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There's a problem with cross-disciplinary learning here. Maths ext2 ppl don't know how to apply mechanics and etc into the real world BECAUSE they don't have to. That's for the physicists. Just out of interest, what school do u go to? In our school, often we can separate like maths-good...

I agree

There's a neat way and a dodgy way to use memory (in maths at least) The neat way (IMHO) is to memorise (i rather the term "learn") the various methods to solve 3u and 4u problems. That way, you're not remembering any crap, so you're not wasting time in your studies, and since the syllabus...

yeah

Sourie, You seem to enjoy saying how dumb other people are, and how dumb the BOS and the HSC is. Are you just trying to sound smart by saying that others are dumb? If so, let me advise you that it's not working (on me, at least), and that people who get cheap thrills from stepping on...

pff... Some people use their memories to get through high school maths, be it extension or not, and some people use their maths skills. Depends which one works out better for them. But if you think maths isn't a good indication of skill because u think it's all memorywork, it means you...

Well it was harder than my expected 3U paper. Anyway, for those who're interested in 7b) WARNING: SPOILER AHEAD! (i) Consider the polynomial: a(x) = (c0) + 2(c1)x + 3(c2)x^2 + ... + (n+1)(cn)x^(n) The integral of a(x) (call this A(x)) is: A(x) = (c0)x + (c1)x^2 + (c2)x^3 + ... +...

nice, josie_is_slut adrenalin rush,..?

f(x) = pi/2

Yeh clue for 7b) polynomials, calculus, binomial.

It was the hardest test, compared with all the past papers I've done. If only it was this hard for 4unit...

There is nothing wrong with 7b(ii)

depends how well others do... they do well: scale less do shit: scale more. That's why you should always hope for a hard test.

Here's a nice elegant way for part (ii) Select any combination of three cards. Of arranging these three cards, there are six ways of doing so, but only one of these will give you decreasing order. Hence P = 1/6 :jaw:

The card question was the silliest question. Part 1 was just a figure the probability of the first card being 4 or greater. Part 2 just needed to figure out internal arrangement. As for geometry, it was predictable as usual: cyclic quads, similar triangles, angle-chasing around the...

This bit is a little iffy... arctan gives values between -pi/2 <= arctan(x) <= pi/2 But arg(z) gives values between -pi <= arg(z) <= pi Clearly you've missed the case arg(z1) = arg(z2) +-pi So I'll fix by starting that b/a = d/c so either 1)arg(z1) = arg(z2) +-pi or 2) arg(z1) =...

O and you're right btw about the algebraic method thing. I just find it stupid proving the extreme case of the triangle inequality (which is btw, another method: "By the triangle inequality, |z1 + z2| <= |z1| + |z2| {equality when z1, z2 point in the same blody direction}")

Elegant? How ironic! Elegance is an easy way to solve a hard problem, not the other way around.

I thought it would've been obvious that if you have length of one side of a triangle equal to the sum of the lengths of other two sides, you have a degenerate triangle, with the two vectors pointing the same way.

Willy, use cos rule: |z1 + z2|^2 = |z1|^2 + |z2|^2 - 2|z1||z2|cosx x = pi - (arg(Z2) - arg(Z1)) show this using diagram since |z1 + z2| = |z1| + |z2| |z1 + z2|^2 = |z1|^2 + |z2|^2 + 2|z1||z2| cosx = -1 x = pi arg(Z2) - arg(Z1) = 0