The value of the car in the year n is given by: V(n)=25000(0.85)^n. We have to determine

a) The purchase price of the car: This is done by replacing n by 0 which gives V = 25000.

b) The annual rate of depreciation: This is equal to the derivative `(dV(n))/(dn)...

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The value of the car in the year n is given by: V(n)=25000(0.85)^n. We have to determine

a) The purchase price of the car: This is done by replacing n by 0 which gives V = 25000.

b) The annual rate of depreciation: This is equal to the derivative `(dV(n))/(dn) = n*25000*(0.85)^(n-1)` , the value of the rate of depreciation is dependent on the year and changes for different years.

c) The car's value at the end of 3 years: The car's value at the end of 3 years is V(4) = 25000*(0.85)^4 = 13050.15

d) The value lost by the car in its first year: The value lost in the first year is V(0) - V(1) = 25000*0.15 = 3750

e) The value lost in the fifth year: This is equal to V(5) - V(6) = 1663.89

f) The number of years for the value to become equal to half of the original purchase price: If the number of years required for this is N, V(N) = 1/2*V(0)

=> 0.85^N = 1/2

=> `N = log 0.5/log 0.85 = 4.265` years.