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Re: 2012 HSC MX1 Marathon what do you mean by a little square?

Re: 2012 HSC MX1 Marathon prove true for n=1, lhs = 1/1^2 = 1 rhs = 2-1/1 = 1 1≤1 .'. true for n=1 assume true for n=k 1/k^2 ≤ 2 - 1/k RTP 1/(k+1)^2 ≤ 2 - 1/(k+1) k>0, .'. k+1 > 1 .'. (k+1)^2 > k^2 .'. 1/(k+1)^2 < 1/k^2 using the assumption, 1/(k+1)^2 < 1/k^2 ≤ 2 - 1/k .'...

108 is in the 2nd quadrant already, but doing 180-108, you get 72, which is in the first quadrant and doesn't make sense since [1-root5]/4 is a negative value. what you could do [what i usually do but i don't know how others do it] is change the value of [1-root5]/4 to make it positive instead...

for questions 1-3 consider the identity sin^2 x + cos^2 x = 1 for 4 change tan to sin/cos for 5 use double angle formula for sin2a and change tan to sin/cos for 6 double angle formula and consider sin(2x + x) instead of sin3x [unless you've learnt the identity for sin3x off by heart]

this conic is a hyperbola branching sideways i.e. x^2 is positive therefore the focus is S(+-ae,0) i think where you got lost is assuming that S could only be +ae, when it could actually be -ae as well since the question doesn't specify that S is +ae and S' is -ae [i.e. it could be that S is...

that's my grade :( i'm going to get 3 new teachers for modern, eco and chem next year

i got 29/30, lost a mark in a locus problem because i left my "a" out when eliminating the parameters -.-

Advanced English - read all my texts and find a couple of related texts for module C and AOS 4U maths - finish off curve sketching 3U - finish off integration Economics - read over all of second topic (Australia's place in the Global Economy) not really doing anything for chemistry, but i...

I know of a few people in my grade as well who ask for notes/help from others yet never returning the favour in any way, but it's mostly a small issue for me as my friends and I tend to share our ideas/notes. I think our school is a lot more competitive than other schools...

your volumes seem a bit too small...did you mean cubic cm? as for the questions you're stuck on: 1. if you think about it, the curved surface is basically your sheet of paper, so therefore its area is the dimensions of your piece of paper, and if you used the same piece of paper for both...

finished yesterday, but i got a headache and sore throat somehow so not feeling as happy as i should be... :(

800-1000 is what i'm told by my teachers for 40-45 minutes. 1300 is possible if you already know the question and such, but it'd be really difficult to pull off that many words without knowing the question beforehand

oh, right, i see what you mean. thanks for the clarification!

hey spiralflex, i'm not completely sure what you mean by the division formula not working...either i've misunderstood what you meant or there's something wrong with my working because i managed to get the answer using the formula. letting the ratio that P divides the interval AB be k:d...

yeah, i just checked with my teacher today, and he said the same thing. thanks for the help!

if t=tan[theta/2], tan[theta]=2t/(1-t^2) cot[theta/2]=1/t .'. you get {[2t/(1-t^2)] - t}/{1/t + [2t/(1-t^2)]} = {[2t-t(1-t^2)]/(1-t^2)}/{[1-t^2+2t^2]/[t(1-t^2)]} = {[2t-t+t^3]/(1-t^2)}/{[1+t^2]/[t(1-t^2)]} = {[t(1+t^2]/(1-t^2)}/{[1+t^2]/t(1-t^2)]} from this you can cancel some of the...

ah, ok, thanks for that! edit: sorry, the constant is actually p. does that change anything?

for what values of p does the equation px^2 - 4x + p = 0 have real roots? the answer is -2<=p<=2, but i was wondering, can p=0? because it just asks for real roots, and if p=0 then x=0, but does that still count?

LOL, it's like an extension from yours :P

that parabola is a sideways version of the parabola x^2=4ay, with the x and y axes interchanged, so therefore your focus is also switched so that it's S(a,0) rather than (0,a) to find the focal length, just do the same as you would with a normal parabola, except this time the coefficient of x...