between 1 and 1 less than the divisor inclusive.
8. a. 10 b. Every whole number divided by 10 will
give a remainder that is a whole number from 0 to 9 inclusive.
9. 3:00 P.M.
Section Quiz: Lessons 4-5 to 4-6
1. D 2. F 3. B 4. G 5. C 6. J 7. B 8. G 9. C
Chapter 4 Enrichment: Perfect Numbers
33,550,336
Chapter 4 Big Ideas
1. Continuous graphs are graphs that are connected by lines or curves. Discrete graphs are graphs that only have distinct points. An example of continuous data is time. An example of discrete data is the number of candles sold.
2. A relation is a function when each domain value is paired with exactly one range value.
3. The input of a function is the independent variable and the output of a function is the dependent variable. The value of the
dependent variable depends on the value of the independent variable.
4. Step 1: Use the function to generate ordered pairs by choosing several values for x. Step 2: Plot enough points to see a pattern for the graph. Step 3: Connect the points with a line or smooth curve.
5. Sample answer: the number of empty seats in a classroom and the number of students seated in the class.
6. First, find the common difference. Then, write a rule to find the nth term.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 131
Chapter 5 Answers
LESSON 5–1
Reading Strategies 1. x + 4y = 9 2. yes;
Each domain value is paired with exactly one range value.
3. no; A constant change of +2 in x does not correspond to a constant change in y.
4. x + y = 4
Review for Mastery 1. no 2. yes 3. yes 4. no 5. no 6. yes 7. D: all real numbers; R: all real numbers 8. D: x ≥ 0; R: y ≥ 0 9. D: {0, 1, 2, 3, 4}; R: {0, 0.5, 1, 1.5, 2} 10.
D: x ≥ 0; R: y ≥ 0
Practice 1. yes 2. Each domain value is paired with exactly
one range value. 3. yes 4. yes 5. A constant change of +1 in x corresponds
to a constant change of −2 in y. 6. −x + y = −4 7. yes 8.
9.
D: x ≥ 0; R: y ≥ 0
x 6 4 2 0 −2
y −3 −1 0 2 3
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 132
Challenge
LESSON 5–2
Reading Strategies 1. x-int: 2; y-int: −3 2. x-int: 3; y-int: −3 3. x-int: −3; y-int: 5
Review for Mastery 1. x-int: 3; y-int: −3 2. x-int: −1; y-int: −2 3. x-int: −2; y-int: 4 4. x-int: 3; the time it took to complete the trip.
y-int: 120; the number of miles left to driven.
5.
6.
7.
Practice 1. 2; 4 2. 4; −3 3. 1; −2 4.
a. x = 3 b. 3 c. y = 2 d. 2
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 133
5.
a. x-int: 5; y-int: 50 b. x-int: the amount of money after 5
weeks. y-int: the amount of money before she makes any withdrawals
Challenge 1. ≈5.8 units 2. ≈4.5 units
3.
−4; 4 4; −4 4; 4 −4; −4 ≈22.6 units
4.
8, 9;
−4, 9; 8, −4.5;
≈50.1 units
LESSON 5–3
Reading Strategies 1. temperature 2. time
3. 123
= 4; 42
= 2; 44
= 1
4. upward is vertical and the y-axis is the vertical axis.
5. people run horizontally and the x-axis is the horizontal axis.
6. − 23
Review for Mastery
1. 1.63 1 .27 0.36 0.181988 1 986 2
− = =−
2. 1.15 1 .63 0.48 0.481989 1 988 1
− −= = −−
3. 2.26 1 .5 1.11 0.5551991 1 989 2
− = =−
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 134
4. 1989 to 1991 5. yes; 1988 to 1989 6. 2
7. − 14
8. 34
9. − 12
10. 1
11. −5
Practice 1. rise 2. run
3. rise to run 4. 13
5. 2 6. − 32
7. zero 8. negative 9. undefined
10. 35 mi/h; 12mi/h; 11mi/h; 39 mi/h; 12
mi/h;
between hours 8 and 10
Challenge
1. Possible answer: 2 ,8
3 ,12
416
2.
They form a straight line.
3. 14
5.
(4, 3), (8, 6)
6. 3 ,4
34
7.
34
8. The fractions are written as yx
and slope
is defined as change in y divided by change in x.
9. yes; (−4, −3) would be written as 34
−−
which equals 3 .4
LESSON 5–4
Reading Strategies 1. 4 2. Possible answer: (−1, 6) 3. 2 4. (−4, 1) and (6, 5)
5. 25
6. 0
7. horizontal
Review for Mastery
1. 34
2. 12
3. 4 4. − 25
5. −2 6. 3
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 135
7. 12
8. − 43
9. 14
Practice
1. 9; 3; 6
2. 1 3 ;2 ( 2)− −
− − 4 ;
4− −1
3. 2 6 ;0 4− −
− 8 ;4
−−
2 4. − 23
5. −2 6. 73
7. 2; the profit increases $2 for every box sold.
8. 32
; For each additional crust made, 1.5
cups of flour are needed. 9. a. 0; b. 0;
2x; 5y; 2; 2 5; 5 −5 −2
c. −5; −2;
−2; −5; 25
−
Challenge 1.
2. Slope of AB = Slope of BC = 2
4. Slope of AB = 2; Slope of BD = 3 5. Points A, B, and C lie on a line if the
slopes of ,AB BC and AC are equal.
6. 3% 7. 300 feet 8. a. 6% b. No; the uphill grade is positive and the
downhill grade is negative. 9. $2
LESSON 5–5
Reading Strategies 1. Divide by 2 when averaging two
numbers. 2. Taking a square root. 3. Yes, the sums of the squares can be
added in any order. 4. (5, 4); ≈ 11.7
Review for Mastery 1. (9, 7) 2. (4, 3)
3. 12, 32
⎛ ⎞⎜ ⎟⎝ ⎠
4. (17, 1)
5. ≈ 3.6 6. ≈ 5.8 7. ≈ 8.2 8. ≈ 9.2 9. ≈ 8.1 10. 10
Practice 1. 2, 8, 1, 3, 10, 4, 5, 2 2. (3, 2) 3. (3, 4)
4. 372
14 311
x
xx
+=
= +=
642
8 62
y
yy
+=
= +=
(11, 2)
5. (1, 2) 6. 4, 1, 5, 1, 3, 4, 9, 16, 25, 5 7. 4.5≈ 8. 6.4≈ 9. 11.3≈ miles
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 136
Challenge 1. (6.5, 1, 2) 2. (–2.5, 5, 4) 3. 4. 5. ≈ 12.4 6. 15
LESSON 5–6
Reading Strategies
1. 8 2. 13
3. –2.5 4. y = 12 – x; no
5. y = 32
x; yes 6. 1 ,2
1,2
12
7. yes 8. 12
Review for Mastery
1. no 2. yes; 43
3. yes; –3 4. yes; – 12
5. yes; 3.5 6. no 7. 40 8. –8.75
9.
y = 2.13x
Practice
1. y = 7x yes yes 7 2. y = 4x – 10 no no n/a
3. y = 52
x yes yes 52
4. 1 ;5
1;5
15
yes yes
5. 4; 2.5; 2 no no
6. k = 1 ;2
y = 1 (8) 42
⎛ ⎞ =⎜ ⎟⎝ ⎠
7. k = 3 ;2
y = 3 45(15)2 2
⎛ ⎞ =⎜ ⎟⎝ ⎠
8.
y = 4x
Challenge 1. a. y = 5t, 40 minutes b. y = 10t, 20 minutes
x
y
z
x
y
z
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 137
c. y = 15t, 13 13
minutes
2. a. y = 15t b. y = 25t c. y = 30t, 3. a. Q = 500 + 30t – 35t b. emptying at 5 gallons per minute c. after 100 minutes, the tank will be empty 4. a. Q = 500 + 30(t + 6) – 35t,
or Q = 680 – 5t b. after 136 minutes, the tank will be
empty
Section Quiz: Lessons 5-1 to 5-6
1. B 2. F 3. B 4. H 5. D 6. F 7. C 8. H 9. D 10. G 11. D
LESSON 5–7
Reading Strategies 1. With a fraction, you have a “rise” and
“run” for graphing. 2. (0, –8) 3. 5; 12 4. –3; 0 5. 1; –4
6. 13
; 3
7.
8.
Review for Mastery
1. y = 14
x + 3 2. y = –5x
3. y = 7x – 2 4. y = 3x – 6
5. y = 12
x + 9 6. y = –x + 3
7. y = –5x + 30 8. y = x – 7
9. y = 43
x + 4
10. y = 2x – 3
Practice
1. 23
; 2 2. –1; 8
3. 3; –6; 6; 6; 11;
–2; 11
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 138
4. y = 2x – 4
5. y = – 12
x + 3
6. y = – 23
x + 2
7. a. y = 25x + 30 b. slope: 25; number of desks per
classroom; y-int 30; number of spare desks
c. 630
Challenge 1. a1 = 3 2. d = 2 3. an = 3 + (n – 1)(2) 4.
x 1 2 3 4 5 6 7
y 3 5 7 9 11 13 15
5.
No, because the domain of the sequence
is restricted to natural numbers: {1, 2, 3, 4, …}.
6. y = 2x + 1 7. a. The slope is the same as the common
difference (m = d = 2). b. The y-intercept is the same as the first
term less the common difference (b = a1 – d = 1).
8. y = –3x + 8; m = d = –3 and b = a1 – d = 5 – (–3) = 8
9. an = 4 + (n – 1)(5); d = m = 5 and a1 = b + d = –1 + 5 = 4
LESSON 5–8
Reading Strategies 1. the slope of the line and one point on
the line 2. the slope; you can substitute the
coordinates of the two given points into the slope formula.
3. y + 10 = 4(x – 3) 4. y = –2x – 1
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 139
Review for Mastery 1.
2.
3.
4.
5.
6.
7. y = –3x + 5 8. y = 14
x + 1
9. y = 4x 10. y = 5x – 3
11. y = 12
x – 1 12. y = –x + 5
Practice 1. C 2. A 3. B 4. y – 8 = 4(x – 3)
5. y + 3 = – 12
(x – 5) 6. y = 5x + 2
7. y = –3x + 12 8. 2; y = 2x + 2
9. 12
; y = 12
x – 6 10. x-int:4, y-int: 10
11. x-int: –1, y-int: 3 12. y = 8x + 30; $1070
Challenge Plot an even number of pairs of points
such that the two points in each pair are collinear but neither point is collinear with any other plotted point.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 140
LESSON 5–9
Reading Strategies Possible answers are given for 1 and 2.
1. y = 3x + 3 2. . y = – 13
x + 3
3. – 13
4. 1
5. –1
6. yes, BC and AC are perpendicular because 1(–1) = –1.
Review for Mastery 1. y = 2x + 1; y = 2x – 3
2. y = – 23
x + 2; y = 32
x + 1
3. 6; – 1
6 4. 4 3;
3 4−
5. y = –x + 11 6. y = 14
x – 3
Practice
1. y = 4; y = 12
x + 3; y = 12
x; y = 2x
2. y – 5 = 6(x + 2); y = –6x; 6x + y = 4; y = 6 3. 0; –2; 0; –2;
The opposite sides have the same slope which means they are parallel. ABCD is a parallelogram because both pairs of opposite sides are parallel.
4. y = x – 4; y = 3; y = –x; y = –3
5. y = 5x + 1; y = 3; y = 15
x; x = 5
6. y = 13
x – 2; x = 2; y – 4 = 3(x + 3);
y = –3x + 9
7. 2 ;5
– 5 ;2
– 3 ;7
25
52
⎛ ⎞−⎜ ⎟⎝ ⎠
= –1 so AB is perpendicular to
.BC ABC is a right triangle because it contains a right angle.
Challenge 1. a. y = 2x + 4 2. a. –3
b. y = – 12
x – 4
3. a. 3
b. y = – 12
x + 72
4. It is a rectangle because it has four right angles.
5–6. x – y = –4; x + y = –4; x – y = 4; x + y = 4
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 141
LESSON 5–10
Reading Strategies 1. 5; 2 – (–3) = 5 2. f(x); f(x) 3. y-axis; opposite 4. rotation (less steep) about (0, –3) 5. reflection about y-axis 6. translation 6 units up
Review for Mastery 1.
translation 3 units up 2. rotation (less steep) 3. translation 5 units down 4. rotation (steeper) 5. h(x) = –5x 6. h(x) = 9x 7. h(x) = –2x + 7 8. rotation (steeper) and translation
2 units down 9. rotation (less steep) and translation
3 units up
10. rotation (less steep) and translation 8 units up.
11. rotation (less steep) and reflection across the y-axis
Practice 1. rotation 2. translation 3. reflection 4.
translation 5 units up 5.
rotation (steeper) about (0, –1) 6.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 142
rotation (less steep) about (0, 0) and translation 7 units down
7. a. The graph will be translated 3 units up. b. The graph will be rotated about (0, 12)
and become less steep.
Challenge
1. y + 3 = 2(x + 2) OR y – 5 = 2(x – 2) 3. y + 1 = 2(x + 2) OR y – 7 = 2(x – 2) 4. In the form y – y1 = m(x – x1), the value of
y1, increased by 2 units. 5. y + 9 = 2(x + 2) OR y + 1 = 2(x – 2)
7. y + 3 = 2(x – 3) OR y – 5 = 2(x – 7) 8. In the form y – y1 = m(x – x1), the value of
x1, increased by 5 units. 9. y + 3 = 2(x + 5) OR y – 5 = 2(x + 1) 10. Translated down 4 units and right 9 units.
Section Quiz: Lessons 5-7 to 5-10
1. A 2. H 3. B 4. G 5. B 6. H 7. D 8. F 9. C 10. G 11. D 12 F
Chapter 5 Enrichment: Coordination
Greatest Common Factor
Chapter 5 Big Ideas
1. To find the x-intercept, substitute 0 for y and solve for x. To find the y-intercept, substitute 0 for x and solve for y.
2. The line of a positive slope rises from left to right. The line of a negative slope falls from left to right. The line of a slope of zero is horizontal. The line of a slope that is undefined is vertical.
3. The slopes of parallel lines are the same. The slopes of perpendicular lines are opposite reciprocals of each other.
4. When the y-intercept b is changed in the function f(x) = mx + b, the graph is translated vertically. When the slope m is changed in the function f(x) = mx + b it causes a rotation of the graph about the point (0, b), which changes the line’s steepness. When the slope m is multiplied by –1 in f(x) = mx + b, the graph is reflected across the y-axis.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 143
Chapter 6 Answers
LESSON 6–1
Reading Strategies 1. number of people; total cost 2. y = 10x + 120; y = 8x + 150
3.
4. (15, 270) 5. For 15 people, the total cost will be
$270 with both catering companies.
Review for Mastery 1. yes 2. no 3. no 4. yes 5. (−1, 3) 6. (2, −3)
7.
(−5, −3)
8.
(3, −3)
Practice 1.
x − y = −2 2x + y = 6
(2) − (4) −2 2(2) + (4) 6
−2 −2 (4) + (4) 6
(8) 6
or
or is not 2.
2x + y = 0 x + 4y = −7
2(1) + (−2) 0 (1) + 4(−2) −7
(2) + (−2) 0 (1) + (−8) −7
(0) 0 (−7) −7
or
or is 3.
(1, −2)
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 144
4.
(−3, 4) 5. 20 miles; $80
Challenge 1.
(−2, 0) and (3, 5) 2.
about (−2.3, 0) and about (2.3, 0)
3.
about (−2.7, 1.7) and (4, 5)
LESSON 6–2
Reading Strategies 1. y = 5 − x or y = −x + 5 2. The first equation is already solved for y. 3. The solution of a system must satisfy
both equations. 4. (6, 4) 5. (−3, 5)
Review for Mastery 1. (2, 3) 2. (7, 9)
3. (−4, 1) 4. (17, 7) 5. (3, 6) 6. (−1, 7)
Practice 1. 3x; 3x; x; 2x; 2; 2; x; 2; 2; 2; 2; 6; 2; 6 2. x − 3; x − 3; 4x; 4x; 4; 4; x; 7; 7; 7; 7; 4; 7; 4 3. (3, 12) 4. (2, 0) 5. (2, −5)
6. a. 20 4526 30
y xy x
= +⎧⎨ = +⎩
b. 2.5 c. $95
Challenge 1. x = 3; y = −1, z = 4 2. x = −1; y = 4, z = −3 3. x = 2; y = 8, z = −5 4. x = −2; y = 13, z = 4
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 145
LESSON 6–3
Reading Strategies 1. Multiply the first equation by 3 and the
second equation by 5 to get common coefficients of −15.
2. 4(9 10 7) 36 40 285(5 8 31) 25 40 155
x y x yx y x y
− = − =⎧ ⎧⇒⎨ ⎨+ = + =⎩ ⎩
3. (1, −3) 4. (10, −10)
Review for Mastery 1. (3, −14) 2. (4, −1) 3. (−6, 18) 4. (−4, −14) 5. (7, 10) 6. (7, −18) 7. (5, −2)
Practice 1. 0y; 3x; 3; 3; 2; 2; 2; 2; 2; 12; 4; 2; 4 2. 2y; −7; 0y; −3; −3; −1; −1; 3; 3; 3; 9; 9; 9;
−2; 2; 2; −1; 3; −1 3. 2; 4; −16; 5; 10; 5; 10; 5; 5; 2; 2; 2; 2; 2;
−10; −2; −2; 5; 2; 5 4. (3, 4) 5. (6, −2) 6. (−8, −1) 7. 3 and −4
Challenge
1. 5 4 8659 6 6 1410
x y zx y z
+ + =⎧⎨ + + =⎩
2. 5x − 2y = 20 3. −21x − 18y = −3780 4. x = 60
5. 2 2854 565
y zy z
+ =⎧⎨ + =⎩
6. y = 140, z = 5 7. sleeping bags: $60; tents: $140;
bug repellant: $5
LESSON 6–4
Reading Strategies 1. inconsistent 2. infinitely many 3. infinitely many
4. consistent, independent 5. one 6. inconsistent; 0; parallel lines 7. consistent, dependent; infinitely many; coincident lines 8. consistent, independent; 1; intersecting lines
Review for Mastery 1. infinitely many solutions 2. no solution 3. no solution 4. inconsistent; no solutions 5. consistent and dependent; infinitely many solutions 6. consistent and independent; one solution 7. consistent and independent; one solution; (0, 8) 8. consistent and dependent; infinitely many solutions
Practice 1. no solution 2. infinitely many solutions 3. infinitely many solutions 4. no solution 5. infinitely many solutions; consistent, dependent 6. no solution; inconsistent 8. They will always have the same amount
of money. The graphs of these equations are the
same line.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 146
Challenge 1. Start −1 −2 −7 −13 −14 −19 − Finish 2. Start −6 −10 −15 −21 −17 −23 − Finish
Section Quiz: Lessons 6-1 to 6-4
1. D 2. G 3. B 4. H 5. D 6. F 7. B 8. G 9. A 10. J
LESSON 6–5
Reading Strategies 1. solid; 2. dotted; below above 3. solid; above
4.
solution; (4, 1) not: (−2, 2) 5.
solution: (−3, −3)
not: (2, 7)
Review for Mastery 1.
2.
3.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 147
4.
5.
Practice 1. no 2. yes 3. no
4.
y ≤ x + 3
5.
y > −3x − 1 6. a. x + y ≤ 8
b.
c. Possible answer: 2 peach, 6 blueberry
or 4 peach, 3 blueberry 7. y ≥ x − 2 8. y < 2x + 4
9. y ≤ − 12
x
Challenge 1.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 148
x Think: y = (x + 3)2 − 4 (x, y)
−6 y = (−6 + 3)2 − 4 = −(3)2 − 4 = 9 − 4 = 5 (−6, 5)
−5 y = (−5 + 3)2 − 4 = −(2)2 − 4 = 4 − 4 = 0 (−5, 0)
−4 y = (−4 + 3)2 − 4 = (−1)2 − 4 = 1 − 4 = −3 (−4, −3)
−3 y = (−3 + 3)2 − 4 = (−0)2 − 4 = 0 − 4 = −4 (−3, −4)
−2 y = (−2 + 3)2 − 4 = (1)2 − 4 = 1 − 4 = −3 (−2, −3)
−1 y = (−1 + 3)2 − 4 = (2)2 − 4 = 4 − 4 = 0 (−1, 0)
2. y ≤ −|x| + 5
x Think: y = −|x| + 5 (x, y)
−3 y = −|−3| + 5 = −(3) + 5 = −3 + 5 = 2 (−3, 2)
−2 y = −|−2| + 5 = −(2) + 5 = −2 + 5 = 3 (−2, 3)
−1 y = −|−1| + 5 = −(1) + 5 = −1 + 5 = 4 (−1, 4)
0 y = −|0| + 5 = 0 + 5 = 5 (0, 5)
1 y = −|1| + 5 = −1 + 5 = 4 (1, 4)
2 y = −|2| + 5 = −2 + 5 = 3 (2, 3)
3. y < 12
x2 − 2
x Think: y = 12
x2 − 2 (x, y)
−4 y = 12
(−4)2 − 2 = 12
(16) − 2 =
8 − 2 = 6 (−4, 6)
−2 y = 1
2(−2)2 − 2 = 1
2(4) − 2 −
2 − 2 = 0 (−2, 0)
0 y = 12
(0)2 − 2 = 12
(0) − 2 =
0 − 2 = −2 (0, −2)
1 y = 1
2(1)2 − 2 = 1
2(1) − 2 =
12
− 2 = −1 12
11, 12
⎛ ⎞−⎜ ⎟⎝ ⎠
2 y = 1
2(2)2 − 2 = 1
2(4) − 2 =
2 − 2 = 0 (2, 0)
4 y = 1
2(4)2 − 2 = 1
2(16) − 2 =
8 − 2 = 6 (4, 6)
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 149
LESSON 6–6
Reading Strategies 1.
2 2 308
x yx
+ ≤⎧⎨ >⎩
2.
3.I = 10 ft, w = 5 ft and I = 11 ft, w = 4 ft
Review for Mastery 1.
Possible Answers: Sol: (1, −3), (−2, −4) not sol: (3, −3), (−3, 2)
2.
Possible Answers: Sol: (−4, 3), (−3, 1) not sol: (3, −3), (2, 2)
3.
4.
5.
6.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 150
7.
Practice 1. yes 2. no 3. yes
4.
a. (−1, 0) and (−3, 2) b. (0, −3) and (4, 0) 5.
a. (0, 0) and (1, 2) b. (1, 0) and (−4, 3)
6.
a. (3, 3) and (4, 4) b. (0, 0) and (0, 3) 7. a.
64 2 20x y
x y+ ≥⎧
⎨ + ≤⎩
b.
c. Any combination represented by the
ordered pairs in the solution region. d. 2 lbs mac. salad, 5 lbs potato salad; 3 lbs mac. salad, 4 lbs potato salad
Challenge 1.
2. The solution is the interior and boundary
of the square formed by the absolute value inequalities.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 151
3.
a. |x| < 3, |y| < 5 b. |x| > 3, |y| > 5 4. Explanations may vary but should give
the following absolute value inequalities: |x − 1| ≤ 2 and |y − 3| ≤ 5
5. | |, where 0
| |x a
ay a
≤⎧ >⎨ ≤⎩
Section Quiz: Lessons 6-5 to 6-6
1. D 2. G 3. A 4. J 5. B 6. J 7. D
Chapter 6 Enrichment: Multiply It Out
1. −6, T 2. −1, 1 3. 11, E 4. 3, R 5. 10, U 6. 1, S 7. 9, Y 8. −8, B 9. −7, O 10. 8, D 11. −11, P 12. 0, V Distributive Property
Chapter 6 Big Ideas
1. Step 1 Solve for one variable in at least one equation, if necessary.
Step 2 Substitute the resulting expression into the other equation.
Step 3 Solve the equation to get the value of the first variable.
Step 4 Substitute that value into one of the original equations and solve.
Step 5 Write the values from Steps 3 and 4 as an ordered pair, (x, y), and check.
2. Use the graphing method when both equations are solved for y or when an estimate is needed.
Use substitution when a variable in either equation has a coefficient of 1 or -1, both equations are solved for the same variable, or either equation is solved for a variable.
Use elimination when both equations have the same variable with the same or opposite coefficients, or a variable term in one equation is a multiple of the corresponding variable term in the other equation.
3. Step 1 Solve the inequality for y (slope-intercept form).
Step 2 Graph the boundary line. Use a solid line for or . Use a dashed line for or < or >.
Step 3 Shade the half-plane above the line for y > or y ≥. Shade the half-plane below the line for y < or y ≤.
4. The graph of a dependent system is the same line. The graph of a consistent system is intersecting lines. The graph of an inconsistent system is parallel lines.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 152
Chapter 7 Answers
LESSON 7–1
Reading Strategies 1. 6 2. 0
3. 8−3 4. 71
b
5. 32 6. 132
7. 1 8. 11,000,000
9. −64 10. − 164
11. 41t
12. 2
3cd
13. 58x
14. 12
Review for Mastery
1. 5; 2•2•2•2•2; 132
2. 10•10•10; 11000
3. 4; 4; 5•5•5•5; 625 4. 45y
5. 8a3 6. 3
29xy
7. 4x
y 8. a
b
9. 2
45yx
10. 4
11. 512 12. 200 13. −8 14. 324
15. 116
Practice 1. 3; 3; 9 2. 2; 2; 2; 2; 16 3. −3; −3; −3; −3; −27 4. −1; −1; −1; −1; −1; −1; −1; −1
5. −1 6. 164
7. 3; 9 8. 2; 3; 1; 3; 127
9. −2; (−2)4; −2; −2; −2; −2; 16; 516
10. 34x
11. 5b2
12. 3
4mn
13. 41
2k
14. f 4 g 15. r 6 s2
16. 14
gram or 0.25 gram
17. 10,000; 1000
Challenge 1. 1; 1; 1; 1; 1; 1; 1; 1; 1 2. 2; 4; 8; 6; 2; 4; 8; 6; 2 3. 3; 9; 7; 1; 3; 9; 7; 1; 3 4. 4; 6; 4; 6; 4; 6; 4; 6; 4 5. 5; 5; 5; 5; 5; 5; 5; 5; 5 6. 6; 6; 6; 6; 6; 6; 6; 6; 6 7. 7; 9; 3; 1; 7; 9; 3; 1; 7 8. 8; 4; 2; 6; 8; 4; 2; 6; 8 9. 9; 1; 9; 1; 9; 1; 9; 1; 9 10. 0; 0; 0; 0; 0; 0; 0; 0; 0 11. For all n, 1n has 1 as its units digit. 12. The pattern is 2, 4, 8, and 6, for n = 1, 2,
3, and 4 and then repeats. 13. The pattern is 3, 9, 7, and 1, for n = 1, 2,
3, and 4 and then repeats. 14. For all n > 0, 5n has 5 as its units digit. 15. If you divide n by 4, then the units digit is
7, 9, 3, or 1, depending on whether the remainder is 1, 2, 3, or 0, respectively.
LESSON 7–2
Reading Strategies 1. 4 × 109 2. 0.0097 3. 4.19 × 1011 4. B 5. A 6. C
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 153
7. D 8. 56,000,000 9. 875,000
Review for Mastery 1. right; 1,000,000 2. left; 0.01 3. right; 10,000 4. positive; 103 5. negative; 10−4 6. positive; 107 7. positive 6.12 × 107 8. negative 4.5 × 10−4 9. positive 4.67 × 106 10. 7451 11. 0.00004231 12. 293,000,000
Practice 1. 0.01 2. 1000 3. 1 4. 0.00001 5. 0.1 6. 10,000,000 7. 5; 5 8. 4; −4 9. 3; −3 10. 226,000 11. 0.00528 12. 0.0476283 13. 482,000 14. 0.029 15. 60,000 16. 4.5 × 103 17. 6.56 × 106 18. 2 × 10−5 19. 2.03 × 10−3 20. 7.2 × 10−3; 9.1 × 10−3; 4.7 × 103;
9.2 × 103; 2.4 × 104; 6.13 × 104 21. 1.1 × 10−3; 4.1 × 10−2; 5.6 × 10−2;
4.2 × 108; 4.5 × 108; 9.2 × 108 22. 7.39 × 106 23. 40,000 sq mi 24. 290,000,000 25. 5,980,000 sq mi
Challenge 1. a. 9.3 × 107 miles
b. 4.84 × 108 miles c. about 5.2 times
2. a. about 39.5 times b. about 7.6 times
3. about 269,000 times 4. 5.9 × 1012 5. a. about 1.18 × 1018
b. about 1.3 × 1010 times
6. about 3.8 × 1028 to 1. 7. about 3.6 × 1051 to 1.
LESSON 7–3
Reading Strategies 1. multiply 2. add 3. Power of a Product 4. Power of a Product; with both properties,
a number is applied to all parts. 5. m24 6. 88
7. 9v10 8. 8
5dc
9. 5832 10. −16y14
Review for Mastery 1. 27 2. 84 • 53 3. 212 • 33 4. m15 n4
5. 68 6. 614
7. 915
8. 8 31
x y
9. 12
10vu
10. yes
11. no; 4h6 12. no; m3 13. 16x10 14. s12 t9
15. 5
2032yx
−
Practice 1. am + n 2. amn 3. an bn 4. 5; 8; 13 5. −2; 7; 5 6. 3; −5; −2 7. a2 b4 8. n 9. r 2 s5 10. 3; 4; 12 11. 0; 5; 0; 1 12. 2; −4; −8; 1; b8 13. 4; 4; 81n4 14. 3; 3; 8x3 15. 3; 4; 12; 12; 3; t15 16. b15
17. 1 18. 161
c
19. 64a6b12
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 154
Challenge 1. 0.00012 2. 0.00048 3. 0.00009 4. 0.0000108 5. 0.00012 6. 0.00048 7. 0.00009 8. 0.0000108 9. 0.00000036 10. 0.000006 11. Multiply the decimals in the same way as
the whole numbers. Count the number of decimal places in each of the three numbers. Find the sum of those numbers. Move the decimal point that many places to the left.
LESSON 7–4
Reading Strategies
1. subtract 2. 48
5⎛ ⎞⎜ ⎟⎝ ⎠
3. Positive Power of a Quotient
4. 144 5. 16625
6. 6481
7. 3
4 5g
f h
8. 18
6ts
9. 10 532
c d
10. 827
11. 4
4xy
12. 14
625g
f
Review for Mastery 1. 25 2. x6 y2
3. ba
4. 3
325
or 8125
5. 18
12xy
6. 6
49mn
7. 3
6ab
8. 4
2xy
9. 94
10. 53
; 53
; 12527
11. 7
3zxy
; 7; 1; 3; 35; 5; 15
12. 2 3c
a b; 1; 2; 3; 4; 8; 12
13. 5
5yx
14. 89
49m
15. 15
1032ba
16. 2
29nm
17. 627
8x 18.
12
416s
r
Practice 1. 4; 81 2. 8; 5; 3
3. 2; 7; −5; 51t
4. 6; 3; 6; 4; 4
3ts
5. 13 41
a b 6. 5
1xy
7. 649
8. 4
423
; 1681
9. 3
34x ;
3
64x 10. 5
4; 25
16
11. 81a4 b4
256c8 12. 3
3278
cb
13. 2
343x yz
; 8 4
12256
81x yz
14. 4n
; 36n ; 2n
15. 1.5; 8 16. 0.2; 2; 2; −1; 2; 2 × 101 17. a. 3.5 × 101
b. $250,000
Challenge 1. 23 × 31 2. 22 × 33 3. 22 × 1131 4. 23 × 32 × 52
5. 2
32 • 32 • 3
= 232
6. 4
2 22 • 3
2 • 3 • 5 =
223 • 5
7. 3
5 22 • 52 • 3
= 3
4 25
2 • 3
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 155
8. 2 3
3 2 22 • 3 • 52 • 3 • 5
= 32 • 5
9. If a prime number base b appear in the numerator (or denominator), it cannot occur in the denominator (or numerator) as well because then the rational number is not fully simplified.
ex: n
mb ab c
= n mb ac
−
10. Every rational number can be written as a quotient whose numerator is 1 or the product of prime numbers raised to positive integer exponents and whose denominator can be written as 1 or the product of prime numbers raised to positive integer exponents, and there are no prime bases common to the numerator and the denominator.
LESSON 7–5
Reading Strategies 1. 3rd or cube 2. 5th 3. 5th; 4th 4. 2 5. 9 6. 2 7. 20 8. 4 9. 27 10. 8 11. 1024 12. 64
Review for Mastery 1. 8 2. 10 3. 1 4. 4 5. 2 6. 7 7. 6 8. 14 9. 3 10. 1 11. 7 12. 5 13. 8 14. 8 15. 4 16. 1 17. 81 18. 1000 19. 4 20. 243 21. 8 22. 32 23. 343 24. 256
Practice 1. B 2. D 3. C 4. A 5. 7 6. 3 7. 1 8. 12 9. 8 10. 9 11. 1 12. 32 13. x8 14. x3y4 15. m4n 16. x2
17. 14 cm
Challenge
Section Quiz: Lessons 7-1 to 7-5
1. C 2. H 3. D 4. H 5. B 6. G 7. D 8. H 9. A 10. F 11. D 12. H 13. B 14. G 15. C 16. F
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 156
LESSON 7–6
Reading Strategies 1. 3; 1 2. The exponents on the variables have a
sum of 4. 3. cubic; binomial 4. −4g2 + 8g + 1 5. −4 6. 2 7. trinomial 8. quadratic
Review for Mastery 1. 8 2. 3 3. 4 4. 6 5. 5 6. 2 7. −6x5 − 5x4 + x3; −6 8. −x3 + 5x2 + 2x; −1 9. 7x2 + 8x −1; 7 10. quadratic binomial 11. cubic polynomial 12. 102 feet
Practice 1. 2; 1 2. 3; 2 3. 5; 4 4. trinomial 5. monomial 6. binomial 7. 3x2 − x + 12; 3 8. −g5 + g4 − 2g3; −1 9. k4 − k3 + k2 + 1; 1 10. quadratic monomial 11. linear binomial 12. quartic trinomial 13. −3; 13; −2 14. 72 in2
Challenge 1. −3xy4 2. 2x4y4 + 3x2y5 3. x5 + x3 + x 4. 2xy + 5xz 5. 2x4 6. 4xyz2 + xyz 7. 4x2 − 3x5 + x 8. x − 3 9. x3y − 3xyz + z 10. 4x2 − 3x2y 11. x2 + y2 + z2 + w2 12. x3 + x2 + x 13. −3 + x + 4x2 14. x4 + x3 + x2
LESSON 7–7
Reading Strategies 1. −3 2. −1 3. −3x and 4x; 7 and −24 4. x2 and −3x2; 8x and 3x; −4 and −2 5. 2x3 + 2x + 7 6. −8x5 + 2x4 7. 9x2 + 16x + 2 8. 2x3 − 2x2 + 3 9. 4x4 + 2x − 2 10. −2x3 + 3x2 − 10x + 4
Review for Mastery 1. no; same variable raised to different
power 2. yes; same variable raised to same power 3. no; different variable raised to same
power 4. 3y2 + 10y 5. 4m4 + 3m 6. 12x5 + 18x4 7. 8x2 + 9x 8. m2 − 2m + 7 9. 10x3 + x2 + 3x + 9 10. 3y5 + 8y4 − 8y3 11. −x2 − 7x 12. 3x3 − 4x + 8 13. 5x4 − x3 + 7x2 + 3 14. 9x3 − 8x 15. 8t4 + 1 16. −2x3 + 4x + 4 17. t3 − t2 − 4t − 6 18. 4c5 − c3 + 8c2 − 7
Practice 1. 4x3 − 6 2. –20p5 − 3p + 14 3. 3m + 6 4. 5y2 + y + 12 5. 6z3 + 4z2 + 5 6. 12g2 + 4g − 1 7. 7x3 + 6x 8. 8k + 1 9. 3s3 + 5s + 20 10. 9a4 + 8a2 11. 9b2 + b − 9 12. 4c3 + 6c 13. w + 8 14. a. 2n + 2 b. 8n + 20 c. 12n + 28
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 157
Challenge 1. 4 2. −4b3 + b − 6 3. 22 − 12d 4. −3g8 − 7g6 − g4 − 7g3 + g2 − g + 16 5. Q(x) = x4 + 1; S(x) = x4 + x3 + 1 6. Q(x) = x3 + x2 + 1; S(x) = x3 + 1 7. Q(x) = x3 + x2 + 1; S(x) = −x3 + x2 8. Q(x) = x5 + x4 + 1; S(x) = x4 + x3 + x2
LESSON 7–8
Reading Strategies 1. 4 2. They have the same exponent on the
same variable. 3. There are no like terms. 4. −6x5 + 12x4 − 3x3 5. 18x3 + 57x2 + 30x 6. 7x2 − 19x − 6 7. 2x5 − 10x4 + 8x3 − 22x2 − 34x + 8
Review for Mastery 1. −10x3y4 2. −8x3y2z2 3. 3x3y3 4. 4; 4; 4x − 20 5. 3x; 3x; 3x2 + 24x 6. 2x; 2x; 2x; 2x3 − 12x2 + 6x 7. 5x + 45 8. −4x3 − 32x 9. 6x4 + 15x3 + 12x2 10. 3x2 − 21 11. 10a4b2 12. −5y3 + 35y2 − 10y 13. x; 4; x2 − x − 20 14. x; 2; x2 + 6x − 16 15. x; 3; x2 − 9x + 18 16. x2 − 5x + 6 17. x2 − 49 18. x2 + 3x + 2 19. x, 3; 2x3 + 10x2 + 20x + 24 20. x, 2; 6x3 + 16x2 + 13x + 10
Practice 1. 20x2 2. 15x3 3. 18y5 4. 15x + 21 5. 8x3 + 28x2 + 12x
6. x2; 2x; 5x; 10; 7. x2; −3x; 4x; −12; x2 + 7x + 10 x2 + x − 12 8. x3; 4x2; 7x; 3x2; 12x; 21; x3 + 7x2 + 19x + 21 9. 8x4; −6x3; 10x; −4x3; 3x2; −5; 8x4 − 10x3 + 3x2 + 10x − 5 10. a. w; b. w + 5; c. w2 + 5w; d. 24 in2;
e. 36 in2
Challenge 1. x + 4 2. x + 1 3. x + 2 4. x + 3 5. x + 6 6. x − 2 7. x − 3 8. x − 4 9. x − 6 10. x + 2 11. x + 3 12. x − 6 13. x − 4 14. x + 6; x + 3 15. x + 1; x − 4 16. x − 2; x − 2 17. 3x − 1 18. x − 7 19. 2x − 3; 3x + 4
LESSON 7–9
Reading Strategies 1. difference of squares 2. perfect square trinomial 3. It will have 3 terms. 4. c4 + 20c2d + 100d2;
perfect square trinomial 5. 4s2 − 9; difference of squares
Review for Mastery 1. yes 2. no 3. yes 4. x2; 14x; 49; x2 + 14x + 49 5. x2; 2x; 1; x2 − 2x + 1 6. 4x2; 40x; 100; 4x2 + 40x + 100 7. x2 − 16x + 64 8. x2 + 4x + 4 9. 49x2 − 70x + 25
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 158
10.difference of squares 11. perfect-square trinomial 12. perfect-square trinomial 13. a2; 49; a2 − 49 14. 4; m2; 4 − m2 15. 4x2; 25; 4x2 − 25 16. x2 − 64 17. 100 − x2 18. 25x2 −4y2
Practice 1. x; x; 5; 5; x2 + 10x + 25 2. m; m; 3; 3; m2 + 6m + 9 3. 2; 2; a; a; 4 + 4a + a2 4. x2 + 8x + 16 5. a2 + 14a + 49 6. 64 + 16b + b2 7. x; x; 10; 10; x2 − 20x + 100 8. y; y; 6; 6; y2 − 12y + 36 9. 9; 9; x; x; 81 − 18x + x2 10. y2 − 14y + 49 11. b2 − 22b + 121 12. 9 − 6x + x2 13. x; 7; x2 − 49 14. 4; y; 16 − y2 15. x; 2; x2 − 4 16. x2 − 64 17. 9 − y2 18. x2 − 1
Challenge 1. 4; 64; −60; −15 2. 36; 4; 32; 8 3. 81; 9; 72; 18 4. 121; 9; 112; 28 5. 9; 169; −160; −40 6. a. The difference (a + b)2 − (a − b)2 is four
times the product ab.; b. 4ab; c. (a2 + 2ab + b2) − (a2 − 2ab + b2) = 4ab
7. (a + b)2 + (a − b)2 = 2a2 + 2b2 8. 4a4 9. 4b4 10. [(a2 + b2) + (a2 − b2)]2
= (a2 + a2 + b2 − b2)2 = (2a2)2 = 22a4 = 4a4
11. [(a2 + b2) − (a2 − b2)]2 = (a2 − a2 + b2 + b2)2 = (2b2)2 = 22b4 = 4b4
Section Quiz: Lessons 7-6 to 7-9
1. C 2. g 3. D 4. F 5. A 6. H 7. A 8. H 9. C 10. F 11. C 12. F 13. B 14. H 15. C
Chapter 7 Enrichment: Good Luck Squares
50 17 31 80 42 99 69 300 18 115 46 91 63
72 94 65 89 550 10 8 121 97 950 59 150 73
39 615 54 78 16 106 32 112 15 88 19 825 29
125 60 215 211 225 117 377 76 9 82 105 52 85
22 800 81 1 12 250 21 47 196 25 13 115 45
815 4 114 37 116 6 53 7 500 101 525 36 68
51 400 43 113 93 58 40 325 650 715 71 11 28
750 49 725 23 111 144 925 64 74 33 315 169 61
86 98 14 625 900 107 256 109 30 100 5 700 56
57 75 83 110 102 415 20 90 915 95 350 108 77
600 70 66 24 450 48 25 425 850 26 104 84 38
34 87 44 92 79 515 35 103 67 96 55 41 27
Chapter 7 Big Ideas
1. For zero exponents, a base of zero gives
you =0 00
0. For negative exponents, a
base of zero gives you − = =1 1
000
nn
.
Since dividing by 0 is undefined, neither value exists.
2. When the exponent is a positive number, move the decimal point that number of places to the right. When the exponent is a negative number, move the decimal point that number of places to the left.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 159
3. An exponential expression is completely simplified if there are no negative exponents, the same base does not appear more than once in a product or quotient, no powers are raised to powers, no products are raised to powers, no quotients are raised to powers, and numerical coefficients in a quotient do not have any common factors other than 1.
4. 1 Multiply the First terms: = 2•x x x . 2. Multiply the Outer terms: =• 2 2x x . 3. Multiply the Inner terms: =•3 3x x . 4. Multiply the Last terms: =•3 2 6 .
Then combine like terms: + +2 5 6x x .
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 160
Chapter 8 Answers
LESSON 8–1
Reading Strategies
1. 1, 2, 4, 7, 14, 28 2. a i a i a i a i a
3. 1, 2, 4, 8, 16 4. a i a
5. 4a2 6. 2m 7. no; 11 and 3 are factors.
8. 2 i 33 9. 2 i 3 i 52
10. 24 i 5
Review for Mastery 1.
22 i 11
2.
23 i 7
3.
34
4. 32 i 11 5. 3 i 52
6. 22 i 3 i 7 7. 1, 2, 4, 7, 14, 28
8. 1, 2, 4, 11, 22, 44 9. 4 10. 5 11. 4 12. 12 13. 2a 14. 3x2 15. y
Practice 1.
32 i 22
2.
3. 53 4. 5 5. 8 6. 12 7. 25 8. 3y2 9. p 10. 6 11. 7y2 12. 16 13. 7
Challenge 1. 3152111 2. 325271 3. 210 4. 24325271 5. 2, 3, 5, and 7 6. 2 and 11 7. relatively prime 8. 3, 5, 17, and 31 9. 21325272 10. 21112 11. 1 12. 3255172315 13. 233471 14. 1
LESSON 8–2
Reading Strategies 1. 4x2(x + 3) 2. 6t(5t3 − 3) 3. p(9p + 1) 4. 4(7r4 − 5r2 − 2) 5. 15p5(2p3 + 3) 6. 2m2(3m6 − 8m − 3)
Review for Mastery 1. 5x(4x − 3) 2. 11a(4a + 1) 3. 12(2y − 3x) 4. (x + 7)(5x + 2) 5. (a + 4)(3a − 2) 6. (4y + 1)2 7. 3x2; 4; 3x2; 4; 3x2 + 4 8. 5a2; 6; 5a2; 6; 5a2 + 6
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 161
9. (3x2 + 2)(7x + 4) 10. (10x2 + 3)(4x − 5)
Practice 1. x; 5 2. 5; m3 3. 5; y5; 2 4. 2y2(5 + 6y) 5. 6t(−2t4 + 1) 6. 3x2(2x2 + 5x +1) 7. 5t(−t + 8) 8. 3x and x + 8 9. (d + 2)(4d + 9) 10. (x − 5)(12 + 7x) 11. 3n2; 12; 3; 3; (n + 3)(n2 + 4) 12. (2x + 5)(x2 + 1) 13. 2y3; 6; y; 2; y; y; y; 2; (y − 2)(2y2 − 3) 14. (m − 3)(4m2 − 5)
Challenge 1. (x + 6)(x + 2) 2. (x − 6)(x − 7) 3. (2x + 7)(x + 3) 4. (5x + 2)(2x + 3) 5. (5x + 4)(x + 3) 6. (2x − 5)(2x − 3) 7. (x + 1)(x2 + x + 1) 8. (x − 2)(3x2 + 4x + 5) 9. (x4 + 3)(x2 + 5x + 7) 10. (6x8 + 7x4 + 3)(x2 + 4x + 8)
LESSON 8–3
Reading Strategies 1. +; + 2. +; − 3. +; + 4. −: − 5. +; − 6. −; − 7. +; − 8. +; − 9. +; + 10. (x − 3)(x − 7) 11. (x + 8)(x − 3) 12. (x + 6)(x + 5)
Review for Mastery 1. 16; 10 Factors Sum
and
and
and (x + 2)(x + 8)
2. 20; −9 Factors Sum
and
and
and (x − 4)(x − 5) 3. (x + 12)(x + 1) 4. (x + 10)(x + 5) 5. (x − 9)(x − 4) 6. −20; 1 Factors Sum
and
and
and (x − 4)(x + 5) 7. −4; −3 Factors Sum
and
and (x + 1)(x − 4) 8. (x − 3)(x + 6) 9. (x − 7)(x + 2) 10. (x − 5)(x + 9)
Practice 1. 3; 2 2. 4; 1 3. 5; 4 4. (x + 7)(x + 3) 5. (x + 6)(x + 5) 6. (x + 8)(x + 2) 7. 6; 2 8. 5; 3 9. 16; 1 10. (x − 9)(x − 3) 11. (x − 4)(x − 11) 12. (x − 8)(x − 5) 13. 10; 4 14. 3; 1 15. 8; 4 16. 12; 2 17. 14; 2 18. 5; 2 19. (x + 3)(x − 5) 20. (x + 2)(x − 10) 21. (x + 6)(x − 8) 22. (x + 3)(x − 4) 23. (x + 1)(x − 3) 24. (x + 1)(x − 2)
17161
108 2
8 4 4
20− 21−1−
− 12−10 2−
− 9 −5 −4
1 9 20 1−
8 2−
1 5 4 −
− 3 −4 1
0 2 2 −
1 0
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 162
25.1; 5
n n2 + 6n + 5 0 02 + 6(0) + 5 = 5 1 12 + 6(1) + 5 = 12 2 22 + 6(2) + 5 = 21 3 32 + 6(3) + 5 = 32 4 42 + 6(4) + 5 = 45
n (n + 1)(n + 5)
0 (0 + 1)(0 + 5) = 5
1 (1 + 1)(1 + 5) = 12
2 (2 + 1)(2 + 5) = 21
3 (3 + 1)(3 + 5) = 32
4 (4 + 1)(4 + 5) = 45
Challenge 1. Distributive Property
2. 1 1 1 1 1,2 3 2 3 6
1 1 1 3 1 2 3 2 52 3 2 3 3 2 2 3 6
×⎛ ⎞ ⎛ ⎞ = =⎜ ⎟ ⎜ ⎟ ×⎝ ⎠ ⎝ ⎠++ = × + × = =×
3.
2
2
1 1 1 16 2 32 3 2 3
(2 1)(3 1)6 3 2 16 5 1
x x x x
x xx x xx x
⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + = + × +⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠
= + += + + += + +
4. Yes; the given polynomial expression is written as a product of polynomials of lower degree. There is no requirement that the coefficients be integers.
5. 1 1155 3
x x⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
6. 1 1287 4
x x⎛ ⎞ ⎛ ⎞+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
7. 4 563 2
x x⎛ ⎞ ⎛ ⎞− −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
8. 5 7123 4
x x⎛ ⎞ ⎛ ⎞− +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠
LESSON 8–4
Reading Strategies
1. 1 i 3
2. 1 i 12; 2 i 6; 3 i 4
3. minus, minus 4. ( x − )( x − ) 5. Possible answer: ( x − ) ( x − ) Outer + Inner
3 −1 1 −12 3 i −12 + −1 i 1 = −37 No 3 −12 1 −1 3 i −1 + −12 i 1 = −15 No 3 −4 1 −3 3 i −3 + −4 i 1 = −13 No 3 −3 1 −4 3 i −4 + −3 i 1 = −15 No 3 −6 1 −2 3 i −2 + −6 i 1 = −12 No 3 −2 1 −6 3 i −6 + −2 i 1 = −20 Yes
6. (3x − 2)(x − 6)
Review for Mastery 1.
Factors Factors Outer + Inner 1; 5 1; 4 1; 4; 5; 1; 9 1; 5 4; 1 1; 1; 5; 4; 21 1; 5 2; 2 1; 2; 5; 2; 12
(x + 2)(5x + 2) 2. (3x + 4)(x + 1) 3. (2x − 7)(x − 3) 4. (2x + 3)(2x + 1) 5. (3x + 5)(x − 4) 6. (5x − 1)(x + 7) 7. −1(2x − 5)(x + 1)
Practice 1. 1; (x + 3)(5x + 2) 2. (2x + 3)(2x + 5) 3. (3x + 5)(x + 4) 4. (3x + 2)(2x + 5) 5. (2x + 1)(4x + 7) 6. (2x + 3)(4x + 1) 7. (4x − 1)(x − 8) 8. (3x − 7)(3x − 2) 9. (3x − 5)(2x − 5) 10. (5x − 2)(x − 4) 11. (7x − 5)(3x − 1) 12. (4x − 3)(3x − 4) 13. (5x + 9)(2x − 1) 14. (3x + 4)(x − 1) 15. (x + 2)(5x − 3) 16. (4x + 3)(x − 3) 17. (2x + 1)(2x − 7) 18. (3x + 4)(2x − 5)
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 163
19. (5x2 + 48x + 27); (5x + 3)(x + 9) 20. −1 (2x − 1)(3x − 4) 21. −1(5x + 2)(4x − 3) 22. 4x + 1
Challenge 1. 4 7 15
1 2 1 5 4 5 0
− − 2. 3 4 32 12 32 3 8 0
−−
−
3. (6x + 1) 4. (8x − 5) 5. (x − 1)(x + 3)(x + 5) 6. (x + 4)(2x − 1)(x − 5)
Section Quiz: Lessons 8-1 to 8-4
1. A 2. J 3. B 4. G 5. D 6. J 7. B 8. H 9. C 10. H 11. B 12. G 13. B 14. F 15. B 16. G
LESSON 8–5
Reading Strategies 1. D 2. P 3. D 4. N 5. P 6. N 7. (4x + 3)(4x − 3) 8. (11x − 2)2 9. (9x + 1)(9x − 1)
Review for Mastery 1. 3x; 10; 60x; 30x ≠ 2ab 2. x; 7; 14x; (x − 7)2 3. 5x; 2; 20x; (5x + 2)2
4. yes; 5x; 9; (5x + 9)(5x − 9)
5. yes; 30 ;x 7; a is not a perfect square
6. yes; 2x; 11; (2x + 11)(2x − 11) 7. (x + 10)(x − 10) 8. (x + y)(x − y) 9. (3x2 + 8)(3x2 − 8)
Practice 1. 5; x; 5 2. 1; 3x; 3x; 1; 3x; 2; 1 3. (x − 9)2 4. (6x + 2)2 5. 6 is not a perfect square
6. 12x ≠ 2(2x i 6).
7. a. x + 4 in. b. 4(x + 4) in. c. 48 in. 8. 3; 3 9. 2p; 7; 2p; 7 10. (t3 + 12)(t3 − 12) 11. (4x5 + y)(4x5 − y) 12. 20 is not a perfect square. 13. the operation between the two squares
is addition.
Challenge 1. 2 2
2 2 3
3 2 2
3 3
a ab ba b
ba ab ba ba ab
a b
+ +× −
− − −+ + +
−
2. a2; b2; a2; b2; = (a2 + b2)(a + b)(a − b) 3. 2 2
2 2 3
3 2 2
3 3
a ab ba b
ba ab ba ba ab
a b
− +× +
− ++ − +
+
4. a. If two polynomials are equal, their corresponding coefficients are equal. The coefficients of a2, ab, and b2 are equal.
b. If t = 0 and u = 0, then r and s are given by undefined expressions. Thus, there are no numbers r, s, t, and u for which a2 + b2 can be factored as (ra + sb)(ta + bu).
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 164
LESSON 8–6
Reading Strategies 1. GCF 2. difference of squares 3. Factor 4 out. 4. (2x + 3)(x − 9) 5. 3(x + 5)(x − 5) 6. 2(3x2 + 1)2
Review for Mastery 1. no GCF; x2 + 14x + 49 is a trinomial; This
is a perfect square trinomial. Method: use (a + b)2.
2. factor out the GCF: 4(x2 − 10); x2 − 10 is a binomial; This is not a difference of squares. Method: Factor out GCF.
3. factor out the GCF: 2(x2 + 4x + 3); x2 + 4x + 3 is a trinomial; This is not a perfect square trinomial; a = 1; Method: Factor out GCF. Then find factos of c that sum to b.
4. 3(x + 10)(x − 10) 5. 4(x + 1)(x − 6) 6. 2(2x − 5)2 7. −7(x + 4)(x − 1) 8. 2(2x + 3)(2x − 3) 9. 10(2x + 3)(x + 1)
Practice 1. yes 2. no; 2n2(n + 4) 3. yes 4. yes 5. no; 2(m + 9)(m + 1) 6. no; 2(p − 4)(p + 3) 7. 5(3g + 2)(3g − 2) 8. 3w(w + 5)2 9. 12y(x − 2)2 10. −3d(d − 10)(d + 10) 11. 2(a + 4)(a − 4) 12. 5(m + 3)(m − 4) 13. (c + 9)(c − 2) 14. 2(x + 3)(x + 1) 15. (f2 + 4)(f − 3) 16. −3(k − 6)(2k − 1) 17. (p4 + m2)(p2 + m)(p2 − m) 18. (a + 1)(a − 1)(2a + 7)
Challenge 1. (x + 2)(x2 − 2x + 4) 2. (4x + y)(16x2 − 4xy + y2) 3. (5x + 6)(25x2 − 30x + 36) 4. (x − 4)(x2 + 4x + 16)
5. (7x − y)(49x2 + 7xy + y2) 6. (9x − 10)(81x2 + 90x + 100) 7. 2(x − 3)(x2 + 3x + 9) 8. 8(3x + y)(9x2 − 3xy + y2) 9. (x + 2)(x2 + 2x + 4)(x3 + 4) 10. 3(x3 + 5)(x + 1)(x2 − x + 1) 11. (x + 3)(x − 3)2(x2 + 3x + 9) 12. (3x + 2)(3x − 2)(9x2 − 6x + 4)
(9x2 + 6x + 4) 13. (2x + y)(2x − y)(4x2 − 2xy + y2)
(4x2 + 2xy + y2)
Section Quiz: Lessons 8-5 to 8-6
1. A 2. J 3. B 4. F 5. A 6. H 7. C 8. G 9. C 10. F 11. C 12. F 13. C 14. H
Chapter 8 Enrichment: A Maze of Multiplication
The picture shaded is an airplane.
1. +2 3x x
2. + +2 8 7x x
3. + +2 2 15x x
4. −22 8x x
5. − 28 8x x
6. − −2 4 21x x
7. − −22 11 6x x
8. − 227 3x x
9. + −215 2 1x x
10. − + +212 18 6x x
11. −2 64x
12. − 245 20x x
13. −14 6x
14. − +2 13 30x x
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 165
15. + − 230 7 2x x
16. −29 49x
17. − 210 4x x
18. − −240 43 6x x
19. −29 4x
20. − −27 32 15x x
Chapter 8 Big Ideas
1. The GCF contains the variable raised to the lower exponent.
2. Step 1 Find the greatest common factor. Step 2 Write each term as a product
using the GCF. Step 3 Use the Distributive Property to
factor out the GCF. Step 4 Check by multiplying. 3. To factor the quadratic trinomial of the
form + +2x bx c , find two factors of c whose sum is b.
4. When c is positive, the trinomial’s factors have the same sign. The sign of b tells whether the factors are positive or negative. When b is positive, the factors are positive, and when b is negative, the factors are negative. When c is negative, the trinomial’s factors have opposite signs. The sign of b tells which factor is positive and which is negative. The factor with the greatest absolute value has he same sign as b.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 166
Chapter 9 Answers
LESSON 9–1
Reading Strategies 1. a = 3, b = −5, c = 2 2. upward 3. D: all real numbers; R: y ≥ −5 4.
x y −2 8 −1 2 0 0 1 2 2 8
5.
6. (0, 0) 7. upward; because a > 0.
Review for Mastery 1. +3; −27 +3; −9; −9; (−27); +18 5; 2; +3; 16; 7; +9; 9 − (−9) = +18 8 − 5 = +3; 43 − 16 = +27; 27 − (9) = +18 quadratic; the second difference are
constant. 2. +1; −8 +1; −4; +4 +1; +6; +10 +1; +22; +16 not quadratic the second differences are
not constant.
3. not quadratic it is linear. 4. all real numbers; y ≥ 0 5. all real numbers; y ≥ −2 6. all real numbers; y ≤ 15
Practice 1. yes; the second differences are constant. 2. yes; it can be written in the form
y = ax2 + bx + c. 3.
x y = x2 − 4 (x, y) −2 y = (−2)2 − 4 = 0 (−2, 0) −1 y = (−1)2 − 4 = −3 (−1, −3) 0 y = (0)2 − 4 = −4 (0, −4) 1 y = (1)2 − 4 = −3 (1, −3) 2 y = (2)2 − 4 = 0 (2, 0)
4. downward 5. upward 6. (−2, −6) 7. minimum: −6 8. D: all real numbers; R: y ≥ −6
Challenge 1, 2, 3 & 5
4. No, the middle terms (a, 3a, 5a) are
different.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 167
6. Yes, the second differences are 2a. 7. 2a 8. a. 2(5) = 10; b. 2(−3) = −6
LESSON 9–2
Reading Strategies 1. x = −2; (−2, 0) 2. x = −1; (−1, −1) 3. x = 3; (3, 14) 4. x = −6; (−6, 20) 5. x = 1; (1, −2) 6. x = −5; (−5, −32)
Review for Mastery 1. x = 3 2. x = 2 3. The axis of symmetry is x = −3. 4. The axis of symmetry is x = 0. 5. −10; 1; 10; 2; 5; x = 5 6. x = 1 7. x = 5; 5; 5; 0; y = 0; (5, 0) 8. (1, 8) 9. The vertex is (−3, −27)
Practice 1. −4 and 0 2. −2 3. no zeros 4. x = 0 5. x = −4 6. x = 5 7. 1; 8; −4; x = −4; (−4, −4) 8. 1; −10; 5; x = 5; (5, 15) 9. 2; −8; 2; x = 2; (2, −11)
Challenge 1. (−6, 0); (−3, −9); (0, 0) 2. 0 = 36a − 6b + c; −9 = 9a − 3b + c; 0 = c 3. 1; 6; 0 4. y = x2 + 6x
5. 2
0 9 318 ; 2 180 9 3
a b cc y xa b c
= − +⎧⎪ = = − +⎨⎪ = + +⎩
6.
2
0 25 50 4 2 ; 3 10
12.25 2.25 1.5
a b ca b c y x x
a b c
= − +⎧⎪ = + + = + −⎨⎪− = − +⎩
LESSON 9–3
Reading Strategies 1. x = −3; (−3, −4); (0, 5); Possible answers:
(−2, −3) and (−1, 0)
2. x = 0; (0, 6); (0, 6); Possible answers:
(−2, 4) and (−4, −2)
Review for Mastery 1. x = −2 2. (−2, −16) 3. (0, −12) 4. Possible answers:
(−1, −15); (1, −7)
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 168
5.
6. 3 seconds 7. 144 feet 8. 6 seconds
Practice 1. x = −1; (−1, −4); −3; Possible answers:
(1, 0) and (2, 5)
2. x = −2; (−2, 18); 10; Possible answers:
(−1, 16) and (1, 0)
3. 16 feet; 1 second; 2 seconds
Challenge 1. 21; 11; 5; 3; 5; 11; 21 3.18; 3.08; 3.02; 3.00; 3.02; 3.08; 3.18 2. (2, 3) 3. x = 2 4. 2 5. −2.27; −2.12; −2.03; −2.00; −2.03; −2.12;
−2.27 6. (−1, −2) 7. x = −1 8. −1 9. a. An equation for the axis of symmetry of
the graph of y = ax2 + bx + c is x = 2ba
− .
b. One approach is to use graphs, and a second approach is to use tables as in Exercises 1 and 5.
LESSON 9–4
Reading Strategies
1. g(x) = 12
x2;
The absolute value of 12
is less than the
absolute value of 4. 2. 4 and −4 have the same absolute value:
4. g(x) will open upward, h(x) will open downward.
3. 2; 4; 1; 3; g(x) = x2 − 7 4. c is negative, so it was translated down
from the origin.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 169
5. 12; g(x) = x2 + 5 is translated up 5 units and g(x) = x2 − 7 is translated down 7 units so they are 5 + 7, or 12, units apart.
Review for Mastery
1.
The graph of g(x) is narrower than the
graph of f(x). The vertex of g(x) is the same as the vertex of f(x).
2.
The graph of h(x) is translated up 5 units
from the graph of f(x). The vertex of h(x) is (0, 5). The vertex of f(x) is (0, 0).
3. The graph of g(x) is narrower than the graph of f(x). The graph of g(x) is opens downward and f(x) opens upward. The axis of symmetry is the same.
The vertex of f(x) is (0, 0). The vertex of g(x) is (0, −1).
4. The graph of g(x) is wider than the graph of f(x). The graph of g(x) and f(x) opens upward. The axis of symmetry is the same. The vertex of f(x) is (0, 0). The vertex of g(x) is translated 4 units up to (0, 4).
5. The graph of h(x) is narrower than the graph of f(x). The graph of h(x) opens downward and f(x) opens upward. The axis of symmetry is the same. The vertex of f(x) is (0, 0). The vertex of h(x) is translated down 1 unit to (0, −1).
Practice 1. f(x), g(x) 2. g(x), h(x), f(x) 3. same; both open upward; f(x): (0, 0); g(x)
down 3 units at (0, −3) 4. g(x) is wider; both open upward; both at
(0, 0) 5. g(x) is narrower; both open upward;
f(x): (0, 0); g(x) up 4 units at (0, 4) 6. same; f(x) opens upward, g(x) opens
downward; f(x): (0, 0); g(x) down 1 unit at (0, −1)
7. a. 400; 256 b. The graph of h2 is a vertical translation
of the graph of h1: 144 units down. c. ball dropped from 400 ft: 5 s; ball
dropped from 256 ft: 4 s
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 170
Challenge 1. a. h;
b. 0, 3, 6, −4; c. vertex moves horizontally d. The parameter h determines a horizontal translation.
2. a. k; b. 0, 2, 5, −3; c. vertex moves vertically; d. The parameter k determines a vertical translation.
3. a. a;
b. 1, 2, −1, 15
;
c. width and direction; d. The parameter a makes the parabola narrower or wider and makes the parabola open up or down.
4. The coordinates of the vertex are (h, k). 5. (−18, 24)
Section Quiz: Lessons 9-1 to 9-4
1. B 2. H 3. D 4. G 5. A 6. G 7. A 8. J 9. B 10. F
LESSON 9–5
Reading Strategies 1. quadratic function 2. y = 5x2 + 7x; f(x) = 5x2 + 7x 3. Because the graph may look like it
intersects a point, but actually only come close to it.
4. −2 and 2
Review for Mastery 1. 0, −3; x = 0
3(0)2 + 9(0); 0 3(0) + 0; 0 0; 0 x = −3 3(−3)2 + 9(−3); 0 3(9) + (−27); 0 27 + −27; 0 0; 0
2. 2; x = 2 (2)2 – 4(2) + 4; 0 4 − 8 + 4; 0 0; 0
3. 0, 3; x = 0 −2(0)2 + 6(0); 0 −2(0) + 0; 0 0; 0 x = 3 −2(3)2 + 6(3); 0 −2(9) + 18; 0 −18 + 18; 0 0; 0
4. about 3.5 seconds 5. about 24.5 seconds 6. about 5.5 seconds
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 171
Practice 1.
x = 2 2.
x = −3 or x = 2 3.
x = −1 or x = 1
4.
no real solution 5. 2 seconds
Challenge 1. 36 ft; Solve 0 = −0.08x2 + 2.88x by
graphing on a graphing calculator. The zeros are 0 and 36.
2. 6 ft; Solve 13.5 = −0.08x2 + 2.88x by graphing y = −0.08x2 + 2.88x − 13.5. The zeros are approximately 6 and 30.
3. yes; no; Two lanes require 24 ft, which is exactly the width that is taller than 13.5 ft. Three lanes require 36 ft; the cars in the outside lanes would hit the tunnel walls.
4. 4 ft
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 172
5. Yes; Evaluate y = −0.08x2 + 2.88x when x = 4 to find y = 10.24. Because an “average” pedestrian is shorter than 10 ft, the walkway will be tall enough.
6. accept
LESSON 9–6
Reading Strategies 1. yes; because 0 × 0 = 0 2. Because you need two or more quantities
being multiplied 3. −2, because −2 + 2 = 0. 4. If abc = 0, then a = 0, or b = 0, or c = 0. 5. x = −5, x = 2 6. x = −4, x = 3 7. x = −2, x = 5
Review for Mastery 1. x − 6; x − 3; 6; 3 2. x + 8; x − 5; −8; 5 3. 3x; x − 7; 0; 7
4. 2x − 3; x + 9; 32
; −9 5. 15
; −2
6. −4; 2 7. −4; 3 8. −5 9. −8, 1 10. −7, 7 11. 0; −6.25 12. 5; −2 13. −3; −7 14. −2; 2 15. −1; 3
Practice 1. a = 0; b = 0 2. 7; −2 3. 5; 1 4. −2; −6
5. 43
; 3 6. x − 5; x − 5; 5
7. x + 1; x + 1; −1 8. 9; 3; x − 9; x + 3; 9; −3 9. x + 3; x + 5; x + 3; x + 5; −3; −5 10. 5; 1 11. 6; −2
12. − 12
; 3 13. 43
; − 12
14. a. −16t2 + 1600 = 0; b. 10 seconds
Challenge
1. −3, −5, 5 2. 52
, 32
, − 32
3. −3, 3, 7 4. 0, −4, 6
5. 0, − 12
, − 53
6. 2, −2, − 112
LESSON 9–7
Reading Strategies 1. x = ±5 2.
3. zero a. The parabola does not cross the
x-axis. b. There is no real number whose square
is negative.
Review for Mastery 1. ±9 2. ±3 3. no solution 4. ±12 5. ±7 6. ±11 7. ±4 8. ±5 9. ±6 10. ±3 11. ±10 12. ±8 13. ±7.07 14. ±3.46 15. ±5.48 16. 6.5 cm 17. 14.6 in.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 173
Practice
1. ;a − a 2. 2; 2; −2 3. 169; 13; 13; −13
4. 900; 30 5. no real solutions
6. 144; 12 7. 100; 100; 10
8. 25;36
56
9. 125; 25; 25; 5
10. ±6 11. ±2
12. ± 15
13. 0
14. no real solutions 15. ±7 16. ±2.83 17. ±5.48 18. ±2.24 19. a. x2 = 225
b. 15 in. by 15 in
Challenge 1. 3 and 4 2. a. 12.25
b. 3.5 > 12 c. 3.5 d. ≈3.428 e. ≈3.464
3. up to 3 places, depending on how the numbers were rounded
4. ≈3.6 5. ≈4.5 6. No; the square root will be an irrational
number which cannot be written as a terminating decimal.
LESSON 9–8
Reading Strategies 1. x = −10 or x = 6 2. x = −9 or x = 19 3. x = −3 or x = −13
4. x = −2 12
or x = −3 12
Review for Mastery 1. 49; 49; x − 7 2. 100; 100; x + 10 3. 9; 9; x + 3 4. x2 + 18x + 81;
(x + 9)2 5. x2 − 16x + 64; (x − 8)2
6. x2 + 5x + 25 ;4
25
2x⎛ ⎞+⎜ ⎟
⎝ ⎠
7. 7; −1 8. −2, −6 9. 9; −7 10. 4; −8
Practice 1. 9 2. 36 3. 16 4. 9; −2, −4 5. 9; 1, 5 6. 1; −4, 6 7. 25; −2, −8 8. x2 − 4x = 5; 4; 5, −1 9. x2 − 4x − 12 = 0; 4; −2, 6 10. a. width = w; length = w + 6
b. w2 + 6w = 91 c. 9 d. w = 7, l = 13
11. a. w2 + 4w = 45 b. w = 5, l = 9
Challenge
1. y = 25
2x⎛ ⎞+⎜ ⎟
⎝ ⎠− 1 2. y =
2112
x⎛ ⎞−⎜ ⎟⎝ ⎠
+ 6
−25
4 −
2114
3. y = 2
2bx⎛ ⎞+⎜ ⎟
⎝ ⎠+ c 4.
2,
2 4b bc
⎛ ⎞− −⎜ ⎟⎝ ⎠
−2
4b
5. a > 0 6. If the parabola opens upward and the
y-coordinate of the vertex is greater than 0, then the parabola is completely above the x-axis. If it is less than 0, then the parabola crosses the x-axis at two points. If it is equal to 0, then the parabola touches the x-axis at one point.
7. a. 0 8. a. (−1, −9) b. 1 b. 2 c. 2
9. a. (4.5, −2.25) b. 2
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 174
LESSON 9–9
Reading Strategies 1. 0 2. 2; −5; 3 3. 2 4. x = 2 or x = 5
5. no real solution 6. x = −2 or x = 53
7. x = 8
Review for Mastery 1. 1; 2; −35; 2; 2; 1; −35; 1; −7, 5
2. 3; 7; 2; 7; 7; 3; 2; 3; − 13
, −2
3. 1; 1; −20; 1; 1; 1; −20; 1; −5; 4
4. 2; −9; −5; −9; −9; 2; −5; 2; − 12
; 5
5. 4; 20; 25; 20; 4; 25; 1 solution 6. 15; 8; −1; 8; 15; −1; 2 solutions 7. 2 solutions 8. no real solutions
Practice 1. 1; 6; 5; 6; 6; 1; 5; 1; −1, −5 2. 1; −9; 20; −9; −9; 1; 20; 1; 5, 4
3. 2; 9; 4; − 12
, −4 4. 1; −3; −18; 6, −3
5. 3; 1; 5; −11; no real solutions 6. 10; 1; 25; 0; 1 7. 64; 2 8. 8, −8 9. −6
10. 4, −8 11. 12
, −5
Challenge 1. a = 2; b = −5; c = −8 2. a = 7; b = 0; c = −56 3. a = 12; b = −16; c = 0 4. a = 0; there is no x2 − term 5. 2 6. 3 7. 1 8. answers may vary; answers may vary 9. answers may vary; answers may vary 10. answers may vary; answers may vary
11. x = 0 12. x = ± ca
13. x = ba
− or x = 0
14. x = 2 4
2b b ac
a− ± −
15. the quadratic formula
Section Quiz: Lessons 9-5 to 9-9
1. B 2. F 3. C 4. G 5. D 6. F 7. A 8. H 9. A 10. H 11. B 12. F
Chapter 9 Enrichment: Factoring Fun
Answer: 2748
1. ( )( )− −2 4x x
2. ( )( )− +2 3x x
3. ( )( )− −1 4x x
4. ( )( )+ −2 1x x
5. ( )( )+ −2 4x x
6. ( )( )− −3 4x x
7. ( )( )+ −3 4x x
8. ( )( )+ −1 4x x
9. ( )( )− −2 1x x
10. ( )( )− −4 4x x
11. ( )( )− +5 3x x
12. ( )( )+ +2 3x x
13. ( )( )− −2 3x x
14. ( )( )+ −5 4x x
15. ( )( )− +2 1x x
16. ( )( )+ −5 1x x
17. ( )( )− +2 4x x
18. ( )( )− +2 5x x
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 175
Chapter 9 Big Ideas
1. If a function has one zero, use the x-coordinate of the vertex to find the axis of symmetry. If a function has two zeros, use the average of the two zeros to find the axis of symmetry.
2. To find the x-coordinate of the vertex, find the axis of symmetry by using zeros or the formula. To find the corresponding y-coordinate, substitute the x-coordinate of the vertex into the function. Then write the vertex as an ordered pair.
3. First, factor. Then, set each factor equal to zero. Finally, solve each resulting equation.
4. Write the equation in the form
= +2x bx c . Find ⎛ ⎞⎜ ⎟⎝ ⎠
2
2b
. Complete the
square by adding ⎛ ⎞⎜ ⎟⎝ ⎠
2
2b
to both sides of
the equation. Factor the perfect square trinomial. Take the square root of both sides. Write two equations, using both the positive and negative square root and solve each equation.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 176
Chapter 10 Answers
LESSON 10–1
Reading Strategies
1. FL, HI, IN, PA, TX 2. Fall
3. 24; 6.25% 4. double line graph
5. circle graph 6. bar graph
7. line graph
Review for Mastery 1. line 2. circle 3. bar 4.
A circle graph is appropriate because it
shows categories as part of a whole. 5. carpool 6. 4 7. 22 8. savings 9. entertainment and clothing 10. $100
Practice 1. Whale Shark 2. 32 years 3. Hound Shark 4. March 5. 4 6. $0.50
7.
It compares each type of pizza to the total
pizzas sold that weekend.
Challenge 1. about 20% 2. about 75% 3. The area of the section for female
survivors is about the same as the area for the section of male survivors.
4. about 1260; The area of the section for males who died is about 10 times the size of the area of the section for females who died.
5-7.
LESSON 10–2
Reading Strategies
1. 3;
11, 11, 15, 16, 20, 28, 28, 48, 49, 49, 55,
56, 62
2. Group 2; 4
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 177
3. 3, 4, 5, 6, 7, 8, 9,10; 4
4. Weights
13 14 15
8 2 0
9 3 0
9 4 1
4 5 6 6 8 9
Review for Mastery 1. 2.
Car Gas Mileage mi/gal Frequency 15-19 5 20-24 8 25-29 12 30-34 7 35-39 2 40-44 1
Practice 1. Rushing Yards
2. a. 2 b. 5 c. 53 d. Period 5; 1 3.
Frequency6 4 4 3 4
4.
5. 4–7, 8–11, 16–19
Challenge 1. 55% 2. 154 3. 18% 4. 980 5. 1075
Low Temp High Temp 9 6
9 9 8 4 2 0 0 9 7 6 3 2 0
3 2 0
3 4 5 6 7 8
8 5 5 6 7 9 0 0 0 0 1 1 3 3 6 6 8 4
Key: 14|2 means 142 lb
stem leaves
3 4 5 6 7 8 9
2 3 8 2 4 4 7 5 8 2 3 4 8 3 6 8 Key: 4|3 means 43
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 178
6. Cumulative Frequency
Cumulative Percentage
22 4% 62 11%
140 26% 225 41% 315 58% 398 73% 486 89% 538 99% 546 100%
7.
8. 40–44 9. 10–34
LESSON 10–3
Reading Strategies 1. the mean 2. because 2 occurs more often 3. when there is an even number of data
values 4. The mode occurs most often; an outlier is
different from the other numbers, so an outlier cannot be a mode.
5. mean: 8, median: 8.5, mode: 3 and 10
Review for Mastery 1. 8, 2, 3, 4, 3, 20, 4; median: 3, mode: 3,
range: 6 2. mean: 5.5, median: 5, mode: 4 and 5,
range: 4 3. mean: 10, median: 10, mode: none,
range: 12 4. 5, 9, 11, 14, 18, 18, 21
5. 5, 9, 14, 18, 21 6. 7. 2, 5, 7, 9, 14, 15 8. 2, 5, 8, 14, 15 9.
Practice 1. 7, 9, 10, 19, 25
mean: 7 9 10 19 25
145
+ + + +=
median: 10 mode: none range: 25 – 7 = 18
2. 2, 3, 3, 5, 5, 5, 5 mean: 4 median: 5 mode: 5 range: 3
3. mean: 11 median: 10.5 mode: 8 and 12 range: 9
4. outlier: 29, increases mean by 4.25, median by 1.5, and range by 18, no effect on mode
5. outlier: 11, decreases mean by 8.5, median by 2.5, no effect on mode, increases range by 28
6a. mean, 44 6b. median, 50, because it is higher than the
mean. 7a. 8, 10, 14, 15, 18, 22, 22, 30, 33 7b. 8, 12, 18, 26, 33 8. Liam 9. Vicki 10. Liam
50 10 15 20 25
100 20 30 40
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 179
Challenge 1. 18 2. 36.5 3. 18.5 4. 27.75 5. –9.75 6. 64.25 7. 73 8. mild 9. 10.
11. 12.
LESSON 10–4
Reading Strategies 1. when there is an outlier in the data set 2. check if the percents add to 100% 3. a survey of 100 people 4. The vertical axis starts and ends just
beyond the data values. Someone might believe there’s been a big increase in money given to schools.
5. The horizontal axis is not equally spaced. Someone might believe a lot of miles were put on very quickly when the car was new.
Review for Mastery 1. a. The vertical scale does not start at
zero. b. Someone may believe there are large
differences between the number of men and women who have enrolled at the school.
c. Possible answer: Someone who may want to show that inequalities exist in admissions.
2. Most of the houses are selling for under $231,000. The mean is not a good descriptor.
3. The sample is biased. People who don’t have e-mail cannot be surveyed.
4. The sample size is too small.
Practice 1. The vertical axis does not start at 0. This
exaggerates the differences in the scores. 2. Lazaro scored more than twice as much
as the other players. 3. Lazaro 4. The vertical axis does not start at 0 and
the categories on the horizontal axis are not at equal time intervals.
5. that the cost of Product X decreased very rapidly
6. sellers of Product X 7. The sectors do not add to 100%. 8. his parents; to make them think he
doesn’t waste his money and that most goes towards school.
9. People eating at an Italian restaurant are more likely to favor Italian food.
Challenge 1. The pictures are different sizes, so the
numbers of some pets are exaggerated. 2. There is the same number of cats as
horses; and there are more dogs and fewer fish than any other animal.
3. A breeder of dogs; to suggest that dogs are the most popular pet.
4. Make the individual pictures the same sizes, or use a standard bar graph.
5. Because it is 3-D, the relative numbers of computers are exaggerated by the volumes.
6. The number of Brand A models is about 4 times greater than the number of Brand C models.
7. The manufacturer of Brand A; to suggest that their models offer significantly more variety.
0 2010 30 40 50 60 70 80
80 120100 140 180160 200 220
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 180
8. Use 2-dimensional bars that only vary by height, not by width and/or depth.
Section Quiz: Lessons 10-1 to 10-4
1. B 2. J 3. B 4. H 5. C 6. J 7. A 8. H 9. D 10. H
LESSON 10–5
Reading Strategies 1. {B, W}
2. a. 14
b. 512
c. 512
d. 34
e. 712
Review for Mastery 1. {1, 2, 3, 4, 5, 6} 2. (A, B, C, D, E)
3. a. 845
b. 79
c. 215
d. 2945
4. a. 99% b. likely c. 792 5. a. 4% b. unlikely c. 1
Practice 1. sample space: {1, 2, 3, 4, 5, 6};
outcome {5} 2. sample space: {A, B, C, D};
outcome {A} 3. impossible 4. as likely as not
5. likely 6. 310
7. 15
8. 2740
9. 3340
10. a. 15% b. 45 11. a. 14% b. 70
Challenge 1. Neither the number of favorable outcomes
nor the number of total outcomes can be negative.
2. because the numerator is always less than or equal to the denominator
3. Possible answer: A probability between 0.0 and 0.25 is a very low probability. A probability between 0.25 and 0.5 is a low probability. A probability between 0.5 and 0.75 is a probability that is better than 50-50 but not close to certainty. A probability between 0.75 and 1.0 is a probability that is considerably better than 50-50.
4. 2 ;13
very poor
5. 4 ,13
about 0.31; not very good
6. 7 ,13
about 0.54; better than 50-50,
but still not very good
7. 11,13
about 0.85, very high
8. 0; as low as it can get 9. 1; as high as it can get. 10. Possible answer: 2n − 6 ≥ −5
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 181
LESSON 10–6
Reading Strategies 1. a. they are the same b. The sum of the numerators equals the
denominator: 2 + 3 = 5. 2. The numerator in the probability is the
same as the first number in the odds ratio. The numbers in the odds ratio sum to the denominator in the probability.
3. 15
4. 45
5. 1:4 6. 4:1
Review for Mastery
1. 12
2. 310
3. 13
4. 56
5. 88% 6. 1:4
7. 49
8. 9899
9. 3:2 10. 121
Practice 1. 50% 2. 30%
3. 16 2 %3
4. 25%
5. 90% 6. 14
7. 0.5 8. 95% 9. a. 5 b. 1
c. 15
10. a. 9 b. 7
c. 79
11. a. 3 b. 2
c. 3 ;2
3; 2
12. a. 25
b. 1115
c. 1:2
Challenge 1. {100, 250, 500, 1000, 5000}
2. 14
3. 316
4. 18
5. 116
6. 1 1 3 1 3, , , ,16 8 16 4 8⎧ ⎫⎨ ⎬⎩ ⎭
7. 1 8. The total of the probabilities for all
possible events must be 1. 9. No; the range of functions can vary widely.
A sum of 1 is only required for a probability function such as the one used here.
10. 916
11. 516
12. 316
13. 38
14. 58
15. 0
LESSON 10–7
Reading Strategies 1. independent 2. dependent
3. independent 4. 14
5. 29
6. 3100
Review for Mastery
1. 1 ;6
1 ;6
136
2. 18
3. 425
4. 320
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 182
5. 1 ;5
1 ;3
115
6. 265
7. 1495
8. 518
Practice 1. independent; the outcome of one roll
does not affect the other rolls. 2. dependent; a marble selected and not
replaced reduces the number to choose from.
3. a. independent
b. 136
4. a. dependent
b. 119
5. a. independent; 7 7%100
=
b. dependent; 1 2.2%45
≈
c. dependent; 7 15.6%45
≈
6. 28 62.2%45
≈ 7. 1 2.2%45
≈
8. 8 17.8%45
≈
Challenge
1. 13
2. 723
3. 769
4. 1146
5. 6n
6. 51n −
7. 30( 1)n n −
8. 1n
9. 11n −
10. 12n −
11. 1( 1)( 2)n n n− −
12. Your chances of drawing a 1 get better until the 1 is drawn. This happens because the number in the denominator
of the probability fraction keeps getting lesser, and the value of the fraction gets closer to 1.
LESSON 10–8
Reading Strategies 1. combinations; 10 2. permutations; 6720 3. permutations; 120 4. combinations; 35
Review for Mastery 1. permutations 2. combinations 3. 30,240 4. 1680 5. 650 6. 720 7. 3060 8. 35 9. 6435 10. 15 11. 10 12. 11,880 13. 462
Practice 1. 3; 3; 9 2. 10; 10; 10; 10; 10,000 3. 4; 2; 2; 16 4. P; 360 5. C; 4 6. C; 15,504 7. P; 40,320 8. 35 9. 24 10. 15,600 11. 210
Challenge 1. 3P3 = 3! = 6
2. 6 23
=
3. n rPr
4. 6 12 3 =i
5. 2n rP
ri
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 183
Section Quiz: Lessons 10-5 to 10-8
1. C 2. J 3. C 4. G 5. A 6. H 7. B 8. H 9. C 10. J 11. C
Chapter 10 Enrichment: Fraction, Decimal, and Percent Mania
Boy do I have problems!
Chapter 10 Big Ideas
1. Find the mean of the middle two data values.
2. The scale of a graph might not start at zero. The sectors of a circle graph might not add to 100%. A sample might not be large enough to represent the entire population. A sample might not be random.
3. Experimental probability is found by conducting an experiment to find probabilities or an event. Theoretical probability is the probability that should occur when an experiment is conducted.
4. If the occurrence of one event does not affect the probability of the other then the events are independent. If the occurrence of one event does affect the probability of the other then the events are dependent. The result of tossing a dime and a nickel is an independent event. A dependent event is choosing a game piece, and then your sister picks another game piece.
5. Permutation; The order of players chosen as starters does matter. When one player is chosen as a starter he/she cannot be chosen for another starting position.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 184
Chapter 11 Answers
LESSON 11–1
Reading Strategies 1. 2 2. 3 3. Possible answer: I divided each term by
the previous term. 4. 162 5. 354,294
6. no; the ratios between terms (1.5, 1.3, and 1.25) are different.
7. −640 8. 18,144 9. 1
Review for Mastery 1. no; there is no common ratio. 2. yes; the common ratio is −2.
3. yes; the common ratio is 1 .2
4. 5; 625, 3125, 15,625 5. −2; −96, 192, −384 6. 1.5; 20.25, 30.375, 45.5625 7. 2; 4, 8, 16 8. 7; −2; 10; −3584 9. −4; 3; 8; −8748 10. 1458 11. −768 12. −885,735
13. 19
Practice 1. 4; 256, 1024, 4096 2. 10; 100,000, 1,000,000, 10,000,000
3. 1 ;2
8, 4, 2
4. −5; 2500, −12,500, 62,500 5. 2048 6. −384 7. 1792 8. 3; 9; 729 9. −2; −2; −8192 10. 1.2; 8; $19.91
Challenge 1. 50 2. −10; −8; −6; −4;
−2; 0; 2; 4; −24 3. 23 + 18 + 13 + 8 + 3 + (−2); 63
4. 12
+ 58
+ 34
+ 78
; 114
or 2 3
4
5. −11.8 + (−13) + (−14.2) + (−15.4) + (−16.6) + (−17.8); −88.8
6. 484 7. −10; −20; −40; −80; −160; −320; −640;
−1280; −2550 8. 3 + (−6) + 12 + (−24) + 48 + (−96); −63
9. 16 + 4 + 1 + 14
; 854
or 211
4
10.
− 3227
+ 6481
+ − 128243
⎛⎝⎜
⎞⎠⎟
+ 256729
+ − 5122187
⎛⎝⎜
⎞⎠⎟
+ 10246561
;
−42566561
≈ −0.649
LESSON 11–2
Reading Strategies 1. always 2. sometimes 3. always 4. never 5.
x y −1 3
2−
0 −3 1 −6 2 −12
Review for Mastery 1. yes 2. no
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 185
3. yes 4. 1.6 feet 5. ≈1601 6.
x y = −4(0.5)x y −2 y = −4(0.5)−2 −16 −1 y = −4(0.5)−1 −8 0 y = −4(0.5)0 −4 1 y = −4(0.5)1 −2
7.
x y = 2(5)x y −1 y = 2(5)−1 0.4 0 y = 2(5)0 2 1 y = 2(5)1 10 2 y = 2(5)2 50
8.
x y = −1(2)x y −1 y = −1(2)−1 −0.5 0 y = −1(2)0 −1 1 y = −1(2)1 −2 2 y = −1(2)2 −4
Practice 1. 10.6 feet 2. yes, as the x-values increase by 1,
the y-values are multiplied by 2. 3. no, as the x-values increase by 1,
the y-values are not multiplied by a constant amount
4. y (x, y)
0.2 (−2, 0.2 )
0.6 (−1, 0.6 )
2 (0, 2) 6 (1, 6)
18 (2, 18)
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 186
5. y = −2 (4)x y (x, y) y = −2(4)−2 −0.125 (−2, −0.125) y = −2(4)−1 −0.5 (−1, −0.5) y = −2(4)0 −2 (0, −2) y = −2(4)1 −8 (1, −8) y = −2(4)2 −32 (2, −32)
6. 18 years 7. 59 years
Challenge 1.
The bases are reciprocals.
The graphs are reflections across the y-axis.
2.
The bases are reciprocals.
The graphs are reflections across the y-axis.
3. The bases are reciprocals of one another and the graphs are reflections of one another across the y-axis.
4. Because 10 and 110
are reciprocals
5. Because 10.25
⎛ ⎞= ⎜ ⎟⎝ ⎠
and 5 and 15
are reciprocals
6. Because 144
xx− ⎛ ⎞= ⎜ ⎟
⎝ ⎠and 4 and 1
4 are
reciprocals
7. Because 5 41.254 5
x xx
−− ⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟
⎝ ⎠ ⎝ ⎠and 5
4
and 45
are reciprocals
LESSON 11–3
Reading Strategies 1a. exponential decay 1b. 1041 birds 2a. exponential growth 2b. 5647 people 3a. exponential growth 3b. $1192.67
Review for Mastery 1. y = 372,000(1 + 0.05)t; ≈$549,613 2. y = 4200(1.03)t; ≈5165 3. y = 350,000(1 − 0.03)t; ≈$291,540 4. y = 1200(0.98)t; ≈1085
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 187
5. A = 17,000(1.03)t; $20,298.89
6. A = 23,00040.021
4
t⎛ ⎞+⎜ ⎟⎝ ⎠
; $26,979.99
7. A = 30(0.5)t; 3.75 g 8. A = 40(0.5)t;2.5 g
Practice 1. y = 270,000(1 + 0.07)t; y = $330,761.61 2. y = 2200(1 + 0.02)t; y ≈ 2478 3. y = 200(1 + 0.08)t; y ≈ $503.63
4. A = 20,000
1+ 0.031
⎛⎝⎜
⎞⎠⎟
(1)(t )
;
A ≈ $25,335.40
5. A = 35,000
1+ 0.0612
⎛⎝⎜
⎞⎠⎟
(12)(t )
;
A ≈ $63,678.89
6. A = 35,000
1+ 0.084
⎛⎝⎜
⎞⎠⎟
4t
;
A ≈ $52,008.16 7. y = 800(1 − 0.02)t; y ≈ 738 8. y = 2300(1 − 0.04)t; y ≈ 1529 9. A = 10(0.5)t; A = 2.5 grams
Challenge 1. A = 10,000(1.0565)t 2.
t 0 1 2 3 A 10,000 10,565 11,161.92 11,792.57
4 5 6 7 12,458.85 13,162.78 13,906.47 14,692.19
3. a. sometime after the end of the third year but before the end of the fourth year
b. sometime after the end of the third year but before the end of the seventh year
4. 7.4 years 5. No; after 10 years, they will have
$17,325.87, which is less than $18,000. 6. 8.3 years 7. 4.1 years 8. 153 years
LESSON 11–4
Reading Strategies 1. exponential 2. linear 3. quadratic 4. quadratic 5. exponential 6. linear 7. linear 8. quadratic 9. exponential
Review for Mastery
1.
linear
2.
exponential 3. exponential 4. quadratic 5. linear 6. exponential; y = 1(4)x 7. linear; y = 3x + 7
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 188
Practice 1.
linear 2.
quadratic 3. exponential 4. linear 5. quadratic 6. a. linear b. y = 0.45x c. $3.60 7. a. exponential b. y = 5(1.2)x c. $17.92
Challenge
2. exponential or quadratic 3. y = −0.251x + 39.469 5. y = 0.002x2 − 0.547x + 46.625 7. y = 46.700(0.987)x
9. Exponential or quadratic; either model is a good fit for the curvature of the scatter plot.
10. −5.711 mi/h; 12.965 mi/h; 4.430 mi/h 11. The exponential model fits best because
it gives decreasing speeds as traffic density increases; the linear model gives negative speeds and the quadratic model gives increasing speeds.
Section Quiz: Lessons 11-1 to 11-4
1. B 2. G 3. C 4. G 5. B 6. J 7. C 8. J 9. D 10. H 11. A
LESSON 11–5
Reading Strategies 1. a. yes b. no c. no d. yes 2. The graph is shifted up 2 units. 3. all real numbers greater than or equal to −5 4. 80 ft/s 5.
Review for Mastery 1. no, x is not under the square-root sign. 2. yes, x is under the square-root sign.
3. x ≥ 85
4. x ≥ 0
5. x ≥ 0 6. x ≥ 5 © Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 189
7. x ≥ 0 8. x ≥ −6 9. x ≥ 0
x ( ) 2f x x=
0 0
2 2
8 4
18 6
32 8
10. x ≥ −3
x ( ) 3f x x= +
−3 0
1 2
6 3
22 5
33 6
11. x ≥ −3
x ( ) 2 6f x x= +
−3 0
−1 2
5 4
15 6
29 8
Practice 1. 36 feet 2. ≈26.8 feet
3. 7 4. 12
5. x + 4; −4 6. x ≥ 0 7. x ≤ 5 8. x ≥ 2 9.
x f(x) −3 3 3 0 0− + = =
−2 −2 + 3 = 1 = 1 1 1 3 4 2+ = =
6 6 3 9 3+ = =
13 13 3 16 4+ = =
10.
x f(x)
0 2(0) 0=
2 2(2) 2=
8 2(8) 4=
18 2(18) 6=
32 2(32) 8=
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 190
11.
x f(x) 1 2 1 1 0− − = 5 2 5 1 4− − = −
10 2 10 1 6− − = −
17 2 17 1 8− − = − 26 2 26 1 10− − = −
Challenge 1. y = x − 10 2. y = 5x
3. y = 13
x− + 2 4. y = x±
5. y = x± − 2 6. y = 4 8x± +
7. The line or curve of each original relation is reflected across the line y = x to create the line or curve of the inverse relation.
8. 1, 2, and 3 9. The graph of y = x2 is a parabola, and the
inverse relation y = x± is the same parabola reflected across the line y = x. The graph of ( )f x x= is the top (positive) half of y = x± .
LESSON 11–6
Reading Strategies 1. Quotient Property of Square Roots 2. Product Property of Square Roots 3. When multiplying powers with the same
base, add the exponents. So x3 i x3 = x6.
4. 26xy y 5. 3 23
x
6. 3 2x7x4 7. 2 22 11x y
8. 2
44 2
3x x
y 9. 52x
Review for Mastery
1. 2 5 2. 10 3
3. 23 6x 4. 79
5. 103
6. 4
335
xy
7. 754
; 25 34
; 5 32
8. 28825
; 144 225
; 12 25
9. 2 29
10. 3 27
11. 5 53
12. 11 23x
13. 3
4m m
n 14. 10 2
7xy
Practice
1. 4 ; 2 2. 64 ; 8 3. 25 144+ ; 169 ; 13 4. 4b − 5. 9 6. 3x +
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 191
7. 6 8. 3; 3 ; 10 3
9. 2 2b ci ; bc 10. 10 5
11. 9 • 10 ; 3 10 12. 7 2y x
13. 110
14. 234
yx
15. 2
4
2125
b
c; 2
215
bc
16. x xy
17. 5x 18. 10
4b
19. 4 23
20. 2113
x x
21. 412 25
x x 22. 2 85 ft; 18.44 ft
Challenge 1−5
1. a. 2 in.
b. 3 in. c. 4 2= in. d. 5 in. 2. The length of each hypotenuse is a
square root expression in which the radicand increases by 1 from the previous hypotenuse.
4. Continue the spiral until you have 12 triangles; the length of the hypotenuse of the 12th triangle will be 13 in.
5. 1n + in.
LESSON 11–7
Reading Strategies 1. They have the same radicand, 6.
2. Keep 6 and subract 5 − 3.
3. Write 2 54 as 6 6.
4. 8 6 5. 3 5−
6. 6 3 2 10x x+ 7. 20 2
8. 21 7y
Review for Mastery
1. yes; 11 2y 2. no
3. no 4. yes; 11− 5. 6 13 6. 5 2
7. 11 5 8. 10 3a
9. 8 x 10. 7 6
11. 16 2i ; 16 2 ; 4 2 ; 5 2
12. 9 3i ; 9 3 ; 3 3 ; 4 3
13. 25 5i ; 25 5 ; 5 5 ; 6 5
14. 12 3 15. 3
16. 4 7 14+ 17. 4 3
18. 4 7x 19. 10y−
Practice 1. 8 2. −2 3. 13; m 4. 7 6 13+ 5. 4 10 6. 6 2 4b b−
7. 5; 4; 9 8. 9; 64; 9; 64; 3; 8; 11 9. 4; 16; 5; 4; 16; 5; 2; 4; 5; 6; 5
10. 8 7 11. 7 3
12. 5 11 13. 9 2 4 3+
14. 4 5 10 2+ 15. 10 3
16. 10 2 5 3+ 17. 3 5 9 2− +
18. 8 6 19. 9 x
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 192
20. 12x 3 21. 16 3t
22. 21 in.m 23. 14 2 m
LESSON 11–8
Reading Strategies 1. FOIL 2. Rewrite the expression so there is no
square root in the denominator.
3. 18 2 4. 5 3 2 6+
5. 58 2 5+ 6. 5 2
7. 4 8. 2x
Review for Mastery
1. 6 2. 5 2
3. 2 22 4. 142
5. 2 63
6. 2 155
7. 4; 8 ; 4 5 2 10+ 8. 2 2 7+
9. 5; 3 ; 5; 3 ; 27 3−
10. 50 13 10+
Practice
1. 9; 3 5 2. 2; 2; 4; 4; 28 3. 100; t2 ; 30 2t 4. 5 2 5. 90 6. 12 14x
7. 3 2 2 3− 8. 2 3 3 2t−
9. 7 5; 5 ; 9 5 5−
10. 5 8 5− 11. 7 + 35
12. 2 10− − 13. 155
14. 33
; 333
15. 4; 22bb
; 10b8b
16. 306
17. 5
18. 17 3+
Challenge
1. Yes; 2( 5 3) 10( 5 3) 22
(28 10 3) ( 50 10 3) 22 0− + + − + + =
− + − + + =
2. 5 3x = − ± ; yes, 5 3x = − + is one of the two roots.
3. No; 24(1 3 2) 4(1 3 2) 17(76 24 2) (4 12 2) 1755 12 2 0
− − − − =− − − − =
− ≠
4. 1 3 2 1 3 22 2 2
x ±= = ± ; yes, 1 3 2x = −
is not one of the two roots. 5. a. conjugates
b. m n p−
6. The quadratic formula includes ± square root; irrational roots will always be conjugates.
7. b2 − 4ac would have to be greater than zero and not a perfect square.
LESSON 11–9
Reading Strategies 1. to clear the radicals 2. You are also checking for extraneous
solutions. 3. x = 64 4. x = −12 5. x = 16 6. x = 9 7. x = 3 8. no solution
Review for Mastery 1. 144 2. 25 3. no solution 4. 100 5. 64 6. 22
7. 8x + ; 6 ; −2 8. 3x + ; 10 ; 7
9. 3 4x − ; x ; 2 10. 3
11. 4 12. 2
Practice 1. 81 2. 36; 18 3. 8; 8; 64; 32 4. 25; 625 5. 8; 66 6. 15; 12
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 193
7. 100 8. 97 9. 40 10. 3 11. 6 12. 4 13. x = 6; x = 3
4 2x x− = − 4 2x x− = − 6 − 4 6 2− 3 − 4 3 2−
2 4 −1 1 2 2 −1 1
The only solution is 6. 14. 36 15. no solution 16. 10
Challenge 1. Their graphs do not intersect. 2. Their graphs do not intersect. 3. Their graphs do not intersect. 4. x − 2 ≠ x − 4 for all values of x 5. x − 2 ≠ x − 6 for all values of x 6. x − 4 ≠ x − 6 for all values of x 7.
10 20 50 100 3.162 4.472 7.071 10 2.828 4.243 6.928 9.899 0.334 0.229 0.143 0.101
200 500 1000 5000 14.142 22.361 31.623 70.711 14.071 22.316 31.591 70.697 0.071 0.045 0.032 0.014
8. In the expression ,x x n− − as n remains constant and x becomes larger, the expression gets closer to 0 in value.
Section Quiz: Lessons 11-5 to 11-9
1. C 2. J 3. A 4. J 5. B 6. F 7. D 8. H 9. B 10. G 11. C 12. J 13. D 14. H
Chapter 11 Enrichment: The Power Of Pi
3.14159265358979323846 1. 81 2. 64 3. 41 4. 625 5. 49 6. 512 7. 256 8. 75 9. 343 10. 625 11. 128 12. 729 13. 387 14. 169 15. 633 16. 52 17. 243 18. 128 19. 864 20. 16
Chapter 11 Big Ideas
1. Exponential growth occurs when a quantity increases by the same rate r in each time period t. Exponential decay occurs when a quantity decreases by the same rate r in each time period t.
2. Linear functions have constant first differences, quadratic functions have constant second differences, and exponential functions have a constant ratio.
3. The graph of ( ) = + 4f x x is a vertical
translation of the graph ( ) =f x x . The graph is translated 4 units up. The graph of ( ) = + 4f x x is a horizontal
translation of the graph ( ) =f x x . The graph is translated 4 units left.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 194
4. An expression containing square roots is in simplest form when the radicand has no perfect square factors other than 1, the radicand has no fractions, and there are no square roots in any denominator.
5. First add 5 to both sides; = 25x . Using the Power Property of Equality you can square both sides of the equation, resulting in an equivalent equation: x = 625.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 195
Chapter 12 Answers
LESSON 12–1
Reading Strategies 1. 5 2. −1 3. −7 4. no 5. yes 6. yes 7. no 8. −3 9. −10 10. 15 11. 2
Review for Mastery 1. 50; 50; 50; yes 2. 24; 24; 24; yes 3. 30; 30; 14; no 4. no 5. yes 6. no 7. 10 8. 3
9. a. 6; b. y = 6x
; c. −2; −3; −6 und.; 6; 3; 2
10. a. −8; b. y = −8x
;
c. 2; 4; 8 und.; –8; −4; –2
Practice 1. 30; 30; 30; yes; xy is constant. 2. 18; 21; 20; no; xy is not constant. 3. no; the equation cannot be written in the
form y = kx
.
4. yes; the equation can be written in the
form y = kx
.
5. a. 12; b. 12; c. −2; −3; −6; und.; 6; 3; 2
6. 10 7. 1 8. a. 20; 10; 25; b. (20)(10); 25(y2); 8; 8
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 196
Challenge
1. 5 13
feet 2. 83 13
pounds
3. Box A: 6 23
feet; Box B: 5 13
feet
4. 11 14
feet
5. No; box B would have to be 11 14
feet
from the pivot point. The board is not long enough.
6. a. L = 74
x; b. x = 47
L
LESSON 12–2
Reading Strategies 1. a. no; b. yes; c. yes 2. It has no variable in the denominator. 3. Find what value makes x − 5 = 0; 5 4. x = 1; y = 2
Review for Mastery 1. 0 2. 5 3. −7 4. x = 0, y = −3 5. x = 0, y = 0 6. x = 3, y = 2 7. x = −5, y = 0 8. x = −2, y = −6 9. x = 3, y = 8 10. x = 1, y = 0
11. x = −2, y = 3
Practice 1. 0 2. −1 3. 7 4. x = 6; y = 0 5. x = −4; y = −5 6. x = 1; y = 4 7. a. x = −2; b. y = 0; c. −1; −2; −4; 4; 2; 1
8. a. x = −5; b. y = −5;
c. −6; −7; −9; −1; −3; −4
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 197
9. a. D: nonnegative values;
R: nonnegative values b.
Challenge 1.
break at x = −1
2.
break at x = 1
3.
y = x − 2.5 and y = 1
x − 2.5
4.
y = x − 3 and y = 1
x − 3
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 198
5.
y = x + 3 and y =
1x + 3
LESSON 12–3
Reading Strategies
1. 424
simplified to 16
and the exponents
were subtracted, making 5
32 .m m
m=
2. b − 9 3. difference of squares 4. (6 − t) became (−1) (t − 6)
5. 21
3t 6.
2
2n
7. 42 7c −
8. 29x
−+
Review for Mastery 1. 0 2. 0; −5
3. −2, 7 4. 21 ; 0
5x
x≠
5. 3 ; 44
x xx
+ ≠−
6. 9 ; 0xx
≠
7. 2 ; 33
xx
≠ −+
8. 1 ; 0, 42
x xx
≠ ≠ −
9. 5x; x ≠ 4
10. x + 3; x − 3; x + 3; 13x −
11. x; x − 4; x − 4; x + 3; 3
xx +
12. x + 3; x − 5; x + 3; x + 2; 52
xx
−+
13. 19x +
14. x + 6
15. 14x +
16. 35x −
17. 15x
−+
18. 12x
−+
Practice 1. 5 2. 0; −3 3. −4, −2
4. x + 4; x − 4; 1x − 4
5. x − 6; x + 6; x − 6; 3x + 6
6. x + 1; x − 1; x + 1; x + 7; x −1x + 7
7. 1x + 2
8. x + 5x − 3
9. xx + 4
10. x − 3; x − 3; − 12
11. − 1x + 4
12. − 1
x + 2
13. a. 6s2
b. ratio for first animal: 38
; ratio for
second animal: 16
; the first animal
would experience greater heat loss because its surface-area-to volume ratio is greater.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 199
Challenge 1.
(3 2)( 3)( 1)( 3)
x xx x
+ −+ −
; x = −1 and y = 3; at x = 3
2.
( 1)( 4)1( 2)( 2)x xx x+ −
− + −; x = 2, x = −2 and y = −1;
none
Section Quiz: Lessons 12-1 to 12-3
1. C 2. J 3. D 4. H 5. A 6. H 7. A 8. F 9. B 10. H 11. D 12. G
LESSON 12–4
Reading Strategies
1. 23
2. 3(x − 2)
3. 5
2ba
4. 3t − 6
5. 112
6. 55
4x
Review for Mastery
1. ( 3)(5 )x
x+ ; (2)
(4) ( 3)x +; 1
10x
2. 3x + 6 3. 23x
4. 206x −
5. 36
xx +
6. 6 243
xx
++
7. 24xx
; 27x
; 328x
8. 12y 9. x2
10. 21
10 25x x− + 11. 5
8 8xx
−+
12. 3 2110x x−
Practice
1. 36x
2. 3
3q
p
3. 2 4
5a a+ 4. 1
2m
5. 6
56cd
6. 2 3
4 4k k
k−+
7. a. x +3 b. 2x +3
c. ( 3)(2 3)(2 2)
x xx x
++ +
d. ≈26%
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 1100
Challenge 1.
2 61
x xx− −+tartS
2 23
x xx
+−
2 4
3xx
−−
2x
x +
2
22 1x x
x+ +
31
xx
+−
2
294
xx
−−
2 3 2x
x x+ +
12x +
Finish
2.
218 93 2x xx
−+
2
3 23 20 12
4x x
x x+ +
−
2276
xx
2
23 1 0 86 3 3 18
x xx x
− −+ −Start
26 24
18 9x x
x−
− 6 3
6 4xx
−+
Finish
22 11 63 2
x xx− −
+
2
28 16
4 4 1x x
x x− +
− +
26 243 2xx
+−
3.
2
2yx
115y
26y
Finish
10y
15xy
56
23xStart
3xy 2
2x y
4.
23
3 10 3x
x x−
− +
212 11 53
x xx+ −−
Start
2
27 30
3 5 2x xx x
+ −+ −
22
4 45 50x
x x+
+ +
2
23 5 2
7 30x x
x x+ −
+ −
2
24 7 15
12 20x x
x x− −
+ +
23 29 1010
x xx+ −
+
24 45 502
x xx+ +
+Finish
2
312 11 5
xx x
−+ −
LESSON 12–5
Reading Strategies 1. Step 4 2. When multiplied with 9x, it gave the
LCD 18x3 3. The numerator would be 2x − 1.
4. 13x −
5. 1112x
Review for Mastery
1. 110x −
2. x − 3
3. 15
4. 110x
5. 55
xx
−+
6. x − 2
7. 40x2 8. 5x(x + 7)
9. 33
⎛ ⎞⎜ ⎟⎝ ⎠
;2
222
xx
⎛ ⎞⎜ ⎟⎝ ⎠
; 25 312x
x+
10. x + 3; x + 2; 22
xx
++
; 12x +
11. 2 143 12
xx
++
12. 16x −
13. 33x +
Practice
1. 2x
2. 16t −
3. 13
4. 23b +
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 1101
5. 9x3y2
6. (a + 4) (a − 4) (a + 5)
7. 13m
8. 45
xx
++
9. 610x
−−
10. 23 31
9x
x x−−
11. a. 11r
b. 23 h or 3 h 40 min3
Challenge
1. 2d 2. 40d
3. 45d 4.
40 65d d+
5. 2
40 65
dd d+
6. ≈49.5 mi/h
7. No; the average of 40 mi/h and 65 mi/h is 40 65 52.5 mi/h.
2+ =
8. 20 mi/h 9. 6.8 mi/h
LESSON 12–6
Reading Strategies 1. x2 ÷ x = x 2. x(x + 2) 3. −x2 − 2x 4. −4 5. Divide, multiply, and subtract with 3x − 4
as the dividend.
6. x + 5 7. x − 5 + 223x +
Review for Mastery
1. 7x2 + 3 2. x2 + 2 + 3x
3. 5x2 + 6x + 1 4. x − 7 5. 3x 6. x + 3 7. x; 6; x2 + 2x; −6x; 12; −6x − 12; x − 6
8. x; 2; x2 + 4x; 2x; 3; 2x + 8; x + 2 + 54x
−+
9. x − 5 10. x + 3 + 22x
−+
Practice 1. 4m; 4m; 3m2 + m 2. 10y3 + 6y − 5
3. 3x3 + 3 − 2x
4. 3w3 − 1w
− 47
2w
5. k + 5 6. x − 7 7. a + 1 8. 2t + 3
9. c + 3 + 24c +
10. k + 2 + 92k −
Challenge 1. a. 9; b. 9 2. a. 0; b. 0 3. a. −5; b. −5 4. a. −6; b. −6 5. a. −3; b. −3 6. P(r) 7. Because the remainder is zero, (x − 1) is
a factor of 2x2 + 3x − 5: (x − 1)(2x + 5). 8. Evaluate Q(r) ; if Q(r) = 0, then the
binomial (x − r) is a factor. 9. a. No; Q(1) = 6(1)4 − 31(1)3 − 49(1)2 +
104(1) + 60 = 90 ≠ 0 b. Yes; Q(−2) = 6(−2)4 − 31(−2)3 − 49(−2)2 + 104(−2) + 60 = 0
LESSON 12-7
Reading Strategies
1. 8x
= 41x −
2. a. no; b. yes; c. yes 3. It eliminates the fractions. 4. 3 is an extraneous solution because it
makes the denominators of the original equation 0, which is undefined.
5. 7 6. −4
Review for Mastery 1. x = −5 2. x = 0, x = 2 3. x = −13 4. x = 0, x = 2 5. x = 0 6. x = 0, x = 6
7. x(x + 2); x = 2 8. x(x + 1); x = 12
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 1102
9. 4x2; x = 4, x = −5 10. (x + 1)(x − 1); x = 2
Practice 1. 8 2. −10 3. 4 4. 3 5. 2 6. 6 7. 3; 4 is an extraneous solution 8. −1 and 6; no extraneous solutions
9. 1 13
h or 1h 20 min 10. 4x
11. 1x
, 14x
12. 1x
+ 14x
= 14
; x = 5
13. 5 and 20
Challenge 1. x = 2, y = 2 2. Possible answer: x = 4, y = 4 3. Possible answer: x = 8, y = 8 4. Possible answer: x = 12, y = 12 5. Possible answer: x = 16, y = 16 6. Possible answer: x = 200, y = 200 7. x = 2n, y = 2n
8. a. 11
+ 11
= 2
b. If x and y are integers with 0 < 1x
< 1
and 0 < 1y
< 1, then 0 < 1x
+ 1y
< 2.
Therefore, 1x
+ 1y
= 2 is only true when
x = 1 and y = 1.
c. Since 0 < 1x
+ 1y
< 2, there are no
positive integers x and y for n > 2. 9. Possible answer: x = 6, y = 6 10. Possible answer: x = 10, y = 10
Section Quiz: Lessons 12-4 to 12-7
1. C 2. J 3. A 4. H 5. B 6. J 7. B 8. H 9. B 10. F 11. B 12. F 13. B
Chapter 12 Enrichment: Experiencing Exponents
Because he always wants a quarterback. 1. H 2. K 3. L 4. U 5. C 6. Y 7. N 8. B 9. W 10. Q 11. T 12. E 13. S 14. R 15. A
Chapter 12 Big Ideas
1. There is an asymptote at x = 3 because when x = 3 the denominator in the function is 0, and dividing by 0 is undefined.
2. A rational expression is in simplest form when the numerator and denominator have no common factors except 1.
3. Step 1 Identify the LCD. Step 2 Multiply each expression by an
appropriate form of 1 so that each term has the LCD as its denominator.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 1103
Step 3 Write each expression using the
LCD. Step 4 Add or subtract the numerators,
combining like terms as needed. Step 5 Factor. Step 6 Simplify as needed. 4. Step 1 Write the binomial and polynomial
in long division form. Step 2 Divide the first term of the
dividend by the first term of the divisor. This is the first term of the quotient.
Step 3 Multiply this first term of the quotient by the binomial divisor and place the product under the dividend, aligning like terms.
Step 4 Subtract the product from the dividend.
Step 5 Bring down the next term in the dividend.
Step 6 Repeat Steps 2–5 as necessary until you get 0 or until the degree of the remainder is less than the degree of the binomial.
© Houghton Mifflin Harcourt Publishing Company Holt McDougal Algebra 1104