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its over............

Re: How many past papers have yous done ? yeh :\

D? To reduce transpiration?

yeh i dont think they come up in 4U

Re: HSC 2013 2U Marathon Write it out again without brackets, group the sines together and then group cosines together

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Re: MX2 Integration Marathon damnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn!

The values of 'a' and 'b' determines the 'fatness' and 'skinniness' of hyperbola A formula: a^2 e^2 = a^2 + b^2 Horizontal hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 a^2 e^2 = a^2 + b^2 Vertical hyperbola \frac{y^2}{a^2}-\frac{x^2}{b^2} = 1 a^2 e^2 = a^2 + b^2 Something to help...

\frac{1}{2}v^2 = xe^{-2x^2} \\ \\ \therefore a = e^{-2x^2} - (4x)(x) e^{-2x^2} \\ \\ \therefore a = e^{-2x^2} - 4x^2 e^{-2x^2} factorise and ur done

i can only think of approximating the solution through halving-the-interval method or newton's method

Re: HSC 2013 2U Marathon (in red) and if u sub in a =0, e^2a = 2a e^0 = 0 1 = 0 trollface.jpg

no sorry

Looks long...

Just do Sydney Grammar, you don't have much time!

Re: MX2 Integration Marathon I give up, my brain has been squished! Can you please your one-two line method? :D

He did a typo, its meant to be \frac{z}{z-i}

Q2. f) (ii) \Delta = (i)^2 - 4(-1-i) = 3 + 4i \\ \\ \therefore z = \frac{-i \pm \sqrt{3+4i}}{2} from part (i), you know the square root of (3+4i), sub in, find z

Try Sydney Grammar

yay!

Ahhh yesss! Thank you, I thought I was going to wait another half an hour....