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Re: HSC 2013 4U Marathon Different co-ordinates for at least 1 corner

Re: HSC 2013 4U Marathon Can you explain what hte question is saying in the first sentence? Is it saying that the integers from 1 to 100 are in pairs? Because if they're are then their sum would be 5050. ======= In the x-y plane, there is an n*n grid of dots in the plane, all with integer...

Yep you are correct haha

Yep :P I do more Physics in the MX2 and MX1 courses, plus I did horrible in a recent task. I still have 12 units though lol

Redo your working for m(PQ), it should turn out to be -1/pq Which means that p=1/q (since -1/pq * 1/pq = -1)

I just officially dropped Physics :') Feels so good

First principles: $The notation$ \ \ f'(a) $is used to show the value of the derivative at$ \ \ a $In this case$ \ \ a=5 \ \ $and by first principles$ f(a) = \lim_{h\ to 0} \frac{f(a+h)-f(a)}{h} The use of x=4.99 and x=5.01 is rather silly, but essentially when you substituted...

Re: MX2 Integration Marathon $By considering$ \ \ t^2=x^2+y^2 $Looking at the x-y plane, for varying t, the curve looking downwards is always a circle, as t varies from$ \ \ -r \rightarrow r \ \ $according to the equation$ z=e^{-t^2} \ \ $where z is the height of the curve at that specific...

re: HSC Physics Marathon Archive Thanks for the tips, I'll keep this in mind if I decide to continue doing Physics, though my decision to drop it is starting to wager in the positive direction. Holy shit man, there was a huge thread on this in the General HSC section, I do not deny and I...

re: HSC Chemistry Marathon Archive $A lawn fertiliser's sulfate content is$ \ \ P \% (w/w) $A sample of$ \ \ m \ \ $grams of fertiliser is analysed, by reacting with excess barium nitrate$ $What is the mass of the Barium Sulfate precipitate expected to form, in terms of the above pronumerals?$

re: HSC Physics Marathon Archive Wow, is this really part of the syllabus? what even

re: HSC Chemistry Marathon Archive You have restated the question for part (ii). I'm asking you to prove (mathematically) that an increase in pressure shifts the reaction to the right (which is why I asked for the equilibrium constant) Reaction Quotient, (Q), is core module right? Or is it...

Re: MX2 Integration Marathon $Prove, without differentiating the Right Hand Side$ \int \frac{dx}{(x+1)(x+2)(x+3)\dots (x+n)} = \ln \left(\prod_{k=1}^{n} (x+k)^{\frac{(-1)^{k-1}}{(k-1)!(n-k)!}} \right) + c $Note that$ \prod_{k=1}^{n} f(k) \equiv f(1) \cdot f(2) \cdot f(3) \dots f(n)

Re: MX2 Integration Marathon $Prove that$ \lim_{n \to \infty} n \int_0^{\frac{\pi}{2}} (1-\sqrt[n]{\sin x}) \ dx = \frac{\pi}{2} \ln 2

re: HSC Chemistry Marathon Archive $The Haber process is given by the chemical equation$ 3H_{2(g)} + N_{2(g)} \leftrightharpoons 2NH_{3(g)} $i) Write down the equilibrium constant for this expression$ $ii) Hence mathematically explain with reference to the reaction quotient$ $ that why an...

Re: MX2 Integration Marathon $Prove without differentiating the Right Hand Side$ \int \frac{dx}{\sqrt{(x-a)(x-b)}} = \ln \left |\frac{2x - (a+b)}{\sqrt{2(a^2+b^2)}} + \sqrt{\frac{4x^2-4x(a+b) - (a-b)^2}{2(a^2+b^2)}} \right | No room for the +c :/

Re: HSC 2013 4U Marathon Yep I see what you mean. I'll give this a go tomorrow.

Re: HSC 2013 4U Marathon Very true, so I can't know that my approximation is correct for an error of 10^{-3}? (which means that is a flaw in the question right?) Or is there a way of showing that |\pi - $(approx)$| < 10^{-3} without knowing the digits of pi?

Re: HSC 2013 4U Marathon Here is the question (fixed): $By considering the binomial expansion of an appropriate quadratic expression$ $Prove that$ \sum_{k=0}^{n} (-1)^k \binom{2n}{k} \binom{2n-k}{2n-2k} = \sum_{k=0}^{2n} \binom{2n}{k}^2 \phi^{2n-k} \tau^k \phi = \frac{1+\sqrt{5}}{2}...

Re: HSC 2013 4U Marathon It may come under inequalities, the way I thought of doing it is considering area underneath 1/(1+x^2) and approximating through trapeziums. Another way that comes to mind is proving the infinite series of pi/4 through integration of tan^{2n} x, then summing the first...