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I think he's just giving you a lot of material to practice on, because that's basically the best way to improve at maths and refine your knowledge. But if you don't really understand the work he's giving to you, it's probably a good idea to get a textbook or a tutor so that you can go through...

Hahaha love the countdown.

Thanks, sounds optimistic :) Anyone else willing to give an estimate?

Got back most of my assessment task 1 results of Year 12, curious as to how I'm progressing :/ Preliminary: (just to fill in the results I haven't gotten back yet) English Advanced: 79% (~60/155) Biology: 94% (3/48) Chemistry: 88% (11/70) Maths 2u: 99% yearly exam mark (25/~110 overall) Maths...

Tbh I don't do PDHPE myself but I know a lot who do - apparently the course is mostly theory, with very little physical activities involved. Though if you do genuinely love PDHPE it's worth giving a shot. Can't say much about Modern History though.

Come from outta nowhere and get dux in the school :P

3) i. Solve y = x^2 - 3x and y = -x^2 + x +6 simultaneously (equate them to each other) 2x^2 - 4x - 6 = 0 x^2 - 2x - 3 = 0 (x-3)(x+1) = 0 x = 3, -1 sub these values into either equation and you'll get the y values for P and Q ii. Area between 2 curves, A = integral of TOP curve MINUS...

Do past HSC questions which are the same topic (not the whole paper), and if your school has them, past papers. Once you've got a hang of the harder questions, then its generally fine to do textbook questions. That's basically what I did and it was smooth sailing from there

Alright, so I basically did it like this: h = b-a/n, where h is the width of each subinterval and n is the number of subintervals = 2-(-2)/6 = 2/3 Drew a table of values with the x coordinates -2, -1, 0, 1, 2 and y values e^-4, e^-1, 1, e^-1, e^-4 Then I used the formula A ~ (b-a)/6...

Hey guys, I'm having trouble with this question: "Use Simpson's Rule and 4 sub-intervals to approximate the area bounded by the curve y = e-x2 , the x-axis, and the lines x = -2 and x = 2." It seems using the formula A = (b-a)/6 {f(a) + f(b) + 4[f(a+b)/2]...} from Cambridge gives me...

For maths, the best ones are probably Cambridge>Fitzpatrick>Jones & Couchman>Maths in Focus>etc.. dunno what to say about Terry Lee though

Maths 2U>Japanese>Maths 3U/Biology>Chemistry>Religion>English If only English wasn't compulsory ahaha. Pretty happy with 2U and Jap, got 99 in both woot.

Yeah this happens at most schools. Honestly if you want to improve your 2u mark all it takes is a good textbook and doing as many practice questions as you need (from easy 2u level to hard 2u level) to understand the stuff. 3u people find 2u waay easier cause the hardest questions they ask can...

Hey guys :) Subjects: Advanced English Mathematics Mathematics Ext 1 Chemistry Biology Studies of Religion I Japanese Beginners Aim: 99+ ATAR to safely get into UNSW Med, but I'd be pretty happy with 97+

My school's starting extension classes this term sort of like a trial. Anyway, good luck guys :D

never mind - i got it now, thanks asianese n = 14

lol nice edit. after that i just expanded the nC5 and nC6 into n!/5!(n-5)! and n!/6!(n-6)! but that's about it help?

Yeah i got past that part, k = 5, k = 6, but the problem is simplifying the coefficients' expression I ended up with something like nC5 x 3^-5 = 2 x nC6 x 3^-6 @deltaforce its the general expression for binomial expansions

"When (3+2x)^n (n is a positive integer) is written out as a polynomial in x, the coefficients of x^5 and x^6 have the same value." i. Find the value of n. ii. Show that t(k+1)/t(k) = 2(15-k)/3k iii. Hence or otherwise, show that the coefficients of x^5 and x^6 are the largest coefficients...

b) Parabola x^2 = 4y P (2p, p^2) 1: y = (x^2)/4 2: y' = x/2 at P (2p, p^2) y' = 2p/2 = p (tangent), -1/p (normal's gradient) 3: y - y1 = m(x-x1) y - (p^2) = -1/p (x - 2p) py - p^3 = 2p - x x + py = p^3 + 2p = equation of the normal c) Equation of normal x +py =...