Q3. - Algebra 1

Question

Jaime bought a car in 2005 for $\mathrm{\backslash (}28500$. By 2008, the car was worth $\mathrm{\backslash )}23700$it. Based on a linear model, what will the value of the car be in 2012?

A. $\mathrm{\backslash (}17300$

B. $\mathrm{\backslash )}18100$

C. $\mathrm{\backslash (}17550$

D. $\mathrm{\backslash )}18475$

Step-by-Step Solution

Answer

The correct option is represented by A. 

Step 1. Determine the formula that can be used.

The standard form associated with the slope and the intercept form related to the line that is straight $y=mx+c$ in which the slope is represented by $m$ and the $y$-intercept is represented by $c$ can be used to write the required equation.

Step 2. Take the help of the obtained method.

Determine the depreciation associated with the car for each year. 

 $$\begin{gathered} \frac{28500-23700}{3}=\frac{4800}{3} \\ =1600 \end{gathered}$$ 

Take the help of obtained standard form represented by $y=mx+c$ in which the required values can be filled by using the obtained depreciation, the initial price of the car, $t$ which represents the time in years and $C$ which will represent the cost.

$C=2850+1600t$

Step 3. Write the conclusion.

The obtained equation is $C=2850+1600t$ so the required value can be determined by using the gap of years between 2005 and 2012 in place of time into the obtained equation so,

$$\begin{gathered} C=28500-1600\times 7 \\ =17300 \end{gathered}$$

The obtained value is the same as in option A.