Question

(1 point) in 2003, the price of a certain automobile was approximately $31,500 with a depreciation of $1,980 per year.
a) write an equation to model the problem. let $p$ be the price of the car and let $t$ represent the number of years after 2003. for example, the year 2005 would be represented by $t = 2$. your answer must include $p=$.
answer:

b) after how many years will the cars value be $15,660? (note: include the units, in this case years.)
answer:

webwork, like mathematics, is case - sensitive so capitals matter. use $p$ and not $p$.

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Explanation:

Step1: Determine the initial - value and rate of change

The initial price of the car in 2003 ($t = 0$) is $P_0=31500$ and the rate of depreciation is $r = 1980$ per year. The general form of a linear - depreciation equation is $P=P_0-rt$. Substituting the values, we get $P = 31500-1980t$.

Step2: Solve for $t$ when $P = 15660$

Set $P = 15660$ in the equation $P = 31500-1980t$. Then, we have the equation $15660=31500 - 1980t$. First, add $1980t$ to both sides: $1980t+15660=31500$. Then subtract 15660 from both sides: $1980t=31500 - 15660$. So, $1980t=15840$. Divide both sides by 1980: $t=\frac{15840}{1980}=8$ years.

Answer:

a) $P = 31500-1980t$
b) 8 years