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I hate those tests. The leading coefficient was negative... proof enough...

I choose what instrument I want to play and then I thrash it. It's a great way to let out stress.

What's the attraction to ivy league schools? Apart from an awesome campus and being able to join the skulls.

I could swear that was my school you were talking about. Except that our exterior is some hundred years older in places...

Best way to get owned: revealing your intentions to a board full of students and teachers.

'To be decided' .... but meant more literally than if it were actually part of my major work. I'll start it soon, I swear...

I agree with you that students should not be penalised for going beyond the syllabus. I also think that you should get a soap box for an avatar, just in the spirit of things.

Martin gave a pretty good explanation here: http://www.boredofstudies.org/community/showthread.php?t=60249

Good call, it certainly saves a lot of time... and space...

I'm doing 4U math, 4U science, 4U english and one thing I have to semi-disagree with on is your grouping science with math. While 'science' and math are sciences, science doesn't have a whole lot to do with math in the HSC. If you can solve a quadratic equation then you're over qualified. A lot...

We use Cambridge for 4U and fitzpatrick for 2U/3U. I'm not a fan of the fitzpatrick 4U textbook, it's way too easy and a lot of his 3U textbook is more difficult! If you can take it I think the cambridge is very beneficial.

1. a) at t=0 ---> M_oe^-kt=10 so M_o=10 at t=140: 5=10e^-140k ln(1/2)=ln(e^-140k) -ln(1/2)/140= k so k = 4.95 x10^-3 to 2d.p. b) After 70 days M = 10e^-70k = 7.07 grams c) 2= 10e^-kt 1/5 =...

Like Euler suggested, the partial fractions technique only works on P(x)/Q(x) where deg P < deg Q I think it has to do with the fundamental theorem of algebra and the fact that if two polynomials of degree n are equal for more than n values of x, then they are equal for all values of x...

Perhaps this is a novel idea, but you could ask all of these questions in the one post?

Just because, in terms of an explanation, it makes more sense. You can see directly why it works and it's an easy expansion. The technique of creating a polynomial with new roots is far more subtle.

1 = a(x² +1) +x(bx +c) = (a +b)x² +cx +a hence a+b=0, c=0, a=1 ---> b=-1 giving you: &int; 1/x - x/(x² +1) dx = lnx - 1/2ln(x² +1) + c = ln [x/&radic;(x² +1)] + c

x = a(x² +9) + (bx +c)(x+1) = (a+b)x² + (b+c)x + 9a + c a+b=0 b+c=1 9a+c=0 solve simultaneously and a= -1/10, b= 1/10, c=9/10 so you get: 1/10 &int; x/(x² +9) + 9/(x² +9) - 1/(x+1) dx = 1/10[(1/2)ln(x² +9) +...

I think the secondary syllabus in the US tends to be a few steps behind ours.

Came to my head first as well but it's the 3U forum and I just tend to think of that as a 4U technique. I don't know about less room for error though...

Motion passed.