At the end of the third year, the worth is 10,479.64 dollars
Explanation:
This problem can be modeled by an exponential function since we have that the car’s value depreciated steadily from year to year, that is, we have a constant ratio. So:
[tex]c(x)=ab^x \\ \\ a:Initial \ value \\ \\ b:constant \ ratio \\ \\ x:Number \ of \ years[/tex]
The automobile was purchased for $35,000, so this is the initial value:
[tex]a=35,000[/tex]
After 7 years it is worth $2,100:
[tex]Substituting \ values: \\ \\ 2,100=35,000b^7 \\ \\ b^7=0.06 \\ \\ b=\sqrt[7]{0.06} \\ \\ b=0.669[/tex]
So our exponential function becomes:
[tex]c(x)=ab^x \\ \\ c(x)=35,000(0.669)^x[/tex]
At the end of the third year, the worth is:
[tex]c(3)=35,000(0.669)^3 \\ \\ c(3)=\$10,479.64[/tex]
At the end of the third year, the worth is 10,479.64 dollars
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