1.) The function f (x) = 12,500 (0.87) ^x models the value of a car x years after it is purchased.

How does the average rate of change between years 11 and 15 compare to the average rate of change between years 1 and 5?

A. The average rate of change between years 11 and 15 is about 2 times the rate between years 1 and 5.

B. The average rate of change between years 11 and 15 is about 3 times the rate between years 1 and 5.

C. The average rate of change between years 11 and 15 is about 13 the rate between years 1 and 5.

D. The average rate of change between years 11 and 15 is about 14 the rate between years 1 and 5.

2. A population of beavers decreases exponentially at a rate of 7.5% per year.

What is the equivalent monthly rate to the nearest hundredth of a percent?

A. 0.65%

B. 0.81%

C. 0.91%

D. 2.37%

1] Given that the value of x has been modeled by f (x) = 12500 (0.87) ^x, then:

the rate of change between years 1 and 5 will be:

rate of change is given by:

[f (b) - f (a) ] / (b-a)

thus:

f (1) = 12500 (0.87) ^1=10875

f (5) = 12500 (0.87) ^5=6230.3

rate of change will be:

(6230.3-10875) / (5-1)

=-1161.2

rate of change in years 11 to 15 will be:

f (11) = 12500 (0.87) ^11=2701.6

f (15) = 12500 (0.87) ^15=1,547.74

thus the rate of change will be:

(1547.74-2701.6) / (15-11)

=-288

dividing the two rates of change we get:

-288/-1161.2

-=1/4

comparing the two rate of change we conclude that:

The average rate of change between years 11 and 15 is about 1/4 the rate between years 1 and 5.

The answer is D]

2] Given that the population of beavers decreases exponentially at the rate of 7.5% per year, the monthly rate will be:

monthly rate = (n/12)

where n is the number of months

=7.5/12

=0.625

This is approximately equal to 0.65%. The correct answer is A. 0.65%