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$The moon's name is actually \emph{Io} (starts with capital i), not Lo.$

$For the first question, the formula in my previous post does not apply, since the region is not one where one of the given curves forms the ``upper boundary'' of the region, and the other the lower boundary. Here is a graph of the situation:$...

$In general, sketch the curves to see what the required region looks like and find the $x$-values of the points of intersections of the curves so you know what to use for the limits of integration. If the function whose curve is further from the $x$-axis (i.e. higher up, if both curves are above...

Re: MX2 2016 Integration Marathon Is this just a typo (and doesn't affect the rest of the working)? (The second fraction on the R.H.S. is just 0.) Ceebs carefully reading through the whole thing haha.

$Note that $\angle BAO = x^\circ$ and $\angle OCB = y^\circ$ (this is because triangles $BAO$ and $BCO$ are isosceles, as they contain two equal sides each (equal radii of the circle $AO$, $BO$ and $CO$) (and base angles of isosceles triangles are equal)), so $x^\circ + x^\circ+y^\circ +...

Re: MX2 2016 Integration Marathon 6 pages...now that is one tedious integral!

Re: MX2 2016 Integration Marathon I doubt this integral would be asked in HSC MX2 these days.

Re: MX2 2016 Integration Marathon $Note that the process would be to take the limit as $u\to 0^+$, so $u\ln \left|\sin u\right|\to 0$ when evaluating the lower limit.$

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread A way to find the area of a parallelogram is the find the perpendicular distance from one vertex to the line through the side opposite, and multiply this by the length of that line (i.e...

$I haven't looked at this proof closely so I don't know whether this affects the proof, but at the start, $z^2 - a$ is not $(z+a)(z-a)$, it is $\left( z + \sqrt{a}\right)\left( z-\sqrt{a}\right)$.$

$Here's how to do (c) (iii) $(\beta )$.$ $Note that the roots to the equation$ $ $z^3 + z^2 + z + 1 + z^{-1} + z^{-2} + z^{-3} = 0\ \ (*)$$ $are precisely the six non-real seventh roots of unity, namely $\mathrm{cis}\left(\pm\frac{2\pi}{7} \right),\mathrm{cis}\left(\pm\frac{4\pi}{7}...

Re: HSC 2016 4U Marathon - Advanced Level I think maybe Sy123 didn't notice your edit to include the complex roots case (or maybe he started typing his reply before the edit).

Re: HSC 2016 4U Marathon - Advanced Level Direct expansion.

Re: HSC 2016 2U Marathon It's a HSC 4U technique of calculating volumes. It's not required for 2U or 3U. (You can find out about it here: http://www.stewartcalculus.com/data/CALCULUS%20Concepts%20and%20Contexts/upfiles/3c3-Volums-CylinShells_Stu%20.pdf )

Re: HSC 2016 2U Marathon Is this from a 2U Paper/Textbook? If it's from 4U, you can easily do it using cylindrical shells.

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread $Since $(1+i)$ is a constant, we just apply the quadratic formula as usual, with $a=1$, $b=-1$, and $c=1+i$.$ $You will need to find the square root of a complex number to simplify the answer. Have you learnt how to do this yet?$

$Let the length of wire (in metres) for the circle be $C\geq 0$ and that for the square be $S \geq0$. Then we must have $C+S =3$ (since we're using a total of 3 metres). We want to find the values of $C$ and $S$ that will minimise the total area.$ $Now, since the perimeter of the square is $S$...

Re: MX2 2016 Integration Marathon A devilishly tedious integral. :p

Re: HSC 2016 4U Marathon - Advanced Level $Try substituting $y=x+\xi$ where $\xi >0$; then the desired result should fall out, recalling that $x\geq1$.$ $In fact, it is clear from your last line; we assume $x<y$, so $x-y<0$, and we can divide out the factor of $(x-y)$ from the last line and...

Re: HSC 2016 4U Marathon - Advanced Level Try considering f(x+a) – f(x), where a is positive and x ≥ 1.