Document 1804562

Chapter 9,
continued
2. G;
5. D;
88.4x 2 2 49.7y 2 2 4390 5 0
Shaded area 5
88.4x2 2 49.7y2 5 4390
Area of large square 2 Area of small square
2
2
}2}ø1
28 5 4y2 2 4x2
y2
y2
88
x2
49.7
28 5 (2y) 2 (2x)
So, mirror B represents a hyperbola.
x2
15}
2}
7
7
y2
x2
The equation }
2}
5 1 describes the possible values of
7
7
x2 1 4y 2 5 16
6.
2
29x 2 2 36y 2
3 (29)
2
9x2 1
9x 1 y 1 8y 5 20
5 2144
y 2 1 8y 5 20
235y 2 1 8y 5 2124
x and y.
2
3. D;
35y 2 8y 2 124 5 0
The light beam originates at (0, 9) and travels towards the
focus, (12, 0).
920
3
5 2}4
m5}
0 2 12
62
3
or
y2250
or
y52
So, the y-coordinate of the uppermost intersection
point is 2.
3
y 5 2}4 x 1 9
3
An equation for the path of the light beam is y 5 2}4x 1 9.
The intersection is the solution of the system:
Chapter 9 Review (pp. 669–672)
1. A parabola is the set of all points in a plane equidistant
from a point called the focus and a line called the directrix.
3
2. The line segment joining the two co-vertices of an ellipse
2
3. The line segment joining the two vertices of a hyperbola is
y 5 2}4x 1 9
is the minor axis.
y
80
} 2 } 5 1 l 5x 2 2 4y 2 5 320
by indicating how wide or narrow the hyperbola is.
2
3
5x 2 4 2}4 x 1 9 5 320
1
2
1
2
2
54
9 2
x 2}
x 1 81 5 320
5x 2 4 }
4
16
2
9
5x 2 2 }4 x2 1 54x 2 324 5 320
11
4
}x 2 1 54x 2 644 5 0
11x 2 1 216x 2 2576 5 0
}}
2216 6Ï2162 2 4(11)(22576)
x 5 }}}
2(11)
2216 6 400
x5}
22
x ø 8.4
or
the transverse axis.
4. The asymptotes of a hyperbola help in drawing the graph
3
When y 5 2}4 x 1 9:
x 5 228
Because x > 0, x ø 8.4.
3
y 5 2}4 (8.4) 1 9 5 2.7
At (8.4, 2.7), light will be reflected to the focus
at (212, 0).
4. G;
The vertices (66, 0) are the closest points to the origin. So,
the shortest possible horizontal distance you could be from
the jet when you first hear the sonic boom is 6 miles away.
5. (26, 25), (2, 23)
}}}
d 5 Ï(26 2 2)2 1 (25 2 (23))2
}}
}
}
}
5 Ï(28)2 1 (22))2 5 Ï64 1 4 5 Ï68 5 2Ï17
1 26 1 2
25 1 (23)
2
24
,}
5 1}
, } 5 (22, 24)
M }
2
2
2 2 2
28
6. (22, 5), (1, 9)
}}
d 5 Ï(22 2 1)2 1 (5 2 9)2
}}
}
}
5 Ï (23)2 1 (24)2 5 Ï9 1 16 5 Ï25 5 5
,}
5 2}, 7
M 1}
2
2 2 1 2 2
22 1 1 5 1 9
1
7. (23, 24) , (2, 5)
}}
d 5 Ï(23 2 2)2 1 (24 2 5)2
}}
}
}
5 Ï (25)2 1 (29)2 5 Ï25 1 81 5 Ï106
M 1}
,}
5 2} , }
2
2 2 1 2 22
23 1 2 24 1 5
1 1
8. (2200, 40) , (30, 2140)
}}}
d 5 Ï(2200 2 30)2 1 (40 2 (2140))2
}}
}}
5 Ï(2230)2 1 (180)2 5 Ï52,900 1 32,400
}
5 Ï 85,300 ø 292
The distance between the skydivers is about 292 yards.
528
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
64
35y 1 62 5 0
y 5 2}
35
y 2 9 5 2}4 (x 2 0)
2
(35y 1 62)( y 2 2) 5 0
Chapter 9,
continued
9. x2 5 16y
17. 3x 2 1 3y 2 5 147
y
4p 5 16
x 2 1 y 2 5 49
(0, 4)
r 2 5 49
2
p54
x
2
The focus is (0, 4).
y 5 24
The directrix is y 5 24.
2
(27, 0)
r57
10. y 5 26x
y
4p 5 26
18. (5, 9)
3
3
p 5 2}2
(
3
22 ,
(5)2 1 (9)2 5 106
)
The standard form of the equation is x 2 1 y 2 5 106.
0
The focus is 1 2}2 , 0 2.
3
3
x
3
x 52
3
The directrix is x 5 }2 .
19. (28, 2)
(28)2 1 (2)2 5 68
The standard form of the equation is x 2 1 y 2 5 68.
The axis of symmetry is
horizontal, y 5 0.
20. (27, 24)
11. x2 1 4y 5 0
(27)2 1 (24)2 5 65
y
x2 5 24y
2
4p 5 24
The standard form of the equation is x 2 1 y 2 5 65.
y51
3
p 5 21
21. 16x 2 1 25y 2 5 400
a 2 5 25
a55
The directrix is y 5 1.
b 5 16
The axis of symmetry is vertical, x 5 0.
Vertices: (5, 0) and
(25, 0)
2
12. Focus: (25, 0)
p 5 25
b54
y
(0, 4)
y2
x2
}1}51
16
25
x
(0, 21)
The focus is (0, 21).
1
(23, 0)
(3, 0)
x
21
(25, 0)
(5, 0)
(0, 24)
Co-vertices: (0, 4) and
(0, 24)
The equation is y2 5 4(25)x, or y2 5 220x.
c2 5 a 2 2 b2 5 25 2 16 5 9
13. Focus: (0, 3)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x
(0, 27)
2
c53
p53
Foci: (3, 0) and (23, 0)
2
The form of the equation is x 5 4py.
22. 81x 2 1 9y 2 5 729
2
The equation is x 5 12y.
p56
2
The form of the equation is y2 5 4px.
2
The equation is y 5 24x.
2
15. x 1 y 5 81
(0, 9)
2
r 5 81
y
a59
b53
}
}
c 5 Ï 72 5 6Ï 2
(23, 0)
2
2
x
(0, 26
2)
(0, 2
x 2 1 y 2 5 40
10 )
x
Foci: 1 0, 66Ï 2 2
23. 64x 2 1 36y 2 5 2304
a2 5 64
2
y
(2
r 5 40
10, 0)
b 5 36
a58
b56
}
x
22
(0, 2
2
(26, 0)
7)
(6, 0)
2
x
c 2 5 64 2 36 5 28
}
c 5 Ï 28 5 2Ï 7
2
r 5 Ï40 5 2Ï 10
(0, 8) y
y2
x2
}1}51
64
36
}
2
(0, 29)
}
(9, 0)
(0, 29)
2
(3, 0)
24
Co-vertices: (63, 0)
22
}
a 5 81
b2 5 9
2)
Vertices: (0, 69)
(29, 0)
x 5 40 2 y
(0, 6
c 2 5 a2 2 b2 5 81 2 9 5 72
r59
2
y
(0, 9)
y2
x2
}1}51
81
9
14. Directrix: x 5 26
16.
(7, 0)
2
The axis of symmetry is
vertical, x 5 0.
2
y
(0, 7)
(0, 22
7)
(0, 28)
Vertices: (0, 68)
Co-vertices: (66, 0)
(22
10, 0)
}
(0, 22
10 )
Foci: (0, 62Ï7 )
Algebra 2
Worked-Out Solution Key
529
Chapter 9,
continued
24. Vertex: (26, 0)
29. Foci: (0, 65)
Vertices: (0, 62)
a52
a2 5 4
c55
c2 5 25
2
2
c 5 a 1 b2
25 5 4 1 b2
b 2 5 25 2 4 5 21
Co-vertex: (0, 23)
2
a56
a 5 36
b53
b2 5 9
y2
x2
An equation is }
1}
5 1.
9
36
25. Vertex: (0, 28)
y2
Focus: (0, 5)
a2 5 64
30. Foci: (69, 0)
c 2 5 25
c55
Vertices: (64, 0)
c 2 5 a2 2 b2
2
c 5 a 1 b2
y2
x2
} 5 1.
An equation is }
1
64
39
4
y2
x2 2 }
51
9
b2 5 9
2
(
22
31. 4x2 1 9y2 1 40x 1 72y 1 208 5 0
10, 0)
(1, 0)
(0, 23)
(21, 0)
2
c 5 a 1 b 5 1 1 9 5 10
x
The conic is an ellipse because B 2 2 4AC < 0 and
A Þ C.
c 5 Ï10
Vertices: (61, 0)
4x2 1 9y2 1 40x 1 72y 1 208 5 0
}
Foci: (6Ï 10 , 0)
4(x2 1 10x) 1 9(y2 1 8y) 5 2208
Asymptotes: y 5 63x.
27. 4x2 2 16y2 5 64
4(x2 1 10x 1 25) 1 9(y 2 1 8y 1 16)
(22
y2
4
}2}51
5, 0)4
y
a54
b2 5 4
b52
5 2208 1 100 1 144
(0, 2)
(24, 0)
a 2 5 16
4(x 1 5)2 1 9(y 1 4)2 5 36
(4, 0)
(0, 22)
(2
5, 0)
}1}51
h 5 210, k 5 24
a2 5 9,
}
a53
c 5 Ï20 5 2Ï 5
b 5 4,
Vertices: (64, 0)
The center is (210, 24).
2
}
Foci: (62Ï 5 , 0)
28. 100y 2 2 36x 2 5 3600
y2
36
x2
100
(0, 2
}2}51
a2 5 36
2
b 5 100
34 )
8 (0, 6)
x
b 5 10
}
}
c 5 Ï136 5 2Ï 34
Vertices: (0, 66)
}
Foci: (0, 62Ï 34 )
3
Asymptotes: y 5 6}5 x
Algebra 2
Worked-Out Solution Key
(0,22
2
(10, 0)
212
c 2 5 36 1 100 5 136
The co-vertices are 2 units above and below the center at
(210, 22) and (210, 26).
y
(210, 0)
a56
b52
The vertices are 3 units left and right of the center at
(213, 24) and (27, 24).
1
Asymptotes: y 5 6}2 x
530
(y 1 4)2
4
(x 1 5)2
9
x
26
c 2 5 16 1 4 5 20
}
A 5 4, B 5 0, C 5 9
B 2 2 4AC 5 0 2 4(4)(9) 5 2144
}
x2
16
y2
2
x
An equation is }
2}
5 1.
65
16
10, 0 )
(2
b53
2
b2 5 81 2 16 5 65
y
(0, 3)
a51
2
81 5 16 1 b2
26. 9x 2 2 y 2 5 9
a 51
c2 5 81
c59
b2 5 64 2 25 5 39
2
a2 5 16
a54
25 5 64 2 b2
34 )
(0, 26)
(210, 22)
22
(210, 24)
(27, 24)
(213, 24) (210, 26)
y
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
a58
x2
An equation is }
2}
5 1.
21
4
Chapter 9,
continued
32. y 2 2 10y 2 8x 1 1 5 0
34. x2 1 y2 1 4x 2 14y 1 17 5 0
A 5 0, B 5 0, C 5 1
A 5 1, B 5 0, C 5 1
2
B 2 2 4AC 5 0 2 4(1)(1) 5 24
B 2 4AC 5 0 2 4(0)(1) 5 0
2
The conic is a parabola because B 2 4AC 5 0.
y2 2 10y 2 8x 1 1 5 0
The conic is a circle because B2 2 4AC < 0, b 5 0, and
A 5 C.
x2 1 y2 1 4x 2 14y 1 17 5 0
y2 2 10y 5 8x 2 1
( y 2 5)2 5 8x 1 24
(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
( y 2 5)2 5 8(x 1 3)
(x 1 2)2 1 ( y 2 7)2 5 36
y2 2 10y 1 25 5 8x 2 1 1 25
h 5 23, k 5 5
The radius is 6.
The vertex is at (23, 5).
h 5 22, k 5 7
4p 5 8
The center is (22, 7).
y
p52
The paprabola opens to the right, so the focus is 2 units
to the right of the vertex at (21, 5).
(22, 7)
y
4
(23, 5)
2
1
35. y2 5 4x lx 5 } y2
4
x
4
2x 2 5y 5 28
33. 9x2 2 y2 2 18x 2 4y 2 5 5 0
21 }4 y2 2 2 5y 5 28
1
A 5 9, B 5 0, C 5 21
B 2 2 4AC 5 0 2 4(9)(21) 5 36
1
2
} y2 2 5y 1 8 5 0
2
The conic is a hyperbola because B 2 4AC > 0.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
9x2 2 y2 2 18x 2 4y 2 5 5 0
9(x2 2 2x) 2 (y2 1 4y) 5 5
9(x 2 2x 1 1) 2 (y2 1 4y 1 4) 5 5 1 9 2 4
2
9(x 2 1)2 2 (y 1 2)2 5 10
(y 1 2)2
}2}51
10
10
}
(x 2 1)2
9
a5}
3
b2 5 10
b 5 Ï 10
x 1 y 2 14 5 0 l x 5 14 2 y
}
Ï10
Vertices: 16}
, 22
3
y
21
10
, 22
3
)
1
When y 5 8: x 5 }4 (8)2 5 16
36. x2 1 y2 2 100 5 0
}
1
y 5 8 or y 5 2
The solutions are (1, 2) and (16, 8).
}
Ï10
10
1
2
} ( y 2 8)(y 2 2) 5 0
When y 5 2: x 5 }4 (2)2 5 1
The center is at (1, 22).
a2 5 }
9
1
2
} ( y2 2 10y 1 16) 5 0
1
h 5 1, k 5 22
(1 2
x
22
(21, 5)
(14 2 y)2 1 y2 2 100 5 0
2
196 2 28y 1 y2 1 y2 2 100 5 0
(22 1
(1 1
10
, 22
3
(1, 22)
(22 2
2y2 2 28y 1 96 5 0
10 )
2(y2 2 14y 1 48) 5 0
)
10 )
x
2(y 2 8)(y 2 6) 5 0
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
The solutions are (6, 8) and (8, 6).
Algebra 2
Worked-Out Solution Key
531
Chapter 9,
37. 16x2 2 4y2
continued
6. (1, 9), (10, 22)
2 64 5 0
}}}
d 5 Ï(1 2 10)2 1 (9 2 (22))2
4x2 1 9y2 2 40x 1 64 5 0
36x2 2 9y2
2
2 144 5 0
}}
5 Ï (29)2 1 (11)2
2
4x 1 9y 2 40x 1 64 5 0
40x2
}
}
5 Ï 81 1 12 1 5 Ï202
2 40x 2 80 5 0
1 1 1 10
9 1 (22)
2
11 7
M}
,}
5 1}
,}
2
2 22
2
40 1 x2 2 x 2 2 2 5 0
40(x 1 1)(x 2 2) 5 0
2
7. y 2 24x 5 0
x 5 21 or x 5 2
y2 5 24x
When x 5 21:
When x 5 2:
16(21)2 2 4y2 2 64 5 0
16(2)2 2 4y2 2 64 5 0
24y2 5 48
The conic is a parabola.
4p 5 24
24y2 5 0
y2 5 224
No real solution
The solution is (2, 0).
y50
p56
The focus is (6, 0).
y
Chapter 9 Test (p. 673)
2
1. (21, 5), (7, 3)
(0, 0)
(6, 0)
x
22
}}
d 5 Ï(21 2 7) 1 (5 2 3)
2
2
}}
5 Ï(28)2 1 (2)2
}
}
}
5 Ï64 1 4 5 Ï 68 5 2Ï17
1
8. x2 1 y2 5 16
}
2
21 1 7 5 1 3
M}
,}
5 (3, 4)
2
2
The conic is a circle with radius r 5 Ï16 5 4.
(0, 4) y
2. (4, 2),(8, 8)
}}
d 5 Ï(4 2 8)2 1 (2 2 8)2
}
1
(24, 0)
21
}
(4
, 0)
(0, 0)
x
}
5 Ï16 1 36 5 Ï52 5 2Ï13
M1 }
,}
5 (6, 5)
2
2 2
418 218
(0, 24)
2
3. (21, 26), (1, 5)
9. 64y 2 x 5 64
}}
d 5 Ï(21 2 1) 1 (26 2 5)
2
2
x2
y2 2 }
51
64
}}
5 Ï(22) 1 (211)
2
2
}
The conic is a hyperbola.
}
a2 5 1
}
5 Ï4 1 121 5 Ï125 5 5Ï5
21 1 1 26 1 5
b 5 64
1
d 5 Ï(2 2 3) 1 (25 2 1)
2
2
5 Ï1 1 36 5 Ï 37
M1 }
,}
5 } , 22 2
2
2 2 12
5
}}}
d 5 Ï(26 2 (23)) 1 (22 2 5)
2
2
}}
5 Ï(23)2 1 (27)2
}
5 Ï9 1 49 5 Ï 58
22 1 5
3
M}
,}
5 1 2}2 , }2 2
2
2
532
Algebra 2
Worked-Out Solution Key
a 59
a53
b2 5 1
b51
Vertices: (0, 63)
5. (26, 22), (23, 5)
2
x
(0, 21)
b58
The conic is an ellipse.
}
26 1 (23)
(8, 0)
29
y2
}}
1
(28, 0)
x2 1 }
51
9
5 Ï(21)2 1 (26)2
}
y
(0, 1)
10. 18x2 1 2y2 5 18
}}
2 1 3 25 1 1
2
Vertices: (0, 61)
4. (2, 25), (3, 1)
}
a51
2
M1 }
,}
5 0, 2}2 2
2
2 2 1
2
2
9
(0, 3)
(21, 0)
1
y
(1, 0)
x
22
Co-vertices: (61, 0)
(0, 23)
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
}}
5 Ï(24)2 1 (26)2
Chapter 9,
continued
11. (x 2 6)2 1 ( y 1 1)2 5 36
}
The conic is a circle with radius r 5 Ï 36 5 6 and center
(6, 21).
y
2
22
x
(6, 21)
12. (x 1 4)2 5 6(y 2 2)
h 5 24, k 5 2
( y 2 2)2
(x 2 8)2
14. } 1 } 5 1
100
81
The conic is an ellipse.
h 5 8, k 5 2
Center: (8, 2)
a2 5 100
a 5 10
2
b 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
The conic is a parabola with vertex (24, 2).
The parabola opens up.
(8, 12)
8
4p 5 6
(21, 2)
3
p 5 }2
(8, 2)
(17, 2)
Focus: 1 24, 2 1 }2 25 1 24, 3}2 2
3
1
x
22
y
(8, 28)
(24, )
1
32
(24, 2)
2
1
21
(x 1 4)2
9
( y 2 5)
15. } 2 (x 1 3)2 5 1
9
x
h 5 23, k 5 5
( y 2 7)2
49
Center: (23, 5)
13. } 2 } 5 1
The conic is a hyperbola.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
h 5 24
The conic is a hyperbola.
k57
a2 5 9
a53
b2 5 1
b51
(23, 5)
(24, 5)
(22, 5)
2
(23, 2)
a2 5 9
a53
b2 5 49
16. Parabola; vertex: (0, 0); directrix: x 5 26
b57
The form of the equation is y2 5 4px
p56
The vertices are 3 units left and right of the center at
(27, 7) and (21, 7).
y
The equation is y2 5 24x.
The form of the equation is (x 2 h)2 5 4p ( y 2 k).
(24, 7)
(21, 7)
24
x
17. Parabola; vertex: (22, 21); focus: (22, 5)
(24, 14)
(24, 0)
1
The vertices are 3 units above and below the center at
(23, 8) and (23, 2).
Center: (24, 7)
(27, 7)
y
(23, 8)
2
x
h 5 22,
k 5 21
p 5 5 2 (21) 5 6
The equation is (x 1 2)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.
Algebra 2
Worked-Out Solution Key
533
Chapter 9,
continued
20. Ellipse; center: (0,0); vertex: (0, 6); co-vertex: (23, 0)
y
2
2
A 5 1, B 5 0, C 5 4
x
The form of the equation is }2 1 }2 5 1.
a2 5 36
b53
b2 5 9
2
B2 2 4AC 5 0 2 4(1)(4) 5 216
The conic is an ellipse because B2 2 4AC < 0 and A Þ C.
x 2 1 4y2 2 6x 2 16y 1 21 5 0
y2
x
The equation is }
1}
5 1.
36
9
1 x2 2 6x 2 1 4(y2 2 4y) 5 221
21. Ellipse; vertices: (21, 4) and (7, 4); foci: (1, 4) and (5, 4)
2
(x2 2 6x 1 9) 1 4( y2 2 4y 1 4) 5 221
(x 2 h)
( y 2 k)
a
b
1 9 1 16
(x 2 3)2 1 4( y 2 2)2 5 4
2
The form of the equation is }
1}
5 1.
2
2
(x 2 3)2
4
} 1 ( y 2 2)2 5 1
The center is the midpoint of the vertices at
25. x 2 1 y 2 1 8x 1 12y 1 3 5 0
21 1 7 4 1 4
,}
5 (3, 4).
1}
2
2 2
A 5 1, B 5 0, C 5 1
h 5 3, k 5 4
B2 2 4AC 5 0 2 4(1)(1) 5 24
The vertices are 4 units from the center, so a 5 4 and
a2 5 16.
The conic is a circle because B2 2 4AC < 0, B 5 0, and
A 5 C.
The foci are 2 units from the center, so c 5 2 and c2 5 4.
c2 5 a2 2 b2
x2 1 y2 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
2
4 5 16 2 b
b2 5 16 2 4 5 12
(x 1 4)2 1 ( y 1 6)2 5 49
( y 2 4)2
(x 2 3)2
The equation is }
1}
5 1.
12
16
26. 4x2 2 9y 2 2 40x 1 64 5 0
A 5 4, B 5 0, C 5 29
22. Hyperbola; vertices: (0, 66); foci: (0, 69)
y2
B2 2 4AC 5 0 2 4(4)(29) 5 144
x2
b
The form of the equation is }2 2 }2 5 1.
a
a56
a2 5 36
c59
c2 5 81
The conic is a hyperbola because B2 2 4AC > 0.
4x2 2 9y 2 2 40x 1 64 5 0
4(x2 2 10x) 2 9y 2 5 264
4(x2 2 10x 1 25) 2 9y 2 5 264 1 100
c2 5 a2 1 b2
4(x 2 5)2 2 9y 2 5 36
b2 5 81 2 36 5 45
y2
36
(x 2 5)2
9
x2
45
y2
4
}2}51
The equation is } 2 } 5 0.
27. y2 2 16y 2 12x 1 40 5 0
23. Hyperbola: vertex: (2, 25); focus: (21, 25);
A 5 0, B 5 0, C 5 1
center: (5, 25)
2
2
(x 2 h)
( y 2 k)
a
b
B2 2 4AC 5 0 2 4(0)(1) 5 0
2}
5 1.
The form of the equation is }
2
2
The conic is a parabola because B2 2 4AC 5 0.
h 5 5, k 5 25
y 2 2 16y 2 12x 1 40 5 0
y 2 2 16y 5 12x 2 40
The vertex is 3 units from the center, so a 5 3 and
a2 5 9.
y 2 2 16y 1 64 5 12x 2 40 1 64
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
( y 1 5)2
(x 2 5)2
The equation is }
2}
5 1.
27
9
( y 2 8)2 5 12x 1 24
( y 2 8)2 5 12(x 1 2)
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
( y 2 3)2
25
}1 }51
534
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
a 5 6,
a
b
24. x 2 1 4y 2 2 6x 2 16y 1 21 5 0
Chapter 9,
continued
29. y 2 2 16x2 1 14y 1 64x 2 31 5 0
33. Upright glass:
A 5 216, B 5 0, C 5 1
B2 2 4AC 5 0 2 4(216)(1) 5 64
The conic is a hyperbola because B2 2 4AC > 0.
y 2 2 16x 2 1 14y 1 64x 2 31 5 0
( 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
( y 1 7)2 2 16(x 2 2)2 5 16
( y 1 7)2
} 2 (x 2 2)2 5 1
16
30. 4x 2 1 y 2 5 16
1
a 5 }2 (6) 5 3, a2 5 9
4x 1 (2 2 x) 5 16
(2104)2 5 4p(45)
2
5x 2 4x 2 12 5 0
10,816 5 4p(45)
(5x 1 6)(x 2 2) 5 0
10,816
4p 5 }
45
5x 5 26 or x 2 2 5 0
10,816
6
An equation is x 2 5 }
y.
45
x 5 2}5 or x 5 2
10,816
When x 5 2}5 : y 5 2 2 1 2}5 25 }
5
6
6
16
4p 5 }
45
2704
When x 5 2: y 5 2 2 (2) 5 0
The solutions are 1 2}5 , }
and (2, 0).
52
6 16
31. x2 1 4y2 2 8y 5 4
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
3 (24)
24y 1 32x 1 8y 1 64 5 0
(x 1 2)(x 1 30) 5 0
x 5 22 or x 5 230
When x 5 22:
When x 5 230:
(22)2 1 4y 2 2 8y 5 4
(230)2 1 4y 2 2 8y 5 4
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
y 5 0 or y 5 2
22
023
23
220
2
5}
51
mXY 5 }
326
23
1 60 5 0
2
325
5}
5 21
mWX 5 }
624
2
2
No real solution
The solutions are (22, 0) and (22, 2).
2
The distance from the vertex to the focus is about 60.1.
1. C;
2 8y 2 4 5 0
1 32x
p5}
ø 60.1
45
TAKS Practice (pp. 676–677)
y2 2 2y 2 8x 2 16 5 0
x
4y2
x2
1}
5 1.
An equation is }
9
9
When x 5 2104 and y 5 45:
4x 2 1 4 2 4x 1 x 2 5 16
2
9
The form of the equation is x2 5 4py.
2
x2 1 4y2
3
1
b 5 }2(3) 5 }2, b2 5 }4
34. Parabola with point (2104, 45)
x1y52ly522x
2
radius 5 1.5, r2 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
32. 2x 1 y 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).
5}
5 21
mYZ 5 }
123
22
522
3
5 }3 5 1
mWZ 5 }
421
}
}
}
}
So, WX and YZ are parallel and XY and WZ are parallel
}
because their slopes are the same. The slopes of WX and
}
} } } }
XY are negative riciprocals, so WX > XY, XY > YZ,
} }
} }
YZ > WZ, and WZ > WX. This means, WXYZ has 4 right
angles.
}}
}
}
WX 5 Ï(6 2 4)2 1 (3 2 5)2 5 Ï22 1 (22)2 5 2Ï2
}}
}}
}
XY 5 Ï(3 2 6)2 1 (0 2 3)2 5 Ï(23)2 1 (23)2 5 3Ï2
Because WXYZ has two pairs of parallel sides, four
right angles, and adjacent sides are not congruent, it
is a rectangle.
2. G;
y
(24, 4)
(2, 4)
(24, 2)
(4, 1)
1
1
(22, 22)
The vertices are at (2, 4),
(4, 1), (1, 22), (22, 22),
(24, 2), and (24, 4).
x
(1, 22)
Algebra 2
Worked-Out Solution Key
535
Chapter 9,
continued
3. B;
9. B;
y
The polygon has one line of
symmetry.
1
3608
5 608 and the
The measure of the central angle is }
6
1
measure of the bisected angle is }2(608) 5 308. To find
the length of the apothem, use a special triangle.
1
x
308
4. H; When a point (x, y) is rotated 1808 about the origin,
the point (x, y) is mapped onto the point (2x, 2y). So,
M(21, 1) l M9(1, 21) and N(1, 3) l N9(21, 23).
y
x
608
4
x
}
So, x 5 4. The apothem is a 5 4Ï3 .
1
1
}
The area of the hexagon is about 166 square units.
1
1
10. J;
x
M9
13
P(blue) 5 }
20
N9
Number of blue marbles 5
P(blue) + Total number of marbles
Point N9 is in Quadrant III.
13
5. B;
The line of symmetry passes through the points (24, 3)
and (2, 23).
+ 100
5}
20
5 65
There are most likely 65 blue marbles in the bag.
23 2 3
26
5}
5 21
m5}
6
2 2 (4)
11. B;
y 2 y1 5 m(x 2 x1)
50 2 35
15
5}
5 0.3 5 30%
Percent of time saved 5 }
50
50
y 2 3 5 21(x 2 (24))
12. H; Because c represents the vertical translation of the
y 5 2x 2 1
The line of symmetry of the hexagon is y 5 2x 2 1.
6. G;
parabola, c must be 0 to be on the origin. So, c 5 0.
13. Let x 5 amount invested in 3% account.
Let y 5 amount invested in 5% account.
4(3 2 x) 2 3(2x2 2 9x 1 10) 5
x 1 y 5 4500
12 2 4x 2 6x2 1 27x 2 30 5
0.03x 1 0.05y 5 186
26x2 1 23x 2 18
23x 2 3y 5 213,500
3 (23)
3x 1 5y 5 18,600
3 100
2y 5 5100
7. D;
The dashed line passes through the points (0, 4)
and (3, 2).
224
2
5 2}3
m5}
320
1. 5x 1 24 5 11 2 2x
2
y 2 4 5 2}3 (x 2 0)
7x 5 213
13
x 5 2}
7
2
2
So, the graph represents the inequality y > 2}3x 1 4.
8. G;
So, $2550 should be invested into the account that pays
5% interest anually.
Cumulative Review Chs. 1–9 (pp. 678–679)
y 2 y1 5 m(x 2 x1)
y 5 2}3x 1 4
y 5 2550
1 24 0 11 2 21 2}
Check: 51 2}
72
72
13
13
65
5x 1 3y 5 220
5x 1 3(0) 5 220
5x 5 220
x 5 24
The x-intercept is 24.
Algebra 2
Worked-Out Solution Key
26
2}
1 24 0 11 1 }
7
7
103
7
103
7
}5}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Oliver saved 30% of the time by driving on the highway.
y 2 3 5 2(x 1 4)
536
3
A 5 }2aP 5 }2(4Ï3 )(6)(8) ø 166.28
N
M
308
2x
a
Chapter 9,
continued
2. {4x 2 7{ 5 13
}
Ïx 2 2 5 x 2 4
6.
x 2 2 5 (x 2 4)2
4x 5 20 or 4x 5 26
x 2 2 5 x2 2 8x 1 16
3
x 5 5 or x 5 2}2
2
x 2 8x 1 16 2 x 1 2 5 0
Check:
{4(5) 2 7{ 0 13
{41 2}2 22 7{ 0 13
{13{ 0 13
0 13
{2}
2 {
x2 2 9x 1 18 5 0
3
(x 2 3)(x 2 6) 5 0
26
13 5 13 x 5 3 or x 5 6
13 5 13 Check:
}
3. x2 2 12x 1 35 5 0
(x 2 7)(x 2 5) 5 0
x 5 7 or x 5 5
72 2 12(7) 1 35 0 0
49 2 84 1 35 0 0
52 2 12(5) 0 0
25 2 60 1 35 0 0
050
4. 2x2 2 5x 1 5 5 0
4x 5 8
(2 )
2 x
050
2 (25) 6 Ï(25) 2 4(2)(5)
2x 5 3
x 5 }}}
2(2)
3
x 5 }2
}
5 6 Ï 215
5 6 iÏ15
5}
x5}
4
4
Check:
Check:
432 2 5 0 3
82503
}
1 5 1 iÏ15 2 1 5 1 iÏ15 2
25
5iÏ 15
5 1 5iÏ15
2}
1500
21 }
22}
4
4
8
2
2}
25 }
1500
4
4
5
4
5iÏ15
4
353
}
}
25
4
}
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
x2 1 5x 1 6 5 3x2 1 x
2x2 2 4x 2 6 5 0
}
1
2 1
2
25
5iÏ 15
5 2 5iÏ15
1}
1500
21 }
22}
4
4
8
5 2 iÏ15 2
5 2 iÏ 15
25 }
1500
2}
4
4
2(x2 2 2x 2 3) 5 0
2(x 1 1)(x 2 3) 5 0
}
}
}5}
(x 1 3)(x 1 2) 5 x(3x 1 1)
050
}
x 5 21 or x 5 3
Check:
}
5iÏ 15
25
5iÏ15
5
}2}2}1}1500
4
4
4
4
21 1 3
3(21) 1 1
2
22
x3 1 3x2 2 18x 2 40 5 0
(x 1 2)(x 1 x 2 20) 5 0
x 5 22 or x 5 25 or x 5 4
Check:
40 5 40 4 1 3(4) 2 18(4) 0 40
64 1 48 2 72 0 40
3
2
40 5 40 6
10
3
5
3
5
3
5
}0}
}5}
x24
x23
9.
2x 2 3
x23
}125}
(x 2 3)1 }
1 (x 2 3)2 5 (x 2 3)1 }
x23 2
x23 2
2x 2 3
x24
x 2 4 1 2x 2 6 5 2x 2 3
(22)3 1 3(22)2 2 18(22) 0 40
28 1 12 1 36 0 40
40 5 40 (25)3 1 3(25)2 2 18(25) 0 40
2125 1 75 1 90 0 40
21
1
21 5 21 2
3
312
}0}
}0}
x3 1 3x2 2 18x 5 40
(x 1 2)(x 1 5)(x 2 4) 5 0
313
3(3) 1 1
21
21 1 2
}0}
050
5.
x
x12
x13
3x 1 1
8.
} 1 } 2 } 2 5iÏ 15 1 5 0 0
}
5 23
22x 5 23
}}
}
252
7. 4 x 2 5 5 3
2
}
Ï6 2 2 0 6 2 4
}
Ï4 0 2
1 Þ 21
Reject x 5 3. The solution is 6.
Check:
}
}
Ï3 2 2 0 3 2 4
}
Ï 1 0 21
x2750
x57
Check:
2(7) 2 3
723
724
723
}120}
3
4
11
4
}120}
11
4
11
4
}5}
Algebra 2
Worked-Out Solution Key
537
Chapter 9,
continued
22. f(x) 5 5(1.4x)
11. y 5 (x 1 1)2(x 2 2)
10. y 5 22x 1 7
The function is an example of exponential growth
because 1.4 > 1.
y
y
1
x
22
The function is an example of exponential decay
because 0.6 < 1.
2
x
22
}
24. f(x) 5 8e22x
The function is an example of exponential decay
because e22 < 1.
13. y 5 4ex
12. y 5 Ï x 1 4 1 3
23. f(x) 5 3(0.6)x
y
25. 3 ln x 2 ln 5 5 ln x3 1 ln 521
x3
5 ln }
5
y
26. log3 4 1 2 log3 7 5 log3 4 1 log3 72
(24, 3)
5 log3(4)(49)
5
x
21
x
21
5 log3 196
27. 5 log x 1 log y 2 3 log z 5 log x5 1 log y 1 log z23
x5y
3x 2 1
15. y 5 }
x2 2 9
14. y 5 ln(x 2 2)
5 log }
3
z
y
y
28. x 5 18, y 5 6
x53
x52
a
1
x
21
x 5 23
16. 2x2 2 20x 2 48 5 2(x2 2 10x 2 24)
5 2(x 1 2)(x 2 12)
17. 6x2 1 7x 2 20 5 (3x 2 4)(2x 1 5)
18. x3 1 8x2 2 4x 2 32 5 (x 1 8)(x2 2 4)
5 (x 1 8)(x 1 2)(x 2 2)
19. f (x) 5 6x 2 1
a
y 5 }x
1
21
29. x 5 5, y 5 215
y 5 }x
a
x
a
65}
18
(215) 5 }5
a 5 108
275 5 a
108
y5}
x
y 5 2}
x
75
30. x 5 6, y 5 9
a
y 5 }x
a
9 5 }6
54 5 a
54
y5}
x
y 5 6x 2 1
(x 2 5)(3)(x 1 7)
x 2 5 3x 1 21
31. } : }
5 }}
x 1 7 x2 2 25
(x 1 7)(x 1 5)(x 2 5)
6x 5 y 1 1
y11
x5}
6
3
x11
Write the inverse as y 5 }
.
6
x11
The inverse function is f (x) 5 }
.
6
5}
x15
2(x 1 4)(x 1 2)(x 2 3)
x14
2x 1 8
32. } 4 }
5 }}
(x 2 3)(x 1 4)
x23
x2 2 x 2 6
5 2(x 1 2) 5 2x 1 4
20. f (x) 5 x3 2 5
7
x23
33. } 1 }
x22
x15
3
y5x 25
x3 5 y 1 5
3}
3}
Write the inverse function as y 5 Ïx 1 5 .
3}
The inverse function is f (x) 5 Ï x 1 5 .
21. f (x) 5 x5
y 5 x5
x 5 y 1/5
Write the inverse function as y 5 x 1/5.
The inverse function is f (x) 5 x 1/5.
538
Algebra 2
Worked-Out Solution Key
(x 2 3)(x 2 2)
7(x 1 5)
x2 2 5x 1 6
7x 1 35
1 }}
5 }}
(x 1 5)(x 2 2)
(x 1 5)(x 2 2)
x 5 Ïx 1 5
1 }}
5 }}
(x 1 5)(x 2 2)
(x 1 5)(x 2 2)
x2 1 2x 1 41
x 1 3x 2 10
5}
2
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
1
Chapter 9,
continued
38. 4x2 2 16y2 2 56x 1 160y 2 268 5 0
34. (28, 5), (24, 21)
A 5 4, B 5 0, C 5 216
}}}
d 5 Ï(28 2 (24))2 1 (5 2 (21))2
B2 2 4AC 5 24(4)(216) 5 256
}
5 Ï(24)2 1 62
The conic is a hyperbola because B2 2 4AC > 0.
}
4x2 2 16y2 2 56x 1 160y 2 268 5 0
5 Ï16 1 36
4 (x2 2 14) 2 16 ( y2 2 10y) 5 268
}
5 Ï52
4 (x2 2 14x 1 49) 2 16 ( y2 2 10y 1 25)
}
5 2Ï13
1
5 268 1 196 2 400
2 1
28 1 (24) 5 1 (21)
212 4
M}
,}
5 }
, }2 5 (26, 2)
2
2
2
2
2
4 (x 2 7) 2 16(y 2 5) 5 64
(y 2 5) 2
(x 2 7) 2
}2}51
4
16
35. (3, 5), (8, 7)
}}
2
h 5 7, k 5 5
d 5 Ï(3 2 8)2 1 (5 2 7)2
Center: (7, 5)
}}
5 Ï(25)2 1 (22) 2
a2 5 16
}
5 Ï25 1 4
a54
2
b 54
b52
c2 5 a2 1 b2 5 20
}
5 Ï29
The vertices are 4 units left and right of the center, at
(3, 5) and (11, 5).
M 1}
,}
5 } , 62
2
2 2 12
318 517
11
y
36. (22, 7), (1, 14)
(7, 7)
}}
d 5 Ï(22 2 1)2 1 (7 2 14)2
(3, 5)
(11,5)
4
}}
5 Ï(23)2 1 (27)2
(7, 3)
}
5 Ï9 1 49
(7, 5)
x
22
}
5 Ï58
1
2 1
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
22 1 1 7 1 14
1 21
M }
,}
5 2}2 , }
2
2
2
39. y2 1 6x 1 4y 1 16 5 0
2
A 5 0, B 5 0, C 5 1
The conic is a parabola because B2 2 4AC 5 0.
37. x2 1 y 2 1 12x 2 4y 1 15 5 0
y2 1 6x 1 4y 1 16 5 0
A 5 1, B 5 0, C 5 1
y2 1 4y 5 26x 2 16
B2 2 4AC 5 24
2
y 1 4y 1 4 5 26x 2 16 1 4
The conic is a circle because B2 2 4AC < 0, B 5 0, and
A 5 C.
x2 1 y2 1 12x 2 4y 1 15 5 0
(x
(x
2
2
1 12x) 1 ( y 2 4y) 5 215
2
1 12x 1 36) 1 ( y 2 4y 1 4) 5 215 1 36 1 4
2
2
2
(x 1 6) 1 (y 2 2) 5 25
h 5 26, k 5 2
(y 1 2)2 5 26(x 1 2)
h 5 22, k 5 22
Vertex: (22, 22)
4p 5 26
3
p 5 2}2
3
7
The focus is 2}2 units left of the vertex, at 1 2}2, 22 2.
r2 5 25, r 5 5
y
Center: (26, 2)
y
6
2
(26, 2)
24
x
(2
7
,
2
22
)
2
x
(22, 22)
Algebra 2
Worked-Out Solution Key
539
Chapter 9,
continued
40. 2x2 1 3y2 1 4x 1 12y 2 14 5 0
42. g 5 games
A 5 2, B 5 0, C 5 3
r 5 rides
2
B 2 4AC 5 24(2)(3) 5 224
1.5g 1 2.5ra30
The conic is an ellipse because B2 2 4AC < 0 and A Þ C.
1.5ga30 2 2.5r
ga}1 2}2 r 2 1 1 }2
5
2
2x 1 3y 1 4x 1 12y 2 14 5 0
21 x 1 2x 2 1 31 y 1 4y 2 5 14
2
2
21 x2 1 2x 1 1 2 1 31 y2 1 4y 1 4 2 5 14 1 2 1 12
An inequality representing the number of games
and rides that can be purchased for $30 or less is
5
ga2}3r 1 20.
2(x 1 1)2 1 3(y 1 2)2 5 28
(y 1 2)2
(x 1 1)2
} 1 }5 1
28
14
}
3
Number of games
h 5 21, k 5 22
Center: (21, 22)
}
a2 5 14
a 5 Ï14
}
Ï
28
b2 5 }
3
28
b5 }
3
Ï
43.
14
}
3
}
The vertices are Ï 14 units left and right of the center, at
}
}
(21 2 Ï14 , 22) and (21 1 Ï14 , 22).
}
Ï
28
The co-vertices are }
units above and below the
3
}
1
Ï 2 1
28
3
)
4
(211
14, 22 )
(21, 22 2
28
3
5x 1 1300 2 2x 5 2500
3x 5 1200
Ï 2
28
y
22
Student
Number
Adult
price + of adults 1 price +
5
+
x
1
2
+
Number
Total
of students 5 income
(650 2 x) 5 2500
}
28
and 21, 22 2 }
.
center, at 21, 22 1 }
3
3
(21, 22 1
(12, 0)
Number of rides
}
c5
The ordered pairs (rides,
games) that use exactly $30
are (0, 20), (3, 15), (6, 10),
(9, 5) and (12, 0).
0 2 4 6 8 10 12 14 r
14
c2 5 14 2 }
5}
3
3
(212
g
21 (0, 20)
18
15
12
9
6
3
0
c2 5 a2 2 b2
28
5
ga2}3 r 1 20
x
14, 22 )
x 5 400
There were 400 adults and 650 2 400 5 250 students at
the basketball game.
44. T 5
F
0.05 0.98
(21, 22)
)
M1 5 TM0 5
41.
y
5
x
x 5 y 1 40
M2 5 TM1 5
2x 1 2y 5 380
2( y 1 40) 1 2y 5 380
5
2y 1 80 1 2y 5 380
4y 5 300
y 5 75
M3 5 TM2 5
x 5 (75) 1 40 5 115
The width of the garden is 75 feet and the length is
115 feet.
G
0.95 0.02
5
F
F
F
F
F
F
F G
GF G
G F G
GF G
G F G
G F G
G F G
M0 5
700
700
0.95 0.02
700
0.05 0.98
700
665 1 14
35 1 686
679
5
721
0.95 0.02
679
0.05 0.98
721
645 1 14
34 1 707
0.95 0.02
0.05 0.98
626 1 15
33 1 726
659
5
741
5
5
659
741
641
759
The matrices M1, M2, and M3 represent the distribution
of customers in company A and company B for the first
three months. Each month, the total number of customers
for company A is declining, while the total number of
customers for company B is increasing.
540
Algebra 2
Worked-Out Solution Key
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2
Chapter 9,
2 2
12 19
18 7
{
{
1
45. Area 5 6}
2
2 2
1
5 6}2 12 19
18 7
continued
1
1
1
{
{
1 2 2
1 12 19
1 18 7
1
5 6}2 [(38 1 36 1 84) 2 (342 1 14 1 24)]
1
5 6}2 (2222)
49.
t
ln s
t
ln s
5.75
6.13
60
70
7.14
20
30
40
50
6.42 6.62 6.83 6.97
80
90
100
7.35 7.65 7.97 8.27
8.27 2 5.75
ø 0.025
m5}
100 2 0
ln s 2 5.75 5 0.025(t 2 0)
ln s 5 0.025t 1 5.75
0.25mi 2
111 units + } ø 6.9 mi2
1 unit
1
10
The data pairs (t, ln s) lie close to a line.
5 111
2
0
2
s 5 e 0.025t 1 5.75
s 5 e5.751 e0.025 2t
The area of the surface of the lake is about
6.9 square miles.
46. h 5 216t2 1 v0t 1 h0
The ball’s initial velocity is v0 5 16 feet per second.
s 5 314(1.025)t
An exponential model for the data pairs (t, s) is
s 5 314(1.025)t.
Let h 5 0 and solve for t to find the time when the ball
hits the ground.
0 5 216t2 1 16t 1 9
}}
216 6 Ï(16)2 2 4(216)(9)
t 5 }}}
2(216)
216 6 28.8
232
tø}
t 5 1.4 or 20.4
Cumulative number of
stamp designs
The initial height of the ball is h0 5 9 feet.
ns
9.0
8.5
8.0
7.5
7.0
6.5
6.0
5.5 (0, 5.75)
5.0
0
0
20
40
(100, 8.27)
60
80
100 t
Years since 1904
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
The ball hits the ground in about 1.4 seconds.
47. The function that gives the sale price of the television
after $50 is subtracted and a 15% discount is applied is
f (t) 5 0.85(t 2 50).
If t 5 $480:
f (t) 5 0.85(480 2 50)
5 0.85(430)
5 365.5
The sale price of the television will be $365.50.
48. p 5 4500
r 5 0.0275
n 5 12
r nt
A 5 P1 1 1 }n 2
0.0275 12t
A 5 45001 1 1 }
12 2
A 5 4500(1.0023)12t
When t 5 5:
A 5 4500(1.0023)60 ø 5165.07
The balance after 5 years will be $5165.07.
Algebra 2
Worked-Out Solution Key
541