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  1. W

    Tips/Tricks for Complex Numbers

    There's a few for finding the roots of complex numbers that will save you a decent amount of time. For one, there's completing the square eg. find the roots of 3+4i 3+4i=4+4i+i^2=(2+i)^2 , hence the roots are +-(2+i) There's also inspection, which often works if |x|,|y| and |x+iy| form a...
  2. W

    Ekman's Compilation of MX2 Questions

    Can I please have a hint for Q18 of integration? I simplified the integrand a bit by using difference of cubes and cancelling some common factors but after that im pretty stuck. Chucking into wolfram Alpha and integral calculator resulted in both timing out
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    Ekman's Compilation of MX2 Questions

    Isn't the solution to Q15 of the polynomials section, 8, not 5?
  4. W

    Mathematical induction

    Prove n^5-n is divisible by 30 for all natural numbers
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    HSC 2016 MX2 Combinatorics Marathon (archive)

    Re: HSC 2016 MX2 Combinatorics Marathon Number of lowest cards of straight (i.e 2-10)=9*4 (assuming ace-2-3-4-5 doesn't count?) Therefore number of straights=9*4*(4^4) P (straights)=9*4^5/52C5 =192/54145
  6. W

    Response Feedback (neutralisation)

    I'll remember that! Thanks :)
  7. W

    Response Feedback (neutralisation)

    yeah it should have been at least three, i've never seen a 2 mark assess question
  8. W

    Response Feedback (neutralisation)

    I don't think it's a very good answer to be honest -you haven't given a clear judgement (which is what the verb assess asks for) by explicitly saying 6M HCl is not good to use here -you should say strong acids should not be used to neutralise strong bases since the reaction is highly exothermic...
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    HSC 2016 MX2 Marathon (archive)

    Re: HSC 2016 4U Marathon Yeah, I realised, after checking in wolfram, which I probably should have done before asking. My bad
  10. W

    HSC 2016 MX2 Marathon (archive)

    Re: HSC 2016 4U Marathon alright, I know I was wrong last time I asked, but just to be sure: cot^-1 k^2/8 or cot^-1 k/8 ?
  11. W

    HSC 2016 MX1 Marathon (archive)

    Re: HSC 2016 3U Marathon Well I got this, but wolfram says 3pi/4. I'm not sure if my working in the previous part is wrong or that the limit should have been pi, cos that would give the answer a)\quad Let\quad \alpha =\tan ^{ -1 }{ (x+1) } ,\quad Let\quad \beta =\tan ^{ -1 }{ (x-1) } \\ \tan {...
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    HSC 2016 MX1 Marathon (archive)

    Re: HSC 2016 3U Marathon Same here, I'm stumped :/
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    HSC 2016 MX2 Complex Numbers Marathon (archive)

    Re: HSC 2016 Complex Numbers Marathon thanks for the help, I finally got it. for part b) it's just summing the converging g.p 1/2 +1/4 +1/8+...+1/2n so by the limiting sum of a g.p S=1/2/(1-1/2) which is 1. Quick Question: What is rep power and what is its significance? Also - do you just...
  14. W

    HSC 2016 MX1 Marathon (archive)

    Re: HSC 2016 3U Marathon It's in the reply above ambility (2nd part)
  15. W

    HSC 2016 MX1 Marathon (archive)

    Re: HSC 2016 3U Marathon Hmmm.. I wonder what was wrong with my solution :/
  16. W

    HSC 2016 MX2 Complex Numbers Marathon (archive)

    Re: HSC 2016 Complex Numbers Marathon :( i thought i worked it out for a sec
  17. W

    HSC 2016 MX2 Complex Numbers Marathon (archive)

    Re: HSC 2016 Complex Numbers Marathon Isn't it meant to be 2cos 2pi r/2n+1 in part i?
  18. W

    If I moved an object in a circle, would I have done any work?

    I'm pretty sure work is only done if there is a force component acting in the DIRECTION of displacement. Since the centripetal force is always perpendicular to the direction of displacement, there is no force acting in the direction of displacement and thus no work is done on an object in...
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    How to learn 2U by self

    If you're going to use a cambridge book (which I highly recommend) to learn 2u, use the 3u yr 11 one instead of the 2u ones. The 3u is by far more challenging than even the yr 12 2u textbook and it contains great extension questions you can attempt to boost your understanding of the topics.
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    HSC 2016 MX1 Marathon (archive)

    Re: HSC 2016 3U Marathon \cos { (2\sin { x) } } =\sin { (2\cos { x } ) } \\ \cos { (2\sin { x) } } =\cos { (\frac { \pi }{ 2 } } -2\cos { x } )\\ $By,\quad the\quad general\quad solution\quad of\quad the\quad cosine\quad function$,\\ 2\sin { x } =2n\pi \pm (\frac { \pi }{ 2 } -2\cos { x }...
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