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Re: HSC 2013 4U Marathon Let G be the centroid of triangle ABC which is inscribed in a circle Produce AG,BG and CG so that they intersect the circle again at points X,Y and Z respectively Show that \frac{AG}{GX}+\frac{BG}{GY}+\frac{CG}{GZ}=3

Re: HSC 2013 4U Marathon S=1+3x+6x^2+10x^3+15x^4+\dots Sx=x+3x^2+6x^3+10x^4+15x^5+\dots S(1-x)=1+2x+3x^2+4x^3+5x^4+\dots Sx(1-x)=x+2x^2+3x^3+4x^4+5x^5+\dots S(1-x)(1-x)=1+x+x^2+x^3+x^4+x^5+\dots S(1-x)^2)=1+x+x^2+x^3+x^4+x^5+\dots=\frac{1}{1-x}$, if $|x|<1 S=\frac{1}{(1-x)^3}$ , if...

Re: HSC 2013 3U Marathon Thread thanks for that , don't know what I was thinking hopefully it makes some sense now

Re: HSC 2013 3U Marathon Thread Let M bet the midpoint of a chord PQ in a circle Two chords AD and BC are drawn so that they intersect at M AB and CD intersect the chord PQ at X and Y respectively Show that M is the midpoint of XY

Re: HSC 2013 3U Marathon Thread S_n=1+\frac{4}{5}+\frac{7}{5^2}+\frac{10}{5^3}+ \dots +\frac{3n-5}{5^{n-2}}+\frac{3n-2}{5^{n-1}} \frac{S_n}{5}=\frac{1}{5}+\frac{4}{5^2}+\frac{7}{5^3}+\frac{10}{5^4}+\dots+\frac{3n-5}{5^{n-1}}+\frac{3n-2}{5^n}...

actually you do need to apply the chain rule d/dx e^(f(x))= e^(f(x)) * f'(x)

yes, it is chain rule, as you can let u=x^2 then d/dx sin(x^2) =d/du sin u * du/dx or dy/dx= dy/du* du/dx, where y=sin(x^2) for the 1st question, you can find it if you can find , say d/dx 2^x if you couldn't differentiate 2^x, think about the definition of log and exponentials (a ^...

+1 If you find mx1 REALLY time consuming, do drop it But it does scale quite well, you don't necessarily have to be able to do everything to get a decent scaled mark

http://www.youtube.com/watch?v=TINfzxSnnIE

some teachers are very fussy about it

it also depends on which area of engineering you're interested in to be honest, if you're finding 2u maths difficult , probably most engineering courses would be challenging

xy=c^2 only holds on the hyperbola the locus found is for R, the midpoint of PQ, which does not necessarily lie on the hyperbola so in general you cannot assume xy=c^2 on this locus dunno if that's clear enough, you could try to sketch it and see what happens

in your last and 2nd last line, did you actually mean multiplication for 53*147^k rather than addition? the way you use the assumption in step2, is to make 47^k ( or 53*147^(k-1), it wouldn't matter ) the subject then substitute it in step3

y=xlnx-x^2 y'=lnx+1-2x y''=1/x-2 =0 at pt of inf, so x=1/2 and sub into y , what you get should be the same as Drongoski's answer of course you should verify the change of concavity

-2a+a=-a , where a=1/2^k

I'd assume you know: the definition of log log a^b = b * log a and the integral of e^(cx+d)

bump

sorry about the messy handwriting

Continuing from spiral's work:

I personally prefer loose leaf paper on a clipboard