The equation of a circle always follows this:
(x - x1) + (y - y1) = r^2
x1 and y1 represent the coordinates as POSITIVE numbers and the r represent the radius
Your answer is (x + 1) + (y - 4) = 3
The trick is to exploit the difference of squares formula,

Set a = √8 and b = √6, so that a + b is the expression in the denominator. Multiply by its conjugate a - b:

Whatever you do to the denominator, you have to do to the numerator too. So

Expand the numerator:






So we have

But √12 = √(3•4) = 2√3, so

Answer:
the correct answer is
D. 80 ft^3
vol. of pyramid = (l× w × h)/3
= (6×8×5)/3
= 80 ft^3
8988989889 IDK hate doing the it but luv getting the answers
54•2=108
36•1=36
108+36=144
14•6=84
144-88=$56
Answer is $56.00