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Answer :
We start with the general exponential depreciation model for the car’s value:
[tex]$$
P(x) = 7000 \times r^x,
$$[/tex]
where [tex]\( P(x) \)[/tex] is the value of the car after [tex]\( x \)[/tex] years, and [tex]\( r \)[/tex] is the annual depreciation factor.
After 2 years the car’s value is given as:
[tex]$$
P(2) = 7000 \times r^2 = 3937.50.
$$[/tex]
To find [tex]\( r \)[/tex], we solve for [tex]\( r^2 \)[/tex]:
[tex]$$
r^2 = \frac{3937.50}{7000} = 0.5625.
$$[/tex]
Taking the square root of both sides gives:
[tex]$$
r = \sqrt{0.5625} = 0.75.
$$[/tex]
Thus, the exponential model that represents the value of the car [tex]\( x \)[/tex] years after purchase is:
[tex]$$
P(x) = 7000 \times (0.75)^x.
$$[/tex]
This matches the first option:
[tex]$$
P(x) = 7000 \times 0.75^x.
$$[/tex]
[tex]$$
P(x) = 7000 \times r^x,
$$[/tex]
where [tex]\( P(x) \)[/tex] is the value of the car after [tex]\( x \)[/tex] years, and [tex]\( r \)[/tex] is the annual depreciation factor.
After 2 years the car’s value is given as:
[tex]$$
P(2) = 7000 \times r^2 = 3937.50.
$$[/tex]
To find [tex]\( r \)[/tex], we solve for [tex]\( r^2 \)[/tex]:
[tex]$$
r^2 = \frac{3937.50}{7000} = 0.5625.
$$[/tex]
Taking the square root of both sides gives:
[tex]$$
r = \sqrt{0.5625} = 0.75.
$$[/tex]
Thus, the exponential model that represents the value of the car [tex]\( x \)[/tex] years after purchase is:
[tex]$$
P(x) = 7000 \times (0.75)^x.
$$[/tex]
This matches the first option:
[tex]$$
P(x) = 7000 \times 0.75^x.
$$[/tex]
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