Recent content by Faera

gah got it. Take 2 squared out as a common factor. *was blind* i didnt see it at first, coz when i divided 64/4, i managed to wind up with 18... its too early in the morning.

hmm. *guesses ramdomly* perhaps... Try to write the 2 squared + 4 squared + 6 squared pattern as sums of squares... eg 2 squared = 1 squared + 1 squared... 4 squared = 2 squared x 4 o.0 or maybe not...

Hey I don't really understand proofs unless they're algebraic when I'm reading online, so what I'll do is print yours of, Xayma, and take a look at it tonight so I have a hard copy with me, (and there's also the bonus of being able to look at circles, not ellipses) and then I'll get back to...

ahh genius. =) thanks!

err... ive got a diagram made up but dont know how to attach.. 0.o

Hmm *kicks self* -.- ive been doing the question WRONG! Blah. Sorry, ND, about that. :( My bad. Thanks for the proof, Xayma. :) *gives cake to all* Well, now I know what circumscribes truly means- and considering i wasted hours of my life on this, I wont be forgetting it any time soon...

Sorry, I dont understand why it can't be correct. I'm not really following. -.-' Perhaps we have to prove that a = (a + b)/2, then?

Why is it obviously not right? circumscribes = Makes a circle going through points A, B, C

Well, we're s'posed to use c/g skill s from 3u to do them, so Im guessing yes. But as to if well get asked them in tests Q3 was asked.. =D But I think the 1st one is a bit too complicated. Not sure about the 2nd.

Mine popped up first, though... =D Oh, and I'll hunt for my past papers book for that, thanks, CM. :)

both. :)

Hey guys- These are out of the 4u Fitzpatrick book, ex 37(a) If you haven't had a go at them you probably should try taking a look at them some time soon because they take aaagggesss to see.. (or it could've just been me). Well, anyway, there are three that I can't get out, any help would be...

uhmm... would this work?: subst x = [tanu]^2 and then after simplifying, subs v = sinu ?

3a. Let angle XAB = angle XAC = x let angle AXC = a, therefore angle AXB = 180 - a in triangle ACX: sinx / CX = sina / AC ....1 in triangle ABX: sinx / BX = sin(180-a) / AB sinx / BX = sina / AB ....2 1 / 2 : BX/CX = AC/BC which means that BX:CX = AC:BC as required...

3c. Let the lines/rays from q's 3a,b, intersect at point R. RO perpendicular PQ (from 3a) Since RO, PQ are the diagonals of QRPO, the QRPO must be a rhombus (since QO and PO are equal, it cannot be a kite). Therefore, PR = OQ and R = vector OP + vector OQ. that is, R = z(0) + i(z(0)) R...