Search results

That wasn't the worst bit, we broke his mini cupboard when we were moving it out. It wasn't aluminium, it was magnesium. You can buy aluminium anywhere.

Re: MX2 2015 Integration Marathon Hey guys can someone solve this integral? \int \frac{dx}{x\sqrt{x^2-1}} For some reason the answers for the paper for this question have +-, can someone explain why?

Re: MX2 2015 Integration Marathon @porcupinetree I = \int \frac{x-\sin{x} \cos{x}}{1-\cos^{2}{x} } dx = \int x\csc^{2}{x} - \cot{x} dx $ Via IBP: $ \int x\csc^{2}{x} = \ln{\sin{x}} - x\cot{x} \therefore I = \ln{\sin{x}} - x\cot{x} - \ln{\sin{x}} +c = -x\cot{x} +c Edit: Paradoxica...

Re: MX2 2015 Integration Marathon This is similar to a previous question you asked. I= \int \frac{x^2-1}{x^4+3x^2+1} dx = \int \frac{1-\frac{1}{x^2}}{x^2 + 3 + \frac{1}{x^2} } dx $ (Divide top and bottom by $ x^2) $ Let u $ = x+\frac{1}{x} $ $ \therefore I= \int \frac{1}{u^2+1} du =...

Re: MX2 2015 Integration Marathon Hey Paradoxica, here are a few integrals for your integration bank: \int \frac{1+x^2}{(1-x^2)\sqrt{1+x^4}} dx \int \frac{(1+x)^3 (x-1)}{(x^2+1)^2\sqrt{x^4+x^2+1}} dx I have a feeling you will be able to solve them :)

Re: MX2 2015 Integration Marathon Im not sure where the substitution comes from but this is what I did: \int \frac{(2-x^2)e^x}{(1-x)\sqrt{1-x^2}} dx = \int \frac{(1-x^2)e^x}{(1-x)\sqrt{1-x^2}} dx + \int \frac{e^x}{(1-x)\sqrt{1-x^2}} dx \frac{d}{dx}...

re: HSC Chemistry Marathon Archive Well you should know that sodium hydroxide is produced via the electrolysis of concentrated brine. So Cell Y has the products of the electrolysis of concentrated brine, hence that is the answer.

Re: MX2 2015 Integration Marathon I know that \tan{x} - \cot{x} = -2\cot{2x} but I thought it looked cooler in its unsimplified form. I am not entirely sure how your new question is solved though. But according to Wolfram, there is no solution for it.

re: HSC Chemistry Marathon Archive It doesn't just go black, it forms a carbon snake.

re: HSC Chemistry Marathon Archive I would talk about both concentrated and dilute aqueous solutions. Its a 5 marker so I guess 3 marks for all three types of electrolysis and the final 2 on equations and the final comparison

Oooo, one question per person or one question overall per unit? Because if it is the latter then we all have to agree on one question to ask.

Calm down guys, there is still 3 days left, carrot will probs give it out the night before

Re: MX2 2015 Integration Marathon I= \int \frac{1}{\sin^{6}{x} + \cos^{6}{x}} = \int \frac{\sec^{4}{x} \cdot \sec^{2}{x}}{\tan^{6}+1} dx $ (Divide top and bottom by $ \cos^{6}{x} ) $ Let u $ =\tan{x} $ $ \therefore I = \int \frac{(u^2+1)^2}{u^6+1} du = \int \frac{(u^2+1)^2}{(u^2+1)(u^4-u^2...

Re: MX2 2015 Integration Marathon ftfy Edit: Here is the solution: I=\int_{0}^{\frac{\pi}{2}} \frac{x\sin{x}\cos{x}}{\sin^4{x}+\cos^4{x}}dx = \int_{0}^{\frac{\pi}{2}} \frac{(\frac{\pi}{2} - x)\sin{x}\cos{x}}{\sin^4{x}+\cos^4{x}}dx \therefore 2I = \frac{\pi}{2} \int_{0}^{\frac{\pi}{2}}...

2014 is by far the hardest paper. If you want a confidence booster, try the 2012 paper.

re: HSC Chemistry Marathon Archive Ooo and if the hsc doesn't have that question then you don't buy me a car... you buy me a house Deal?

Not really... iirc, james ruse has a ~95% cohort average for extension 2 math. They have ~90% average for chemistry. And the cohort sizes for both subjects are 100+ people. (Might need soulful to correct these because this is what I had heard before trials, but I doubt that the averages changed...

re: HSC Chemistry Marathon Archive First time for everything buddy

re: HSC Chemistry Marathon Archive Although the dot point says to have one specific chemist, meaning you have to find an actual chemist who has to be working. I have heard that you need one male and one female chemist who have different roles. In terms of the chemical principles, you can say...

Whoops, I keep on getting confused by them