The volume of a cone varies jointly as the square of its radius and its height. (V=k*r^2*h) If the volume of a cone is 9pi inches when the radius is 3 inches and its height is 3 inches, find the volume of a cone when the radius is 5 inches and the height is 9 inches. Express the answer in terms of pi. Given answer is 75 pi. How do I work this? Please help? Anyone?
first you need to calculate k you have V=k*r^2*h so 9pi = k*3^2*3 so what's k ?
that's where I am having the problem.
If I could figure out the k, I'd be halfway to figuring this out. The other half is trying to figure out how to express the answer in terms of pi
because I have another question exactly like it, and i feel like I am overlooking something incredibly simple
\[9 \pi = k*3^2*3\] \[k = \frac{ 9 \pi }{ 3^2*3 }\]
to express the answer in terms of pi, you have to keep k in terms of pi
how do I do that?
you just keep the pi and simplify the rest so \[k = \frac{ 9 }{ 9*3 } \pi\]
can you simplify it ?
k = 1/3 π
correct
hmmm....
you didn't understand ?
actually am trying to figure out how to plug the same value into another equation. want to see if the k i got from this answer will also be the same in the other question
actually for a cone k is always 1/3 pi but that's only for a cone you can use this k to find the volume of the other cone mentioned in this question "radius is 5 inches and the height is 9 inches"
you're a life saver! i wish I could give you 2 medals for this question.
since the other question is also about a cone, then yes, I can do the math and move on. i simply could not figure out how to manipulate the math to make pi behave
so you are able to do it now ?
let me see. V=1/3 *r^2*h my next question is ?π=1/3*4^2*3
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1/3π*4^2*3 1/3π*16*3 wouldn't the 1/3 and the 3 cross cancel each other out and leave me with 16π?
yeah correct
thankee-sai!
glad to help! :)
now let's see if I can manage to pull off a somewhat passing grade on this test :/
i hope you can
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