Seth has just decided to replace his computer. his old computer cost him $1,433 when he bought it exactly seven years ago. seth paid for it with his credit card, which has an apr of 11.70%, compounded monthly. he made no other purchases with the card and paid off his balance after two and a half years of making identical monthly payments. the computer consumed about $0.79 of electricity every day. in total, what percentage of the lifetime cost of the computer did the electricity make up? (assume that two out of the seven years were leap years, and round all dollar values to the nearest cent.)

The solution would be like this for this specific problem:

Cost of the computer:

i = 0.117 / 12 
i = 0.00975 

= monthly payment 
= (i * P * (1 + i)^n) / ((1 + i)^n - 1) 
= (0.00975 * $1433 * (1 + 0.00975)^30) / ((1 + 0.00975)^30 - 1) 
= $55.32 

Electricity cost:

= $0.79 * 365.25 * 7 
= $2019.83 

Electricity Percentage: 

= $2019.83 / ($2019.83 + $1659.60) 
= 54.89% 

Electricity: 

= ($0.79 * 365 * 5) + ($0.79 * 366 * 2) 
= $2020.03 

Electricity Percentage: 

= $2020.03 / ($2020.03 + $1659.60) 
= 54.898%

In total, the electricity made up 54.898% of the lifetime cost of the computer.

for the points! : )

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