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Re: HSC 2013 2U Marathon $By considering triangle PBA, show that$ \ \sin\alpha = \frac{x^2 - 16}{4x}

Re: HSC 2013 2U Marathon Yep that's correct.

Re: HSC 2013 2U Marathon You have to adjust y = 3x - 1 into the form 3x - y - 1 = 0, and then use the formula. This way, a = 3, b = -1 and c = -1.

Re: HSC 2013 2U Marathon $The shortest distance between 2 lines is when the gradients are parallel.$ y = x^2 +3x +5 \ \fbox{1} y = 3x - 1 \ \fbox{2} m_2 = 3 $Find the tangent of$ \ \fbox{1} \ $at the point where the gradient is 3:$ y'_1 = 2x + 3 = 3 \therefore \ x = 0, \ $and after...

Re: HSC 2013 2U Marathon KL = KP because its just been folded over. I'll adjust my solution to make it more clear.

Re: HSC 2013 2U Marathon If you 'unfold' the sheet from L, P will replace L. From there it is basically given that KL (or KP) is 6 + x

Re: HSC 2013 2U Marathon QK = 6 - x KP = 6 + x $Since the sheet of paper has been folded over from P (now L),$ \ KP = KL \therefore \ KL = 6+x $By Pythagoras,$ \ (6 + x)^2 - (6 - x)^2 = QL^2 QL^2 = 36 + 12x +x^2 - 36 + 12x - x^2 \therefore \ QL^2 = 24x

Re: HSC 2013 2U Marathon Only 2/3

Re: HSC 2013 2U Marathon Back to the marathon guys? $Solve the equation$ \ 2\ln(x\sqrt{3}) = \ln(6 - 7x).

Either way, he will get the highest internal mark while everyone else (except for the person coming last) will have their marks adjusted, conforming to the gaps between each rank. Since the gap between 1st and 3rd is only 1 mark, you won't be too far off. If you do beat him in externals, yes...

Not sure about the cut off for last year but looking at the raw mark database, the estimated cut off is 83. In general it will vary around this point so I doubt 2012 was too far off. You can definitely beat him/her, provided you perform very well externally. Besides 1 mark will probably not...

Re: HSC 2013 2U Marathon Yep. Well done!

Re: HSC 2013 2U Marathon Yeah that's true. So what could you do with the curve using that general idea?

Re: HSC 2013 2U Marathon Haha nope..

Re: HSC 2013 2U Marathon Drawing a graph might help you figure out what to do.

Re: HSC 2013 2U Marathon $Find the shortest distance between the$ $curve$\ y = x^2 + 3x + 5 \ $and the line $y = 3x - 1.

Yes. Those raw marks are all real, not estimates. It is only the band 6 cutoff which is an estimate. However, with some subjects it is generally difficult to achieve the raw marks listed due to the tight nature of the marking criteria.

Re: HSC 2013 2U Marathon I guess I'm just tired :L

Re: HSC 2013 2U Marathon M = 2\alpha^2 L^2 = (-\alpha - 2\alpha)^2 = 9\alpha^2 \frac{L^2}{9} = \alpha^2 \frac{2L^2}{9} = 2\alpha^2 = M

Re: HSC 2013 2U Marathon Haha no problem. Hmm that boundary would also include the numbers in between 3 and 4 (i.e. 3.1, 3.2 etc.) so I don't think its right.