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  1. nightweaver066

    good job haha. and it was the last place you were going to search anyways hows your holidays?

    good job haha. and it was the last place you were going to search anyways hows your holidays?
  2. nightweaver066

    ahahaha yeah you have been tough hiding for the past year

    ahahaha yeah you have been tough hiding for the past year
  3. nightweaver066

    What's so surprising LOL Sorry i don't want to use credit :D

    What's so surprising LOL Sorry i don't want to use credit :D
  4. nightweaver066

    Past student answers

    It's pretty outdated but i don't think there's anything more recent released online. http://arc.boardofstudies.nsw.edu.au/go/hsc/std-packs/
  5. nightweaver066

    Integration

    It's best to ask your maths teacher/head teacher.
  6. nightweaver066

    HSC 2012-2015 Chemistry Marathon (archive)

    Re: HSC 2012 Chemistry Marathon Your first question. I don't have time so i'll just run over a few points. - Petroleum is a finite resource and we are consuming it quicker than we're excavating it - Experts predict it will run out in the next 50 years - Oxidation-reduction reactions are able to...
  7. nightweaver066

    CHEM PRAC HELP!!

    Jacaranda should have them. If not, write one up yourself, post it up and we'll check.
  8. nightweaver066

    I just solved this. Can you? (MX2 students should have a go)

    E(X)=\lim_{R \to \infty}\int_{0}^{R} x\frac{e^{-x/\beta}x^{\alpha - 1}}{\Gamma (\alpha) \beta^{\alpha}} dx = \frac{1}{(\alpha - 1)!\beta^{\alpha}} \times \lim_{R \to \infty}e^{-\frac{x}{\beta}}x^{\alpha}dx $After applying tabular integration to the integral with the limit,$ \lim_{R \to...
  9. nightweaver066

    I just solved this. Can you? (MX2 students should have a go)

    Working is a bit long, but i combined the x and x^(alpha - 1), took out the constants (the T(alpha) and beta thing), considered the integral seperately, applied tabular integration, deduced what the integral would result in: \alpha!(-1)^{\alpha + 2}\beta^{\alpha + 1} Subbed that back in to the...
  10. nightweaver066

    HSC 2012-2015 Chemistry Marathon (archive)

    Re: HSC 2012 Chemistry Marathon I think you have as well. "Petrochemical feedstocks" = petroleum. Maybe reattempt the question with a focus on catalytic/thermal cracking? (Go through the process and provide balanced chemical equations) Good attempt though if it was simply ways of attaining...
  11. nightweaver066

    HSC 2012-2015 Chemistry Marathon (archive)

    Re: HSC 2012 Chemistry Marathon Oops.. Main point was gotten across though :P
  12. nightweaver066

    HSC 2012-2015 Chemistry Marathon (archive)

    Re: HSC 2012 Chemistry Marathon ^^^ Check it out if you guys need it. :) New question:
  13. nightweaver066

    HSC 2012-2015 Chemistry Marathon (archive)

    Re: HSC 2012 Chemistry Marathon Use [IMG] link [./IMG] without the dot NH_4Cl $ is the product of ammonium hydroxide and hydrochloric acid$\\NH_4OH_{(aq)} + HCl_({aq)} \rightarrow NH_4Cl_{(aq)}\\$As it is a product of a weak base and strong acid, the product must be acidic, i.e. it forms an...
  14. nightweaver066

    HSC 2012-2015 Chemistry Marathon (archive)

    Re: HSC 2012 Chemistry Marathon Since it says justify, you should include that the esterification process is very slow. For the last point you made, mention how the reactants and products are volatile. Other than that, good :)
  15. nightweaver066

    Physics Marathon 2012

    $Ideas to Implementation$ 1. E = \frac{V}{d} = \frac{320}{0.016} =20000V/m 2. d = u_x t t = \frac{d}{u_x} $Using $ y = u_yt + \frac{1}{2}a_y t^2 = \frac{u_y}{u_x}d + \frac{a_y}{2u_x^2}d^2 \equiv ax^2 + bx + c \therefore $ Parabolic trajectory$ 0.008 = \frac{a_y}{2(3.7 \times 10^7)^2}d^2...
  16. nightweaver066

    HSC 2012-2015 Chemistry Marathon (archive)

    Re: HSC 2012 Chemistry Marathon Bump
  17. nightweaver066

    I just solved this. Can you? (MX2 students should have a go)

    Just ended up with \alpha \beta (-1)^{\alpha} :/
  18. nightweaver066

    Inverse Trig. Differentiation

    Sorry that i skipped steps lol. What i did was expand the [(2x - 1)/3]^2, combine the two terms on the bottom to have a common denominator, took out 1/root(9) from the bottom, moved it to the top and changed it to 3.
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