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ME1-96 ME2-94 All the while I was aiming for 100 and 98.

In all honesty, the concepts that 3 and 4 unit encounter are really quite simple. Perhaps this is because I have finished both the ME1 and ME2 courses. Having said that, I didn't do too well - I have always had a bad exam record.

Lol, I've done that before too.

Damn, MATH1151 doesn't offer this =[

-1/4 my friend

For engineering, I would assume the entire ACPW section as well as integration by substitution. Combinatorics is quite redundant and as is circle geometry...

http://community.boredofstudies.org/14/mathematics-extension-2/236297/complex.html

B/Com (Actuarial Studies)/ B/Sci (Advanced Mathematics) (Quantitative Risk)

Saw some hotties whilst attending UNSW info day - are there any Russian imports to this uni?

Mah-deh - If you have any problems, you could always call me up homie. Either way, you're a clever kid and will be fine.

Flashback bump. Look how far we've come xD

\text{In one of the questions above, you proved that} \,\, z_{1}^{2}+z_{2}^{2}+z_{3}^{2}=z_{1}z_{2}+z_{1}z_{3}+z_{2}z_{3} \,\, \text{where the points} \,\, z_{1}, z_{2} \,\, \text{and} \,\, z_{3} \,\, \text{were the vertices of an equilateral triangle.} \\\\ \therefore...

You've posted this question already...although this disguises it quite nicely.

This is a nice solution.

Maybe - I'll be there for O-Week and if I see a congregation of weirdos I'll know what's up ^^

Divide the equation by \sqrt{3}\cos 4t

You were very close though, keep at it :)

\frac{1}{2}\left [ \frac{\cos\frac{\theta}{2}\left ( \cos\frac{\theta}{2}-i\sin\frac{\theta}{2} \right )}{1-sin^{2}\frac{\theta}{2}} \right ]=\frac{1}{2}\left [ \frac{\cos\frac{\theta}{2}\left ( \cos\frac{\theta}{2}-i\sin\frac{\theta}{2} \right )}{\cos^{2}\frac{\theta}{2}} \right ] \\\\ =...

:O Does not! Gella eats forbidden banana ^^

Yeah, neg rep queen - you suck