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Who wouldn't want a free BBQ!?

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I'm on campus. So I'm basically there instantly.

Re: 2012 HSC MX1 Marathon D: Ext 2 is very fun!

That's about right, but "If you're look at something "stationary", you're looking for Lo" isn't really the best wording lol. If the question says "this is how long you see it in your FOR, how long is it in the objects FOR?" Then you are looking for Lo. Though I can never remember which Lv and Lo...

Re: 2012 HSC MX1 Marathon Haha would come in handy.

1. $The first step would be to convert $ 30 000kmh^{-1} $ into a fraction of c.$ \\\\ 30 000kmh^{-1}= \frac{75}{9}kms^{-1} \\\\ = \frac{75}{2 700 000}c \\\\ = \frac{3}{108 000}c $Now use the length contraction formula. $ l_v=l_o\sqrt{1-\frac{v^2}{c^2}} \\\\ l_v = ?, l_o = 30m...

Re: 2012 HSC MX2 Marathon z = 0, 1

Re: 2012 HSC MX2 Marathon I used sum of roots to find the final root, then subbed in x=-1/2 to find d. But either way works. P(a\cos\theta,b\sin\theta) $ and $ Q(a\cos\phi,b\sin\phi) $ lie on the ellipse $ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1. \\\\ $Find the equation of line $PQ

Re: 2012 HSC MX2 Marathon $Let $ z = x+ iy, \\\\ 3|z-1|^2 = |z+1|^2 3(x-1)^2 + 3y^2 = (x+1)^2 + y^2 \\\\ 3x^2-6x+3+3y^2=x^2+2x+1+y^2 \\\\ 2x^2 -8x +2 +2y^2 = 0 \\\\ 2(x-2)^2+2y^2=6 \\\\ (x-2)^2+y^2 = 3 \\\\ |x +iy-2|^2=3 \\\\ $As $ x+iy =z, |z-2|^2=3 $New question... $ p(x) = 2x^3...

Re: 2012 HSC MX2 Marathon $The polynomial $ 2x^3 - x^2 + 5 = 0 $ has roots $ \alpha, \beta, \gamma. \\\\ $Find the polynomial with integer coefficients whose roots are $ \alpha^3, \beta^3, \gamma^3

Re: 2012 HSC MX2 Marathon \alpha $ and $ \beta $ are the roots of $ z^2 - 2z +2 = 0. \\\\ $If $ \cot\theta = x + 1, $ prove that$ \\\\ \frac{(x+\alpha)^n - (x + \beta)^n}{\alpha - \beta} = \frac{\sin n\theta}{(\sin\theta)^n}

I have no idea who it is either.

Well I was tossing up with USyd and UNSW, but UNSW gave me accommodation back at the start of november, then I got the AAA scholarship, and then I got another one for $5000 per year for the duration of my course. So, I went with UNSW :D

UNSW - they give me scholarships :p

Re: 2012 HSC MX2 Marathon Haha your welcome! Goodnight.

Re: 2012 HSC MX2 Marathon Yep, real coefficients so the roots come in conjugate pairs. And also, when you do the s^2 + c^2 = 1, remember when you take the sqrt its plus/minus.

Re: 2012 HSC MX2 Marathon It is very close lol, but not quite there. The solutions are z = \frac{-1}{2} \pm \frac{\sqrt{3}}{2}i, z = \frac{-1}{4} \pm \frac{\sqrt{15}}{4}i

Re: 2012 HSC MX2 Marathon Lol yeah what you said didn't make much sense :p

Re: 2012 HSC MX2 Marathon I don't think that's right. You should have: \cos\theta = -\frac{1}{4} \\\\ sin^2\theta = 1 - \frac{1}{16} \\\\ \sin\theta = \pm\frac{\sqrt{15}}{4} $And with $ \cos\theta = \frac{1}{2} $ we get $ \sin\theta = \pm\frac{\sqrt{3}}{2}