Answer:
Volume of the right pyramid = 288 m²
Step-by-step explanation:
Volume of the pyramid = 
From the ΔAOB,
By Pythagoras theorem,
AB² = AO² + OB²
(6√2)² = AO² + (6)²
72 = AO² + 36
AO = √(36) = 6 m
Since base of the pyramid is a square so area of the base = (Length × Width) = (side)²
Now volume of the pyramid = ![\frac{1}{3}[(Length)(width)]\times height](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B3%7D%5B%28Length%29%28width%29%5D%5Ctimes%20height)
= 
= 288 m²
Therefore, volume of the right pyramid is 288 m².
<span>(7y2 + 6xy) – (–2xy + 3)
7y^2 + 6xy +2xy +3
7y^2 +8xy +3
8xy + 7y^2 +3</span>
Since , the relation is linear .
Let equation is y = mx + c .
Putting , x = 0 and y = 32 .
32 = c .......( 1 )
Also , putting x = 100 and y = 212 .
We get :
212 = 100m + c .......( 2 )
Comparing equation 1 and 2 .
100m = 212 - 32
x = 1.8
Therefore , y = 1.8x + 32 .
Hence , this is the required solution .