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Re: 2012 HSC MX1 Marathon What about now: <a href="http://www.codecogs.com/eqnedit.php?latex=\textup{It is known that the series }1@plus;\frac{1}{2}@plus;\frac{1}{4}@plus;\frac{1}{8}... \textup{ converges to 2}. \\ \textup{By comparing the series }...

Thanks.....

Talk about things like ice core samples, testing air and comparing to 'clean' locations etc. You can talk about the fact that only recently technology has developed to allow accurate readings therefore we have no real figures to compare to. I'd probably also mention that as sulfur dioxide and...

Re: 2012 HSC MX1 Marathon <a href="http://www.codecogs.com/eqnedit.php?latex=\textsl{Prove that the series } 1@plus;\frac{1}{2}@plus;\frac{1}{3}@plus;\frac{1}{4}@plus;... \textsl{ diverges}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\textsl{Prove that the series }...

I don't have time to post a solution but I'll outline my method and maybe someone else can help/you can do it. 1. Prove for n=1 2. Assume true for n=k (3^k>k^2) 3. We need to prove that 3^(k+1)>(k+1)^2. This is the same as proving that 3^(k+1)-(k+1)^2>0 4. Split 3^(k+1) into 3(3^k) and then...

You made a mistake in your first line. It should be the integral of (x+2)^2 not (x=2)

<a href="http://www.codecogs.com/eqnedit.php?latex=V=\pi\int_{0}^{2}y^2dx\\ V=\pi\int_{0}^{2}(x@plus;2)^2dx \\V=\pi\int_{0}^{2}x^2@plus;4x@plus;4dx\\ V=\left [ \frac{x^3}{3}@plus;2x^2@plus;4x \right ]^{2}_{0}\\ V=\frac{56\pi}{3}" target="_blank"><img...

Re: HSC 2012 Chemistry Marathon The properties of Technetium 99m make it very useful for its applications as a medical tracer in medicine. As it undergoes gamma decay (maybe write equation, but its only 3 marks), it can be placed inside the body effectively 'tagging' the required area. For...

Re: 2012 HSC MX2 Marathon Still doesn't work...

Re: 2012 HSC MX1 Marathon That is fantastic :). I'm going to rep you for that.

Re: 2012 HSC MX1 Marathon Yep. That's what me and a few friends got but we don't have the actual answer so I can't tell you 100%. I'm pretty confident though. How did you get it so fast, my method was quite complicated... If you don't want to type it all out, can you just post a general...

Re: 2012 HSC MX1 Marathon This is an interesting question: Consider a circle with equation <a href="http://www.codecogs.com/eqnedit.php?latex=x^2@plus;y^2=10" target="_blank"><img src="http://latex.codecogs.com/gif.latex?x^2+y^2=10" title="x^2+y^2=10" /></a>. From the point P (-3,4), two...

Wait what? Did you argue like hell for that?

Depends on the marker. Its right but I know a few teachers that frown on that sort of thing even though there's nothing wrong with it. I'm guessing you can so it in the HSC but maybe someone else can confirm. Honestly though, if you've proven that LHS=x=RHS, just reverse the steps you used to...

<a href="http://www.codecogs.com/eqnedit.php?latex=u=x^3-1 \\ \frac{du}{dx}=3x^2\\ du=3x^2dx\\ \therefore \int 3x^2\sqrt(x^3-1)=\int \sqrt(u)du\\ =\frac{2(u)^\frac{3}{2}}{3}\\ =\frac{2}{3}(x^3-1)^\frac{3}{2}@plus;C" target="_blank"><img src="http://latex.codecogs.com/gif.latex?u=x^3-1 \\...

I think in this case they didn't get 100% raw mark. For example, one of my friends got 83/84 this year which was reported as 100 but when they calculate state ranks, they go to decimal places so someone that got 84/84 would have beaten him.

Re: HSC 2012 Chemistry Marathon A- Citric acid Analyse the relationship between the position of elements on the periodic table, and the acid-base behaviours of their oxides.

With a lot of these, the easiest way is to graph the original function and then test a few points. For example, with y=x^2-1, it is clear that for every value of x, y=x^2-1 will be one less than y=x^2. Therefore you just shift the graph down one unit. When you do a few of them, you will be able...

Yep. That's how you do it. First you have to establish that the point P is directly above the focus and then you can just use the focus directrix definition.

I think you'll find that if T is on the directrix and the x axis, due to symmetry of the ellipse, the points P and Q both lie on the line x=ae (This is from memory so I might be wrong). Have a go using that fact :)