A car purchased for [tex]$\$10,000$[/tex] depreciates under a straight-line method in the amount of [tex]$\[tex]$750$[/tex][/tex] each year. Which equation below best models this depreciation?

A. [tex]y = 10000 - 750x[/tex]
B. [tex]y = 10000x + 750[/tex]
C. [tex]y = 10000 + 750x[/tex]
D. [tex]y = 10000x - 750[/tex]



Answer :

Certainly! Let's analyze the problem step-by-step:

1. Initial Value: The car is purchased at a value of [tex]$10,000. This means that at the starting point (i.e., at \( x = 0 \) years), the value of the car \( y \) is $[/tex]10,000.

2. Depreciation Rate: The car depreciates by [tex]$750 every year. This implies that the value of the car decreases by $[/tex]750 for each year [tex]\( x \)[/tex].

3. Formulating the Equation:
- We need to express the value of the car [tex]\( y \)[/tex] after [tex]\( x \)[/tex] years.
- Initially, the value is [tex]$10,000. - For each year \( x \), we subtract $[/tex]750 from the initial value.

4. Mathematical Representation:
- The initial value is subtracted by [tex]\( 750 \times x \)[/tex] (since $750 depreciation multiplied by the number of years [tex]\( x \)[/tex]).

5. Putting It All Together:
- The equation that models the car's depreciation can be written as:
[tex]\[ y = 10000 - 750 x \][/tex]

6. Choosing the Correct Option:
- Among the given options, the equation that matches our derived formula is:
[tex]\[ \boxed{y = 10000 - 750 x} \][/tex]

Thus, the best equation that models the depreciation of the car is:
A. [tex]\( y = 10000 - 750 x \)[/tex]