A new car is purchased for 23900 dollars. The value of the car depreciates at 9.25% per year. To the nearest tenth of a year, how long will it be until the value of the car is 6000 dollars?

9514 1404 393

  14.2 years

Step-by-step explanation:

The value is described by the exponential function ...

  y = (initial value)·(decay factor)^x

where y is the value after x years.

The decay factor is (1 - annual depreciation) = 1 - 0.0925 = 0.9075, so we want to find x when ...

  6000 = 23900·0.9075^x

  6000/23900 = 0.9075^x . . . . . . . . . . . . . . divide by 23900

  log(6000/23900) = x·log(0.9075) . . . . . . . take logarithms

  x = log(6000/23900)/log(0.9075) ≈ 14.2396

It will be about 14.2 years until the value of the car is $6000.

the correct functions are

h(x) = 3x + 2  

g(x) = 2^x

step-by-step explanation:

i proceed to calculate the different values ​​of k (x)

case 1)  k(x) = g(x) ∘ h(x)

g(x) ∘ h(x)=g(h(x))

k(x) = [2]^[3x + > is the option  4. k(x) = 2^(3x + 2 )

case 2)  k(x) = g(x) + h(x)

k(x) = [2^x]+[3x + 2]=2^x+3x+> is the option  1.) k(x) = 2^x + 3x + 2

case 3)k(x) = h(x) ÷ g(x)

k(x) =  [3x + 2]/[2^x]=3x/(2^x)+2/(2^x)=3x*(2^-x)+2^(1-x)

is the option  3.)   k(x) = (3x)2^(-x) + 2^(-x + 1 )

case 4)k(x) = g(x) - h(x)

k(x) =  [2^x]-[3x + 2]=2^x-3x-> is the option  2.) k(x) = 2^x - 3x - 2 

case 5)  k(x) = g(x) × h(x)

k(x) = [2^x]*[3x + > is the option  5.) k(x) = 2^x(3x + 2)

case 6)k(x) = h(x) ∘ g(x)

h(x) ∘ g(x)=h(g(x))

k(x)=3*[2^x]+> is the option  6. k(x) = 3(2^x) +2