since the diameter of the base of the cylinder is 6 feet, then its radius is half that, or 3 feet.
![\bf \textit{volume of a cylinder}\\\\ V=\pi r^2 h~~ \begin{cases} r=radius\\ h=height\\[-0.5em] \hrulefill\\ r=3\\ h=9 \end{cases}\implies V=\pi (3)^2(9)\implies V=81\pi](https://tex.z-dn.net/?f=%5Cbf%20%5Ctextit%7Bvolume%20of%20a%20cylinder%7D%5C%5C%5C%5C%0AV%3D%5Cpi%20r%5E2%20h~~%0A%5Cbegin%7Bcases%7D%0Ar%3Dradius%5C%5C%0Ah%3Dheight%5C%5C%5B-0.5em%5D%0A%5Chrulefill%5C%5C%0Ar%3D3%5C%5C%0Ah%3D9%0A%5Cend%7Bcases%7D%5Cimplies%20V%3D%5Cpi%20%283%29%5E2%289%29%5Cimplies%20V%3D81%5Cpi)
True. Math right to left would be absurd. Please Mark Brainliest!!!
Answer:
A,D,E
Step-by-step explanation:
A) (5)+3=8
D)12x5=60
E) 12+5=17
Answer:
It's $7
Step-by-step explanation:
7+7+7= 21 and add two $16 tickets and that's $53
Lets find that out writing equations and solving them, a multiple of 1/6 is something like this (1/6)x, so we have
(1/6)x > 3/6
and
<span>(1/6)y < 4/6
</span>lets solve both equations:
<span>(1/6)x > 3/6
</span>x > 6(3/6)
x > 3
<span>(1/6)y < 4/6
</span>y < 6(4/6)
y < 4
So the number must be between 3 and 4, which is obvious, lets try with 3.5 then, that is 3 5/10 or 35/10 = 7/2 in fractional form, and lets try it out:
(1/6)(7/2) = 7/12
finally we compare with the original fractions:
1/6 < 7/12
2/12 < 7/12
So, it comply with being greater than 1/6, now lets compare with 4/6
7/12 < 4/6
<span>7/12 < 8/12
</span>therefore is also smaller than 4/6 and hence 7/12 is a multiple of 1/6 between 3/6 and 4/6
