if it has a diameter of 8, that means its radius is half that, or 4.
![\bf \textit{volume of a cone}\\\\ V=\cfrac{\pi r^2 h}{3}~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=4\\ h=5 \end{cases}\implies V=\cfrac{\pi (4)^2(5)}{3}\implies V=\cfrac{80\pi }{3} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill \stackrel{using~\pi =3.14}{V= 83.7\overline{3}}~\hfill](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cone%7D%5C%5C%5C%5C%0AV%3D%5Ccfrac%7B%5Cpi%20r%5E2%20h%7D%7B3%7D~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Ar%3D4%5C%5C%0Ah%3D5%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Ccfrac%7B%5Cpi%20%284%29%5E2%285%29%7D%7B3%7D%5Cimplies%20V%3D%5Ccfrac%7B80%5Cpi%20%7D%7B3%7D%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0A~%5Chfill%20%5Cstackrel%7Busing~%5Cpi%20%3D3.14%7D%7BV%3D%2083.7%5Coverline%7B3%7D%7D~%5Chfill%20)
Answer:
JKLM is similar to ABCD
Step-by-step explanation:
Answer:
-17
Step-by-step explanation:
4.5*6=27
so -44++27
If we add the equations it looks like
-5y + 8x + 5y + 2x = -18+58
so 10x=40
so x=40/10=4
now let's replace x by 4 in the second equation
5y +2*4=58
5y=58-2*4=58-8=50
so y=50/5=10
so (x, y) = (2, 10)
3)
82 + p >= 150
p>= 150 - 82
p>= 68
answer A.
4)
x/-10 > 6
x > -60
answer
x > -60
5)
5z >= - 75
z >= -75/5
z >= -15
answer
z >= -15
