A bank loaned out $21,500, part of it at the rate of 15% annual interest, and the rest at 9% annual interest. The total interest earned for both loans was $2,025.00. How much was loaned at each rate?

So the portion of the loan lent at 4% is 21000(0.5955) = $12,505.5

And the portion of the loan lent at 15% is 21000(0.4045) = $8494.5

Step-by-step explanation:

This problem can be solved in several ways. The way I like is as follows.

Ask yourself at what interest would $ 21,000 yield and interest of $ 2,215 in one year ? The equation to solve is:

21000(P) = 2215

P = 0.1055 or 10.55 %

This number is a weighed average of the 4% yield and the 15% yield so the next equation to solve is

10.55 = 4(1 – X) + 15X

Where X is the proportion of the loan given at the indicated interest. So

10.55 = 4 – 4X + 15X

6.55 = 11X

X = 0.5955

1-X = 0.4045

So the portion of the loan lent at 4% is 21000(0.5955) = $12,505.5

And the portion of the loan lent at 15% is 21000(0.4045) = $8494.5

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

step-by-step explanation:

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