Answer:
Option A. one rectangle and two triangles
Option E. one triangle and one trapezoid
Step-by-step explanation:
step 1
we know that
The area of the polygon can be decomposed into one rectangle and two triangles
see the attached figure N 1
therefore
Te area of the composite figure is equal to the area of one rectangle plus the area of two triangles
so
![A=(8)(4)+2[\frac{1}{2}((8)(4)]=32+32=64\ yd^2](https://tex.z-dn.net/?f=A%3D%288%29%284%29%2B2%5B%5Cfrac%7B1%7D%7B2%7D%28%288%29%284%29%5D%3D32%2B32%3D64%5C%20yd%5E2)
step 2
we know that
The area of the polygon can be decomposed into one triangle and one trapezoid
see the attached figure N 2
therefore
Te area of the composite figure is equal to the area of one triangle plus the area of one trapezoid
so

Answer:
Remove all perfect squares from inside the square root. ... I think it's about eighth or ninth grade. ... so if you have the cube root of the square root of (x-5) =2, you get ((x-5)^(1/2))^1/3 = 2, power to power requires multiplication, so (x-5)^1/6 = 2, ...
Missing: 176 xy
The formula of the midpoint of HE:

We have H(0; 0) and E(2a; 2a). Substitute:
Answer:
18, -3
using the formula for midpoint M=(x1+x2)/2 ,(y1+y2)/2