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Use congruent triangles. (SAS) Then use that to deduce that the sides in the circle are equal and thus must be the radii of the circle, making the point the centre.

I would find the whole thing then half it, given that it's easier to integrate with the equation.

hell yeah, I always fist pump after getting a tough question. The exam supervisors think I'm high and shit.

inb4 enigma Anyways, let the side of the base be s and A be the surface area of the block you'll need \frac{dV}{dt} = \frac{dV}{ds} . \frac{ds}{dA} . \frac{dA}{dt} $it's given that$ \ \ \frac{dV}{dt} = -12 \\ \\ \\ V = s.s.\frac{s}{2} = \frac{s^3}{2} \\ \\ \\ A = s^2 + s^2 +...

Are you just required to sketch the inverse function, in that case you just need the equation of the inverse equation and not worry about the point of intersection with y = x $The inverse of$ f(x) = tan(\frac{x}{2}) \ \ \ \is \\ \\ \\ f^{-1}(x) = 2tan^{-1}(x) , $since the domain of the...

Mate I know someone who failed 2 of his 2U test (including the half yearlies) He got his game on real hard for the rest of the year, ended up in the 70s for the CSSA trials, then walked away with a B5. What I would advice is get some help from your peers and not just from your teacher, and of...

Re: HSC 2015 4U Marathon 0<x<1 \Rightarrow e^x < 3 \\ \\ \therefore \int_0^1 x^{\alpha}e^x < \int_0^1 3x^{\alpha} = \frac{3}{\alpha+1}

Re: MX2 2015 Integration Marathon I_0 = 2/3 when you multiply top and bottom by (2n+2)(2n)....(4)(2) you realise there are n+1 2s to factorise

yes it's basically getting the required mark. However some courses would have lecturers with strict marking criteria to reduce the amount of students attaining HDs.

Re: MX2 2015 Integration Marathon could be accelerated :P or it could be a uni student in disguise.

I don't really understand the first one lol. For the 2nd one Let the Midpoint of AB be M construct OM (where O is the center) and you'll see that angle OMA is a right angle as a radius bisects a chord at right angles. From triangle OMA, use Pythagoras to find OM (i got 10 mm), you'll see...

Find dy/dy using implicit differentiation (product rule) then sub the point (2,1) to obtain the gradient of the tangent then use the point gradient formula

2013 3u hsc :d

Re: MX2 2015 Integration Marathon $Let$ \ \ \ x = tan\theta \ \ \ \ \Rightarrow \int_{0}^{\frac{\pi}{4}} {\ln(1+tan\theta)} = I \\ \\ \\ = \int_{0}^{\frac{\pi}{4}} {\ln(1+tan(\frac{\pi}{4}-\theta)) = \\ \\ \int_{0}^{\frac{\pi}{4}} \ln\frac{2}{(1+tan\theta)} = \int_{0}^{\frac{\pi}{4}}...

Wut Sent from my C5303 using Tapatalk

ahhh crap...my rust is pretty evident :) EDITED!!

$since w is the cube root of unity,w^3 = 1 \ \ \ \ \ \ 1+w+w^2 = 0$\ \ and \ \ \ w = cis(\frac{2\pi}{3}) \\ \\ 1+2w+3w^2 = 1+w+w^2 +w+2w^2 = w+2w^2 = w^2+w^2+w+1-1 \\ \\ \\ = w^2-1 = cis(\frac{4\pi}{3}) - 1 = \frac{-1-i\sqrt{3}}{2} - 1 = \frac{-3-i\sqrt{3}}{2} and you do the similar working...

What trig substitution made you use partial fraction afterwards? $if you let$ \ \ x = 4\tan\theta \\ \\ \Rightarrow 16 \int sec^3 \theta \ \ d\theta From there you use IBP (integration by parts) then you sub it back in. (i dunno if that's the best method though lol)

2y+ x = 2log3 + log 2 = log3^2+log 2 = log 9 + log 2 = log18 Therefore that's incorrect

If you ever encounter a tough question and you feel that it would potentially burn your time (e.g you were like" woah wth is this??") best to just leave it and attempt ALL the other questions that you're comfortable with before returning to that question (and the others that you had trouble with...