Recent content by frenzal_dude

It has to be an error in the textbook, another reason to think this is because they even forgot to put the end bracket for the 2(1+cos(..... Thanks for your help, I went through your working out and it seems fine, they obviously forgot to put "=a(t)" before the "2|cos(PItT)|"

do you reckon it's an error in the textbook?

I need to find the absolute value of b(t): b(t)=a(t)[1+e^{-j2\pi tT}] Here is the answer in the textbook: \therefore \left | b(t) \right |=\left | a(t) \right |\sqrt{2(1+cos(2\pi tT)}2|cos(\pi tT)| However I got a different answer when working it out, and can't understand how they got to that...

dude did u read what i wrote? I didn't need help on the question, I got it right assuming D was closer to the origin (1,3) (which is totally fine). I was just wondering why my answer didn't match the answer in the book. There was nothing in the question that said the letters need to go around...

I know it's an easy question, the only reason i posted it here was coz i was confused why the book had a different answer, both answers are correct depending on where you consider D(x,y) to be. btw I'm a maths tutor so that's why I put this question up.

I'll be 24 in June, did my HSC in 2005. I got the question right if you consider D to be close to the origin, but strictly speaking the letters should be in alphabetical order around the paralellogram. And d/w about the complex plane stuff, I did all that in my engineering degree.

I got the right answer now. Thanks nigga20101, I put D(x,y) in the wrong place when drawing my parallelogram, ofcourse the letters have to go in order. My first answer is also correct if you don't worry about the order of the letters around the parallelogram.

ohexploitable you said that (-3,5) is the right answer, can you pls show your working? Can you also show me why my working out is wrong?

mAD will be the same as mCB since they are parallel lines. So then we can easily find AD since we have point A and the gradient. Likewise for mBD, the gradient will be the same as mAC since they are parallel lines and we have point A, so with that point and knowing the gradient we know the line.

How did you get your answer? I tried to find the equation of the two lines which run through D(x,y) and used simultaenous equations to find D, I can't see where I went wrong in my working out.

ABCD is a parallelogram. The coordinates of A,B and C are (-1,4), (4,6) and (2,7) respectively. Find the coordinates of D. I got D(1,3) but the answer says: D(-3,5) Working out: mAD = (7-6)/(2-4) = -1/2 AD: y = (-1/2)x + 7/2 mBD = (7-4)/(2+1) = 1 BD: y = x + 2 Find the intersection which...

minor point? I think it's a major point since without using the brackets you get 2 totally different expressions! In the exam I doubt the marker is going to think 'oh I know what the student meant', I'm pretty sure the student would get 0 for writing the expression without brackets.

everyone please remember to use brackets, what you have just written is this: x+\frac{2}{x}-1 If you meant to write that then that's ok, but if you meant to write this: \frac{x+2}{x-1} then you need to write it like this: (x+2)/(x-1)

thanks Drongoski, haha u beat me by 2 mins!

This is actually an easy question, can't believe noone's given the answer yet: If the circle touches the x and y axes, then the centre is going to be (h,k) where h=k. The point where it touches the y axis is (0,k), the point where it touches the axis is (h,0), subbing these along with (1,2)...