So in finding the area of the square where the part of the yard that is not covered, first is to calculate the are of both then subtract the two and the are for the circle is pi*r^2 so it means pi*(4x)^2 then for the are of square the formula is A^2 and its (20x)^2, and i think the best answer would be 20X^2-(pi*4x^2)
10s^2-15s+20 because 5(2s^2)=10s^2, 5(-3s)=-15s, and 5(4)=20.
Assuming we're solving for x, all that needs to be done is to add 5 on both sides.
x - 5 >= -2
+5 +5
x >= 3
ANSWER
The y-intercept is (0,-11)
The x-intercepts are:
(-1,0) and (11,0)
EXPLANATION
The given function is
![f(x) = {x}^{2} - 10x - 11](https://tex.z-dn.net/?f=f%28x%29%20%3D%20%20%7Bx%7D%5E%7B2%7D%20%20-%2010x%20-%2011)
To find the y-intercept, we put x=0 into the function.
![f(0) = {0}^{2} - 10(0) - 11](https://tex.z-dn.net/?f=f%280%29%20%3D%20%20%7B0%7D%5E%7B2%7D%20%20-%2010%280%29%20-%2011)
![f(0) = - 11](https://tex.z-dn.net/?f=f%280%29%20%3D%20%20-%2011)
The y-intercept is (0,-11)
To find the x-intercepts, we put f(x)=0.
![{x}^{2} - 10x - 11 = 0](https://tex.z-dn.net/?f=%7Bx%7D%5E%7B2%7D%20%20-%2010x%20-%2011%20%3D%200)
![(x + 1)(x - 11) = 0](https://tex.z-dn.net/?f=%28x%20%2B%201%29%28x%20-%2011%29%20%3D%200)
This implies that,
![x = - 1 \: or \: x = 11](https://tex.z-dn.net/?f=x%20%3D%20%20-%201%20%5C%3A%20or%20%5C%3A%20x%20%3D%2011)
The x-intercepts are:
(-1,0) and (11,0)
we have given y interms of x.
we have given that add 2 to x.
which means ![y=x+2](https://tex.z-dn.net/?f=%20y%3Dx%2B2%20)
so we have
.
so when x
.
.
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