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Unfortunately I don't :( ... I was considering doing it, but it would mean that my accelerated 2U - which I did this year - would have gone to waste. Oh well, more time for 'fun' subject like English and SOR [jks, I absolutely hate English - words cannot describe] haha Thank you very much for...

Hey everyone! :) I've seen people at school with these 'PDF documents' which have a range of questions from past HSC exams that are organised into sections by topic. I was just wondering if anyone had one which they would be willing to share with me? It would be greatly appreciated...

Thank you for your input 'HAX0R', however i'm questioning the validity of your previous statement considering you are only in year 7. I'm intrigued! why you think Binomial Theorem is the easiest? What specific parts of motion do you find hard and why? Cheers

Ahhh! Haha ok cheers! I'll do a 'bit' of extra work on binomials haha

Hello everyone! :D I was just after some personal opinions regarding what topics people found hardest in 3U? :) Thanks in advance for your input (if there is any that is. If there isn't, then that's awks for me :P haha )

Same here man!! I did heaps of practice papers and I was getting above 90% in almost every one and had an average of 93% during the year!! But, I get into the exam room and I can't think straight! Fuarr! Lost marks in Q11 as well :/ (just like in the trials hahaha) True that! It's over!

Oh my! I'm in a very similar position to you! I made THE most retarded mistakes in my life! No joke... We just had to screw up this test didn't we?! Fuarrrr So frustrating knowing that you could have gone much better!

Re: HSC 2013 2U Marathon Good luck to everyone tomorrow!! :D :D I hope you all do very well!

Phew! At least I didn't get 10! :D haha

MAJOR TYPO! haha It should be "...then a/d + b/d = c/d" NOT "a/d + b/d = c/a"

Soz, after going over it again, there is a problem, but besides that it's fine... I'll put up how I would have done it in a sec

The triangle is a right angle triangle.

Re: HSC 2013 2U Marathon

Re: HSC 2013 2U Marathon Yeah, the answer is 5^x.ln5... There are a few ways which you can do it.

Re: HSC 2013 2U Marathon 1 + (sinx)^2 + (sinx)^4 + (sinx)^6 +... = 1/(1-(sinx)^2) = 1/(cosx)^2 = (secx)^2 Therefore, the integral of "1 + (sinx)^2 + (sinx)^4 + (sinx)^6 +..." = the integral of (secx)^2 = tanx Therefore the answer is tan(π/4) - tan(0) = 1

Re: HSC 2013 2U Marathon Yeah, that's correct - then just use the table and test the values around the 'possible point of inflection' to confirm (or not to confirm) that it is a POI. The reason why you need the table is because in some cases f''(x) may equal 0, but there is no POI. For...

Re: HSC 2013 2U Marathon Hmmm... I agree with asianese "so trixxy". One question, is this some sort of geometric progression? Bit stumped lol :P

Re: HSC 2013 2U Marathon I'm curious - What other functions also satisfy the above 'restrictions' besides f(x) = lnx?

Re: HSC 2013 2U Marathon π = pi sin(2π-x) = -sin(x) sin(π/2 - x) = cosx ∴ sin(2π-x)/sin(π/2 - x) = -sin(x)/cosx = -tanx