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your going into yr11 next year correct?? im prob not the best person...but i would not suggest you begin ext2 together with ext1...you could get an idea of the work if you have the cambridge 4u book...currently im doing the maths in focus hsc course (maths in focus is good for a foundation but...

hehe yep i plan too =D i had a look at 4u n really different to 3u i was really surprised =O...btw sumone post a question to keep this thread going thnx =D

im not an accelerant i have just finished ext1 course from doing the maths in focus n cambridge books i want to be ranked first till yr 12 for maths =D...n no better to learn by yourself (at least thats wat i think)

no...wat makes him think im an accelerant?? o.O

huh?? i dont quite get u...

thanx templar =D

im having difficulty with this question: Find real numbers p, q and r such that: p^2 + q^2 + r^2 = 1 p + q + r = 3 p + 2q - r = 5 Thanx in advance =D

Yep! Good Work =D although you could have done it much simpler: Alternatively: x^3 + 12x^2 + 30x + 4 = 0 (x+4)^3 + 30x + 4 = 48x + 64 (x+4)^3 - 18x -60 = 0 (x+4)^3 - 18(x+4) + 12 = 0 This is in the form u^3 + cu + d = 0 Where, u = x + 4 .'. u^3 - 18u + 12 =0 Dont forget to post up a...

Yep i used the triangle <--- makes my life so much easier with the t-substitution btw i challenge you to be the first one to answer my question =D

1st post yay! =D

i go there for physics n maths seems good evrything is results focused...im doing the 4 week trial n i thinks its good but the location is too far from my house so i can only go on sats...i would highly recommend it =D

haha ok my solution: (1-t^2)/(1+t^2) + [t/√(1+t^2)]^2 = (1-t^2)/(1+t^2) + (t^2/1+t^2) = 1/1+t^2 is this ok gurmies?? =D ok i shall post up another question (easy): Use a suitable substitution to change x^3 + 12x^2 + 30x + 4 = 0 into a cubic equation of the form u^3 + cu + d = 0

go for it ext1 maths wasnt as hard as i thought it to be i never thought i would consider taking ext2 (which i am now) after doing the ext1 course. You'll do good...93 in SC = top 3-4% in state good luck

haha ok i'll limit it to 3u then =D gurmies u wanna post up a new question?

For gurmies' question: Sn = {n[2a + (n-1)d]}/2 AND Tn = a + (n-1)d So rearranging Sn : Sn = {n[a + a + (n-1)d]}/2 Sn = [n(a + Tn)]/2 2.Sn = n(a + Tn) (2.Sn)/n = a + Tn .'. Tn = [(2.Sn)/n] - a Then substituting Sn = 17n - 3n^2 Tn = {[2(17n - 3n^2)]/n} - a Tn = [2(17-3n)] - a Tn...

haha np bro =D

hey my ans: first finding the volume of the liquids: Liquid A = 2.4/1.5 Liquid A = 1.6 mL Liquid B = 0.5/0.75 Liquid B = 0.67mL (use 2/3 mL for accuracy) Now working out the concentration of the liquids: Liquid A = (1.6/250) x 100 Liquid A = 0.64% Liquid B = [(2/3)/250] x 100 Liquid...

thnx for the post michael i found it useful =D

Hey guys i need help with these two questions (they from cambridge 3u) Q1. Consider the equations 12x^2 - 4xy + 11y^2 = 64 and 16x^2 - 9xy + 11y^2=78 (a) by letting y = mx show that 7m^2 + 12m - 4 = 0 (b) hence, or otherwise, solve the two equations simultaneously Q2. (a) Find x...

ok since x = -b/2a (axis of symetry) .: -2 = -b/2a .: -4a = -b .: b = 4a using the general form of a parabola y = ax^2 + bx + c we sub in wat we worked out above to get: y= ax^2 + 4ax + c then using the two points specified we can make the two equations: 3= 4a-8a+c = -4a+c (subbing in...