Question

(1 point)
a new car costs $33000. the cars value will depreciate over time, which means it will lose value. for tax purposes, depreciation is usually calculated linearly.
(a) if the car is worth $26400 after 3 years, write a formula for the value of the car after n years. note: webwork is case - sensitive here, so if you use the variable n in your answer you must keep it capitalized.
p =
(b) suppose you want to sell the car when its value reaches 15400. find the number of years after buying it this will occur. if your answer is not a whole number, please round your answer up.
n = years

Explanation:

Step1: Find the rate of depreciation

The initial value \(P_0 = 33000\). After \(N = 3\) years, \(P=26400\).
The formula for linear depreciation is \(P=P_0+kN\), where \(k\) is the rate of change.
Substitute \(P_0 = 33000\), \(N = 3\), and \(P = 26400\) into \(P=P_0+kN\):
\(26400=33000 + 3k\)
Solve for \(k\):
\(3k=26400 - 33000=-6600\)
\(k=\frac{-6600}{3}=-2200\)

Step2: Write the formula for part (a)

Using \(P=P_0+kN\) with \(P_0 = 33000\) and \(k=-2200\), the formula is \(P = 33000-2200N\)

Step3: Solve for \(N\) in part (b)

Set \(P = 15400\) in the formula \(P = 33000-2200N\)
\(15400=33000-2200N\)
\(2200N=33000 - 15400\)
\(2200N=17600\)
\(N=\frac{17600}{2200}=8\)

Answer:

a. \(P = 33000-2200N\)
b. \(N = 8\) years