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May 23, 2008
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Moderator, Male, from Melbourne

Armitage2Riewoldt was last seen:
Mar 19, 2018 at 1:10 PM
    1. SaintsSeptember
    2. SJ
      Nah I'm still in the habit of using the drop-down menu on the bottom, and MyBigfooty. I must press that link hundreds of times per day.
    3. SJ
      Nah that's it. Cheers. I never go to boards via that way so thanks.
    4. SJ
      How do you know how many posts there have been on your board?
    5. esaint66
      Not really it's just a few people are so it would be nice to know :)
    6. esaint66
      Hey mate, Are you on saintsational ?
    7. MC_Bomba
      have a micky new year.
    8. ssfc0203
      R u gutted?
    9. Drummond
      Good luck with getting the job mate. :)
    10. Fender
    11. A•|dan
      Lol kewlies. I'm finishing grade 2 soon. :)
    12. A•|dan
      lol nah i was joking. so wassup g
    13. A•|dan
      uh.. another hint please?
    14. A•|dan
      oh hiya ther, just noticed i got this. now who could this be? hmm
    15. holllywood
      hahaha at methosd 2 paper.

      hardest thing ive ever done.
    16. Ben the Gooner
      Ben the Gooner
      I think so. It was only worth two, so one for the correct rule (chain rule or product rule) and one for the correct values.
    17. Ben the Gooner
      Ben the Gooner
      1)a) dy/dx=5(6x-5)(3x^2-5x)^4
      1)b) f'(0)=1 (lolx at the slight misprint in the question stem!)
      2) Negative hyperbolic graph: x-int at (1,0), y-int at (0,-2)
      Vertical asymptote x=-1, horizontal asymptote y=2
      3) x={-2pi/9, 2pi/9} over [-pi/2,pi/2]
      4)a) Show that k=pi/2 by integration such that A=1
      4)b) Pr (X<1/4 | x<1/2) = [2-sqrt(2)]/2
      5) C= 0.5log(6)
      6)a) Mode=3 (0.4 > {0.3,0.2,0.1})
      6)b) Probability=0.3*
      {*Derived from Binomial Distribution, where n=2, r=2, p=Pr(X=x):
      Pr(X=2)=(2 C 2)[Pr(X=x)]^2[1-Pr(X=x)]^0 = [Pr(X=x)]^2
      => Probability = 0.1^2+0.2^2+0.3^2+0.4^2 = 0.3}
      7) Markov Chain: Pr(CCD) + Pr(CDC) + Pr(DCC) = 0.336
      8)a) dom f' = R/{1,2}
      8)b) Modulus graph...stationary points (local maxima) (-1,4), (2/3 , 17/27)
      Non-inclusive endpoint at (1,0), flip the given curve such that y>=0
      9)a) y=[4000sqrt(3)]/[3(x)^2]
      9)b) Show that A=[4000sqrt(3)]/x + [(sqrt(3)x^2)/2]
      9)c) Minimum** Surface Area at x=cuberoot(4000) =10cuberoot(4) cm (looking for tangible evidence that VCAA overlooks root simplification...)
      {**The graph for A(x) is an oblique curve whose only stationary point is a local minimum at x=10cuberoot(4), hence the minimum surface area occurs at x=10cuberoot(4)}
      10)a) Inverse f: (-1,infinity)->R, inverse f(x) = 0.5loge(x+1)
      10)b) Graph of y=x, x>1 with non-inclusive endpoint at (-1,-1)
      10)c) (-2x)/(2x+1)*** => a=-2,b=2,c=1
      {***We need to evaluate f[-inverse f(2x)] => -inverse f(2x) = -0.5loge(2x+1) = 0.5loge[(2x+1)^(-1)] = 0.5loge[1/(2x+1)]
      f[-inverse f(2x)] = e^2[0.5loge[1/(2x+1)]]-1 = 1/(2x+1) - 1 = [1-2x-1]/[2x-1] = (-2x)/(2x-1), QED}

      From another forum.

      Apparently there was a misprint in the first question. :eek:
    18. Ben the Gooner
      Ben the Gooner
      I don't know about the answers. I'll have a look around.

      My graph was sort of a cross between y=x and y=e^x

      I'd be happy with 2-3/5 for that last one. My inverse was right, and I may scrape one for the graph or the working for the last question.
    19. Ben the Gooner
      Ben the Gooner
      Pretty good. I struggled a bit with that question, but realised that the height wasn't xcm, so I got it out in the end. That last question (f(^-1(x))) was a shocker.
    20. ssfc0203
    21. Ben the Gooner
      Ben the Gooner
      How was the exam?
    22. Blind Blink
      Blind Blink
      well your username used to be mick13 or something, so I didn't think this was you. hmph.
    23. Blind Blink
      Blind Blink
      ummmmm nath?
    24. wheels4
      haha what happened to that one ?
    25. wheels4
      oh yees, i joined two days before you. bragging rights now :P haha
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