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Re: HSC 2012 Marathon :) How? In the original function, x can not be 0.

Hey guys, for fun's sake and ofcourse the sake of the HSC, I've decided to start a new marathon thread (after the last one died). Answer the questions then post your own :). To encourage MX1'ers of any level of ability to take part, please rate your questions in terms of difficulty. To start...

lol, i live in GWS and its completely fine. No where near as bad as to classify us 'disadvantaged'. meh, no one minds an advantage for themselves.

LOL, I completely forgot about the formula for the sum of the first n terms. I did it this way. Let the first term be a, common ratio be b. a+ab+ab^2+ab^3+ab^4=77 (1) ab^5+ab^6+ab^7+ab^8+ab^9=-2464 (2) From (2): b^5[a+ab+ab^2+ab^3+ab^4]=-2464 i.e. 77b^5=-2464 b=-2 From (1)...

The reason I couldn't give two shits about being behind.

lol dude, one question. When you wrote this...were you really serious?

^ what umat would you need?

you're going to be at a great disadvantage compared to the others who have already taken two years of chem. not only do you have to learn high school chem but at the same time you're gonna need to do the chem in med.

yeh I was just thinking about what Riproot said. should have done chem mate

^ I was kidding, I mean no one is expected to be able to derive the formula for centripetal acceleration in uniform circular motion let alone do it in an exam lol. But yeh, in year 12, there are people who simply memorise formulas for say maximum height reached by a projectile when they can do...

I knew about the 'simpler' way (the way deswa just did it) I just couldn't be bothered thinking backwards lol. To be honest, using algebra would be the dopie way to do it.

ah yeh true, its pretty damn simple that way too lol.

There are some errors in the sentences, it's not too bad though.

that actually requires more thinking LOL.

You can actually doing it using elementary methods. Just use the substitution u=x^2 then decompose the fraction.

Just slightly correcting realisenothing. The factorised form of the equation is actually k(x-alpha)(x-beta) where k is some constant since obviously many parabolas can be drawn passing through alpha and beta.

LOL, when in doubt, derive the formulae.

^ oh is it lol, that's cool then. That's why I love physics right now. No memorisation involved yet.

It's a matter of opinion which one is the hardest topic lol. I honestly loved the topic because it didn't involve much memorisation at all. Electricity looks a lot different. It's gonna be a tonne more theory.

^damnit, I tried making a table but it didnt work. The ones on the left represent the amount sabrina has at the 1st, 2nd, 3rd exchange and the ones on the right representing the amount samantha has. x and y is how much they had initially.