Check the picture below.
thus, if one side on the square is x - x², then the area for the square is (x - x²)(x - x²), or (x - x²)²
now... what is the "x" value for the region, let's check their intersection.
![\bf \begin{cases} y=x\\ y=x^2 \end{cases}\implies x=x^2\implies 0=x^2-x\implies 0=x(x-1) \\\\\\ x= \begin{cases} 0\\ 1 \end{cases}\\\\ -------------------------------\\\\ \displaystyle \int\limits_{0}^{1}\ (x-x^2)^2\cdot dx\implies \int\limits_{0}^{1}\ (x^2-2x^3+x^4)\cdot dx \\\\\\ \cfrac{x^2}{2}-\cfrac{2x^4}{4}+\cfrac{x^5}{5}\implies \left. \cfrac{x^2}{2}-\cfrac{x^4}{2}+\cfrac{x^5}{5} \right]_{0}^{1}\implies \left[ \cfrac{1}{2}-\cfrac{1}{2}+\cfrac{1}{5} \right]-[0]\implies \cfrac{1}{5}](https://tex.z-dn.net/?f=%5Cbf%20%5Cbegin%7Bcases%7D%0Ay%3Dx%5C%5C%0Ay%3Dx%5E2%0A%5Cend%7Bcases%7D%5Cimplies%20x%3Dx%5E2%5Cimplies%200%3Dx%5E2-x%5Cimplies%200%3Dx%28x-1%29%0A%5C%5C%5C%5C%5C%5C%0Ax%3D%0A%5Cbegin%7Bcases%7D%0A0%5C%5C%0A1%0A%5Cend%7Bcases%7D%5C%5C%5C%5C%0A-------------------------------%5C%5C%5C%5C%0A%5Cdisplaystyle%20%5Cint%5Climits_%7B0%7D%5E%7B1%7D%5C%20%28x-x%5E2%29%5E2%5Ccdot%20dx%5Cimplies%20%5Cint%5Climits_%7B0%7D%5E%7B1%7D%5C%20%28x%5E2-2x%5E3%2Bx%5E4%29%5Ccdot%20dx%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bx%5E2%7D%7B2%7D-%5Ccfrac%7B2x%5E4%7D%7B4%7D%2B%5Ccfrac%7Bx%5E5%7D%7B5%7D%5Cimplies%20%5Cleft.%20%5Ccfrac%7Bx%5E2%7D%7B2%7D-%5Ccfrac%7Bx%5E4%7D%7B2%7D%2B%5Ccfrac%7Bx%5E5%7D%7B5%7D%20%5Cright%5D_%7B0%7D%5E%7B1%7D%5Cimplies%20%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D-%5Ccfrac%7B1%7D%7B2%7D%2B%5Ccfrac%7B1%7D%7B5%7D%20%5Cright%5D-%5B0%5D%5Cimplies%20%5Ccfrac%7B1%7D%7B5%7D)
Answer:
SSS or SAS
Step-by-step explanation:
Remark
Since the two diagonals bisect each other, you have created two sets of equal lines. If that is the case, then you could prove the equality of the triangles 2 different ways.
1) The angles at the center are vertically angles which makes them equal. Since they are equal, the triangles are congruent by SAS. Notice that the angle is between the 2 equal sides.
2) Since CD and AB are marked as being equal, the triangles are congurent by SSS.
Answer:
B. Place the compass at P and draw an arc of length OP across the circle.
Below is the diagram for help.
Step-by-step explanation:
We are given with an already constructed circle of diameter 
and 
as radius.
To draw a regular hexagon(six-sided), we need to make first cut on the circle of an arc equal to radius of the circle taking any point of the diameter as center
So here, We will consider 
as a center and take an arc of length and make a cut on the circle.
This would be the first step and James need to repeat the same steps by considering the newly formed cut as center each time.
Answer:
B.
Step-by-step explanation:
The vertical line test is what I used to test this.
