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yeah I just realised it all failed sigh.......

x = 60 lol BEcause If you look at the angles at the bottom corners , they both equal to 80. There fore you have a cyclic quadrilateral. Since the the top chords subtend equal angles. there fore the angles subtended from the left chord are equal making x = 60 Sorry If I'm not clear

Re: MX2 Integration Marathon for 1<x<2 1/2<(1-1/x)<1 0<(1-1/x)^n<1 Divide the whole inequality by n Integrate the whole thing with respect to x with limits 0 to 1 0<(integral)<1/n Then apply the limit as n approaches infinity to get the result

Re: MX2 Integration Marathon Lol what steps did you take?

Re: Does anyone here play chess? Lol do you play in NSWJCL?

Re: MX2 Integration Marathon Use integration by parts (lol) I = xsin(x^2) - integral (2xcos(x^2)) let u = x^2 for the integral final answer should be I = xsin(x^2) - sin(x^2) + C P.S: How do you use the latex lol

Re: MX2 Integration Marathon Let u = x^2 then integration by parts There fore I = 1/2 (x^2.e^x^2 - e^x^2)+C

Re: HSC 2013 2U Marathon

Re: MX2 Integration Marathon I'm new to this so... so you use that f(x) = f(a-x) then you add them together. 2I = sin^-1(x-pi/4) then of course divide by 2

What course do you want to do?

Man it's annoying sometimes when you've spent a month preparing for an exam and you don't get the mark you expected.....:mad2:

Unless other schools set them at a later date, it's a bit unnecessary.

Pretty Good overall, though there were some tough questions. Like, who would have thought that rates of change would combine with parametric.

But but..... M + 5/6 can't be an integer...just sayin

The Q there is not an integer.

Did you guys managed to get the MI in Q 14 a? I couldn't factorise a 6 out :(

Re: 2013 CSSA 4U Maths thoughts? lol surprisingly easy haha

Yeah lol I didn't get how that graphs question is worth 7 marks ... I just talked about the problems associated with the nitrogen and sulfur oxides

The CSSA chem trials wasn't that bad lol

Haha I wouldn't realise it's you though, since there are many other maths genius like you and the fact that I wouldn't recognise your profile. Now how did you find me?