Search results

Re: 2012 HSC MX2 Marathon good stuff, someone revised this thread

average bro

I was also considering purchasing one, yet there are so many better and more difficult papers + answers available, and solutions are available everywhere online. If it was my call i would say it is not beneficial.

yeh that's what he means. I don't think the range is able to be solved by HSC means, but prove me wrong :)

go to an english tutor

Re: HSC 2012 Marathon :) well done, now post a new question

Re: HSC 2012 Marathon :) oh god i cbfd latexing for that haha, ill let some other latex leet do that one

here's a good question from my test today if you want to have a crack at it, its not very hard if you find the trick. <a href="http://www.codecogs.com/eqnedit.php?latex=the~equation~x^3-kx^2@plus;k^2x-1=0~has~roots~a,b,c\\ find~and~equation~with~the~roots~\frac{a}{bc},\frac{b}{ac},\frac{c}{ab}"...

Re: HSC 2012 Marathon :) YES WOOOOOOOOOO!! now you post a question :D :D :D

Re: HSC 2012 Marathon :) oh no you copied the question wrong originally :( but that is how you do it, except if you can, dont switch radians for degrees

Re: HSC 2012 Marathon :) ahah maybe your form of simple is my form of hard lol. I mean more conceptually difficult then actually process difficult, and more because of the fact that i have never seen anything like it and that in a test I would have tried substitution for 10 minutes and gave...

Re: HSC 2012 Marathon :) fuark that is a crazy question ok next one, find,<a href="http://www.codecogs.com/eqnedit.php?latex=\lim_{x\rightarrow -5}\frac{\sqrt{20-x}-5}{5@plus;x}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\lim_{x\rightarrow -5}\frac{\sqrt{20-x}-5}{5+x}"...

Re: HSC 2012 Marathon :) i think I have it, is it this? <a href="http://www.codecogs.com/eqnedit.php?latex=\frac{log_{2}e}{1@plus;log_{2}e}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\frac{log_{2}e}{1+log_{2}e}" title="\frac{log_{2}e}{1+log_{2}e}" /></a>

Re: HSC 2012 Marathon :) so so so close, just check one of your signs

Re: HSC 2012 Marathon :) 2^log(x) diverges as x-->0, so the integral cannot be evaluated by HSC means, this is just going out on a limb, im not really sure lol edit: shit it converges ahhhhhhhh

Re: HSC 2012 Marathon :) other then that last step where i missed the fraction, can you show where my substitution working has gone wrong deswa or timske? also, wolfram alpha confirmed the result as 1/ln(2) as well.

Re: HSC 2012 Marathon :) nope, you guys have the root3 part correct hahah

Re: HSC 2012 Marathon :) nope, go again :) post your working and i will check where you have gone wrong.

Re: HSC 2012 Marathon :) i am sure that is the answer

Re: HSC 2012 Marathon :) <a href="http://www.codecogs.com/eqnedit.php?latex=\int_{0}^{1}2^xdx\\ let~u=2^x\\ du=ln(2)2^xdx\\ at~x=1,u=2\\ at~x=0,u=1\\ =\int_{1}^{2}\frac{1}{ln(2)}\\ =(x)ln(2)~dunno~how~to~do~numbered~box~brackets\\ =ln(2)" target="_blank"><img...