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Oh k sorry about that. So when z = x+iy (x and y are real numbers) and i = sqrt(-1) z has 2 componants to it, the real part and the imaginary part. so Re(z) = x Because Re(z) means the real part of the complex number z. The imaginary part is the part in front of the i. So Im(z) is the...

$let z = x+iy$\\ \therefore z(bar) = x-iy\\ \frac{1}{z(bar)-i}\\\\ =\frac{1}{x-iy-i}\\\\ =\frac{1}{x-i(y+1)} \\\\ = \frac{x+i(y+1)}{x^2 + (y+1)^2} - $by realising the denominator\\\\ \therefore Im(\frac{x+i(y+1)}{x^2 + (y+1)^2})\\\\ = \frac{(y+1)}{x^2+(y+1)^2} = 2\\\\ \therefore (y+1) =...

Thanks for that mate, I have another question from Patel 8. A truck of mass 2000kg starts to climb an incline of angle given by theta = tan^-1(1/10). The total resistive force is 2000N Find the retardation it experiences

Question 2 from Exercise 7A I'm hopeless at Mechanics so I'm doing lots of questions to get better. The combined air and road resistance of a car in motion is proportional to v^2, where v is its speed. When the engine is disengaged the car moves down an incline making an angle sin^-1(1/30) with...

http://www.hsccoaching.com/Resources/2008Ext2.pdf question 3c? The way he did it was by working out In + I(n-1) on the left side then subrating I(n-1) at the end. Quite smart actually

he said that n was only a constant integer, that means that sin(npi/2) can be -1, 1 or 0, ie if n =2 its zero? same with cos, isn't cos(npi/2) = 1 when n=0 or minus 1 when n=2?

Oh yeah. ... I knew that, I was just testing that everyone else knew that.

Isn't it quicker for this equation to prove 1/w = conjugate(w) Then use the fact that a poly with real coefficients has roots occuring in conjugate pairs?

I do get what you're saying but (a-x) TIMES f(a-x) = (a TIMES f(a-x)) MINUS (x TIMES f(a-x)) you split the a minus x into two separate integrals.

Because f(a-x) = f(x) ie 2\int_{0}^{a}xf(a-x)\\\\ f(a-x) = f(x)\\\\ $(therefore you can replace f(a-x) with f(x))$\\\\ = 2\int_{0}^{a}xf(x)\\\\ = I I don't know how to explain it any clearer. excuse the < br / > I can't get rid of them

You split the (a-x) function. Because it is (a-x)*(f(a-x)) it equals a*(f(a-x)) - x*(f(a-x)) Do you get that? I'll give you an example without variables. (5-2) *(f(x)) = 5*(f(x)) - 2*(f(x)) Do you understand?

because f(a-x) = f(x) as originally stated, and as stated in the proof.

Pretty sure this is correct. I did skip a few steps but it's all correct as far as I know. I = 2\int_{0}^{a}xf(x)\\\\ \int_{0}^{r}f(x) = \int_{0}^{r}f(r-x)\\\\\therefore I = 2\int_{0}^{a}(a-x)f(a-x)\\\\ I = 2\int_{0}^{a}af(a-x)\, - 2\int_{0}^{a}xf(a-x)\\\\ I = 2\int_{0}^{a}af(a-x)\, - I\\...

hey, hopefully this shows you why it is true. the inv function of a+1 = a-1 inv function of x^2 = root(x) and below is the inv. function of 1/x \\y = 1/x\\ (inv. function)\\x = 1/y\\ xy = 1\\ y = 1/x

damnit, didnt check that, there goes 2 marks :(

part v) I got from p to q then q to s, because I did what he said ^^^ replaced the 2.6 with 1.1 in the original equation then differentiated again (but you just replace it in the derivative lol)

Re: Man that was a hard paper! It wasnt ridiculously hard, but the questions were very unconventional.

Don't forget Megadeth/Slayer a week or two back, it was fantastic. Screamfest is EASILY the best heavy metal festival lineup Australia has ever seen.

Even if it says "find the maximum turning point" and theres only one, check it anyway, the max could still be at one of the endpoints, but I'm pretty sure they wouldn't do that, you should still check it anyway.