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  1. sharky564

    HSC Chemistry Exam Solutions / Thoughts

    I mean same, but x(x+10^-7)/(0.20-x), and the resulting quadratic has coeff for x 4.3E-7 x as opposed to 3.3E-7 x
  2. sharky564

    HSC Chemistry Exam Solutions / Thoughts

    There was a question which you had to determine the pH of a solution given the Keq was something of the order 10^-7, and the pH was like 10.smth. To do that q accurately, you had to have the initial concentration of [OH-] in pure water to be 10^-7, otherwise your quadratic is a bit off. I don't...
  3. sharky564

    MX2 Integration Marathon 2021

    \displaystyle \int_0^{\frac{\pi}{2}} \dfrac{\ln \left ( a^2 \sin^2 (x) + b^2 \cos^2 (x) \right )}{a^2 \sin^2 (x) + b^2 \cos^2 (x)} \; \mathrm{d}x = \dfrac{\pi}{ab} \ln \left ( \dfrac{2ab}{a + b} \right )
  4. sharky564

    BoS Maths Trials 2019

    no, not quite, but close also, if i did maths all the time, i'd probably bomb out in my atar, so we'll see
  5. sharky564

    BoS Maths Trials 2019

    Everyone who's got experience in oly maths can confirm international maths olympiad is much harder than 4u maths. 4u maths - requires maybe a month or so to get good at oly maths - years of training required, thousands of hours doing maths
  6. sharky564

    MX2 Integration Marathon 2021

    That last section which results in a dilogarithm is entirely 4U-able, and you may see that if you try the Feynman in a slightly different way.
  7. sharky564

    My solutions to the 2019 Mathematics Extension 2 Paper

    Oh right, that makes sense. I'll edit that.
  8. sharky564

    My solutions to the 2019 Mathematics Extension 2 Paper

    No because that swapping is taken into account in 2^4 - 2
  9. sharky564

    My solutions to the 2019 Mathematics Extension 2 Paper

    No, what you're saying is you're either choosing any two of the 4 possible spots for one digit (e.g. you're choosing the last two digits to be 1 if your initial 2 digits are 2, 1) and iterating over all possible ordered pairs of numbers, meaning you also count the time you choose the first two...
  10. sharky564

    My solutions to the 2019 Mathematics Extension 2 Paper

    That counts codes like 2211 twice?
  11. sharky564

    My solutions to the 2019 Mathematics Extension 2 Paper

    Thanks! I've replaced the images with the pdf.
  12. sharky564

    My solutions to the 2019 Mathematics Extension 2 Paper

    Of course, I can make mistakes, so any errors found would be great. EDIT: Newest version now a file on google drive as my last edit caused it to become too large: https://drive.google.com/open?id=1U_buMWY-R6i_ti3HunPg9TbSj_o74zFR
  13. sharky564

    BoS Maths Trials 2019

    We're actually identical twins uh what?
  14. sharky564

    Maths Extension 2 predictions/thoughts?

    yo, if anyone could chuck us the paper, that'd be great
  15. sharky564

    BoS Maths Trials 2019

    Considering I had contributed to the paper, it obviously couldn’t be me.
  16. sharky564

    HSC Standard/Advanced Paper 1 Exam Thoughts

    For 1984, the question asked about re-understanding loneliness, so I did a very conceptual essay on how there are different ways in which Winston is lonely: - he is living in a society where the only expected loyalties are to the Party, so he is lonely with his beliefs - he is living in a...
  17. sharky564

    help on abstract perms and combs

    See if it can :), you might have another approach to it.
  18. sharky564

    MX2 Integration Marathon 2021

    After a partial evaluation of the integral, you should get the original integral, which you can use for symmetry. See how that's possible. (note that this is quite cancerous to do, so you need to have some persistence)
  19. sharky564

    MX2 Integration Marathon 2021

    I mean, you have to show why that sum is \frac{\pi^2}{12}, but there are many ways of doing that in 4U, so it's basically done. That was also my method :).
  20. sharky564

    help on abstract perms and combs

    In general, the number of ways of choosing a cards from b consecutive cards such that there are at least c cards between any pair of cards is \binom{b+c-ac}{a}, which isn't too difficult to prove using double-counting.
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