Christian purchased a new car. The value of the car is depreciating at a
constant rate of 12% per year. The function V = 18,000(1 - .12)^t can be
used to model the value of the car in dollars after t years.
A. Explain what 18,000 represents in the equation of the function. (2 points)

B. What will be the value of the car in 6 years? (2 points)
C. Christian also considered buying a car that costs $20,000 and depreciates
by 15% each year. Which car will have more value in 6 years? (4 points)

WILL GIVE BRAINLIEST!

18000 represents the price of the new car.

The value of the car in 6 years can be found by substituting 6 for the variable t, as follows:

In 6 years, it will be worth around $8359.27.

The equation for the car worth $20000 is this:

In 6 years, the other car will be worth $7542.99, so the car worth $18000 will have more value in 6 years.

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

total of apples - take away four

6-4 = 2

answer is 2 apples