Area of the square base = 5^2 = 25 cm^2
Area of one of the triangular side = 1/2 * 5 * 8 = 20 cm^2 and there are 4 sides
So the surface area of while pyramid = 4 * 20 + 25 = 105 cm^2
Volume of the cone = (1/3) pi r^2 h
= (1/3) pi * 5^2 * 8
= 209.44 ft^2
2 ways: Easy and hard
Hard=A
Easy=B
A: 1/2x+4
work from there so we do fun stuff with it
make something that can be simplified so
1/2x+4 times (2/2)=x+8
now square the whole thing and put the result in a square root thingie
(x+8)^2=x^2+16x+64

multiply the whole thing by 4/4 and put
![\sqrt{16} [\tex] on top so then [tex] \sqrt{x^2+16x+64}](https://tex.z-dn.net/?f=%20%5Csqrt%7B16%7D%20%5B%5Ctex%5D%20on%20top%20so%20then%20%0A%5Btex%5D%20%5Csqrt%7Bx%5E2%2B16x%2B64%7D%20)
times

=

=

to solve it, factor out the 16 in the square root and then square root 16 to get 4
then it will be (4 times square root of equation)/4=square root of equatio
factor square root of equation and square root it and get x+8
divide by 2 to get 1/2x+4
B: 1/2x+4
put stuff that cancels out
1/2x+3x-3x+4+56-56
move them around
3 and 1/2x-3x+60-56
or
2x-3x+1 and 1/2x+30-20+30-36
then just add like terms to solve
Answer:
A
Step-by-step explanation:
The triangles aren't similar.
The triangles share a side so we cant use anything involving SSS SAS.
The triangles has only known congruent angle so we cant use AA.
The answer is B
the steps are these
interchange the x and y variables
x=5y-8
solve for y so we get
y = x+8
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