Answer:
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Step-by-step explanation:
Step-by-step explanation:
Y=4(5)-9
=11
you must follow the formula given
Circle: x^2+y^2=121=11^2 => circle with radius 11 and centred on origin.
g(x)=-2x+12 (from given table, find slope and y-intercept)
We can see from the graphics that g(x) will be almost tangent to the circle at (0,11), and that both intersection points will be at x>=11.
To show that this is the case,
substitute g(x) into the circle
x^2+(-2x+12)^2=121
x^2+4x^2-2*2*12x+144-121=0
5x^2-48x+23=0
Solve using the quadratic formula,
x=(48 ± √ (48^2-4*5*23) )/10
=0.5058 or 9.0942
So both solutions are real and both have positive x-values.
First we find the zeroes so we don't take the integral of negative bits
4x-x²
x(4-x)
zeroes at x=0 and x=4
it opens down
so the part we are interested in is the bit between x=0 and x=4
![\int\limits^4_0 {4x-x^2} \, dx =[2x^2- \frac{1}{3}x^3]^4_0=(32- \frac{64}{3})-(0)= 10.6666666666](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E4_0%20%7B4x-x%5E2%7D%20%5C%2C%20dx%20%3D%5B2x%5E2-%20%5Cfrac%7B1%7D%7B3%7Dx%5E3%5D%5E4_0%3D%2832-%20%5Cfrac%7B64%7D%7B3%7D%29-%280%29%3D%20%2010.6666666666)
or aout 10 and 2/3
C is answer
