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I think the website as a whole (not just post-Exam threads) may have declined in popularity drastically over the years. A good resource is Google Trends, which provides data of relative search popularity on Google over time for any phrase you're interested in. The search term "Bored of Studies"...

$\noindent Let $\alpha$ be the smallest angle and $\beta$ the largest. The smallest angle is the one opposite the shortest side and the largest angle is the one opposite the longest side. So by the cosine rule, we have$ $$\begin{align*}\cos \alpha &= \frac{16^{2} + 19^{2} -5^{2}}{2\times...

$\noindent a) Since $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we have $\cot \theta = 0$ iff $\cos \theta = 0$, which is when $\theta = \frac{\pi}{2} + n\pi$, for some $n\in \mathbb{Z}$.$ $\noindent b) Either $\sin \theta = 0$ or $\cos \theta = \frac{1}{2}$. The former equation has...

$\noindent For the first one, let $I = \int \frac{3x^{3} - 2}{x^{2} + 4}\, \mathrm{d}x$. Then$ $$\begin{align*}I &= 3\int \frac{x^{3}}{x^{2} + 4}\, \mathrm{d}x - \int \frac{2}{4 + x^{2}}\, \mathrm{d}x \\ \Rightarrow I &= 3\int \frac{x^{3}}{x^{2} + 4}\, \mathrm{d}x - \tan^{-1}...

Re: ATAR Notes vs Bored of Studies Wow, that is a lot of online users (on ATAR Notes). It does seem to be their highest ever though so far. Do you think it'll continue to increase? They do cater for other states as well though, so maybe that boosts their usage?

Note that for the second question, it is a hyperbola (with eccentricity 2), and its equation can be written in the form ((x – h)^2)/(A^2) – ((y – k)^2)/(B^2) = 1, where (h, k) is where the hyperbola is "centred". Since 4U students study conics, they should be able to get this answer and bypass a...

In fact, since the line y = -1 is a line of the form y = const., we can immediately write down the distance of P(x, y) to this line as being |y+1| without need for calculation. (In general, distance of P(x, y) to the line y = C would be |y – C|.)

Its name is pretty much taboo on BOS (has been for a long time I think), reason being (I'm guessing) that it's a sort of rival site to BOS.

Did you ask your teacher why you've been losing marks for such apparently trivial things?

They can probably find out your name from your student number without needing to memorise it (probably not supposed to though).

Incidentally, here's the course song for Harvard's Stat 110: .

I guess so, but it's probably safer to use the algebraic method.

$\noindent Yes pretty much, except you don't need a graph to show $k^{2} > 2k+1$ for $k\geq 4$, it can be shown algebraically: $k^{2} = (k+1)(k-1)+1 > 2k+1$, since $k-1>2$ and $k+1>k$.$

$\noindent Using implicit differentiation / chain rule, differentiating both sides wrt $x$, we have$ $$\begin{align*}y + xy' + \frac{1}{1+\left(x+y\right)^{2}}\cdot \left(1+y'\right) &= 0.\end{align*}$$ $\noindent Note that at $x=0$, we have $y=1$, so at this point (substituting $x =0,y=1$...

Re: HSC 4U Integration Marathon 2017 $\noindent Equation $(\star)$ follows from using $\int_{-b}^{b}f(x)\,\mathrm{d}x = \int_{-b}^{b}f(-x)\,\mathrm{d}x$.$ $\noindent To help show $(\star \star)$, recall or prove that $\int_{0}^{\frac{\pi}{2}} \frac{1}{1+\tan^{\alpha}x}\,\mathrm{d}x =...

Re: HSC 4U Integration Marathon 2017 $\noindent Another method is as follows (we can replace $2016$ with any non-negative even integer $N$ and $\pi$ in the integrand with any positive real constant $a$, so we will do this). $\noindent Let $I:= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}...

Re: HSC 4U Integration Marathon 2017 What's your final answer? Can't seem to spot it. Edit: Oh found it.

You should probably do subjects you think are best for your ATAR if you want to get in to UNSW Medicine.

Easily possible

Re: HSC 4U Integration Marathon 2017 This is of course a proof that 22/7 > pi.