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With a triangle the angles place restrictions on the sides which is why all that is required to two triangles to be similar is for their angles to be equal. In quadrilaterals, unlike triangles, there are so such restrictions on the lengths of the sides so more evidence is required to call two...

son of a bitch I could have sworn I saw 26/30 yeh, I need some good fucking sleep..

noice m8

farrkkk, gun brah. I wish I was that good in english.

bahahahhaha, well i dont have the paper right now dude but remind me in the next couple days or something and ill defs do it lol.

yeh thats good :)

yeh thats good :)

wait a second dude, I think did do that LOOOLL...I'm not kidding.

bahahhahahahhaahahahahhahahhahahhahaha lol.

oh yeh i am.. sorry, braindead tday

aren't u tonnes good? i remember a couple pages back your ranks were fucking gun.

you have to be trolling..

oh yeh guys, got 96% on my recent 3U maths exam. I always do stupid shit that stops me from getting 100%. This time I copied one of the trigonometric expressions onto my answer sheet incorrectly. AND AFTER THAT, I copied the expression that I copied incorrectly from the question booklet...

Re: 2012 HSC MX2 Marathon wasnt sure..

C=(P+kv^3)t t=d/v => C(v)=(P+kv^3)(d/v)=Pd/v + kdv^2 C '(v)=-Pd/v^2 + 2kdv To minimise C we set C '(v)=0, => 2kdv^3=Pd v^3=P/2k v=cuberoot(P/2k) show that its a minimum

Re: 2012 HSC MX2 Marathon can you treat 'i' as a constant?

soo much yess!!!!!!

That's a strange school you go to..how many assessments have you even done for them to even think of giving you an atar estimate?

Carrotsticks, can you please put the solution to this one?

Re: 2012 HSC MX2 Marathon ^ cylindrical shells would be easier there