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Hi guys, I've been revising over a bit of space and have come across questions asking to calculate the size of a rocket's thrust when they give you they velocity of the exhaust gas and the rate a which it burns. Eg. A rocket with an exhaust velocity of 1.3 × 10^3 ms^–1 burns fuel at a rate of...

Re: 2012 HSC MX2 Marathon \textup{Prove that the equation of the chord joining }P(acos\theta, bsin\theta) \textup{ and } Q(acos\alpha, bsin\alpha)\textup{ of the ellipse } \\ \frac{x^2}{a^2} + \frac{y^2}{b^2}=1 \textup{ is } bcos\frac{\theta + \alpha}{2}x + asin(\frac{\theta + \alpha}{2})y =...

multiplying by x/x is the same as multiplying by 1. ie it does not change the answer, only rearranges the terms. the purpose of this is to eliminate the fractions.

Re: 2012 HSC MX1 Marathon is the answer 108 u^2?

lol what school? i believe most are like that

anecdote brah

having a go at it. where's the question from?

beaten to it :/

lol w^4 = (w^3) x w but w^3 = 1 therefore w^4 = 1 x w = w

wow! how approx how many words is 11 pages of your writing? did u think a lot of it was waffle? great marks though

Re: 2012 HSC MX2 Marathon let z1 = 1 + 2i and z2 = 3 +i therefore arg(z1) = arctan(2) and arg(z2)=arctan(1/3) we want to prove arctan(2) - arctan(1/) = pi/4 ie arg(z1)-arg(z2)=pi/4 arg(z1)-arg(z2)=arg(z1/z2) where z1/z2 = (1+i)/2 arg( (1+i)/2 ) = pi/4. therefore...

Re: The Official Tennis Thread andy murray vs michael llodra. 2012 australian open, round 3, 2nd set, score 5-2, murray serving for the set; best game in tennis ever. http://www.youtube.com/watch?v=YzjGPOY4zO4

Re: 2012 HSC MX2 Marathon -.- oh yea :/

Re: 2012 HSC MX2 Marathon wait, i know when you divide complex numbers you subtract the arguments, does it work by the same token if you subtract two complex numbers you divide the arguments? is that what you did in the first step or am i missing something??

Re: 2012 HSC MX2 Marathon lol "You must spread some Reputation around before giving it to Carrotsticks again."

Re: 2012 HSC MX2 Marathon oh okay, thanks for the method! ill use it if that type of question comes up :)

Re: 2012 HSC MX2 Marathon for q2, thanks heaps man; perfect diagram understood everything :) you're a king. i was drawing |z-2i| = k as a circle with centre (0,-2) instead of (0,2) ie |z+2i|=k. i think it makes the question a lot harder? true? anyway thanks a lot!

Re: 2012 HSC MX2 Marathon Also: Determine the values of k for which the simultaneous equations |z-2i| = k and |z - (3+2i)|=2 have exactly 2 solutions. thanks

Re: 2012 HSC MX2 Marathon Sorry guys, simple question but i dont really understand it :( If z = 1-√3i Find a possible value for n (n>1) such that arg(z) = arg(zn)

this man is a genius