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sorry, dont have a scanner lol. but the question's on p93 q16 and it links to the diagram on p90 q7. maybe someone else who has the book can scan it...

it was from yr 12 cambridge 3u

you can try and post hte questions up here if you want. someones gonna answer it eventually.

yeh thats all that was specified although it did come with a diagram of a small ball on the hypotenuse of a right angled triangle and and distance from that ball to the point on the hypotenuse that meets the horizontal was called x. it also gave a hint about resolution of forces whatever that means.

bump... anyone q

are you sure this is right? the inverse sine bit seems a little dodgy. wouldnt that be the answer if it was 5 / √[(25/4) - (x - 1/2)²]} instead of 5 / √[(x - 1/2)² - 25/4]}. or maybe im missing something. can you please explain that step.

you serious? what did you do then?

there could be more than one answer when you convert cartesian equation to parametric. the easiest way is to either make something the subject or use trig identities. for 7, you could even have y=2t-4 and x=t

A smooth piece of ice is projected up a smooth inclined surface. Its distrance up the surface at time t seconds is x = 6t - t^2. At what angle @ should the surface be inclined to the horizontal to produce these equations. the answer is 11 degrees and 47 minutes. how would you approach this...

To construct cartesian equations from parametric equations, you gotta try to eliminate the parameter. 1. x=t^2 - 2t (1) y= t^2 + 2 (2) y-x = 2+2t so t = (y-x-2)/2 substitute this into eqn (1) or (2) 2. x=t<SUP>2</SUP>+t y=(t<SUP>2</SUP>/2) - 3t 2y=t^2 - 6t x-2y = 7t...

ok thanks, i get it now

Prove that a cyclic quadrilateral has the maximum area of all quadrilaterals with the same side lengths in hte same order. (by letting sides be a,b,c,d and letting P be the included angle of a,b and Q be included angle of c,d) I got up to the part where: dA/dP = absin(P+Q)/2sinQ = 0 P+Q =...

thanks for that...by the way, how do you type up maths?

How do you prove Simpson's rule. I remember learning it in 3u but never actually proved it? Any ideas guys?

partial fractions r the easiest i reckon. you just use it when both numerator and denominator are polynomials and the bottom is factorisable. when the degree of the top is bigger than or equal to the bottom, you gotta do polynomial division before carrying out partial fractions.

really? Slicing seems pretty easy for simple functions though. What are the advantages of each?

Find the volume of the torus with inner radius 3 and outer radius 6. This was all that was asked in the question. What the hell is a torus? Did they miss out some sort of equation or something?

equation of PQ (using 2 point formula) is: y = (p+q)x - apq it passes through (0, -a) so subbing it in equation gives: -a = 0 -apq pq = 1 to find PS and QS, you can either use distance formula or use definition of parabola (ie, distance from focus to point on parabola is equal to...

yeh i got clintmysters answer as well let x = u^2 dx = 2udu ∫ 2u/[u(1+u^2)]du = 2∫ 1/(1+u^2)du = 2 tan<SUP>-1</SUP>(u) + c =2 tan<SUP>-1</SUP>(sqrtx) + c btw, clintmyster, i dont think u can take out e^x as its not a constant. so its: let u = (e^x) du=(e^x)dx ∫...

yeh, its a typo. check the edit. its meant to be pi/2 as opposed to pi