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sunsphere314
- Newbie | Next Rank: 10 Posts
- Posts: 3
- Joined: Mon Mar 21, 2011 1:56 pm
now we can solve quadratics by allowing r to be variable ==> r^2+hr-180=0
from the condition given in the problem h=3 and r^2+3r-180=0,
r(1,2)=(-3 +- sqr{9+ 4*180})/2=(-3 +- 27)/2; r(1)=12 and r(2)=-15
we can have only positive radius, hence r=12 and diameter 2r=24
e
sunsphere314 wrote:Would someone please explain?
A cylinder has a surface area of 360 pi, and is 3 units tall. What is the diameter of the cylinder's circular base?
a) 8 b)12 c)16 d) 20 e)24
Thank You!












