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Document 1804607
Chapter 6,
continued
10 times as massive: 10k 5 ks 6
64. Q 5 3.367*h3/2
10 5 s6
Q 5 3.367(20)(5)3/2
6}
6Ï 10 5 s
Q ø 753
The flow rate of the weir is about 753 cubic feet
per second.
65. a. V 5 s 3
V 5 (16)3
61.5 ø s
Reject the negative value, 21.5. The river’s speed must
be about 1.5 meters per second to transport particles that
are 10 times as massive as usual.
100 times as massive: 100k 5 ks6
100 5 s6
The volume of the cube is 4096 cubic millimeters.
6}
6Ï 100 5 s
b. Tetrahedron: V 5 0.118x 3
4096 5 0.118x 3
34,712 ø x 3
3}
Ï34,712 ø x
33 ø x
Edge length: about 33 millimeters
Octahedron: V 5 0.471x 3
4096 5 0.471x 3
8696 ø x
3}
3
Ï8696 ø x
21 ø x
Edge length: about 21 millimeters
Dodecahedron: V 5 7.663x 3
4096 5 7.663x 3
535 ø x 3
3}
Ï535 ø x
8øx
Edge length: about 8 millimeters
Icosahedron: V 5 2.182x 3
4096 5 2.182x 3
1877 ø x 3
3}
Ï1877 ø x
12 ø x
Edge length: about 12 millimeters
c. No; the icosahedron has the greatest number of
faces but not the smallest edge length. Of the three
polyhedra with equilateral triangles for faces, the
icosahedron has the smallest edge length, but the
dodecahedron’s faces are regular pentagons, so its
edge length is smaller.
66. An equation for mass m in terms of speed s is m 5 ks6,
where k is a constant.
At speed of 1 meter per second: m 5 k(1)6 5 k
Twice as massive: 2k 5 ks 6
2 5 s6
6}
6Ï 2 5 s
61.1 ø s
Reject the negative value, 21.1. The river’s speed must
be about 1.1 meters per second to transport particles that
are twice as massive as usual.
324
Algebra 2
Worked-Out Solution Key
62.2 ø s
Reject the negative value 22.2. The river’s speed must be
about 2.2 meters per second to transport particles that are
100 times as massive as usual.
Mixed Review for TAKS
67. D;
2x(4x 1 1) 2 (7x 1 3)(x 2 4)
5 8x 2 1 2x 2 7x 2 1 28x 2 3x 1 12
5 x 2 1 27x 1 12
68. F;
a2 1 b2 5 c 2
42 1 b2 5 82
16 1 b2 5 64
b2 5 48
b ø 6.9
Perimeter ø 8 1 4 1 6.9 5 18.9
The trench is about 18.9 meters long.
Lesson 6.2
6.2 Guided Practice (pp. 421– 423)
1. (51/3 + 71/4)3 5 (51/3)3 + (71/4)3
5 5(1/3 + 3) + 7(1/4 + 3)
5 5 + 73/4
2. 23/4 + 21/2 5 2(3/4 1 1/2) 5 25/4
31
3
3. }
5}
5 3(1 2 1/4) 5 33/4
31/4
31/4
201/2 3
4. }
5
51/2
1
2
F 1 205 2 G 5 (4
1/2 3
}
)
1/2 3
5 23 5 8
5. S 5 km2/3
5 8.4(9.5 3 104)2/3
5 8.4(9.5)2/3(104)2/3
ø 8.4(4.49)(108/3)
ø 17,506
The sheep’s surface area is about 17,506 square
centimeters.
4}
4}
4}
4}
6. Ï 27 + Ï 3 5 Ï 27 + 3 5 Ï 81 5 3
3}
Ï250
7. }
3} 5
Ï2
Î2502 5 Ï125 5 5
}
3
}
3}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
V 5 4096
Chapter 6,
8.
continued
Î34 5 ÏÏ34 + ÏÏ88 5 ÏÏ2432 5 Ï224
}
5}
5}
5}
5}
}
}
5}
}
5}
}
5}
}
5
3}
3}
3}
3}
13. (165/9 + 57/9)23 5 (165/9)23 + (57/9)23
5 1625/3 + 527/3
3}
9. Ï 5 1 Ï 40 5 Ï 5 1 Ï 8 + Ï 5
3}
1
161/3 + 52/3
16 + 5
16 + 5
161/3 + (52)1/3
}
5}
5/3
7/3 +
1/3
2/3
3}
5 Ï 5 1 2Ï5
3}
3}
3}
3}
5 (1 1 2)Ï 5 5 3Ï5
10.
3}
3}
Ï27q
5 Ï 3 (q
9
3
Î
5 Ï 3 + Ï (q
}
)
3 3
5}
2
3
5 3q
16 + 5
3
4001/3
5}
32,000
}
5
12. }
5 2x
3x1/2y1/2
13.
3
5 10
5
(x2)5 x 2
x
Ï
Ï
x10
}
}
}
5
5
}
} 5 }
y
y5
Ï5 y5
Ï5 y5
3/4
6xy
(1 2 1/2) (3/4 2 1/2)
}
11.
)
3 3
y
}
}
81/3 + 501/3
5}
32,000
5 2x1/2y1/4
}
3}
Ï50
5}
16,000
}
Ï9w5 2 wÏw3 5 3w 2Ï w 2 w 2Ï w
}
133/7
13
5 (3w 2 2 w 2)Ï w 5 2w 2Ï w
}
}
3}
3}
}
}
3}
3}
6.2 Exercises (pp. 424– 427)
15. Ï 20 + Ï 5 5 Ï 20 + 5 5 Ï 100 5 10
Skill Practice
16. Ï 16 + Ï 4 5 Ï 16 + 4 5 Ï 64 5 4
}
3}
1. No, 2Ï 5 and 2Ï 5 are not like radicals because they do
4}
4}
4}
4}
the radicand has no perfect nth powers as factors and any
denominator has been rationalized.
3. 53/2 + 51/2 5 5(3/2 1 1/2) 5 52 5 25
4. (6
5. 3
1/4
)
9
(2/3 + 1/2)
56
+ 27
1/4
1/3
56
5 (3 + 27)1/4 5 811/4 5 3
(1 1 4/5)
6. }
59
924/5
5 99/5
801/4
7. }
5 801/4 + 51/4
521/4
5 (80 + 5)1/4
3}
4}
}
4}
5 312/3 + 312/4
5 34 + 33 5 37 5 2187
5}
Î
Î Î
}
5}
Ï64
5 64
} 5 Ï 32 5 2
19. }
5} 5
2
Ï2
}
}
}
}
Ï1
1
Ï3
3
1
}5 }5}
20. }
} 5 }
} 5
5
75
25
Ï25
Ï 75
4}
4}
4}
4}
Ï36 + 9
Ï324
Ï36 + Ï9
21. }
5}
5}
4}
4} 5
4}
Ï4
Ï4
Ï4
}
}
}
4
4
4
Ï128 67/8
Ï8 + Ï16
}
22. }
5}
8 } + 7/8
8}
8}
6
Ï6
Ï2 + Ï3
Î3244 5 Ï81 5 3
}
4
4}
}
1281/4 + 67/8
5}
6
5 161/4 + 251/4
(1282)1/8 + (67)1/8
5 }}
5 2 + 251/4
6
(1282 + 67)1/8
52+5
1/2
1 2
4}
4}
18. (Ï 3 + Ï 3 )12 5 1 31/3 + 31/4 212
5 4001/4
73 21/3
8. }3
5
4
4}
13
5 Ï16 + Ï 4 5 2Ï4 5 2Ï 2
2. A radical expression with index n is in simplest form if
2/3 1/2
7}
Ï371,293
17. Ï 8 + Ï 8 5 Ï 8 + 8 5 Ï 64 5 Ï 16 + 4
}
not have the same }
index. The expression 2Ï5 has an
3
index of 2 and 2Ï 5 has an index of 3.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
135/7
13
1
13
}
14. }
5 13(3/7 2 5/7) 5 1322/7 5 }
5}
2/7 +
5/7
5/7
}
F1 2 G
5}
6
7 3 21/3
7 [3 + (21/3)]
7 21
4
}
5 }4
5 }4
5 }7
4
1 2
1 2
(4,586,471,424)1/8
5 }}
6
5}
Ï1331
1
112/5
113/5
9. }
5 11(2/5 2 4/5) 5 1122/5 5 }
+}5}
11
112/5 113/5
114/5
4(69,984)1/8
5}
6
10. (123/5 + 83/5)5 5 [(12 + 8)3/5]5
5 (963/5)5 5 963 5 884,736
120(22/5 1 2/5)
1
12022/5 + 1202/5
11. }}
5}
5}
5 73/4
723/4
723/4
723/4
(641/3)5/3 + (641/3)2/3
645/9 + 642/9
12. }
5 }}
3/4
43/4
4
47/3
5}
43/4
}
8
2Ï
69,984
23. C;
4}
5}
3
4}
5 419/12
5 4 + 47/12
4}
4}
4}
5 3Ï 16 + Ï 2 + (26Ï 5 )
4}
4}
5 3 + 2 + Ï2 + (26Ï5 )
4}
5 236Ï 2 + 5
4}
5 236Ï 10
7/3 2 3/4
54
4}
4}
3Ï 32 + (26Ï5 ) 5 3Ï16 + 2 + (26Ï 5 )
}
}
}
}
}
24. Ï 72 5 Ï 36 + 2 5 Ï 36 + Ï 2 5 6Ï 2
6}
6}
6}
6}
6}
3}
25. Ï 256 5 Ï 64 + 4 5 Ï 64 + Ï 4 5 2Ï 4 5 2Ï 2
5 4 + (47/2)1/6
5 4 + (27)1/6
5 4(2 + 26)1/6 5 8 + 21/6
Algebra 2
Worked-Out Solution Key
325
Chapter 6,
3}
7}
7}
5 26Ï 2 1 2 + 2 + Ï2
5 26Ï 2 1 4Ï 2
7}
3}
4}
4}
4}
4}
4}
5 (12 2 28)Ï2
5 10 + 4 + Ï2
4}
4}
5 216Ï 2
5 40Ï 2
3}
Ï36
3
3
}
29. }
4} 5 4}
Ï16 + 9
Ï144
3
5}
4}
4}
Ï16 + Ï9
4}
3
Ï9
}
5}
4} + 4}
2Ï 9 Ï9
4}
3Ï 9
5}
4}
2Ï 81
}
3Ï 3
5}
6
4}
30.
4}
4}
6}
6}
}
}
6}
}
6}
6
4}
4}
5 10Ï 2 2 16Ï 2
4}
5 (10 2 16)Ï 2
5}
6}
4}
5 26Ï 2
3}
3}
3}
3}
40. 5Ï 48 2 Ï 750 5 5Ï 8 + 6 2 Ï 125 + 6
3}
3}
3}
3}
5 5Ï 8 + Ï 6 2 Ï125 + Ï6
3}
3}
5 5 + 2 + Ï6 2 5 + Ï6
3}
3}
5 10Ï 6 2 5Ï 6
3}
5 (10 2 5)Ï 6
3}
5 5Ï 6
5}
9 +9
}
}
31. }
5}
5} + 5} 5
Ï243
Ï27 Ï9
9(1/3 1 1/5)
5}
3
Ï9
1/3
1/5
3}
3}
41. The radical expressions 2Ï 10 and 6Ï 5 are not like
radicals because they don’t have the same radicand.
3}
3}
Therefore, 2Ï10 1 6Ï5 cannot be combined.
42. To make the denominator a perfect cube, you must
98/15
multiply the numerator and denominator of the fraction
by y so that the value of the fraction does not change.
5}
3
(9 )
Î Î
1/2 16/15
3
3
x
y
}2 5
}
}
}
5}
3
x+y
}
Ï3 xy
Ï3 xy
}
5}
} 5 }
2
y +y
y
Ï3 y 3
316/15
5}
3
43. x1/4 + x1/3 5 x(1/4 1 1/3) 5 x7/12
5 31/15
44.
( y4)1/6 5 y(4 + 1/6) 5 y 2/3
45.
Ï81x 4 5 Ï34x4 5 Ï34 + Ïx4 5 3x
15 }
5 Ï3
6}
6}
6}
32. 2Ï 3 1 7Ï 3 5 (2 1 7)Ï 3 5 9Ï 3
3}
2Ï 5
3
1
1
3
33. }Ï 5 2 }Ï 5 5 } 2 } Ï 5 5 }
5
5
5
5
5
3}
1
2
3}
5}
5}
5}
34. 25Ï 2 2 15Ï 2 5 (25 2 15)Ï 2 5 10Ï 2
4}
34}
3 4 } Ï7
1
14}
35. }Ï 7 1 }Ï 7 5 } 1 } Ï 7 5 }
8
8
8
2
8
1
3}
2
3}
3}
3}
3}
36. 6Ï 5 1 4Ï 625 5 6Ï 5 1 4Ï 125 + 5
3}
5 6Ï 5 1 4Ï125 + Ï5
3}
3}
5 6Ï 5 1 4 + 5Ï 5
3}
3}
5 6Ï 5 1 20Ï5
3}
5 (6 1 20)Ï 5
3}
5 26Ï 5
326
4}
4}
4}
6}
Ï1296
Ï64
3}
Ï36
5}
2
4}
5 2 + 5Ï 2 2 8 + 2Ï2
Ï3
5}
2
3}
4}
4}
5 2Ï625 + Ï 2 2 8Ï16 + Ï2
Î814 5 ÏÏ814 + ÏÏ1616
}
}
5}
4}
39. 2Ï 1250 2 8Ï 32 5 2Ï 625 + 2 2 8Ï 16 + 2
3}
Ï36
1
}
}
}
28. 3 } 5 }
3} + 3} 5 3} 5
6
6
Ï6 Ï36
Ï216
3}
4}
5 12Ï 2 2 28Ï 2
4}
6}
4}
4}
4}
5 10Ï256 + Ï 2
Ï9
4}
5 12Ï 2 2 7 + 4Ï2
4}
5 10Ï256 + 2
3}
4}
4}
5 12Ï 2 2 7Ï 256 + Ï2
5 10Ï512
4}
4}
4}
38. 12Ï 2 2 7Ï 512 5 12Ï 2 2 7Ï 256 + 2
4}
4}
Ï36
7}
5 (26 1 4)Ï 2 5 22Ï2
27. 5Ï 64 + 2Ï 8 5 5 + 2Ï 64 + 8
3}
7}
7}
5 6Ï 2
Î
7}
7}
3}
3}
Ï1
7}
7}
5 Ï216 + Ï 2
3}
7}
3}
5 Ï216 + 2
}
7}
5 26Ï 2 1 2Ï128 + Ï2
3}
5 Ï432
4}
7}
37. 26Ï 2 1 2Ï 256 5 26Ï 2 1 2Ï 128 + 2
3}
26. Ï 108 + Ï 4 5 Ï 108 + 4
Algebra 2
Worked-Out Solution Key
4}
4}
4}
4}
2
46. }
5 2x3/2
x23/2
y4/3
x2/5y
(2/5 2 1) (1 1 1/3)
23/5 4/3
}
47. }
5
x
y
5
x
y
5
x3/5
xy21/3
}
}
}
3
3
5
3
Ïx15 Ï (x )
x5
x15
48. 3 }
5}
} 5 }
} 5 }
2
3 6
3
2
3
y6
Ïy
Ï( y ) y
3}
6 } 23
1
1
}
49. (Ï x 2 + Ï x4 ) 5 }
3 5 2/3
3}
6}
2
4
(
x
+
x4/6)3
(Ïx + Ïx )
Î
1
1
1
x
5}
5 }4
5}
(2/3 1 4/6) 3
4/3 3
[x
]
(x )
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3}
continued
Chapter 6,
}
continued
}
}
Îx + Ïx 5
Îx + Ïx 5
Î x + Ïx5
50. }
} 5 }
}
} 5 }
}
}
16
16
Ï25 + Ïx
Ï25x
Ï25 + Ï (x8)2
3 }
3 }
3 }
}
3}
}
1/3
5x
}
5 (3 2 2)x 2Ï x
}
x(1/3 1 5/2)
x17/6
5}
5}
8
5x
5x8
5 x2Ï x
4}
6
}
4}
2
4}
5 (2xy 1 3y)Ï 2x2
5 2(x ) 1 2(2x
3
1
52.
53.
54.
,x
}
5 2(7x
1/4
}
Ï49x 5 Ï49x x 5 Ï49x + Ï x 5 7x x
4
2Ï
5 14x
}
12x 2y 4y 2z12
Ï4 12x2y6z12 5 Ï4 }
}
4 4 12
4
y z +Ï
12x2y2
5Ï
}
34
2 2
5 yz Ï12x y
}
5 24x
68.
}
55.
Ïx
8
}
5 Ïx4xyz8
}
c 5 5x1/3
}
5 Ïx4z8 + Ï xy
5}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
5}
23 Ïx4
23Ïx4
23Ïx 4
23
56. }
5}
+}
5}
5}
5}
5}
5}
5}
6
10
6
4
x2
Ïx Ïx
Ïx
Ïx
Î Î
Î Î
}
57.
58.
3
x
y
x 3 + y2
3
}4 5
3
}
3}
}
Ïx3y2
3 2
Ïx3 + Ï
y
Ï3 y 6
Ï3 y6
3
}
3 2
xÏ
y
y +y
}
}
5
3
}
5
3
20x 3y 2
9xz
9xz + xz
Ï4 + 5x y z
}
5
}
Ï9x2z4
}
Î
5 }2 (4x1/3)(3x1/3)
5 12x1/3
5 }2 + 12x(1/3 1 1/3)
1
9x z
5 1 2}3 2 }2 2Ï x
1
}
Ï4x y + Ï5z
}
4 2
5
5
}
Ï9x2z4
}
2x2yÏ 5z
2xyÏ 5z
3xz
3z
7}
2
4}
6
5}
72. (x6y2)20.75 5 x6 + (20.75)y2 + (20.75)
1
5 x24.5y21.5 5 }
x4.5y1.5
14 }
11
Ïx
y0.75
y3/2
3
1
1
3
61. } y3/2 2 } y3/2 5 } 2 } y3/2 5 }
4
4
4
2
4
5 x21.25y0.75 5 }
1.25
2
x
62. 27Î y 1 16Î y 5 (27 1 16)Î y 5 9Î y
3 }
3 }
63. (x 4y)1/2 1 (xy1/4)2 5 x(4 + 1/2)y1/2 1 x2y(1/4 + 2)
2 1/2
5x y
2 1/2
1x y
1
x0.3
73. }
5 x(0.3 2 1.5) 5 x21.2 5 }
x1.2
x1.5
74. (x5y23)20.25 5 x5 + (20.25)y23 + (20.25)
5}
60. 3Ï x 1 9Ï x 5 (3 1 9)Ï x 5 12Ï x
1
}
1
71. y20.6 + y26 5 y[20.6 1 (26)] 5 y26.6 5 }
y6.6
7}
2
x25/14
(25/14 2 1)
5}
5 x11/14 5
x 5x
3 }
}
70. x 0.5 + x 2 5 x(0.5 1 2) 5 x2.5
Ïx Ïx
Ïx + Ïx
Ïx
} }
}
59. }
7}
7} 5 7} + 7} 5
Ïx7
Ïx 5 Ïx5 Ïx2
x6/4 + x2/7
x(6/4 1 2/7)
5}
5}
x
x
5}
1
5 2}6Ï x
5}
5}
2
2
4}
6
5 6x2/3
}
}
1 }
1 }
1
1
69. C; 2}Ï 4x 2 }Ï 9x 5 2} + 2Ï x 2 } + 3Ï x
6
6
6
6
20x4y2z
}
2 4
}
4 2
3 }
1
5 3x1/3 1 4x1/3 1 5x1/3
y
}
20x3y2 + xz
}
5}
A 5 }2 bh
}
5}
5}
} 5 }
}
2
4
2
}
4}
6
1
P5a1b1c
}
5 x 2z 4 Ï xy
}
c2 5 a2 1 b2
c 5 5Ï (x1/3)2
}
5}
5 35x1/2
1/4
}
yz + Ï x z 5 Ï x yz
5
5 35x(1/4 1 1/4)
1 10x
c 5 Ï 25x2/3
}
3 5
5 (7x1/4)(5x 1/4)
1/4
c2 5 25x2/3
3
6y
5 2xy2Ï
}
) 1 2(5x )
c2 5 9x2/3 1 16x2/3
}
3
1/4
c2 5 (3x1/3)2 1 (4x1/3)2
3
3
8x3y6 + Ï
6y
5Ï
2
A 5 *w
1/4
5 (14 1 10)x1/4
48x 3y7
Ï3 4x 3y5 + Ï3 12y 2 5 Ï3 }
3
5 Ï8 + 6x3y6y
}
}
}
5 2x(3 1 2/3)
67. P 5 2* 1 2w
}
4
}
5 x3(2x 2/3)
5 2x11/3
1/8
}
}
)
2/3
5 2x 1 4x
51. Sample answer: x
5
2/3
3
5}
6 }
5x 5Î x
}
A 5 *w
66. P 5 2* 1 2w
1
+x
}
5}
31/6 5
30/6
1/6
7/8
}
4}
2
5 2xyÏ 2x 1 3yÏ2x
x231/6
5x
4}
4
4
65. yÏ 32x 1 Ï
162x2y4 5 yÏ16 + 2x4x2 1 Ï
81 + 2x2y4
5}
5}
5
5
1
5x
}
}
5/2
5x
x(17/6 2 8)
}
5 3x2Ï x 2 2x2Ï x
Ïx + Ïx
x +x
5}
5}
8
8
5
}
}
64. xÏ 9x3 2 2Ï x5 5 xÏ 9x2x 2 2Ï x4x
2 1/2
5 2x y
y20.5
75. }
5y
y0.8
(20.5 2 0.8)
21.3
5y
1
y
5}
1.3
76. 10x0.6 1 (4x0.3)2 5 10x0.6 1 16x0.3 + 2
5 10x0.6 1 16x0.6 5 26x0.6
Algebra 2
Worked-Out Solution Key
327
continued
77. 15z 0.3 2 (2z0.1)3 5 15z0.3 2 8z 0.3 5 7z 0.3
85. d 5 1.9[1 5.5 3 1024 2*]1/2
}
x5Ï3
(5Ï3 2 2Ï3 )
}
78. }} 5 x
x2Ï3
}
5 1.9[(5.5 3 10
}
5 x3Ï3
ø 1.9(0.23) 5 0.44
x2:
x: 2
80. }
5}
5 x(2: 2 2:/3) 5 x4:/3
:/3
x2:/3
x
}
}
The optimum pinhole diameter is about 0.44 millimeter.
86.
}
81. x2yÏ2 1 3x2yÏ2 5 4x2yÏ2
82. a.
1
b. 2x + 2x 1 1 5 }
16
1
2x 1 (x 1 1) 5 }
16
3
9
}x 5 243
3 5 243 + 9x
1
3
243
22x 1 1 5 }
16
} 5 9x
1
81
22x 1 1 5 }4
1
}2 5 9 x
9
22x 1 1 5 224
922 5 9 x
2x 1 1 5 24
} 5 9x
x
a2 1 b2 5 c2
}
x2 1 x2 0 (xÏ2 )2
}
2x2 0 x2 + (Ï2 )2
2x2 5 2x2 2x 5 25
5
2
x 5 2}
2.5120.77
2.512m1
87. a. }
5}
5 2.5120.74 ø 1.98
2.512m2
2.5120.03
Altair is about 2 times fainter than Vega.
2.5121.25
2.512m1
b. }
5 2.5120.48 ø 1.56
m2 5 }
2.512
2.5120.77
Deneb is about 1.6 times fainter than Altair.
(4x)x 1 2 5 64
4x + (x 1 2) 5 64
4
x2 1 2x
5 64
4
x2 1 2x
5 43
x 2
x
1
2
x 5 22
c.
2.5121.25
2.512m1
c. }
5}
5 2.5121.22 ø 3.08
2.512m2
2.5120.03
Deneb is about 3 times fainter than Vega.
}}
V0Ï (V0)2 1 2gh0
2
x 1 2x 5 3
}
V0Ï (V0)2
(x 1 3)(x 2 1) 5 0
3V
4:
}F
5 r3
5 57.5(32)2/3
5 11(5.9 3 104)2/3
ø 57.5(10.08)
5 11(5.9) (104)2/3
ø 11(3.27)(108/3)
2/3
The bat’s surface area
ø 11(3.27)(464.16)
is about 580 square
ø 16,696
centimeters.
The human’s surface
area is about 16,700
square centimeters.
L
84. v 5 8.8 }
A
Î
g
4
b. S 5 km2/3
Î
(V0)2
V 5 }3:r 3
89. a.
Problem Solving
}
V0 + V0
5}
5}
5}
g
g
x 5 23 or x 5 1
ø 580
}}
V0Ï (V0)2 1 2g(0)
88. d 5 }} 5 }}
g
g
x2 1 2x 2 3 5 0
83. a. S 5 km2/3
)(100)]
5 1.9(0.055)1/2
}
}
}
( } })
79. (xÏ2 )Ï3 5 x Ï2 + Ï3 5 xÏ6
1 2
* 5 10 cm 5 100 mm
1/2
24
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 6,
}
7
1.4 3 10
5.5 3 10
5 8.8 }3
}
ø 8.8Ï0.25 3 104
}
5 8.8Ï 2500 5 8.8(50) 5 440
The velocity of the jet is about 440 feet per second.
Î4:3VF5 r
}
3
}
b. S 5 4:r 2
3V
F
1Î}
4: 2
}
5 4:
3
2
3V 2/3
5 4: 1 }
F
4: 2
(3V)2/3
5 4: }
2/3
(4:)
5 (4:)(1 2 2/3)(3V)2/3
5 (4:)1/3(3V)2/3
c. Balloon with volume V:
S1 5 (4:)1/3(3V)2/3
Balloon with volume 2V:
S2 5 (4:)1/3(3 + 2V)2/3 5 (4:)1/3(2)2/3(3V)2/3
5 (4:)1/3(4)1/3(3V)2/3
3}
3}
5 Ï 4 (4:)1/3(3V)2/3 5 Ï4 S1
3}
The balloon with twice as much water has Ï4 , or
about 1.59 times the surface area of the other balloon.
328
Algebra 2
Worked-Out Solution Key
Chapter 6,
continued
n}
90. Sample answer: Ï xm
n 5 3, m 5 5, x 5 21:
n 5 4, m 5 5, x 5 21:
11. (x6y 4)1/8 1 2(x1/3y1/4)2 5 x3/4y1/2 1 2(x2/3y1/2)
}
(21) 5 21
Ï}
4}
4
, but 21 has no
Ï(21)5 5 Ï21}
3
5
real 4th root; Ï{(21) {5 1
5
4
n 5 3, m 5 4, x 5 21:
n 5 2, m 5 4, x 5 21:
}
(21)4 5 1
Ï3 }
Ï(21)4 5 1
}
}
}
7Ï7
Ï75
3Ï7 1 4Ï 7
}
12. }}
5
} + }
}
}
Ï75 Ï75
Ï75
3/2
7 + 7 + 75/2
71 1 3/2 1 5/2
75
5}
5 }5 5 1
5}
5
5
7
7
7
3
}
Absolute value is needed when n is even and m is odd.
3
}
3
}
}
Ïx
2Ïx 4
2x2
2x2Ï x
Ïx
2Ï x + Ïx 3
}
13. }
5
5 }6
} 5 }
}
} 5 }
} + }
7
8x8
4x
Ï82 + x14 + x 8x Ïx Ï x
Ï64x15
5}
}
}
}
5}
5}
91. A;
22 2 3
5}
5 24xy2Ï2x
5}
5 }8
slope 5 }
24 2 4
28
15. Three labels A, B, C, are added to the graph to indicate
Choice A: 25x 1 8y 5 14
the three right triangles.
8y 5 5x 1 14
5
7
a2 1 b2 5 c2
(8 2 2)2 1 82 5 c2
62 1 82 5 c2
100 5 c2
10 5 c
For right triangle B
a2 1 b2 5 c2
42 1 82 5 c2
80 5 c2
}
4Ï 5 5 c
5
slope 5 }8
92. G;
25 a 26x 1 3 a 15
26x 1 3 a 15
6x a 8
26x a 12
4
x a }3
x q 22
4
22 a x a }3
}
1. 363/2 5 (Ï 36 )3 5 63 5 216
1
1
1
1
2. 6422/3 5 }
5}
5}5}
}
642/3
(Ï3 64 )2 42 16
8
For right triangle C
a2 1 b2 5 c 2
22 1 42 5 c 2
20 5 c 2
}
2Ï 5 5 c
1. f(x) 1 g(x) 5 22x 2/3 1 7x2/3 5 (22 1 7)x 2/3 5 5x2/3
2. f(x) 2 g(x) 5 22x2/3 2 7x2/3 5 (22 2 7)x2/3 5 29x2/3
5}
4. (232)2/5 5 (Ï 232 )2 5 (22)2 5 4
5. x4 5 20
3. The functions f and g each have the same domain: all real
6. x5 5 210
4}
numbers. So, the domains of f 1 g and f 2 g also consist
of all real numbers.
4. f(x) + g(x) 5 3x(x1/5) 5 3x(1 1 1/5) 5 3x6/5
5}
x 5 6Ï 20
x 5 Ï 210
x ø 62.11
x ø 21.58
7. x6 1 5 5 26
3}
f (x)
3x
5. } 5 }
5 3x(1 2 1/5) 5 3x4/5
g(x)
x1/5
3}
6. The functions f and g each have the same domain: all real
8. (x 1 3)3 5 216
x6 5 21
x 1 3 5 Ï 216
x 5 6 Î21
6 }
x 5 Ï 216 2 3
x ø 61.66
x ø 25.52
4}
9. Ï 32 + Ï 8 5 Ï 32 + 8 5 Ï 256 5 4
}
4
6.3 Guided Practice (pp. 429–431)
3. 2(6253/4) 5 2[(Ï 625 )3] 5 2(53) 5 2125
4}
B
4
Lesson 6.3
4}
4}
A
The perimeter of the right triangle is
}
}
}
10 1 4Ï 5 1 2Ï5 5 10 1 6Ï 5 .
Quiz 6.1–6.2 (p. 427)
4}
2
C
For right triangle A
y 5 }8 x 1 }4
25 a 26x 1 3
5}
5 2xy2Ï 2x 2 6xy2Ï 2x
5
25
}
5
5
14. y 2Ï 64x6 2 6Ï
2x6y10 5 y2Ï 32 + 2x5x 2 6Ï
2x5xy10
Mixed Review for TAKS
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
5 x3/4y1/2 1 2x2/3y1/2
}
3}
10. (Ï 10 + Ï 10 )8 5 (101/2 + 101/3)8
5 F 10
G
(1/2 1 1/3) 8
5 (105/6)8
numbers. So, the domain of f + g also consists of all real
f
numbers. Because g(0) 5 0, the domain of }g is restricted
to all real numbers except x 5 0.
7. r(m) + s(m) 5 (1.446 3 109)m20.05
5 (1.446 3 109)(1.7 3 105)20.05
ø (1.446 3 109)(0.55)
20/3
5 10
3}
20
5 Ï 10
3}
5 Ï 1018 + 102
}
3
(106)3 + 102
5Ï
3}
5 106Ï 102
3}
5 1,000,000Ï 100
ø 791,855,335
The white rhino has about 791,855,335 heartbeats over
its lifetime.
8. g( f (5))
9. f ( g(5))
f(5) 5 3(5) 2 8 5 7
g(5) 5 2(5)2 5 50
g( f (5)) 5 g(7)
f ( g(5)) 5 f (50)
5 2(7)2
5 3(50) 2 8
5 98
5 142
Algebra 2
Worked-Out Solution Key
329
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