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Answers to Math
Applications
Math Applications for this
chapter are on pages 422–429.
The notes and solutions shown
below accompany the suggested
applications to assign with this
lesson.
3. a. 125 + 0.4x
b. 125 + 0.4(300) = 125 +
120 = 245
c. 25 + 0.6x
d. 25 + 0.6(200)
= 25 + 120 = 145
6. a. a 2
= $750,000(1.1)
= $825,000
b. a 3
= $825,000(1.1)
= $907,500
$905,000 < $907,500, so
Mari did not earn a bonus.
c. a 5
= $750,000(1.1) 5–1 =
$750,000(1.4641)
= 1,098,075
d. yes; a 2
would now equal
$900,000 instead of
$825,000 and would
increase each subsequent
number.
e. a n
= 1.1(a n–1
)
7. a. For each 1 cent change
in earnings per share, the
closing share price changes
0.68. The expression is
0.68x.
b. 0.68(9) = $6.12
15. a. Each term is the product of
the previous two terms, so
a n
= a n–2
• a n–1
b. 0.3 = 0.5(a n–2
)
0.6 = a n–2
1
24
1
16
1
36
1
40
2 de
3 7
f
396 Chapter 9 Sequences and Series