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Anyone Have Solutions to Edexcel C3 Jan 2010 Paper? watch

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    Does anyone have the solutions to the paper yet?? If so, Can you post them , I think its past the time limit of when the discussion can begin.
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    There's one in the maths exams forum
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    How did it go you guys? Last year it was **** easy, so I'd imagine they made it pretty hard this time.
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    (Original post by gdunne42)
    There's one in the maths exams forum
    You'd think you could link it to make our lives easier but no -_-
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    Yeah I do.

    X = come back later unless you want 65 warning points.

    K = put this in your head and keep it there.

    Y = remember it for future, answers will be up later, wait patiently.

    Therefore applying the chain rule, d/dX x dK x d/Y = go have a kebab for now and come back later.



    Also, as an above posted said, you will know when it's time. It might be time right now, but come back later to be on the safe side. :dance:
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    (Original post by SomeStudent)
    Last year it was **** easy, so I'd imagine they made it pretty hard this time.
    yeah you could say that :emo:
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    It was hard

    Did anyone get that weird logs question? xDD
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    (Original post by neilpatel8)
    Does anyone have the solutions to the paper yet?? If so, Can you post them , I think its past the time limit of when the discussion can begin.
    http://www.thestudentroom.co.uk/show...158161&page=32

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    (Original post by neilpatel8)
    Does anyone have the solutions to the paper yet?? If so, Can you post them , I think its past the time limit of when the discussion can begin.
    Some TSR user posted this solution a while ago.

    http://www.facebook.com/l.php?u=http...4c4d8a4732b0e6
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    (Original post by john1234567)
    yes :awesome:
    genious... pure genious...
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    (Original post by FallenPetal)
    genious... pure genious...


    3^xe^{7x+2} = 15

    3^xe^{7x}e^2 = 15

    ln(3^xe^{7x}e^2) = ln(15)

    ln(3^x) + ln(e^{7x}) + ln(e^2) = ln(15)

    xln(3) + 7x + 2 = ln(15)

    x(ln(3) + 7) = ln(15) - 2

    x = \frac {ln(15) - 2}{ln(3) + 7}
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    (Original post by john1234567)


    3^xe^{7x+2} = 15

    3^xe^{7x}e^2 = 15

    ln(3^xe^{7x}e^2) = ln(15)

    ln(3^x) + ln(e^{7x}) + ln(e^2) = ln(15)

    xln(3) + 7x + 2 = ln(15)

    x(ln(3) + 7) = ln(15) - 2

    x = \frac {ln(15) - 2}{ln(3) + 7}
    Ah. Well the good thing is that I got it to the bit before you need to factorise out the x :woo:

    The bad news is I ran out of time...

    muchos thankos
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    (Original post by FallenPetal)
    Ah. Well the good thing is that I got it to the bit before you need to factorise out the x :woo:

    The bad news is I ran out of time...

    muchos thankos


    i had been struggling with it for around fifteen minutes until i got to that stage, and then i noticed you could factorise the x and i was so happy that i was about to get five more marks
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    I wish I could kill logs.
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    (Original post by john1234567)


    i had been struggling with it for around fifteen minutes until i got to that stage, and then i noticed you could factorise the x and i was so happy that i was about to get five more marks
    Ditto.

    It took me about 10 minutes to remember you can treat ln3 as a real number :P
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    (Original post by john1234567)
    yeah you could say that :emo:
    aw you guys shouldn't worry. Last January 2009 C3/S1/M1 were very hard and they lowered the grade boundaries a lot! And then Jun09 C3 paper was so easy it was unbelievable :p: though the grade boundaries were bit high, but still. So don't worry, this summer you will get a better paper
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    (Original post by FallenPetal)
    It was hard

    Did anyone get that weird logs question? xDD
    the weird logs question wasn't that bad lol
 
 
 
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