I can tell you how to solve this problem.
You can find the areas using both the values as the radiuses (find the are of a circle).
Subtract the smaller area from the larger area and there is your answer.
Hope this helped.
Good Luck!
Well you see here, you need to take attest pre-calculus to answer this question. I am in 6th grade.
The value of x such that f(x) = g(x) is x = 3
<h3>Quadratic equation</h3>
Given the following expressions as shown
f(x) = x^3-3x^2+2 and;
g(x) = x^2 -6x+11
Equate the expressions
x^3-3x^2+2 = x^2 -6x+11
Equate to zero
x^3-3x^2-x^2+2-11 = 0
x^3-3x^2-x^2 + 6x - 9 = 0
x^3-4x^2+6x-9 = 0
Factorize
On factorizing the value of x = 3
Hence the value of x such that f(x) = g(x) is x = 3
Learn more on polynomial here: brainly.com/question/2833285
#SPJ1
11 5/8 + 9 1/2= 21 1/8
3/4 - 2/5= 7/20
4 5/6 - 2 1/2= 2 1/3
I think these are right, I apologize if they are wrong.
