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Re: 2012 HSC MX1 Marathon <a href="http://www.codecogs.com/eqnedit.php?latex=\textup{A function is }y=f(x)\textup{ is given in parimetric form by }x=tan\theta \textup{ and }\\y=4sin2\theta\textup{ for }\frac{-\pi}{2}<\theta<\frac{\pi}{2}.\textup{ State the domain and range of the function and...

how do u know water will act as an acid the second reaction? couldnt it act as a base and the reaction be HCO^-_{3(aq)} + H_2O_{(l)} \rightarrow H_3O^+_{(aq)} + CO_3^{2-} ???

Re: 2012 HSC MX2 Marathon The other question was posted so for (c) The degree of the remainder is always one less than the degree of the divisor. Therefore it is of degree 1. <a href="http://www.codecogs.com/eqnedit.php?latex=\\P(3)=3 \textup{ and }P(4)=4\\\\\textup{Let...

Re: 2012 HSC MX2 Marathon \textup{For }\int^a_0 f(a-x)dx\\\textup{Let }u=a-x \therefore du=-dx\\\textup{also when }x=a\textup{ , }u=0\\\textup{and when }x=0\textup{ , }u=a\\\therefore \int^a_0 f(a-x)dx=-\int_{a}^{0}f(u)du\\=\int_{0}^{a}f(u)du\\=\int_{0}^{a}f(x)dx=LHS\\\\Q.E.D.

\\\Delta = (-2)^{2}-4(2\times-1)\\=4+8\\=12\\\nless 0 ??? \textup{The way to do it is prove that }2k^{2}-2k-1> 0 \textup{ for } k> \frac{1+\sqrt{3}}{2}\textup{ and }\\k<\frac{1-\sqrt{3}}{2}. \textup{ This satisfies the statement only for }k\geq 2.\textup{ You can then sub in }k=0\textup{ and...

Re: HSC 2012 Chemistry Marathon The Ziegler-Natta catalyst allows surface polymerisation to occur which results in a tightly packed polymer, but the low pressure is definitely a contributing factor as well. If you look at LDPE, the extremely high pressure results in the radicals backbiting and...

Re: HSC 2012 Chemistry Marathon LDPE is conducted in an environment with pressure equivalent to approximately 3000 atmospheres, and with extremely high temperatures. The high pressure atmosphere leads to haphazard branching in the propagation process, leading to the low density. HDPE on the...

Hey guys i need help with this question: \\ \textup{(a) Show that the equation } x^{3}-6x^{2}+9x-5 = 0 \textup{ has only one real root } \alpha. \\ \textup{(b) Determine the two consecutive integers between which } \alpha \textup{ lies.} \\\textup{(c) By considering the product of the roots of...

Re: 2012 HSC MX2 Marathon thanks champ; i was just having trouble with part (ii). it turned out to be pretty easy hahaha. +rep

Re: 2012 HSC MX2 Marathon past catholic CSSA trial

Re: 2012 HSC MX2 Marathon ... this better work

Re: 2012 HSC MX2 Marathon http://community.boredofstudies.org/attachment.php?attachmentid=24460&d=1329975443 heres the question again

Re: 2012 HSC MX1 Marathon the answer is 19. i got it by expanding and it didn't take as long as i expected, but is there a way of doing it without expansion? using binomial theorem or something?

Re: 2012 HSC MX1 Marathon Find the term independent of x in the expansion (x+1 + x^{-1} ) ^{4 without expanding the whole bracket.

can you explain the flaw in my reasoning mr carrotsticks?

so what's the answer to OP's question ? :S

i got \frac{75\pi}{2} u^{2} ??? if the argument is \frac{\pi}{4} then it will be \frac{3}{4} of a circle, but because its \pm\frac{\pi}{4} there will be 2 three quarter circles of radius 5. so i just did 2\times\frac{3}{4}\times\pi\times25 = \frac{75\pi}{2} u^{2}

thanks man!! waiting for the answers now ay!! haha repped

parti (i) son

im not sure, is it in the syllabus? did you learn it from uni physics or something? should i bother learning it? thanks heaps anyway