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    3x^-1/2 ii) -12x^-1 +c

    when a=5 sum from u2 to u4 is 25
    when a=6 sum from u2 to u4 is 0

    r=2 d=3.5 difference is 768-33=735

    Angle is 66.5~

    y=sin(2x)

    graph is translated (0, -3)

    logs simplify to -3m

    log question x=2.14

    area of B is 1/2r^2(theta)-1/2a^2sin(theta)
    a is 6root2cm~

    trap rule A=22.8 V=1140

    sub 4.8 into x, y is> 4.4 therefore wont fit

    area according to model is 24 V=1200
    sub 5+h and 5 into f(x) simplify to get 8+h
    as h-->0 m=8
    tangent is y=8x-25
    area with axes is 39.0625
    simplify logs to get straight line form with gradient b and y intercept log a
    a~4.17 b~0.19

    therefore y=4.17*10^0.19t

    last question sub 4 into w viruses: 260.
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    What was the question to which people got the answers 25 and 0? What did it ask you to do?
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    So disappointed- i know how to do all the questions that were on that paper except for 1 marker but when doing the paper, i made too many mistakes, just had no concentration. Even with the question asking for the equation of the sine curve on the graph i wrote y=sinx- missed the transformation completely...guess i'm retaking c2 next year.
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    Also. My integrated area was massive. 144cm^2. I redid the integration 100 times but I don't see my mistake.


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    (Original post by ScalyJake)
    I also got this. Because the cubed was for the entire log, so you didn't bring it down to the front.
    Yeah if you plugged in values for a and m into the calculator it worked out right
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    Was the equation for the sin graph y=sin(2x) ????????
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    Can we all agree there will be low grade boundaries as it was so hard this paper
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    (Original post by Lucuniexet)
    Was the equation for the sin graph y=sin(2x) ????????
    Yeah
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    (Original post by tom476zf)
    yes got this not sure is its right though ?
    Got same (-3m) pretty sure its right
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    (Original post by Mr Moon Man)
    Yeah
    Oh **** I wrote theta. Fml so many stupid mistakes


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    (Original post by Qwertykeyboard15)
    Can we all agree there will be low grade boundaries as it was so hard this paper
    Personally, i didn't find it hard at all. I still ****ed up that paper- though this wasn't due to difficulty, but to...i don't even know
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    (Original post by Crozzer24)
    The log graph was awful, felt impossible to draw a best line of fit, ended up with y intercept of 0.6 and gradient 0.2 which seems way off
    I got exactly the same
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    I put -m^3
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    I got y intercept of 0.58
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    (Original post by Lucuniexet)
    for the simplifying log question i got -3m
    is this familiar to anyone?!
    *****t i just left it as -1^3m. do you think ill get marks for that
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    (Original post by WuMyster)
    Also. My integrated area was massive. 144cm^2. I redid the integration 100 times but I don't see my mistake.
    I got 144 as well but the answer was 24 apparently which makes a lot more sense as they should be similar but I don't know what I did wrong

    I presume you were meant to x the fraction by the x values first and then integrate it.
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    Answer was -m^3 because the cube was for the whole log
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    Am I the only one who got 20 and 0
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    Did anyone get 5roo4/18 for q.3 b ??
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    (Original post by Akhawais)
    Answer was -m^3 because the cube was for the whole log
    you times indices across brackets giving you 3m
 
 
 
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