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There was no ambiguity whatsoever about the question. The word ONE was not a metaphor for, nor was it alluding to TWO related texts.

Well, it all stems from the fact that roots are equally spaced out around a circle. If you take the first case, using De Moivre's Theorem, [cis(pi/4)]^(1/3) = cis(pi/12). This gives you the starting root. For the other two, just add and subtract 2kpi/3 [to maintain equal angular spacing]...

According to ATAR calculators, the two subjects I have coming up (Chemistry + Economics) will be major determinants of my ATAR. I'm in the same boat as you, genuinely cannot muster the motivation to do anything.

This is silly. The aura surrounding the students who did two related texts is obviously elevated...blah blah blah they can adapt two different texts...what a crock of shit. I had to cut my prepared essay (yes I prepare) and did ONE related text as if it were in an essay for two. No I did not...

Pardon me if I was too vague. He could have posted with a follow-up question. As he didn't, I thought it sufficed.

I'm \,\, going \,\, to \,\, assume \,\, it's \,\, z^{3}=\frac{1+i}{\sqrt{2}} \,\, because \,\, this \,\, expression \,\, yields \,\, workable \,\, values. \\\\ z^{3}=\frac{1+i}{\sqrt{2}}=cis\left ( \frac{\pi}{4} \right ) \\\\ z = cis\left ( \frac{\pi}{12}+\frac{2k\pi}{3} \right ), \,\, k=0,\pm 1...

Yeah, I got the induction and barely got that rectangles question...I winged the first part of the probability but don't think i'll get any marks for it. I worked backwards and deduced that W = p/(1-q^n), but didn't note that it arose from GP limiting sum.

Hey, you should change your ATAR aim to 99.8+ to suit those individual subject aims :hat:

Why FYL? I suppose test WAS simple...oh well =[

Same here. If lucky, maybe like 97.

I think it was sqrt[2]-2/sqrt[3]

I differentiated 3 times for this...

http://www.speedyshare.com/917426113.htmlHere you go.

I was told you lose the mark once. If you don't add it in another question, you won't get pinged.

(i) Using the exterior angle of the triangle inscribed, @ + arg(z+1) = arg(z-3) ===> @ = arg(z-3) - arg(z+1) = pi/3(ii) This question is quite complex and requires you to recall some circle properties. Alternatively you could use inverse trig by letting z = x+iy, which will eventually give you...

LOL xD

22C7 x 15C6 x 9C5 x 4C4 = 22!/(4! x 5! x 6! x 7!)

Okay, what you have done is essentially split the fraction in a mathematically incorrect manner (I think). Recall that A/(B+C) =/= A/B + A/C. What you have there is an integral in the form f'(x)/f(x) which is always ln|f(x)| + C.

Can't see a flaw with lolokay's solution.

Lol, there's never a cue for such obscenities :P