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The HSC sucks and getting it done in one year at the end of highschool is the quickest way to put this crap behind me.

yeah, 4 subjects and at least 10 units and you're sweet.

I semi-meditate. I like to chill out, relax and harness my chi. A lot of people sit there cramming up until the minute before the exam but I prefer to put everything away at least 15-25 mins prior in order to calm myself. Also, before math exams I'll sometimes listen to mozart to increase my...

Yeah, bar our average of 43% :p

If you're trying to clamber around the upper UAI ranges where students put a lot of effort in, the HSC can be quite tough.

∫ cosec²x dx = ∫ cot²x.sec²x dx = ∫ (tanx)^-2.sec²x dx let u = tanx ---> du = sec²xdx = ∫ u^-2 du = -1/u = -1/tanx = -cotx +c

I assume you want to know how to prove them? Working backwards: ∫ secx.tanx dx = ∫ (cosx)^-2.sinx dx let u=cosx ---> du = -sinxdx = - ∫ u^-2 du = -(-1/u) = 1/cosx = secx + c ∫ cosecx.cotx dx = ∫ (sinx)^-2.cosx dx let u=sinx...

There are certainly 4 dimensions. Beyond that... it depends on how many you need to make your theory work I geuss.

:p, you're missing out man. There are some interesting ideas in that stuff.

I expect he's refering to the ideas in recent theoretical physics, eg: http://www.damtp.cam.ac.uk/user/gr/public/qg_ss.html http://relativity.livingreviews.org/Articles/lrr-1998-1/download/lrr-1998-1.pdf So naturally I think you will agree that there are 11 dimensions ah... even though...

Haven't you heard???

The rate of change of area with respect to time: dA/dt = 4 The rate of change of area with respect to radius: A = πr² hence dA/dr = 2πr -----> dr/dA = 1/(2πr) dr/dt = dA/dt . dr/dA = 4. 1/(2πr) = 2/(πr) so given that dr/dt = 2/(πr) ---> when r=10, dr/dt...

The first three pages are a little too easy, they don't really include any extended questions from those topics. The last page is more interesting but even then it's fairly medium level difficulty.

I've pretty much done Complex, Graphs, Polynomials, Conics, Integration and about half of Volumes.

Find the locus of the point P(x,y) that is equidistant from the line y=-1 and the point (0, 1)

You're right, you use the Acos(x-a) transformation. You'll just end up with Acos(2x-a) instead so that you have to halve all the values you calculate (or if you have a range you'll be going through double the range).

For Q3. Solve the equation: 5cos@ + 12cos@ - 13 = 0 I think you may have mistyped it but that's just 17cosθ = 13 cosθ = 13/17 and solve as per usual. Is there a square, a sinθ or a 2θ in there somewhere? For Q4. Solve the equation: 4cos@ + 3sin@ = 1 use...

b) Hence find the equations of any quadratics that pass through the origin and are tangent to both y= -2x - 4 and to y = 8x -49 I've never tried one like this but I'll give it a shot. If the quadratic passes through the origin then y = ax² +bx + c, where c=0 y= -2x-4, gradient...

1. a) Show that the tangent to P: y = ax^2 +bx + c with gradient m has y-intercept c - (m-b)^2/4a y' = 2ax +b m=2ax +b x = (m-b)/2a when the gradient of the tangent is m y= a([m-b]/2a)² +b[m-b]/2a +c = (m² -2mb +b²)/4a + (2mb -2b²)/4a +c...