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a = 1/pi, r = 1/pi S(infinity) = (1/pi)/(1-1/pi) S(infinity) = (1/pi)/[(pi-1)/pi] S(infinity) = 1/(pi-1) And that is where I give up :)

hmmm, a guy in yr 11 at my school finished 4u in year 9 and just finished a mathematics course at uni.

Ah, please do. Post it here :)

haha yep. You're right. lol woops :) I will go edit it now. -_- *sleeps*

Personal pronouns. When can you use them? My english teacher; a head marker for HSC said that you should avoid where possible and only use them if you see no other way in module B (personal response for critical study). However, a friend of mine at Hornsby Girls (normally rank 1 in english)...

Also ffx is WAY too long to properly analyse. My friend is doing it. But personally I think that is a very bad choice.

Ah yes. I need to go back and study this topic. I never actually did it. But yes, this is the best way to do it. Sorry.

Re: 回复: Re: ~~Mathematics Revising Game~~ Ok got it: 10bi) d@/dt = pi/90 (2 degrees) @ = pi(t)/90 + C @ = pi(t)/90 ii) tan@ = x/10 x = 10tan@ x = 10tan(pi(t)/90) iii) v = (pi/9)sec^2(pi(t)/90) v = pi/9 * 1/(cos30)^2 v = 4pi/27 km/s v = 1600pi/3 km/h Right, a question: (random catholic trial...

Re: 回复: Re: ~~Mathematics Revising Game~~ I prob shouldn't have looked at the answers. They aren't correct... If @ = pi(t)/90 then when t = 0, @ = 0. However when t = 0, @ = pi/2 as we have been told. Therefore answers wrong :) EDIT: Hence the hint? Or am I just wrong? EDIT: NEVERMIND

Your answer is better. I never bothered to learn all that stuff. This is just the way I do maths :) Derive the formulas in the test :)

Using quadratic formula? Copy and paste the parts you don't get. I appologise to anyone who got confused, I made a minor multiplication error.

x = [-24 + root(24^2 - 4*4*k)]/8 or x = [-24 - root(24^2 - 4*4*k)]/8 These are the roots. Now, one has to equal double the other. Let root(24^2 - 4*4*k) = m Therefore: [-24 + m]/8 = [-24 - m)]/4 -24 + m = -48 - 2m m = -8 or [-24 + m]/4 = [-24 - m]/8 -48 + 2m = -24 - m m = 8 Therefore...

Yep it was grouping symbols. My bad.. Should have made it clearer.

Re: 回复: Re: ~~Mathematics Revising Game~~ Root[(cos^2(x) + sin^2(x) + cot^2(x))*sin^2(x)] = 1 LHS = Root[(cos^2(x) + sin^2(x) + cot^2(x))*sin^2(x)] = Root[(1 + cot^2x)sin^2(x)] = Root(sin^2(x) + cos^2(x)) = Root(1) = 1

Woops =S Sorry must have made mistake =S Should i post another? EDIT: I am pretty sure it works... =S EDIT: Yeah I couldn't do it on computer is what I meant. My question: It is relatively simple, are you sure you aren't messing up the grouping symbols?

9bi) t = 0 to t = 10 ii) t = 25 to t = 30 iii) t = 10 iv) t = 30 v) t = 20 vi) ...How? Question: Prove that: Root[(cos^2(x) + sin^2(x) + cot^2(x))*sin^2(x)] = 1

39/51 * 26/50 * 13/49 = 2197/20825 = 8.443 * 10^-7 (4s.f.) Question = find d(sec^2(x))/dx. EDIT: Thanks to syriangabsta for pointing out my mistake.

I realise this. Reread what I was talking about.

Look at the question... It is pretty obviously meant to be a multiplication. I think you should check the question again TearsofFire, you appear to be mistaken.

Hmm, correct me if I am wrong. But minor error. 2^x = 5 log2(5) = x etc. Not ln2(5)