any of you geeks can help me on how to solve this shit?

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I see brother, well I will see if I get the same ,IRONm@N wrote:This is somewhat easy, but you could make a lot of mistakes.
I got
r = sqaure root (m/p*pi)
oh sorry, I meantabdisamad3 wrote:I see brother, well I will see if I get the same ,IRONm@N wrote:This is somewhat easy, but you could make a lot of mistakes.
I got
r = sqaure root (m/p*pi)
that is what i got as well thanks for your time bro,IRONm@N wrote:oh sorry, I meantabdisamad3 wrote:I see brother, well I will see if I get the same ,IRONm@N wrote:This is somewhat easy, but you could make a lot of mistakes.
I got
r = sqaure root (m/p*pi)
r = cube root (m/p*pi)
not sqaure root
we do that as well sxb but this is just a math lesson we have go through, Iam not big fan of algebra my self,Monk-of-Mogadishu wrote:In my agricultural engineering course we use the same kind of equations but we plug in real data for the variables, for example our r, p, and m would all relate to a measurement of some kind with an exact number. That's the only kind of algebra I can do, when the variables and numbers actually mean something. If you give me a bunch of letters that mean shit I'll walk away, I hate algebra/trig/calculus unless its applied to a situation.![]()
well you might be right sxb,,I gotta re-calculate and see what I did wrong,IRONm@N wrote:no its
r = cube root (2m/3p*pi)
in the brackets you get
m = (3*pi*r^2/4)*2pr
then after simplifying the 2 with the 4 it becomes
m = 3*pi*pr^3/2
2m = 3pi*r^3
r^3 = 2m/3pi*p
r = cube root (2m/3p*pi)
forget about the old one, this is should be right.
so you got a solution?Addoow wrote:The answer is wrong.