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WELL, he's entitled to express his opinion. mm... *chuckles*

should be do-able with slices at the ends, the slices are not annular, but 'cresent' shaped so might mess things up a little.

try it if you have time, I will post the solution when I figure it out :p (I haven't tried it myself yet)

you can ask the teacher for the raw assessment marks. definitely before it gets sent in, I am not sure about during the course though, but my teacher is happy to let us know. just make sure you keep a good gap in the assessment marks between you and those who you predict will do bad, and...

theres are better ways to do this: The volume is just rotating the area bounded by y=sqrt(R^2 -x^2) y=r about the x axis. you can either do this by method of cylindrical shells or annular slices. some necessary working are omitted in the following. shells: dV = 2*Pi*(2*sqrt(R^2...

necessary working may have been omitted: consider the component of gravity parallel to the inclined plane. that would have a magnitude of (mg/30) and direction down the plane. the total acceleration when the car is rolling down with the engines off would be : a = F/m = (kv^2 - mg/30)/m...

priorities(the order to look for in general) 0.)obvious definite integrals eg even and odd functions. 1.)standard form 2.)direct substitutions ( f'(g(x)*g'(x) integrals) 3.)inverse substitutions ( x = arcsin(t) etc) 4.)Complete square and/or partial fractions 5.)By Parts. 6.)T or other...

General --property of rational roots of Polynomials over Z. --Polynomials with multiple roots. --find sum of the roots to the power of n. --Using the fundamental theorem of algebra to prove things. --Solving quartics with symmetric coefficients. I am making a wild guess that there will...

where was that quote from?

Thats right!!! the t formula, thanx for reminding me.

I think we can generally trust our teacher's pace

Follow your instincts when things are unfamiliar

sqrt(3) * COSX - SINX = 1 The good method, as mentioned earlier by wogboy: [sqrt(3) / 2]*cos(x) - [1/2]*sin(x) = 1/2 cos(Pi/6)cos(x) - sin(Pi/6)sin(x) = 1/2 Cos(Pi/6 + x) = 1/2 Pi/6 + x = 2n(Pi) + Pi/3 OR 2n(Pi) - Pi/3 x = 2n(Pi) + Pi/6 OR 2n(Pi) - Pi/2 The Stupid...

hmm.. I will post the proof on simpson's theorem ( last year's 4 b) tomorrow.

above is sarcasm at best, especially the "not me ne way" part.

All textbooks are good, My school uses Arnold/Arnold and the Coroneos supplement(this one is really nasty). It's not that much of being smart, perseverance solves the question: don't stop when the mess seems to overwhelm you.

for locus problems.. it might help if you know what it looks like before you start. 99% of the time you will get another conic(perhaps degenerate) that is a quadratic curve: ax^2 + bxy + cy^2 + dx + ey + f = 0

blame it on selective and grammar school brain draining :P

well.... let's call x(1) p instead. then x(2) = p+2c solving we get y(1) = c^2/p y(2) = c^2/(p+2c) the locus of the mid point has parametric form (p as parameter) x = p+c y = [c^2/p + c^2/(p+2c)]/2 notice p + 2c = x+c p = x- c so eliminating p y = [c^2/(x-c) +...

last year my school had 1 person taking 4 unit :P.