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This is going to be so hard to see: Ok. y = x/root2x = x(2x)^1/2 y' = (2x)^1/2 + 1/2(2x)^-1/2 . 2 . x = root 2x + x/ root 2x sub 4 in: = root 8 + 4/root8 = 12/ root8 = 3root 8/2 what did i do wrong??

lol. how did i manage to get 8 / root 2. exactly the reverse. how would you work out this answer without simplifying it first. if anyone is bothered that would be much appreciated.

care to explain?

i think liverpool is the worst culprit. have you been down to westfield?, there is literally drug dealer after drug dealer lined up outside.

apparently it was asked in a past cssa paper.

lol, you guys are funn-ay. i wish i could take part.

Does UNSW offer a similar course? i was interested in this field.

does it use the same radical?

i can't seem to find the answers in any textbook.

side BX=CX this can be proved by considering part a) which by alternate angles Angle ABD = Angle ACD. If these two base angles are equal, then the sides making the isosceles triangle are also equal. i hope it makes sense.

a) For the first one, since Angle ABC= Angle BCD, and the lines are parallel. Then by alternate angles, Angle ABD = Angle ACD b) By SAS. side: one side is shown to be equal, angle is the same (vertically opposite), and the other side but i'm not bothered to load the page again. Re: since angle...

what's the question?

i'm out.