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Harrizon [31]
2 years ago
13

Donna has a $300 loan through the bank she is charged a simple rate The total interest she paid on the loan was $63 As a percent

age what was the annual interest rate on her loan
Mathematics
1 answer:
klasskru [66]2 years ago
3 0

The annual interest rate is 21%

<h3>How to determine the annual interest rate?</h3>

The given parameters are

Loan Amount, P = $300

Interest, I = $63

Number of years, T = 1

The annual interest rate is calculated as

I = PRT

Substitute the known values in the above equation

63 = 300 * R * 1

Evaluate the product

300R = 63

Divide through by 300

R = 21%

Hence, the annual interest rate is 21%

Read more about simple interest at:

brainly.com/question/20690803

#SPJ1

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Match each interval with its corresponding average rate of change for q(x) = (x + 3)2. 1. -6 ≤ x ≤ -4 1 2. -3 ≤ x ≤ 0 -4 3. -6 ≤
MrMuchimi
The average rate of change of a function f(x) in an interval, a < x < b is given by
\frac{f(b) - f(a)}{b - a}

Given q(x) = (x + 3)^2

1.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -4 is given by \frac{q(-4)-q(-6)}{-4-(-6)} = \frac{(-4+3)^2-(-6+3)^2}{-4+6} = \frac{1-9}{2} = \frac{-8}{2} =-4

2.) The average rate of change of q(x) in the interval -3 ≤ x ≤ 0 is given by \frac{q(0)-q(-3)}{0-(-3)} = \frac{(0+3)^2-(-3+3)^2}{0+3} = \frac{9-0}{3} = \frac{9}{3} =3

3.) The average rate of change of q(x) in the interval -6 ≤ x ≤ -3 is given by \frac{q(-3)-q(-6)}{-3-(-6)} = \frac{(-3+3)^2-(-6+3)^2}{-3+6} = \frac{0-9}{3} = \frac{-9}{3} =-3

4.) The average rate of change of q(x) in the interval -3 ≤ x ≤ -2 is given by \frac{q(-2)-q(-3)}{-2-(-3)} = \frac{(-2+3)^2-(-3+3)^2}{-2+3} = \frac{1-0}{1} = \frac{1}{1} =1

5.) The average rate of change of q(x) in the interval -4 ≤ x ≤ -3 is given by \frac{q(-3)-q(-4)}{-3-(-4)} = \frac{(-3+3)^2-(-4+3)^2}{-3+4} = \frac{0-1}{1} = \frac{-1}{1} =-1

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