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n3EE!DDITIONAL!NSWERS
BEGINNINGONP!!
(x 2 3)2
4
Chapter Test (p. 673) 24. ellipse, }}}} 1 ( y 2 2)2 5 1
2
2
25. circle, (x 1 4) 1 ( y 1 6) 5 49 26. hyperbola,
10.1 Mixed Review (p. 689)
84.
85.
y
y2
(x 2 5)
2
}}}} 2 } 5 1 27. parabola, ( y 2 8) 5 12(x 1 2)
4
9
( y 2 3)2
(x 1 1)2
28. ellipse, }}}} 1 }}}} 5 1 29. hyperbola,
4
25
( y 1 7)2
2
}}}} 2 (x 2 2) 5 1
16
y
2
6
21
1
x
21
x
Chapter 10
10.1 Skill Practice (pp. 686–687)
3.
long-sleeve
short-sleeve
long-sleeve
short-sleeve
long-sleeve
short-sleeve
M
L
XL
4.
86.
M long-sleeve
M short-sleeve
L long-sleeve
L short-sleeve
XL long-sleeve
XL short-sleeve
jam
white, jam
margarine
white, margarine
jam
wheat, jam
margarine
wheat, margarine
87.
y
y
15
2
24
x
25
x
white
88.
89.
y
y
wheat
1
5.
corn
green bean
potato
corn
green bean
potato
corn
green bean
potato
chicken
fish
pasta
6.
cherry
oak
pine
stained cherry
painted cherry
unfinished cherry
stained mahogany
painted mahogany
unfinished mahogany
stained oak
painted oak
unfinished oak
stained pine
painted pine
unfinished pine
57. To find the number of permutations of n distinct objects,
the first event has n ways of occurring, the second event
has (n 2 1) ways of occurring since one of the n objects was
used for the first event, the third event has (n 2 2) ways of
occurring since one more of the n objects was used for the
second, and so on until you reach the last event which only
has 1 way of occurring. The fundamental counting principle
indicates that to find the number of ways that all the events
can happen, multiply each of the number of ways that the
individual events can occur. For n distinct objects it would
be n p (n 2 1) p (n 2 2) p … p 1, which is the same as n!.
21
1
x
21
x
n!
10.2 Skill Practice (pp. 694–695) 42. nC0 5 }}}}}
5
(n 2 0)! p 0!
n!
n!
n!
n!
n!
}}} 5 } 5 1 43. nCn 5 }}}}} 5 }}} 5 }}} 5
n! p 1
n!
(n 2 n)! p n!
0! p n!
n! p 1
n!
n!
r!
} 5 1 44. nCr p rCm 5 }}}}} p }}}}}} 5
n!
(n 2 r )! p r! (r 2 m)! p m!
n!
n!
}}}}}}}}} ; nCm p n 2 mCr 2 m 5 }}}}}} p
(n 2 r )! p (r 2 m)! p m!
(n 2 m)! p m!
(n 2 m)!
n!
}}}}}}}}}}}}} 5 }}}}}}}}}
[(n 2 m) 2 (r 2 m)]! p (r 2 m)!
(n 2 r )! p (r 2 m)! p m!
n!
n!
n!
45. nC1 5 }}}}} 5 }}}} ; nP1 5 }}}}
(n 2 1)! p 1!
(n 2 1)!
(n 2 1)!
n!
n!
46. nCr 5 }}}}} ; nCn 2 r 5 }}}}}}}}}} 5
(n 2 r )! p r!
[n 2 (n 2 r )]! p (n 2 r )!
(n 1 1)!
n!
n!
}}}}} 5 }}}}} 47. n 1 1Cr 5 }}}}}}} 5
r! p (n 2 r )!
(n 2 r )! p r!
[(n 1 1) 2 r]! p r!
(n 1 1)!
n!
}}}}}}} ; nCr 1 nCr 2 1 5 }}}}} 1
(n 1 1 2 r )! p r!
(n 2 r )! p r!
n!
n!
n!
}}}}}}}}} 5 }}}}} 1 }}}}}}}}} 5
[n 2 (r 2 1)]! p (r 2 1)!
(n 2 r )! p r!
(n 2 r 1 1)! p (r 2 1)!
n! p (n 2 r 1 1)
(n 2 r 1 1) p (n 2 r )! p r!
ADDITIONAL ANSWERS
mahogany
stained
painted
unfinished
stained
painted
unfinished
stained
painted
unfinished
stained
painted
unfinished
chicken, corn
chicken, green bean
chicken, potato
fish, corn
fish, green bean
fish, potato
pasta, corn
pasta, green bean
pasta, potato
n!r
(n 2 r 1 1)! p r p (r 2 1)!
}}}}}}}}}} 1 }}}}}}}}}} 5
n! p (n 2 r 1 1)
n! p (n 2 r 1 1) 1 n! p r
n! p r
}}}}}}} 1 }}}}}} 5 }}}}}}}}}} 5
(n 2 r 1 1)! p r!
(n 2 r 1 1)! p r
(n 2 r 1 1)! p r!
n! p (n 2 r 1 1 1 r )
n! p (n 1 1)
(n 1 1)!
}}}}}}}} 5 }}}}}}} 5 }}}}}}}
(n 2 r 1 1)! p r!
(n 2 r 1 1)! p r!
(n 1 1 2 r )! p r!
AA43
Chapter 9,
continued
}
Ï10
10
a2 5 }
9
a5}
3
b2 5 10
b 5 Ï10
36. x2 1 y2 2 100 5 0
x 1 y 2 14 5 0 → x 5 14 2 y
}
}
1
Ï10
Vertices: 16}
, 22
3
2
(22 1
21
10
, 22
3
196 2 28y 1 y2 1 y2 2 100 5 0
2y2 2 28y 1 96 5 0
y
(1 2
(14 2 y)2 1 y2 2 100 5 0
10 )
(
10
, 22
3
(1, 22)
)
11
(22 2
2(y2 2 14y 1 48) 5 0
)
2(y 2 8)(y 2 6) 5 0
x
y 5 8 or y 5 6
When y 5 8: x 5 14 2 (8) 5 6
When y 5 6: x 5 14 2 (6) 5 8
10 )
The solutions are (6, 8) and (8, 6).
37. 16x2 2 4y2
34. x2 1 y2 1 4x 2 14y 1 17 5 0
2
4x 1 9y 2 40x 1 64 5 0
A 5 1, B 5 0, C 5 1
36x2 2 9y2
B 2 2 4AC 5 0 2 4(1)(1) 5 24
The conic is a circle because B2 2 4AC < 0, b 5 0, and
A 5 C.
2
(x2 1 4x) 1 ( y2 2 14y) 5 217
(x2 1 4x 1 4) 1 ( y2 2 14y 1 49) 5 217 1 4 1 49
2
(x 1 2) 1 ( y 2 7) 5 36
2 144 5 0
2
4x 1 9y 2 40x 1 64 5 0
40x2
2 40x 2 80 5 0
40 1 x2 2 x 2 2 2 5 0
x2 1 y2 1 4x 2 14y 1 17 5 0
2
2 64 5 0
2
40(x 1 1)(x 2 2) 5 0
x 5 21 or x 5 2
When x 5 2:
When x 5 21:
2
2
16(2)2 2 4y2 2 64 5 0
16(21) 2 4y 2 64 5 0
The radius is 6.
2
24y2 5 0
24y 5 48
h 5 22, k 5 7
y2 5 224
The center is (22, 7).
y50
No real solution
y
Chapter 9 Test (p. 673)
(22, 7)
1. (21, 5), (7, 3)
4
}}
d 5 Ï(21 2 7)2 1 (5 2 3)2
x
22
1
35. y2 5 4x → x 5 } y2
4
2x 2 5y 5 28
1
1
4
}}
5 Ï(28)2 1 (2)2
}
}
}
5 Ï64 1 4 5 Ï68 5 2Ï17
M1 }
,}
5 (3, 4)
2
2 2
21 1 7 5 1 3
2
2 } y2 2 5y 5 28
2. (4, 2),(8, 8)
1
2
} y2 2 5y 1 8 5 0
1
2
} ( y2 2 10y 1 16) 5 0
}}
d 5 Ï(4 2 8)2 1 (2 2 8)2
}}
5 Ï(24)2 1 (26)2
}
1
2
} ( y 2 8)(y 2 2) 5 0
y 5 8 or y 5 2
1
When y 5 8: x 5 }4 (8)2 5 16
1
}
}
5 Ï16 1 36 5 Ï52 5 2Ï13
M1 }
,}
5 (6, 5)
2
2 2
418 218
3. (21, 26), (1, 5)
}}
d 5 Ï(21 2 1)2 1 (26 2 5)2
}}
When y 5 2: x 5 }4 (2)2 5 1
5 Ï(22)2 1 (211)2
The solutions are (1, 2) and (16, 8).
5 Ï4 1 121 5 Ï125 5 5Ï5
}
}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The solution is (2, 0).
}
M1 }
,}
5 0, 2}2 2
2
2 2 1
21 1 1 26 1 5
594
1
Algebra 2
Worked-Out Solution Key
n2ws-09b.indd 594
6/27/06 11:16:48 AM
Chapter 9,
continued
10. 18x2 1 2y2 5 18
4. (2, 25), (3, 1)
}}
y2
d 5 Ï(2 2 3) 1 (25 2 1)
2
2
x2 1 }
51
9
}}
5 Ï(21)2 1 (26)2
}
The conic is an ellipse.
2
}
5 Ï1 1 36 5 Ï37
1
2 1 3 25 1 1
2
2
2 1
5
2
M } , } 5 } , 22
2
a 59
a53
b2 5 1
b51
(0, 3)
(21, 0)
1
x
Co-vertices: (61, 0)
(0, 23)
}}}
d 5 Ï(26 2 (23)) 1 (22 2 5)
2
2
11. (x 2 6)2 1 ( y 1 1)2 5 36
}}
5 Ï(23)2 1 (27)2
}
The conic is a circle with radius r 5 Ï36 5 6 and center
(6, 21).
}
5 Ï9 1 49 5 Ï58
1
(1, 0)
22
Vertices: (0, 63)
5. (26, 22), (23, 5)
}
y
2
26 1 (23)
22 1 5
3
M}
,}
5 1 2}2, }2 2
2
2
9
y
2
6. (1, 9), (10, 22)
22
}}}
d 5 Ï(1 2 10)2 1 (9 2 (22))2
x
(6, 21)
}}
5 Ï(29)2 1 (11)2
}
12. (x 1 4)2 5 6(y 2 2)
}
5 Ï81 1 12 1 5 Ï202
1 1 1 10
9 1 (22)
h 5 24, k 5 2
2
11 7
M}
,}
5 1}
,}
2
2 22
2
The conic is a parabola with vertex (24, 2).
The parabola opens up.
7. y2 2 24x 5 0
4p 5 6
y2 5 24x
3
The conic is a parabola.
p 5 }2
4p 5 24
Focus: 1 24, 2 1 }2 25 1 24, 3}2 2
3
p56
1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The focus is (6, 0).
y
(24, )
y
2
1
32
(24, 2)
(0, 0)
1
(6, 0)
21
x
22
x
( y 2 7)2
(x 1 4)2
13. } 2 } 5 1
49
9
The conic is a hyperbola.
8. x2 1 y2 5 16
h 5 24
}
The conic is a circle with radius r 5 Ï16 5 4.
(0, 4) y
(24, 0)
21
1
a2 5 9
a53
b2 5 49
b57
The vertices are 3 units left and right of the center at
(27, 7) and (21, 7).
(4, 0)
(0, 0)
k57
Center: (24, 7)
x
y
(24, 14)
(0, 24)
2
(24, 7)
2
9. 64y 2 x 5 64
x2
y2 2 }
51
64
The conic is a hyperbola.
a2 5 1
2
b 5 64
a51
b58
2
(28, 0)
29
(0, 21)
y
(21, 7)
(27, 7)
(0, 1)
(8, 0)
x
(24, 0)
2
x
24
Vertices: (0, 61)
Algebra 2
Worked-Out Solution Key
n2ws-09b.indd 595
595
6/27/06 11:16:58 AM
Chapter 9,
continued
2
2
( y 2 2)
(x 2 8)
14. } 1 } 5 1
100
81
20. Ellipse; center: (0,0); vertex: (0, 6); co-vertex: (23, 0)
The conic is an ellipse.
h 5 8, k 5 2
Center: (8, 2)
a2 5 100
a 5 10
b2 5 81
b59
The vertices are 10 units above and below the center at
(8, 12) and (8, 28).
The co-vertices are 9 units left and right of the center at
(17, 2) and (21, 2).
y
(8, 12)
y2
2
x
The form of the equation is }2 1 }2 5 1.
a 5 6,
a2 5 36
b53
b2 5 9
2
a
b
y2
x
The equation is }
1}
5 1.
36
9
21. Ellipse; vertices: (21, 4) and (7, 4); foci: (1, 4) and (5, 4)
(x 2 h)2
( y 2 k)2
a
b
The form of the equation is }
1}
5 1.
2
2
The center is the midpoint of the vertices at
21 1 7 4 1 4
,}
5 (3, 4).
1}
2
2 2
8
h 5 3, k 5 4
(21, 2)
The vertices are 4 units from the center, so a 5 4 and
a2 5 16.
(8, 2)
(17, 2)
The foci are 2 units from the center, so c 5 2 and c2 5 4.
x
22
c2 5 a2 2 b2
4 5 16 2 b2
(8, 28)
b2 5 16 2 4 5 12
( y 2 5)2
15. } 2 (x 1 3)2 5 1
9
(x 2 3)2
(23, 5)
(24, 5)
Center: (23, 5)
a53
b2 5 1
b51
y
(23, 8)
h 5 23, k 5 5
22. Hyperbola; vertices: (0, 66); foci: (0, 69)
(22, 5)
1
The vertices are 3 units above and below the center at
(23, 8) and (23, 2).
16. Parabola; vertex: (0, 0); directrix: x 5 26
2
The form of the equation is y 5 4px
p56
The equation is y2 5 24x.
17. Parabola; vertex: (22, 21); focus: (22, 5)
The form of the equation is (x 2 h)2 5 4p ( y 2 k).
h 5 22,
a
c2 5 81
c59
2
2
2
c 5a 1b
b2 5 81 2 36 5 45
y2
x2
The equation is }
2}
5 0.
45
36
23. Hyperbola: vertex: (2, 25); focus: (21, 25);
center: (5, 25)
(x 2 h)2
( y 2 k)2
a
b
The form of the equation is }
2}
5 1.
2
2
h 5 5, k 5 25
k 5 21
p 5 5 2 (21) 5 6
2
The equation is (x 1 2) 5 24( y 1 1).
18. Circle; center: (0, 0); passes through (25, 2)
x2 1 y2 5 r2
(25)2 1 (2)2 5 r2
29 5 r2
The equation is x2 1 y2 5 29.
19. Circle; center: (1, 24); radius 5 6
h 5 1, k 5 24
The equation is (x 2 1)2 1 ( y 1 4)2 5 36.
596
x
x2
b
a2 5 36
a56
2
(23, 2)
y2
The form of the equation is }2 2 }2 5 1.
The vertex is 3 units from the center, so a 5 3 and
a2 5 9.
The focus is 6 units from the center, so c 5 6 and
c2 5 36.
c2 5 a2 1 b2
36 5 9 1 b2
b2 5 36 2 9 5 27
(x 2 5)2
( y 1 5)2
The equation is }
2}
5 1.
27
9
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The conic is a hyperbola.
a2 5 9
( y 2 4)2
The equation is }
1}
5 1.
12
16
Algebra 2
Worked-Out Solution Key
n2ws-09b.indd 596
6/27/06 11:17:04 AM
Chapter 9,
continued
24. x 2 1 4y 2 2 6x 2 16y 1 21 5 0
29. y 2 2 16x2 1 14y 1 64x 2 31 5 0
A 5 1, B 5 0, C 5 4
A 5 216, B 5 0, C 5 1
2
B2 2 4AC 5 0 2 4(216)(1) 5 64
B 2 4AC 5 0 2 4(1)(4) 5 216
2
The conic is an ellipse because B 2 4AC < 0 and A Þ C.
The conic is a hyperbola because B2 2 4AC > 0.
x 2 1 4y2 2 6x 2 16y 1 21 5 0
y 2 2 16x 2 1 14y 1 64x 2 31 5 0
1 x 2 6x 2 1 4(y 2 4y) 5 221
2
2
(x2 2 6x 1 9) 1 4( y2 2 4y 1 4) 5 221
1 9 1 16
( y2 1 14y) 2 16(x2 2 4) 5 31
( y2 1 14y 1 49) 2 16(x2 2 4x 1 4) 5 31 1 49 2 64
(x 2 3)2 1 4( y 2 2)2 5 4
( y 1 7)2 2 16(x 2 2)2 5 16
(x 2 3)2
4
( y 1 7)2
} 2 (x 2 2)2 5 1
16
} 1 ( y 2 2)2 5 1
25. x 2 1 y 2 1 8x 1 12y 1 3 5 0
30. 4x 2 1 y 2 5 16
x1y52→y522x
A 5 1, B 5 0, C 5 1
2
4x 2 1 (2 2 x)2 5 16
B 2 4AC 5 0 2 4(1)(1) 5 24
2
The conic is a circle because B 2 4AC < 0, B 5 0, and
A 5 C.
2
2
4x 1 4 2 4x 1 x 2 5 16
5x 2 2 4x 2 12 5 0
2
x 1 y 1 8x 1 12y 1 3 5 0
(x2 1 8x) 1 ( y2 1 12y) 5 23
(x2 1 8x 1 16) 1 ( y2 1 12y 1 36) 5 23 1 16 1 36
(x 1 4)2 1 ( y 1 6)2 5 49
26. 4x2 2 9y 2 2 40x 1 64 5 0
A 5 4, B 5 0, C 5 29
B2 2 4AC 5 0 2 4(4)(29) 5 144
The conic is a hyperbola because B2 2 4AC > 0.
4x2 2 9y 2 2 40x 1 64 5 0
4(x 2 10x) 2 9y 5 264
2
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
4(x2 2 10x 1 25) 2 9y 2 5 264 1 100
4(x 2 5)2 2 9y 2 5 36
(x 2 5)2
9
y2
4
}2}51
(5x 1 6)(x 2 2) 5 0
5x 5 26 or x 2 2 5 0
6
x 5 2}5 or x 5 2
When x 5 2}5 : y 5 2 2 1 2}5 25 }
5
6
The solutions are 1 2}5 , }
and (2, 0).
52
6 16
31. x2 1 4y2 2 8y 5 4
y2 2 2y 2 8x 2 16 5 0
x2 1 4y2
3 (24)
2 8y 2 4 5 0
24y2 1 32x 1 8y 1 64 5 0
x2
1 32x
1 60 5 0
(x 1 2)(x 1 30) 5 0
x 5 22 or x 5 230
A 5 0, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(0)(1) 5 0
The conic is a parabola because B2 2 4AC 5 0.
y 2 2 16y 2 12x 1 40 5 0
y 2 2 16y 5 12x 2 40
y 2 2 16y 1 64 5 12x 2 40 1 64
( y 2 8)2 5 12x 1 24
( y 2 8)2 5 12(x 1 2)
2
28. 25x 1 4y 1 50x 2 24y 2 39 5 0
A 5 25, B 5 0, C 5 4
B2 2 4AC 5 0 2 4(25)(4) 5 2400
The conic is an ellipse because B2 2 4AC < 0 and A Þ C.
25x 2 1 4y 2 1 50x 2 24y 2 39 5 0
25(x2 1 2x) 5 4( y2 2 6y) 5 39
25(x 1 2x 1 1) 1 4( y2 2 6y 1 9) 5 39 1 25 1 36
2
25(x 1 1)2 1 4( y 2 3)2 5 100
(x 1 1)2
4
16
When x 5 2: y 5 2 2 (2) 5 0
27. y2 2 16y 2 12x 1 40 5 0
2
6
( y 2 3)2
25
}1 }51
When x 5 22:
2
When x 5 230:
2
(230)2 1 4y 2 2 8y 5 4
2
4y 2 8y 5 0
4y 2 2 8y 1 896 5 0
4y(y 2 2) 5 0
41 y2 2 2y 1 224 2 5 0
(22) 1 4y 2 8y 5 4
y 5 0 or y 5 2
No real solution
The solutions are (22, 0) and (22, 2).
32. 2x 2 1 y 2 1 2x 2 5 5 0
x2 1 y2 2 2x 2 3 5 0
2y2
2850
2( y2 2 4) 5 0
y2 5 4
y 5 62
When y 5 62: (62) 2 2 x 2 1 2x 2 5 5 0
2x 2 1 2x 2 1 5 0
2(x 2 2 2x 1 1) 5 0
2(x 2 1) 2 5 0
x51
The solutions are (1, 2) and (1, 22).
Algebra 2
Worked-Out Solution Key
n2ws-09b.indd 597
597
6/27/06 11:17:09 AM
Chapter 9,
continued
x 2 1 y 2 2 12x 2 6y 2 40 5 0
2x 2 2 y 2 2 10x 2 8y 1 44 5 0
222x 2 14y 1 4 5 0
33. Upright glass:
2
radius 5 1.5, r 5 2.25
An equation of the water’s surface with the glass upright
is x2 1 y2 5 2.25.
Tilted glass:
The water’s edge forms an ellipse.
2
7
2
y1}
2 6 2 1 ( y 2 3)2 5 85
1 2}
11
11
2
7
64
y2}
1 2}
11
11 2
896
121
49
121
1 ( y 2 3)2 5 85
4096
121
} y 2 1 } y 1 } 1 y 2 2 6y 1 9 2 85 5 0
4y2
1}
An equation is }
5 1.
9
9
49y 2 1 896y 1 4096 1 121y 2 2 726y 2 9196 5 0
170y 2 1 170y 2 5100 5 0
170 ( y 2 1 y 2 30) 5 0
170( y 1 6)( y 2 5) 5 0
y 5 26 or y 5 5
Because y < 0 for vertex D, use y 5 26:
34. Parabola with point (2104, 45)
The form of the equation is x2 5 4py.
When x 5 2104 and y 5 45:
(2104)2 5 4p(45)
10,816 5 4p(45)
10,816
7
4p 5 }
45
2
42
2
44
x 5 2}
(26) 1 }
5}
1}
5}
54
11
11
11
11
11
10,816
The vertex D should be (4, 26) to make ABCD
a rhombus.
y.
An equation is x 2 5 }
45
10,816
2. The form of the parabola’s equation is x 2 5 4py.
4p 5 }
45
When x 5 24 and y 5 212.5:
2704
45
p 5 } ø 60.1
(24)2 5 4p(12.5)
(24)2
The distance from the vertex to the focus is about 60.1.
The focus is 11.52 inches below the vertex of the
parabola. The focus is above the bottom edge of the
reflector, because 11.52 < 12.5.
1. Partial credit would be given, because the solution was
without error but incomplete. Part of the question was
not answered.
2. Full credit would be given, because the solution is
complete and correct.
Standardized Test Practice (pp. 676–677)
3.
y
B (40, 80)
A
(235, 35)
Distances between the points:
}}
2
}}}
}
BC 5 Ï(23 2 (25))2 1 (5 2 (24))2 5 Ï85
}}}
}
CD 5 Ï(25 2 4)2 1 (24 2 (25))2 5 Ï81
}}
x
}
Slope of AO:
}
AB 5 Ï(6 2 (23)) 1 (3 2 5) 5 Ï85
2
O (0, 0)
240
1. A(6, 3), B(23, 5), C(25, 24) and D(4, 25)
(55, 25)
20
}
DA 5 Ï(4 2 6)2 1 (25 2 3)2 5 Ï68
The vertex D(4, 25) can be moved so that
}
CD 5 DA 5 Ï85 .
This can be done by drawing two circles around the }
points A(6, 3) and C(25, 24), each with a radius of Ï85 ,
and finding their intersection, where x > 0 and y < 0.
The equations of the circles are:
(x 2 6)2 1 ( y 2 3)2 5 85
(x 1 5)2 1 ( y 1 4)2 5 85
35 2 0
m5}
5 21
235 2 0
1
2}
51
m
}
Midpoint of AO:
M1 }
,}
5 2} , }
2
2 2 1 2 22
}
Perpendicular bisector of AO:
235 1 0 35 1 0
1
35 35
y2}
5 1 x 2 1 2}
2
22
35
35
35
35
y5x1}
1}
2
2
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
p5}
5 211.52
(4)(212.5)
Standardized Test Preparation (p. 675)
598
2
Substitute for x.
1
a 5 }2 (6) 5 3, a2 5 9
3
9
1
b 5 }2 (3) 5 }2, b2 5 }4
x2
7
x 5 2}
y1}
11
11
y 5 x 1 35
Algebra 2
Worked-Out Solution Key
n2ws-09b.indd 598
6/27/06 11:17:14 AM