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Because it's true.

Hope it was easier than last years trials. Everyone screwed that one up.

Using that identity, it becomes S 1/2 (1 - cos4x) = ....

Speaking to students, I think most would have been prepared to do bonded if offered, but there are also many (like me) who would not have accepted a bonded place, and gone to another uni/degree instead.

It's one of the frustrating features of CSSA papers. Questions 7 and 8 can be some of the easiest in the paper.

I think you may have forgotten to take the negative reciprocal for the gradient for the normal.

You could collapse it into single trig function using auxiliary angle method.

ln (y-1) - ln2 = t + lnt antilogs: y - 1 = e(ln2 + t+ lnt) = e^ln2 . e^t . e^lnt y - 1 = 2. e^t .t So y = 2te^t +1

True. That's why my assessment results for 3u were comparatively poor; I kept on screwing up in early questions.

No, far from it. You'll find >15 people will have gotten 100 aligned.

3u 92%, 100 aligned 4u 95%, 99 aligned

Medicine is actually very enjoyable, and makes all the effort you put in getting into the course worthwhile. :)

Extend BD to centre O. OD is R - .1. OA is R. Use pythag in triangle OAD.

It should work out. Remember that a = k/t^3. You should get two equations: one that gives you c = 5, and the other with c and k, getting you k.

1. Angle PAX = angle XBA (alt segment) Also XAB is isos (equal chords on equal arcs) So angle xab = angle xba = angle pax 2. angle dbe = angle bcd (alt segment) = angle abc (|| lines) angle bde = angle bac (ext angle cycl quad) AA //

The answer involves proving triangles ANY and BMX are congruent by SAS, which you get through isos trianges and alt segment theorem. Could someone try to write it out, because I'm a little busy atm.

Yes, it's a nice trick. If anyone has Cambridge year 11, see ex. 12B extension for some questions using it. Here are are a few: Differentiate: sqrt(x)sqrt(x+1)sqrt(x+2) sqrt(x)(x-1)^2/(x+1). x^ (1/x) The various log laws do come in handy.

Doing a uni degree is better, but you can always take a year off to travel/work. I think Lexi was saying its better to do something during the year than do nothing and solely study for UMAT. It'll keep you focused, and plus there's the chance you may miss out again, and so you can try for...

Looks right. It can also be done like this: sin3x = -sinx sin3x = sin-x Then use general solutions for the sina = sinb case.

Let width be x, length 2x, and the other dimension y. Use the surface area: 4x^2 + 6xy = 300. Solve for y. Then find an expression for the volume by multiplying the 3 dimensions. Maximimum by differentiation.