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my vote goes to Gordo

forget t results S = integral sign S root(sinxcos^3x) = S root(sinx(1-sin^2 x)cosx) = S root(sinxcosx - sin^3 x.cosx) = S root(sinx (cosx - sin^2 x.cosx)) = S root(sinx (cosx - (1 - cos^2 x) cosx) = S root(sinx (cosx - cosx + cos^2 x)) = S root(sinx(cos^2 x) = S cosx root(sinx) =...

haha yeah I know, I just thought it slightly weird that the minus of something could equal its reciprocal.. stupid, but yeah.. hahahhaah sorry for wasting all of your time

i divided by -1 is obviously (- i) correct? hmm.. if i = root(-1) = (-1)^1/2 then i/-1 = ((-1)^1/2)/(-1) = (-1)^1/2 / (-1)^1 ie = (-1)^-1/2 = 1 / i so... 1 / i = (- i) ??? how does this work.. hmm.. just a thought.

ahhh ok so I'm just an idiot thanks a lot I'm still a bit hazy though not sure what I did wrong really... ahh this is just diggin myself deeper could you explain what I screwed up? why can't I use the binominal probability with this..? hmm.. I hate probability.. thanks

hmm.. not sure how that came about withoutaface tell me where you got your first line from... and what did I do wrong..?

I maintain that my one's better

hey I've got a really annoying question that I found in Jeff Geha's book, and since I don't understand what he's up to, I thought I'd post my solution here and see what you guys think.. The question : in a certain box it is known that there are two black marbles and the rest are red. Four...

I've got an answer for your first one n0 = 2, so it isn't all positive integers assume true for n = k k^5 - k = 30I (where I is some integer) = k(k^4 - 1) = k (k^2 - 1)(k^2 +1) = k(k-1)(k+1)(k^2+1) = 30I now for n = k+1 = (k+1)(k)(k+2)((k+2)^2 + 1) [from expanded form]...