SOLUTION
The price of the car = $12,000
The depreciate by 10%
![\begin{gathered} \text{ The depreciating value for the first year } \\ 12,000*((10)/(100))^1 \\ \text{Then} \\ 12,000*0.1 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/q15dpje3r5kt090qwceksjj12omgsuqb14.png)
Then
![12,000-12,00(0.1)](https://img.qammunity.org/2023/formulas/mathematics/college/5v8ipptqmek1e505q55r28sfp7jejcxzad.png)
Then
![\begin{gathered} 12000(1-0.1) \\ 12,000(0.9) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/krbtgjqql0dyoyi9n2md7lxqqdrao2cgj4.png)
For the first year the depreciating value will be
![12,000(0.9)](https://img.qammunity.org/2023/formulas/mathematics/college/2rgpj89zxdtoj5wdec5hktwq7gwyiby0tb.png)
Base on the number of years, the exponential equation will be
![\begin{gathered} f(x)=12,000(0.9)^x \\ \text{where } \\ x=\text{ number of years } \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i5ge6ngeubvioujbm5qbaz75rm0ux4joag.png)
Therefore
The exponential equation that represent the value of the car is
F(x)=12,000(0.9)^x
The price of the car in 5 yeras will be obtain by substituting x=5 into the equation above
![\begin{gathered} f(x)=12,000(0.9)^x \\ \text{where x=5} \\ f(x)=12,000(0.9)^5=7085.88 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/qzo0t1a5wsuj0xibb9u5oeraehrf9be7is.png)
The car will worth $7085.88 after 5 years
Similarly, The for 12 years we have x=12
![f(x)=12,000(0.9)^(12)=3389.15](https://img.qammunity.org/2023/formulas/mathematics/college/sdi10qd3kjv6mn5q6q0cyc6cw639fhntaz.png)
The car will worth $3389.15 after 12 years