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Sagot :
Given that price of the car is $12000 and depreciates is 10%.
After one year the value of the car is
[tex]=12000-\frac{12000\times10}{100}[/tex][tex]=0.9\times12000[/tex]After two years the value of the car is
[tex]=\text{the value of car at first year-}\frac{the\text{ value of the car at first year }\times10}{100}[/tex][tex]=0.9\times12000-\frac{0.9\times12000\times10}{100}[/tex][tex]=0.9\times12000\times\frac{100}{100}-\frac{0.9\times12000\times10}{100}[/tex][tex]=\frac{0.9\times12000\times100-0.9\times12000\times10}{100}[/tex][tex]=\frac{0.9\times12000(100-10)}{100}[/tex][tex]=\frac{0.9\times12000\times90}{100}[/tex][tex]=0.9\times0.9\times12000[/tex][tex]=0.9^2\times12000[/tex]proceeding this way, we get the car value f(x) for x number of years is
[tex]f(x)=12000\times0.9^x[/tex]Hence the required equation is
[tex]f(x)=12000\times0.9^x[/tex]
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