1. samuel bought a cement mixer for $54,205. the value of the cement mixer depreciated at a constant rate per year. the table below shows the value of the cement mixer after the first and second years: year 1 2 value (in dollars) 47,158.35 41,027.76 which function best represents the value of the cement mixer after t years? f(t) = 47,158.35(0.87)^t f(t) = 54,205(0.13)^t f(t) = 47,158.35(0.13)^t f(t) = 54,205(0.87)^t

Step-by-step explanation:

Given: The initial value of cement mixer (A) = $54,205

After one year, the value of mixer = $47,158.35

The constant rate of depreciation (b)=

The exponential function represent the value after depreciation of t years is given by :-

Hence, the function best represents the value of the cement mixer after t years will be

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

6 hours 48 min

step-by-step explanation:

68÷5=13.6

13.6×30=408min=6 hours 48 min.

hope to and hope that my answer is brainliest.(follow me )