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  1. mahmoudali

    HSC 2013-14 MX1 Marathon (archive)

    Re: HSC 2013 3U Marathon Thread $i) Find the probability that in$ \ \ (n+k) \ \ $tosses of a fair coin, there will be$ \ \ n \ \ $heads$ $ii) Hence explain why$ \sum_{k=0}^n \frac{\binom{n+k}{n}}{2^{n+k}} = 1
  2. mahmoudali

    lmfaooo nakruf strong representation of the DIAMONDA

    lmfaooo nakruf strong representation of the DIAMONDA
  3. mahmoudali

    sy the <diamond> where is the diamond trollface.jpg

    sy the <diamond> where is the diamond trollface.jpg
  4. mahmoudali

    sy123 ibn sina :D

    sy123 ibn sina :D
  5. mahmoudali

    HSC 2013-14 MX1 Marathon (archive)

    Re: HSC 2013 3U Marathon Thread my first method was totally wrong i realised after 2 hours :D total number of ways they can be arranged is 7! =5040 number of ways all couples sit together = 3!2!2!2!2! =96 number of ways 2 couples sit together but other 2 random = 5!2!2! =480 number of ways...
  6. mahmoudali

    HSC 2013-14 MX1 Marathon (archive)

    Re: HSC 2013 3U Marathon Thread $$$$$$$ Hurry the hell up sy make another question that we can all benefit from not only the 4Uers $$$$$$$ :D
  7. mahmoudali

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon i skipped so many steps because i cbfed showing it using this latex bs wastes to much time and mathewYan can you please tell me where i made mistakes?
  8. mahmoudali

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon too easy sy :D
  9. mahmoudali

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon cosec^2 x -1 = cot^2 x \ \ \ $simple substitution into f(x) $ \ \ \int_{\frac{-\pi}{4}}^0 \ \ \frac{1}{\sqrt{cot^2x}} dx \ \ \ \ \int_{\frac{-\pi}{4}}^0 \frac{1}{|cotx|} dx \ \ = \ \ \ int_{\frac{-\pi}{4}}^0 |tanx|\ \ = \ \ $(-ln(cosx)) _\frac{-\pi}{4}^0$...
  10. mahmoudali

    HSC 2013-14 MX1 Marathon (archive)

    Re: HSC 2013 3U Marathon Thread i) \int_0^{\pi /2} \sin^{2k} x \cos x \ dx = \frac{1}{2k+1} \int_0^{1} \ u^{2k} \ \ du = \left(\frac{x^{2k+1}}{2k+1}\right)_0^1 = \frac{1}{2k+1} ii) \int_0^{\pi /2} \cos^{2N+1} x\ dx $ can go fuck itself with its binomial 7 marker bullshit $
  11. mahmoudali

    HSC 2013 Maths Marathon (archive)

    Re: HSC 2013 2U Marathon i) $ point R(p,q) sub into$ \ \ y= \frac{1}{x} \ \ = q= \frac{1}{p} ii) \frac {dy}{dx} = \frac{-1}{x^2} \ \ $sub x= p to get gradient$ \ \ M=\frac{-1}{p^2} $y-y_1= m (x-x_1)$ \ \ y-q= \frac{-1}{p^2}(x-p) \ \ y- \frac {1}{p} = \frac{-x}{p^2} +\frac{1}{p}...
  12. mahmoudali

    hey fag SY123 BRAH

    hey fag SY123 BRAH
  13. mahmoudali

    Integration

    dont you just: [ln(x-x^3)] from -2 to -e and you get ln(6) -ln(-e+e^3)= ln(6/(e(-1+e^2))) = ln(6/-1+e^2) - ln(e) hence that girls solution ln(6/e^2 -1) -1 im pretty sure she didnt want the u substitution method but its better that you show your supreme ways sy ;)
  14. mahmoudali

    flop lol how gay

    flop lol how gay
  15. mahmoudali

    Wooooww when did this wikipedia thing come out is it another technological advancement?

    Wooooww when did this wikipedia thing come out is it another technological advancement?
  16. mahmoudali

    Fuck didnt work

    Fuck didnt work
  17. mahmoudali

    Yes yes no yes yes

    Yes yes no yes yes
  18. mahmoudali

    go spam frost again

    go spam frost again
  19. mahmoudali

    :spin:you w0t m8

    :spin:you w0t m8
  20. mahmoudali

    you better accept it

    you better accept it
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