Answer:
Option C. 7,000 − 36x
Explanation:
Let
y ----> depreciated value of the car
x---> rate of depreciation
t ----> the time in months
we know that
The linear equation that represent this situation is
y=7,000-xt
For
t=3 years
Convert to months
t=3*12=36 months
substitute
y=7,000-x(36)
y=7,000-36x