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Re: Australian Maths Competition 2013 ye it is probably 126. Nevertheless this is what I did, probably I missed something haha http://imgur.com/a/fsfVN

Re: Australian Maths Competition 2013 If its asking for the maximum number of chords without intersection, not number of ways, then I think its 125, idk though, 75% sure its wrong.

the vertex is actually (1,3). So the focal length is 3 units so you move 3 units right of the vertex (since negative y^2=4ax) to get the directrix which is x=4

You made a mistake when solving k^2+k-2>0, it should be k>1 or k<-2

Ye heard 17/30 was the cut off for HD too. But don't worry I'm pretty average at chem and found it pretty easy to okay-ish difficulty, I bet you'll do good haha. So ye good luck tomorrow!

finally got 3g to play pokemon

Ah okay thankyou!

please help, an explanation if possible. Thanks

Anyone did the ANCQ yesterday?

http://community.boredofstudies.org/263/space/253170/how-did-aether-model-violate-principle-relativity.html found a thread, may be helpful idk

It's the American calculus course

Re: HSC 2016 2U Marathon haha ye remember it from this youtube vid, funny indeed lol

Re: MX2 2016 Integration Marathon I= \int \tan ^4x\cos 2x $ d$x \\ $ applying IBP; $ u=\tan ^4x \Rightarrow $ d$u=4\tan ^3x\sec ^2x, \\ $ d$v=\cos 2x \Rightarrow v=\sin x\cos x\\ \\I=\tan ^4x\sin x\cos x-4\int \tan ^3x\sec ^2x \sin x \cos x$ d$x \\ = \sin ^2x\tan ^3x-4\int \tan ^4x$ d$x \\...

Re: MX2 2016 Integration Marathon Woah that's actually pretty helpful, thanks :)

Re: MX2 2016 Integration Marathon I =\int \frac{\sqrt{1+x^2}}{x} $d$x \\ $ let $ u=\sqrt{1+x^2} \Rightarrow $ d$x=\frac{\sqrt{1+x^2}}{x}$ d$u \\ I = \int \frac{u^2}{u^2-1}$ d$u$ $ = \int 1+\frac{1}{u^2-1}$ d$u \\ $ let $\frac{A}{u+1}+\frac{B}{u-1}=\frac{1}{u^2-1} \\ $ by expanding and...

oh yeah soz

$recall that the formula to find the volume of a curve rotated about the x-axis from $ a $ to $ b$ is : $ \pi\int_{a}^{b} y^2 $ d$x. $ Here $ y = \cos \pi x $ and the limits are from 0 to 1, so plugging in the values and integrating, you get $: \\ V=\pi\int_{0}^{1} \cos^{2}( \pi x) $ d$x \\ =...

yeah InteGrands answer is right

Re: MX2 2016 Integration Marathon since no one did this, I gave it a try; I=\int\frac{\sin x-\cos x}{(\sin x+\cos x)\sqrt{\sin x\cos x+\sin^2 x\cos^2 x}}$ d$x \\ $ let $u=\sin x\cos x \Rightarrow $ d$u=(\cos ^2x-\sin ^2x)$ d$x\\ \therefore $ d$x=\frac{ du}{(\cos x+\sin x)(\cos x-\sin x)}...

I think it's a quote haha