From the Figure :
Point A is (-3 , -3)
Point B is (6 , 6)
We know that, The Mid Point of Two Points (x₁ , y₁) and (x₂ , y₂) is given by :
![\mathsf{\implies [(\frac{x_1 + x_2}{2})\;,\;(\frac{y_1 + y_2}{2})]}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20%5B%28%5Cfrac%7Bx_1%20%2B%20x_2%7D%7B2%7D%29%5C%3B%2C%5C%3B%28%5Cfrac%7By_1%20%2B%20y_2%7D%7B2%7D%29%5D%7D)
![\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{-3 + 6}{2})\;,\;(\frac{-3 + 6}{2})]}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20Midpoint%5C%3Bof%5C%3BLine%5C%3BAB%5C%3Bis%5C%3B%5B%28%5Cfrac%7B-3%20%2B%206%7D%7B2%7D%29%5C%3B%2C%5C%3B%28%5Cfrac%7B-3%20%2B%206%7D%7B2%7D%29%5D%7D)
![\mathsf{\implies Midpoint\;of\;Line\;AB\;is\;[(\frac{3}{2})\;,\;(\frac{3}{2})]}](https://tex.z-dn.net/?f=%5Cmathsf%7B%5Cimplies%20Midpoint%5C%3Bof%5C%3BLine%5C%3BAB%5C%3Bis%5C%3B%5B%28%5Cfrac%7B3%7D%7B2%7D%29%5C%3B%2C%5C%3B%28%5Cfrac%7B3%7D%7B2%7D%29%5D%7D)

Answer:
SAS is the answer in my opinion
Answer:
(3+d)(9-3d+d^2)
Step-by-step explanation:
a^3 + b^3 = (a+b)(a^2-ab+b^2)
a= 3
b= d
I assume you're supposed to establish the identity,
cos(A) cos(2A) cos(4A) = 1/8 sin(8A) / sin(A)
Recall the double angle identity for sine:
sin(2<em>x</em>) = 2 sin(<em>x</em>) cos(<em>x</em>)
Then you have
sin(8A) = 2 sin(4A) cos(4A)
sin(8A) = 4 sin(2A) cos(2A) cos(4A)
sin(8A) = 8 sin(A) cos(A) cos(2A) cos(4A)
==> sin(8A)/(8 sin(A)) = cos(A) cos(2A) cos(4A)
as required.