Recent content by MOP777

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 :(