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here's my attempt at number one... I = \int \frac{1 - \cot^{n-2}(x)}{\tan(x) + \cot^{n-1}(x)} \, \mathrm{d}x &= \int \frac{\cot(x) - \cot^{n-1}(x)}{1 + \cot^{n}(x)} \, \mathrm{d}x \begin{align*}\textup{Let } u = \cot(x) \textup{, so } \mathrm{d}u &= -(1 + \cot^2(x)) \...

\begin{align*}z^5 &= i\\&= \text{cis} \left ( \frac{\pi}{2} \right )\\&= \text{cis} \left ( \frac{\pi}{2} + 2k\pi \right )&&&k\in\mathbb{Z}\\z&=\text{cis} \left ( \frac{\pi}{2} + 2k\pi \right )^{\frac{1}{5}}\\&=\text{cis} \left ( \frac{\pi}{10} + \frac{2k\pi}{5} \right )\\\end{align*}...

tywebb posted the link a while back (with an alternate solution and generalisation to Q8 b) iirc)

Oh lol Terry Lee's book makes quite a fuss about it anyway the solutions are here

well you know that y = \sin(x) so you can say \mathrm{d}y = \cos(x) \, \mathrm{d}x use the slicing method and you should get it :-) edit: you have to put 2\pi \int_{0}^{\frac{\pi }{2}} x(1-sinx)dx (no spaces)

hmm.. try using: \left ( 1 + x \right )^m\left ( 1 + \frac{1}{x} \right )^n &= x^{-n}(1+x)^{m+n} and comparing coefficients

hmm.. unsure of the actual question but I think this is the gist of it: 1. \begin{align*}\frac{\mathrm{d} }{\mathrm{d} x} \left ( \frac{\ln (x)}{x} \right ) &= \frac{x\times\frac{1}{x} - 1\times\ln(x)}{x^2}\\&=\frac{1-\ln(x)}{x^2}\end{align*} 2. \begin{align*}\therefore \int\frac{1 - \ln...

your friends are incorrect see page 18 http://www.boardofstudies.nsw.edu.au/syllabus_hsc/pdf_doc/science_revised_support.pdf

is it the seventh root of 7?

o) \begin{align*}\int \frac{\left ( \sqrt{x} - 1 \right )^2}{5\sqrt{x}} \, \mathrm{d}x &= \frac{1}{5} \int \frac{x - 2\sqrt{x} + 1}{\sqrt{x}} \, \mathrm{d}x\\&=\frac{1}{5} \int x^{\frac{1}{2}} - 2 + x^{-\frac{1}{2}} \, \mathrm{d}x\\&=\frac{1}{5}\left ( \frac{2}{3}x^{\frac{3}{2}} - 2x +...

are you sure? think about it – if you look at it from the top with magnets on either side the coil would shrink from a rectange into a line, still remaining perpendicular to the magnetic field

Aren't the sides of the coil always perpendicular to the magnetic field?

In general, most acids are weak, however you should know that: hydrochloric (HCl), nitric (HNO3), sulfuric (H2SO4), hydrobromic (HBr), and hydroiodic (HI) acids are all strong. Strong bases you should know: NaOH, KOH, LiOH, RbOH, and CsOH. A combination of electronegativity & the size of the...

it's in jacaranda :P

simple: Zn2+(aq) + 2H2O(l) ⇌ Zn(OH)2(aq) + 2H+(aq) not simple: Zn(H2O)42+(aq) + H2O(l) ⇌ Zn(H2O)3OH+(aq) + H3O+(aq)

just use \begin{align*} blah &< 2blah\\ 2blah &> blah\\ etc. \end{align*} when in latex (the asterisks are to prevent auto-numbering)

\begin{align*}|4x - 1| &> 2\sqrt{x(1-x)}\\(4x-1)^2 &> 4(x(1-x)) \qquad (\textup{Both sides are greater than zero})\\16x^2 - 8x + 1 &> 4x - 4x^2\\20x^2 - 12x + 1 &> 0\\(10x - 1)(2x - 1) &> 0\\\implies x < \frac{1}{10} \textup{ or } x>\frac{1}{2}.\end{align*}\\\textup{But } \sqrt{x(1-x)} \textup{...

Re: 2012 HSC MX2 Marathon correcto \\\textup{By considering the area under the curve,}\\\\y=\frac{1}{1+x^2},\\\\\textup{between 0 and 1 as the limit of a sum of rectangles,}\\\textup{evalutate,}\\\\\lim_{n\to \infty} \sum_{k=1}^{n} \frac{n}{n^2 + k^2}.

Re: 2012 HSC MX2 Marathon :jump:

except some of the H+ ions provided by the HCl will react with OH- ions to form water...