Search results

$\noindent 17. (a) The height of the ladder is clearly $2\sqrt{\lambda^2 -\alpha^2}$, by Pythagoras. So $B$ is $\left(0, 2\sqrt{\lambda^2 -\alpha^2}\right)$. So note that the midpoint $P$ is $\left(\alpha,\sqrt{\lambda^2 -\alpha^2}\right).$ $\noindent For (b), I'm not sure what they mean, of...

Haha yeah. Meant tedious like having to solve simultaneous equations etc.

Guessing she meant all of them haha. They're mainly just tedious / dull.

Ah ok, good :).Feeling a bit lazy to write out my method now, maybe wil later, unless someone else posts their solution and is similar to mine.

Is the answer: 20??

What about Gödel's Incompleteness Theorem?

There might be a better way, but keep in mind that it is possible to tediously find the values of sin (and hence cos and then tan) of all integer angles from 1° to 90°. This is described for example at this page...

$\noindent We'll assume $x \in \left[0,2\pi\right)$ (if it's not, convert it to the equivalent angle in this range). We have$ $$\begin{align*} |z|^2 &= (1-\cos x)^2 +\sin^2 x \\ &= 1-2\cos x + \cos^2x +\sin^2 x \\ &= 2-2\cos x \quad (\text{using the Pythagorean trig. identity}) \\ &= 4...

$\noindent Using partial fractions, the primitive is $\frac{1}{2}\ln \left| \frac{1+x}{1-x}\right|$, $x \neq \pm 1$ (if $x=\pm 1$, the argument of the log is either undefined or 0, neither of which is permissible).$ $\noindent So your statement is essentially just saying that for $|x|<1$, this...

Re: MX2 2016 Integration Marathon Yeah, this question is similar to that Q., I just changed it slightly to yield the infinite sum for a different number. :) It's not too hard a Q. since it's a guided one, but I put this in mainly to give a HSC-style Q. since we haven't had one of those for a...

Yeah, just an algebraic rearrangement. It made no difference here though as u was just 0, as Nailgun noted.

Re: MX2 2016 Integration Marathon I suppose they give us "guidance" in the sense that they tell us what the inequality is we're supposed to prove, whereas seanieg89 wanted people tackling his question to come up with relevant inequalities etc. themselves.

Re: MX2 2016 Integration Marathon $\nondent \textbf{A NEW QUESTION}$ $\noindent Let $x\geq 0$. Let $n$ be a positive integer.$ $\noindent (i) Show that$ $$-x^n \leq \frac{1}{1+x} -\left(1 - x + x^2 -x^3 +\cdots +(-1)^{n-1} x^{n-1} \right) \leq x^n$.$ $\noindent (ii) Hence, use integration...

Re: MX2 2016 Integration Marathon I thought he was being serious.

Ah right (didn't bother to read the actual question haha).

$\noindent The rearranged form is $a = \frac{2(s-ut)}{t^2}$. Since you got the answer right (according to the given answers), I suppose you just forgot to write the brackets?$

Re: HSC 2016 4U Marathon No, 'major' just means the longer one, 'minor' the shorter one. It's just that for most of these HSC Ellipse Q's, they place the longer one on the horizontal axis. The 'semi-' means half, so semi-major axis is half the major axis, like how a radius is half a...

$\noindent For the first one, we want $x^2 \leq 9\Longleftrightarrow -3\leq x \leq 3$. So the domain is $\{ x\in \mathbb{R}: -3\leq x\leq 3\}$.$ $\noindent For the second, the domain is just $\mathbb{R}^+$ (the set of all positive reals).$ $\noindent For the third, the domain is all real $x$...

Generally speaking you do need to memorise chemical equations for all the relevant reactions in the HSC Chemistry course, since you usually need to include them in your answers to get marks (there are usually marks associated for including relevant equations).

It's probably better to just stick to Advanced. Standard is often said to be 'easy', but that actually doesn't mean it's easier to get a higher ATAR with.