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%