Question 1125094: At the beginning of 2010 Chasity's car was worth $10,000, but the value of her car decreases exponentially. She notices that the value of her car decreases by 20% every 3 years.
a. What is the 3-year growth factor for the value of Chasity's car?
b. What is the 1-year growth factor for the value of Chasity's car?
c. Write a function
f
f
that determines the value of Chasity's car (in dollars) in terms of the number of years t since the beginning of 2010.
f(t)=
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! car was worth 10,000 in 2010.
value of the car decreases by 20% every 3 years.
formula to use is f = p * (1 + r) ^ n
f is the future value
p is the present value
r is the interest rate per time period.
n is the numgber of time period.
p is the present value = 10,000
f is the future value = 10,000 - 20% of 10,000 = 8,000
n is equal to 3 years.
you want to find r, which is the average growth factor each year.
formula becomes 8000 = 10,000 * (1 + r) ^ 3
divide both sides of this equation by 10,000 to get 8000 / 10,000 = (1 + r) ^ 3
take the cube root of both sides of the equation to get .8 ^ (1/3) = 1 + r
subtract 1 from both sides of the eqution to get .8 ^ (1/3) - 1 = r
solve for r to get r = -.0716822333.
that means that the car loses 7.16822333% of its value each year.
the 1 year growth factor is therefore 1 - .0716822333 = .9283177667.
the 3 year growth factor is .8 which is equal to .9283177667 ^ 3.
the general formula to use is f = p * (1 - .0716822333) ^ t
f is the future value
p is the present value
r is the interest rate per year.
t is the number of years.
since the present value is 10,000 in 2010, then the formula becomes:
f = 10,000 * (1 = .0716822333) ^ t.
t becomes the number of years since 2010.
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