Question 10 A bank loaned out $29,000, part of it at the rate of 13% annual interest, and the rest at 4% annual interest.

The total interest earned for both loans was $2,195.00. How much was loaned at each rate?

$11,500 was invested at 13%.

$17,500 was invested at 4%

Step-by-step explanation:

This is a simple interest problem.

The simple interest formula is given by:

In which E is the amount of interest earned, P is the principal(the initial amount of money), I is the interest rate(yearly, as a decimal) and t is the time.

After t years, the total amount of money is:

In this question:

Loans totaling 29,000.

P was invested at 13%

29000 - P was invested at 4%.

First investment:

Principal P.

Interest 13% = 0.13.

One year, so t = 1.

So

Second investment:

Principal 29000 - P.

Interest 4% = 0.04.

One year, so t = 1.

So

The total interest earned for both loans was $2,195.00.

This means that

So

So we solve the following system:

$11,500 was invested at 13%.

29000 - 11500 = 17500

$17,500 was invested at 4%

ik how do do this just give me a few minutes and i will give it to

let's call the solutions x₁ = (5 - 2√7)/3 and x₂ = (5 + 2√7)/3

then the equation is (x-x₁)(x-x₂) = 0

if you fill that in you get a lot of numbers which you can recognize as ax²+bx+c=0.

step-by-step explanation:

you can see that a=1 (there is just an x²)

b = (-5 -2√7 + 2√7 - 5)/3 = -10/3

c = -2√7)/3 ) * (-5+2√7)/3) = -1/3

so your answer is 3x^2 – 10x – 1 = 0