Question

The present cost of a car is $9,000. the cost of the car depreciates at the rate of $1000 per year. part

Answer 1

A. f(t)=9000-1000t; Since 9000 is the number we start on, that is written as the constant. And since the value decreases by $1000 each year, we can put the $9000 it costs currently and will decrease by $1000 per year (t), this can be written as the function f(t)=9000-1000t

B. $6000; Just plug 3 into the equation to find the answer.

f(t)=9000-1000(3)

f(t)=9000-3000

f(t)= 6000

It costs $6000 after three years

C. $3500; To solve this just subtract 500 from 9000 to get 8500, and plug this in as the constant in our function.

f(t)=8500-1000(5)

f(t)=8500-5000

f(t)=3500

The answer is $3500