Search results

How would I describe my HSC... well, it was: U07B4TYMkhc

Shorter is better. Shorter gives the sense you're more direct and to the point rather than rambling on...

You continue in what you like.

I know that 1 + ... + (n-1) = \frac{n(n-1)}{2} because that's the sum of the first n-1 numbers. And I know that: \frac{n(n-1)}{2} = \frac{n!}{2!(n-2)!} because if you simplify the RHS, you get the LHS And I know that \frac{n!}{2!(n-2)!} = C(n,2) = \; ^{n}C_{2} because that's like... the...

You'll do fine. Anyways, I was looking at some of the stuff in MATH2130: Higher Methods for Differential Equations. LOL i am so dead: "By letting y = [x; t]T we can transform a non-autonomous system (1) in Rn to an autonomous system in Rn+1:" what what what what what

Nice!

Ah well, don't worry about it. It's over now. Just focus on next sem and work hard from Day 1. Now you know uni isn't as easy as you might've thought before.

Ah okay. Well, I'm a guy so I wouldn't really know about all this... formal dress whatnot. But yeah, if you say so... i'm sure someone more well versed in this will come along soon and help you out :p

Remember that Arts is hard to do well in. Lots pass, but few get HDs. It's not like say... maths where if you do super well, you know you'll get a good mark.

yeah you'll need to ask soemone else for fins1613. i'm not touching finance ever again. YAY! But yes, maths I can help you with.

oh we can crop them out then :p

And you can do it. If you want, I'm thinking some group study sessions next sem. want in?

SnowFox obviously did well taking photos if no one really noticed. EDIT: ...and the Librarian. they'll get a natural and non-posed feel to the pictures

I'm going to uni tomorrow. gonna do mathsoc stuff

dont tell me you invited a gazillion people :p

You're not going for the evening gown-esque wear?

Re: UWS Chatter Thread but you aren't just anyone you're a bos-er NOW GO (H)D IT UP!

(also answer my text, woman!) or i'll get you that lovely red dress you looked -so- nice in :p hehehehehehehe

You have to prove: e^{C(n,2)} > n! Okay, now we note that that is equivalent to: C(n,2) > \textup{ln}(n!) So if we prove that, we prove what we need to. We know that: \textup{ln}(n!) = \textup{ln}(1) + ... + \textup{ln}(n) We use our inequality, x > \textup{ln}(1+x). So for...

uni.books@gmail.com ? You made a whole new account for selling books? oh you