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Re: 2012 HSC MX2 Marathon Yep What? lol

Her name is Twilight... this isn't going to well...

Re: 2012 HSC MX2 Marathon You could do it that way but with \cos\theta = \frac{1}{4}, you aren't going to get a very exact answer. Whats the relationship between \cos\theta $ and $ \sin\theta?

Watching the first one now!

Re: 2012 HSC MX2 Marathon Sounds good. Just need need to find the value of \sin\theta now.

Yeah your right, I missed the 1 on the end, I will correct it.

Yeah they do a bit...

Where do I find this "My little pony"?

Re: 2012 HSC MX1 Marathon Haha oh spiral you genius! I've probably done things just as bad. You be surprised how many stupid mistakes people make.

And when I mean difference, I mean big difference lol. Probably means that you have a lot of misconceptions about evolution.

Re: 2012 HSC MX2 Marathon $If $ z = \cos\theta + i\sin\theta, $ prove that $ z^n + z^{-n} = 2cos n\theta $ and hence solve$ \\\\ 2z^4 + 3z^3 +5z^2 + 3z^2 + 2 = 0

There is a difference between abiogenesis and evolution.

Lies!

Cos I was trying to do it in my head lol, next time I use paper first...

You have to do it all by text, http://community.boredofstudies.org/showthread.php?t=234259 should help get you started.

Wow d.... last time i do math on latex, going to do it on paper first from now on. Was such a good feeling when I came up with the right answer :rolleyes2:

You're going to need to show us the rest of the question for us to do (iii) C. $A double root at $ x = 1 $ means $ p(1) = 0 $ and $ p'(1) = 0 p(x) = 2x^3 - (2a + 1)x^2 + (2 + b)x - 1 \\\\ p'(x) = 6x^2 -2(2a+1)x + (2+b) \\\\ p'(1) = 6 - 4a - 2 + 2 + b = 0 \\\\ 6 - 4a + b = 0 \\\\ p(1) = 2 - 2a...

It's essentially what it is, distance from the origin.

For the first one, just remember modulus is basically a 2 dimensional absolute value, so the same way |1-6| = 5, |6-1| = 5, the modulus of complex numbers also works like that.

$If $z = \frac{1}{w-1} $ then $ |z| = |\frac{1}{w-1}| \\\\ = \frac{|1|}{|w-1|}. \\\\ $As $|1| = 1, \\\\ |z| = \frac{1}{|w-1|}