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Re: HSC 2013 3U Marathon Thread does this mean only 1 replacement occurs, and from the second draw on they stay out of the bag? terribly ambiguous question. i'm going to assume you mean you replace them all. i) (1/3)^2 * (2/3)^2 * 4C2 = 24/81 ii) 4 are drawn, for the sum to be greater than 9...

Re: HSC 2014 4U Marathon - Advanced Level \text{Given that m & n are both positive integers, show that }\\ 17m^{2} + 4n^{2} \text{ and } 17n^{2} + 4m^{2} \text{ cannot both be perfect squares}

but circle geo is 3U so the angle in a semi-circle theorem is outside of the scope of this course isn't it?

Re: HSC 2014 4U Marathon Yes this is exactly what i meant. Don't the perpendicular bisectors meet at the circumcenter?

You need to get surface area in terms of volume so that you can find dA/dV I'd rather not spoon feed, tell me if you still can't get it :)

Re: HSC 2014 2U Marathon i was unaware this was 2u

Re: HSC 2014 2U Marathon oh that's 2U? well then.

Re: HSC 2014 2U Marathon substitution is a 3U method. i'm actually not sure how to do this question using 2U methods.

Re: HSC 2014 4U Marathon is it sufficient to consider the construction of perpendicular bisectors of each of the sides of the triangle formed by the 3 points, and noting that for any 3 points, the 3 perpendicular bisectors all meet at one location, meaning there is only one possible center for...

never allowed to use it lol

wicked excited

what happened to the B? shouldn't it be 2yy' = 2Ax + B

i did mine this morning. i thought mine was really easy tbh but i guess we'll see when the results come back! a lot of people found it quite difficult, but it was a lot easier than most of the past papers i've done (sydney grammar, independent etc.)

Re: HSC 2013 3U Marathon Thread \\ i)\text{ } x = \pm1 \text{, } y = \pm\frac{\pi}{2} \\ ii)\text{ } x \neq 0 \text{, } y = \pm\frac{\pi}{2}

Re: MX2 Integration Marathon one application of IBP with dv/dx = x^m & u = (1-x)^n gave a solutions of (n/m)(same integral but m is now m+1, n is now n-1). By repeating the process n times, you get what i've got in my last post.

Re: HSC 2014 4U Marathon ok i'm not 100% sure what constitutes letting z = x+iy Am i allowed to simultaneously solve arg(z-3-4i)-arg(z-2+2i)=\frac{2\pi}{3} \\ |z-3-4i| = |z-2+2i| Thus finding the equation of the circle?

Re: MX2 Integration Marathon great question. not sure how valid my solution is. i did IBP once, then said pattern follows such that... I = \frac{n!m!}{(n+m)!}\int_{0}^{1}x^{n+m}dx

Re: HSC 2014 4U Marathon I = \frac{2\ln(\sec\alpha + \tan\alpha)}{3} + C \\ \text{Where } \sin\alpha = \frac{2}{3}\left(\cos\theta + \frac{1}{2}\right) cbb trying to simplify it LOL

he's talking about the HSC papers not CSSA EDIT: with 55/70, yes you're on track for an e4 definitely

from memory it was about 52-53/70 in 2012 and last year i remember someone saying 48/70 was the bottom e4