you take out a php 20,000 loan at a 5% interest rate if the interest is compounded annually how much will you after 10 years?​

the future value is 32,577.89

Step-by-step explanation:

The computation of the amount after 10 years is shown below:

As we know that

Future value = Present value × (1 + rate of interest)^time period

= 20,000 × (1 + 0.05)^10

= 32577.89

Hence the future value is 32,577.89

answer: the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.

step-by-step explanation:

the given rate of interest i= 85%

in calculating the monthly payment for a loan with an interest rate compounded monthly, the value of the interest rate per period we use in the formula must be in decimal form.

to convert it in decimal, we divide it by 100, we get

hence, the value should be used for i in the formula of calculation the monthly payment , the interest rate per period is 0.85.

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