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depends on what you want.. if you want a polynomial, it's the lagrange polynomial (check wiki) t = 11[(x-1)(x-2)(x-3)]/6 - 6[(x-1)(x-2)(x-4)]/2 + 3[(x-1)(x-3)(x-4)]/2 - 2[(x-2)(x-3)(x-4)]/6 works out to be t = x^2 - 2x + 3 if you know it's a quadratic, you have t = ax^2 +bx + c in...

yeah .. it's easy. Read your router manual if you use one

tails, white tie

Probably won't get you into Law at Usyd.. but Uni shouldn't be a problem. Your subjects are fine... high 90s are quite achievable with those subjects (provided one gets high band 6s for each subject)

they could but there's no official policy on that

well, near 0, the numerator is roughly log(N)*x....

the limit is log(N) doesn't look like 2 unit

Anything fancy is probably bad.

Yours is wrong... it's sqrt(4-3)/2 = 1/2 OP: foci are at (+/- 1,0)

Stare at it til you figure out how to do it

Take a proper summer holiday?

the december camp is slightly less intense than the april one. don't have that sheet .. but it's usually a mix between AMC like questions and AIMO. It's more raw aptitude that they are looking for rather than knowledge. That said many people does have substantial background. occasionally there...

Dodgy solution.. here's probably what the author of that solution was thinking and what was wrong with that.. the author was trying to calculate P(atleast 1 not chosen) by P(1 chosen) + P(2 chosen) + ... + P(n-1 chosen) The precise and correct interpretation is P(exactly 1 chosen) +...

You learn quite a bit... before you go to the camp you get a set of 30ish questions so you have a feel of the standards required. you start each morning with a 4 hour AMO/IMO like test.. except the first day, which is an AIME (multiple choice, 999 choices for eqach question). then classes run...

serial transmission works by sending data bit by bit over the same wire.. parallel transmission sends a few bits at the same time thorugh different wires. generally serial ports require some additional hardware or software flow control. parallel connection are generally cheaper.... and they...

Done correctly -> yes. In particular, with inequalities.. they usually want you to prove that LHS < RHS where LHS and RHS depends on some parameter(s) say x. so the statement they want you to prove is for all x, LHS(x) < RHS(x) to prove it by contradiction, you would first assume that...

integral [0 to y] x^(4n+2) = 1/(4n+3) y^(4n+3) assuming y <=1 allows you to conclude that it is less than 1/(4n+3)

helps if you post the questions. for the 2nd one.. the key is to notice that 2 - 1/(n+1) > 2 - 1/n + 1/(n+1)^2 it's not that hard to prove by rearranging, noting that n(n+1) > (n+1)^2. you can also prove the inequality by noting that the sum is less than 1+ integral[1 to n] 1/x^2 dx by...

helps if you post the question. You integrate the result from part 2 with limits 0 to y. the condition >= 0 is to ensure the integral does not flip signs andthe condition <= 1 is there to allow you to estimate the last bit.

+1 There's a good reason for a higher UAI for syd and UNSW, especially when UNSW is at a relatively less convennient location than UTS.