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Yes, that is awesome Ngai, wouldn't that save me time in an exam! I really have to learn little tricks like that to make the questions quicker and easier, I've been learning a few clever substitutions that help with conics at the mo. Thanks for those solutions...and for using sub and...

lol, mildly humourous......I lived in NZ for 12 years, so I'm too used to australians being stupid :/ but this is probably more true.

oh a fair while, Question 9 probably 15-30 mins, and Question 8 took me a touch over an hour because I thought I was doing it wrong because it looked so damn messy, but eventually I kept getting the same thing, parametrically and cartesian, so I just carried on and it turned out to work, as ugly...

Ok now Question 9; (I always use "//" as "all over") angle between two lines; tan @ = | m1-m2 // 1+m1m2 | at P: dy/dx = dy/d@ . d@/dx = b cos @ . 1/-asin@ m1= -b/a . cos@/sin@ at PS: m2 = bsin@-0 // acos@-ae at PS' : m3 = bsin@ // acos@+ae therefore angle between tangent and...

Question 8 is a ripper, so is question 9 for that matter...but I'll give you the jist; Q8: edit: I implicitly differentiated to get the gradients and used the line formula to get PR and QR... let R be (x0, y0) P be (x1, y1) Q be (x2, y2) PR is 2xx1 + 3yy1 = 30...

Yeah actually I only bought them recently too, its just convinience factor; also that you can use some as a bootable drive so if you have other work to do...

k^2/4 . (k+1)^2 + (k+1)^3 =(k+1)^2.k^2/4 + (k+1)^2.4(k+1)/4 =(k+1)^2.k^2/4 + (k+1)^2.(4k+4)/4 =(k+1)^2.(k^2+4k+4)/4 You have got to be kidding me, that's it....lol, can't believe that was it. *sigh* I need rest. Thanks slide.

Estel....no.... Slide, good idea, it's probably my fault, should have asked the complete question, I have a worked answer but one line is tripping me out, I think i know WHAT theyve done but no clue HOW they've done it; QUESTION: Use Mathematical Induction to prove: 1^3 + 2^3 + 3^3 +...

read further...its not simply a matter of them giving you LHS and RHS then equating etc; you have deduced LHS, you know what RHS *SHOULD* be, so you manipulate LHS till it looks like RHS...ffs, has no one done mathematical induction!?! edit: My apologies, I don't mean to sound like an...

no, I know they're equal, but its MI, I have to work the LHS out to look exactly like the RHS...like trig identities...I don't think you can simply put them on the one side and show that it equals 0... As in, I've only shown you guys what its supposed to look like, which is the (k+1)^2/4 ...

how on earth do you get; (k+1)^3 + k^2/4 . (k+1)^2 to equal: (k+1)^2 /4 . (k+2)^2 EDIT: if you can't read that; then: 3 2 2 (k+1) + k . (k+1) 4 = 2 2 (k+1) . (k+2) 4

:laugh: I assume to add it you need to upload the image somewhere and then attach it using the IMG code. What is your question exactly?

*annoyed* which is what I wrote with working....unless you were clarifying...

Just make sure you know how to write a good prac report; sources of error, experiment controls etc, conclusion relates back to hypothesis...

word to that. It wasn't anything major anyway, just draw the graph and you can see it.

Are you sure, Zenger? if you draw the function and rotate it around the x axis and then around y=5, you'll find they look different.... rotating the function y=sqrt x about y=5 is the same as rotating the function y=(sqrt x)-5 about the x axis. So our new function is y=(sqrt x)-5 and you...

You can, as long as its mathematically legitimate you can. Infact you can use any mathematically legitimate techniques. However, why use 4u techs in a 2u paper, chances are you will complicate it for yourself and create more room for simple algebraic error etc.

Done complex numbers, almost done conics. Assessment in 2 weeks :/. Conics I'm cool with so far, hardly any problems at all. Complex numbers I dont think are that difficult, there are some nuts out questions I have seen, but I will need to study it for at least a week in advance since I...

That's pretty, I like it.

You just need to know the formulae and know how to apply them and you should be fine. These are slightly different from the book, but its the same, and I think it looks nicer. Trapezoidal rule: (h/2)[f(a)+f(b)-2(f(x1)+f(x2)+...+f(xn))] ; where h is the width of one "strip", a is the first...