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I hear some of us are worrying about the lack of queries / questions coming from HSC students. Considering that this is March (and the psychology that ext2 students are not posting dumb questions because they're afraid ppl will think they shouldn't be doing the course), I reckon this is pretty...

think that was a past HSC question...around question 6'ish...can't remember it's been so long...

[to OLDMAN] slanging match? umm...yeh oldman's soln is the usual approach and should always work unless the question is really a bitch. In summary the cookbook is as follows: 1) Differentiate 2) Sub in 3) repeat (2) until the solution comes out. but this one has an alternative...

1) Let A_1, A_2, ..., A_n represent the nth roots of unity w_1, w_2, ..., w_n. Suppose P represents z such that |z| = 1 (btw, w_1 is omega subscript 1, etc) i) Prove w_1 + w_2 + ... + w_n = 0 ii) Show that |PA_i|^2 = (z-w_i)(z(bar) - w_i(bar)) (for all i = 1, 2, ..., n) iii) Hence prove...

Ahh yes the first 2 were boring the one today was quite funny actually - although it was only for 1/2 hour (btw scheduled for 2 hours so yay to the freetime) and the communication tuts are cool - it's jus socialising...

Because "OLDMAN" is complaining about the lack of problems, I've decided to cause one myself... Prove that x^3 + 3px^2 + 3qx + r has a double root if and only if: (pq - r)^2 = 4(p^2 - q)(q^2 - pr) Enjoy :p

u're right; i wouldn't dare to write that solution in a real exam cos this was the first time i used that method to solve a conics problem and i was surprised it worked too :D

where/what/when? Umm UNSW/med/now

Idunno if my solution is the same as OLDMAN's, but anyway: When comparing z to 1/z, what's not important here is arg(1/z) = -arg(z) What's more important is that |1/z| = 1/|z| So If z1, z2, z3 represented by A, B, C, (origin O as usual), we have cyclic quad OABC (without loss of...

uhh...surviving how bout u? im supposed to be studying joints of the arm atm, aiya...

Here's a funky solution: Prove the case is true for a circle x^2 + y^2 = a^2 (which shouldn't be very hard at all, and only involves euclidean geometry). Here it is below, anyway: In a circle, angle QPR is 90', so you'll find triangle QPT is right angled at P. If QT and the tangent at...

Yet another way: completing the square (cos^2@)(x^2) + (sin2@)x + 1 = 0 (cos^2@)(x^2) + (2sin@cos@)x + sin^2@ + cos^2@ = 0 (xcos@ + sin@)^2 + cos^2@ = 0 (xcos@ + sin@ + icos@)(xcos@ + sin@ - icos@) = 0 x = (-sin@ - icos@)/cos@, (-sin@ + icos@)/cos@ = -tan@ +- i

Didn't realise the new address for this forum...sorri

I'll just write down the method. It's just basically co-ordinate geometry. 1) Find the equation of the normal at P(whatever, whatever) 2) Find the points X, Y (remember all points on the x axis has y = 0, and all the points on the y axis has x = 0) 3) Find the distances of PX and PY...

Well, after 2 boring hours on the train to canberra I've managed to come up with: Xk = integer part of (3^N)/2 this is the theoretical maximum because of the fact that each weighing gives you either L, M, R (base 3 information), and that you don't know whether it's heavier or lighter (divide...

Well you can't solve anything unless you have an equal sign somewhere so I'll assume z^2 - i = 0 so z^2 = i |LHS| = |RHS| so |z^2| = 1 |z| = 1 let z = cis(x) so z^2 = cis(2x) = i = cis(pi/2) 2x = pi/2 + 2k*pi x = pi/4 + k*pi (k = 0,1) so z = cisx = cis(pi/4), cis(pi/4 - pi) = (+-)...

hey how are the maths tutors out there? are they worth the money? Do they just brush up on school stuff or do they teach extra stuff that gives u some sort of an edge? hehe me jus doin a market survey

It's prolly: latuus rectum x = ae, of the general ellipse x^2/a^2 + y^2/b^2 = 1 Sub in x = ae: you have (ae)^2/a^2 + y^2/b^2 = 1 y^2/b^2 = 1 - e^2 y^2 = b^2(1-e^2) for ellipse, we have b^2 = a^2(1 - e^2) (1 - e^2) = b^2/a^2 y^2 = b^2(b^2/a^2) so y = (+-)(b^2/a) the distance PQ =...

If that is true then all the people on the merit list did more than 11 hours study per day, 7 days a week!

whoa ok whenever you make words from some but not all of repeated letters such as this, you've got to case bash. Here we have 5 different letters: H,O,N,G,K we have pair O,N,G Notice that you can have 2 pairs at max (since we have 5 letter words). So: case1: no repeated letters: 5...