Spencer estimates that his new car that costs $24,380 would be worth $13, 500 after two years and $8,230 after five years. a.) What will be the annual rate of depreciation for the first two years? b.) What will the percent be for the five years?

22.31%, 13.25%

Step-by-step explanation:

Given:

Spencer estimates that his new car that costs $24,380 would be worth $13, 500 after two years

And would be worth $8,230 after five years.

Question asked:

What will be the annual rate of depreciation for the first two years ?

What will the percent be for the five years ?

Solution:

Cost of asset (new car) = $24,380

Scrap value of the car after two years = $13, 500

We have to find rate of depreciation for first two years.

So, useful life will be = 2 years

First of all we will find annual depreciation, then rate of depreciation.

Annual depreciation = 5440

Now,

Therefore, annual rate of depreciation for the first two years is 22.31%.

Similarly, we will find rate of depreciation for 5 years.

This time useful life, we will consider 5 years:

Scrap value = $8,230 after five years (given)

Annual depreciation = 3230

Therefore, annual rate of depreciation for the five years is 13.25%

show us the picture

step-by-step explanation:

show the !

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