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Thanks everyone. This has really been helpful =)

Hi, I am doing the HSC next year and i really need a band 6 in english to achieve my goal. So here is my question, how do you get a band 6 in english? Is it witchcraft, do you have to know every Shakespearian play? Do you constantly have to read books like moby dick and stuff? How is it done?

Is there a list that tells you which codes correspond to which subjects?

Re: distinguished acheievers list Im also in it for maths woot woot ;) Congrats to anyone who makes the list =)

Instead of thinking of it like that, divide it into real and imaginary, as such \frac{1}{z} = \frac{x-iy}{x^2+y^2} = \frac{x}{x^2+y^2} - \frac{y}{x^2+y^2} i Then its clear by the definition of a conjugate what the result is. (Sorry jetblack2007 for the intrusion)

This is generally what you do, but of course if the shape has points D, C, B and A it is conventional to start with the first one alphabetically ie A, then just go clockwise around the shape.

The | | is the modulus, and we know |z| = \sqrt{x^2+y^2} So |z - (2+2i)| = |x-2 + i(y-2)| =\sqrt{(x-2)^2 + (y-2)^2} Also |z-(6+6i)| = |x - 6 + i(y - 6)| = \sqrt{(x-6)^2 + (y-6)^2} \therefore 3\sqrt{(x-2)^2 + (y-2)^2} = \sqrt{(x-6)^2 + (y-6)^2} ie (3x - 6)^2 + (3y - 6)^2 = (x-6)^2 + (y-6)^2...

Ah, okay ;) But if you are having trouble with the question, with the criterion you have given i can tell you that if the coefficient of the polynomial is an and s|an for n=0,1,2,... then s | P(x).

the symbol | means "divides", so 5 | 10 means 5 divides 10 (making 2) or x | y means x divides y. It also means that y is a multiple of x ie y = kx. Also x is a divisor of y. All these statements are equivalent and when working on these problems its useful to consider which one to work with...

My suggestion is that you get yourself a copy of Cambridge mathematics 4 unit and do the vectors part in the complex numbers chapter. In order to succeed in 4 unit math you must not only learn how to apply a method (like simply addition of vectors) but you have to understand the method and...

Could you say which rule regarding polygons you are talking about?

fixed ;) or more simply from = \frac{1+cos\theta-isin\theta}{1+cos\theta} = \frac{1+cos\theta}{1+cos\theta} - \frac{isin\theta}{1+cos\theta} = 1 - \frac{isin\theta}{1+cos\theta} $Let$\,\, t=tan\frac{\theta}{2}, \,\, theta \neq \pi \,\, $Then$\,\, z = 1 - \frac{i\cdot \frac{2t}{1+t^2}}{1+...

I have not touched over complex number vectors in quite some time, but its seam to me that ||z| - |w|| = |z|-|w|, so we must only show |z| - |w| \leq |z+w| ie |z| \leq |z+w| + |w| Construct the vectors z and w. Complete the parallelogram to get z+w. From the triangle inequality, |z+w| +...

z^5 = -1 z^5 = r^5\left(\cos5\theta + i\sin\theta5\right) -1 = \cos\pi + i\sin\pi \therefore r^5\left(\cos5\theta + i\sin\theta5\right) = 1\left(\cos\pi + i\sin\pi\right) equating terms r^5 = 1 \implies r = 1 cis 5\theta = cis \pi Generalising, cis 5\theta = cis\left(2\pi k + \pi)...

Do we put our school results in from this year or last year. Im in year 10 so would i put my school results in from year 9? Thanks

Do we have to fill in our school results or does the school do that?

What do ya think a 94/120 will get me =S

Thankyou sooo much =)

Thankyou sooo much =)

Hi, Does anyone know where i can find some preliminary physics yearly exams? Thankyou