Search results

These notes are EXCELLENT! Very detailed and simple to understand :) Highly recommended!!!

Re: HSC 2012 Marathon :) But the *hint* was a fraction.. it didn't say you had to use one

Re: HSC 2012 Marathon :) An alternate solution with a different substitution: I= \int \frac{dx}{\sqrt{x}\sqrt{x+1}} \\ $Let u$=\sqrt{x+1} \Rightarrow \sqrt{x}=\sqrt{u^2-1} \\du=\frac{dx}{2\sqrt{x+1}} \\ =2 \int \frac{du}{\sqrt{-1+u^2}} \\= 2ln(\sqrt{u^2-1}+u)+C \\ =2ln(\sqrt{x}+\sqrt{x+1}) + C

Your last comment lol

lmao^

my bad!!

Let's try a different approach $The domain of $y=sin^{-1}(x^2)$ is equal to the range of its inverse function, so let's find the inverse$ \\ x=sin^{-1}(y^2)\\ sinx=y^2\\y=\sqrt{sinx} \\ $We know the values of $sinx$ range from $ -1\le x \le1,$ however as it is in a square root y must be $\ge0.$...

Why the hell is that 3 marks lol

It really depends on the person imo

roflmfao

Re: HSC 2012 Marathon :) lol'd

From what I've heard and looked at so far, and from what my teacher has told me, it is only the HSC examination that contributes to your ATAR. If you can't sit your exam for whatever reason, they use the mark you got in the trial. As far as I know there are no ranks/moderated assessments marks...

I'm pretty sure the HSC Exam is the only thing that counts towards our ATAR

For those of you still wondering about sin. $Graph cosx from $-2\pi<x<2\pi \\ $Now graph sinx from $-2\pi<x<2\pi You're welcome

Lol no it's in the 3u prelim course

Carrotsticks just did...

That's 3u

God help the creative writers 50 years from now....everything will be a cliche.

Been following gunners since '02. They have so much potential but for some reason the squad always cracks under pressure

they're all y=x ?