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SOR1 definitely

Is it b^2 / 4 by any chance?

I wished the hsc was fixed... like everyone does the same assignments and exams

4unit integration? Na, i've only read parts of it.

It's worth $0.05 when there's no ink. (unless you're going to wear on your finger/ear/body or just show off)

Ok homos, we get it.

He prob wants someone to post the solution step by step. I'm definitely not going to do that

Yea..no way I'm typing everything back into wolfram to re-check. I probably made a mistake whilst entering the whole thing BTW, you sound like you've finished 4unit.

i duno why but i always think your tring to say that your the best

Is the answer for the gradient question -18? or +18? That question's fucking tedious

The set of complex numbers lie on right branch of x^2/16 -y^2/(9/16) = 1

Lol terry lee pg 64...how did I know. Terry lees solution is much better

That's an interesting question... Expanding that would give, 1-(sum roots)-(sum of two roots at a time) -......+(product of roots) I think it would eventually give 1+(n-1)= n. Lol this is because I saw that sum of non real roots = -1 and product of roots = 1

Uhhh it is? Don't think so...send me the link That uses the theorem |S'P-SP| = 2a for the hyperbola

You mean, http://goo.gl/a99V?

The points O, I, Z and P on the Argand Plane represent the complex numbers 0, 1, z and z + 1 respectively, where z = cosθ + i sinθ is any complex number of modulus 1, with 0 <θ <π . (i) Explain why OIPZ is a rhombus. (ii) Show that (z-1)/(z+1) is purely imaginary. (iii) Find the modulus...

Oh god. Noones going to type all of the proof. a) is x(x1)/a^2 - y(y1)/b^2 = 1 b) PS/PS' = (x1 -a/e) / (x1+a/e) TS/TS' = (ae-a^2/x1) / (ae + a^2/x1) =(x1 -a/e) / (x1+a/e).

Or you can go cos(pi/7) = 2cos^2 pi/14 - 1 and do the same for sinpi/7. The 1 will cancel out

So in other words, you wanted her to stay because you liked her? i.e. found her attractive I don't believe in good and bad teaching styles because every person has a different perspective of the teachers method

It's funny though...that question was quite specific on the lead-acid cell.