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Answer 32 Questions

Content type
User Generated
Subject
Linear Algebra
School
Orange Coast College
Type
Homework
Rating
Showing Page:
1/4
1
1) Not Given
2) Answer: B
( )
( )
0
1
1
01
6
11
1
66
f
f
==
==
3) Answer:
1a
Explanation: As f increases by increasing value of x, this implies that a must be greater than 1.
Domain:
( )
,−
Range:
( )
0,
4) Answer: C
5) Not Given
6) Answer:
Domain g:
( )
,
Range:
( )
0,

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2
7) Answer: B
8) Answer: A
( )
0.08
1
t
P t ke=+
( )
0 1 29000
28999
Pk
k
= + =
=
( )
0.08
0.08
1 28999 72500
72499
28999
1 72499
ln
0.08 28999
11.45389278
t
t
P t e
e
t
t
= + =
=

=


=
9) Answer: C
10) Answer: B
( )
( )
0.8 16
3499.82608 3500
3500
16
1 18
N
e
+
==
11) Answer: D
( )
( )
0.14 9
15912.99089 159139 56100Se
==
12) Answer: D
To find k:
( )
6
6
9
7000 500
7000
14
500
1
ln 14
6
0.43
k
k
e
e
k
=
= =
=

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Unformatted Attachment Preview
1 1) Not Given 2) Answer: B 1 =1 60 1 1 f (1) = 1 = 6 6 f ( 0) = 3) Answer: a 1 Explanation: As f increases by increasing value of x, this implies that a must be greater than 1. Domain: ( −,  ) Range: ( 0,  ) 4) Answer: C 5) Not Given 6) Answer: Domain g: ( −,  ) Range: ( 0,  ) 2 7) Answer: B 8) Answer: A P ( t ) = 1 + ke0.08t P ( 0 ) = 1 + k = 29000  k = 28999 P ( t ) = 1 + 28999e0.08t = 72500 72499 28999 1  72499  t = ln   0.08  28999   t = 11.45389278  e0.08t = 9) Answer: C y0 = 900, k = 0.116 P ( t ) = 900e0.116t 10) Answer: N (16 ) = B 3500 = 3499.82608  3500 −0.8 16 1 + 18  e ( ) 11) Answer: D S ( 9 ) = 56100e−0.14(9) = 15912.99089  15913 12) Answer: D To find k: 7000 = 500e6 k 7000  e6 k = = 14 500 1  k = ln (14 )  0.439 6 3 Now, P (8) = 500e( 0.439)8  16757.61563 13) Answer: B 14) Answer: D 15) Answer: C 16) Answer: D 17) Answer: C 7 ln y − 13ln x = 7v − 13u 18) Not Given 19) Answer: When base is ‘e’ 20) Answer: B 21) Answer: B F ( 31) = 320 − 40 log 5 (125 ) = 320 − 40 log 5 ( 53 ) = 320 − 120 = 200 22) Answer: C 3 − r = ...
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