Search results

yeh it is 24abc i forgot the extra 4

2

my final answer is 24abc but i did it really fast so i might of made an error. All I did was factorise using sum and difference of 2 cubes

Re: 2012 HSC MX1 Marathon Well of course, you ended your Proof with qed and open square, Must be a solid proof

Re: 2012 HSC MX1 Marathon well then suffer all the silly mistakes, (1/3 > 1/2)

Re: 2012 HSC MX1 Marathon it's alright, just lay off those cookies till after hsc

Re: 2012 HSC MX1 Marathon you ridiculing carrotsticks

Re: 2012 HSC MX1 Marathon those cookies make you say the darnest things

and besides this question is in terry and cambridge too, its a typo in this paper, but the quickest and best method to do this question is use properties of geometry and not to use algebra

its properties of geometry, if we are given the diagonals are equal, there are only two shapes that this occurs, due to pythegoras, which means the vertice angles will be 90 degrees. I also confirmed this was wrong using algebra method (which is very long) and i ended up with tan inverse of...

limited spots left (2-4 max). Only did revision of last year stuff with a few extras. If you want to join, next week will be deadline, otherwise you will be too far behind.

part (e) is wrong....geometrically speaking if |z_{1}+z_{2}|=|z_{1}-z_{2}| then that refers to the diagonals being equal when you do vector addition. Now there are only two shapes that this occurs for, which is the square and rectangle, both of which have 90 degree vertice angles. Now...

Re: 2012 HSC MX1 Marathon \!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\!\! LHS =tan^{4}x \\ = (tan^{2}x)^{2} \\ = (sec^{2}x -1)^{2} \\ = sec^{4}x- 2sec^{2}x + 1 \\ = sec^{2}x(sec^{2}x-2) +1 \\ = sec^{2}x(tan^{2}x - 1) +1 \\ = sec^{2}xtan^{2}x - sec^{2}x + 1 \\ = RHS

Re: 2012 HSC MX1 Marathon yeh its bad

bumpity bump

Re: 2012 HSC MX1 Marathon Because 10C5 picks 5 people to go At table one, but then those 5 people Still need to be rearranged 4! Ways Around the first table as well

Re: 2012 HSC MX1 Marathon Yeh a combo of 3am and iPhone spell check

Re: 2012 HSC MX2 Marathon Yeh I have proved that question numerous ways, Including the good old fashion gradient AB = gradient BC way, whIch doesn't involve arguments, but I Feel If you understand thE argument method then you should Use it, besides the question said they were concyclic points...

Re: 2012 HSC MX1 Marathon This should be the only method you use for this Sought of question, and if you don't knOw it then Your teacher is bad lol. I went through a chord of Contact question involvIng the hyperbola with my Class the other day and if you didnt know this method That...

^