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Re: HSC 2017 Maths (Advanced) Marathon It can be done using only high school algebra as 1729 did, but admittedly it probably wouldn't come up in a HSC paper (not without any guidance anway). You can find out more about weighted averages at the Wikipedia article for it...

Re: HSC 2017 Maths (Advanced) Marathon Well done. :)

Re: HSC 2017 Maths (Advanced) Marathon $\noindent Well what you've done here is use the same weighted average for each $\overline{x}_{j}$ ($j=1,2,\ldots , K$). This is because you let them all be equal to $\sum_{i =1}^{n}\alpha_{i}x_{i}$ (so you gave each of them necessarily the same set of...

I think so.

However, a few years ago, 'The president of the Board of Studies, Tom Alegounarias, said it was ''a substantial exaggeration'' to suggest HSC physics was more of a history than a science.' (Though he 'conceded there was ''more language in it and less scientific and mathematical analysis''.')...

[Source: http://www.theaustralian.com.au/national-affairs/education/feminised-physics-a-formula-for-failure-says-michelle-simmons/news-story/b26b46725b1d0a97e57585aabea20f37 ] Thoughts?

Re: HSC 4U Integration Marathon 2017 I guess you can post your LaTeX attempts (or do it in the LaTeX Practice Thread) and people may give LaTeX tips then (since then we can see what's wrong with the code, if anything).

Re: HSC 4U Integration Marathon 2017 Still learning what? LaTeX, or Integration?

Yeah, but they weren't merged at the time trecex1 posted his link, they only got merged after I posted the link to this one.

He was probably meaning to refer to this: http://community.boredofstudies.org/1178/general-discussion-2017-hsc/359272/what-has-everyone-done-holidays.html .

General isn't good for ATAR I think (or for learning maths), especially compared to the other levels of HSC maths. You should do consider doing 4U.

$\noindent Replace $x$ with $x+2$. The answer is $(x+2) + 2y -4=0$, i.e. $x+2y -2 = 0$.$

$\noindent Let $P$ be parametrised as $\left(-2t, -t^{2}\right)$ ($t\in \mathbb{R}$). The tangent at $P$ has slope $t$, so has equation $y = -t^{2} + t\left(x + 2t\right)$, i.e. $y = tx +t^{2}$. The intersection points with the tangent and the parabola $x^{2} = 4y$ thus satisfy $x^{2} =...

Once we enter the complex world, the square root function is multi-valued (in fact, double-valued) (except when z = 0, in which case sqrt(z) has only one value). (Actually also for real positive numbers, it'S multivalued). We need to specify something called a "branch" to make the square root...

Some languages (e.g. Latin) generally scale very well as well.

$\noindent If you're after $\sin \theta$ and $\cos \theta$ in terms of $t\equiv \tan \frac{\theta}{2}$, these are$ $$\sin \theta = \frac{2t}{1+t^{2}}, \quad \cos \theta = \frac{1-t^{2}}{1+t^{2}}.$$

$\noindent Do you want $\sin \theta$ and $\cos \theta$ in terms of $T\equiv \tan \theta$? For $\theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, these are$ $$\sin \theta = \frac{T}{\sqrt{1+T^{2}}}, \quad \cos \theta = \frac{1}{\sqrt{1+T^{2}}}.$$ $\noindent Also, $\mathrm{d}\theta =...

Note that arctan(t) + arctan(1/t) is only equal to pi/2 if t := a/b > 0. If t < 0, then it's -pi/2. The result in the question in the case t < 0 is similar, but with a minus sign in front of one of the trig. functions.

Try searching up their annual school reports. Here's some James Ruse ones: http://www.jamesruse.nsw.edu.au/about/reports . E.g. The 2015 report for James Ruse says its median ATAR that year was 99.15.

$\noindent Let $r$ be the radius of the circle $\mathcal{C}$ with diameter $CF$. Once we find this, we are pretty much done (answer will be $CE = 2r-2 = 2(r-1)$). Consider $\triangle OAE$, where $O$ is the centre of $\mathcal{C}$. Then Pythagoras' Theorem here gives $r^{2} = (r-2)^{2} +...