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    (Original post by Ayaz789)
    http://pastpapers.download.wjec.co.uk/s12-0975-01.pdf
    I silll am quite sure im right :/
    All you do on that one is e^3
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    (Original post by Salamon16)
    No
    Idek , anyone got the unofficial ms
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    (Original post by Mango88)
    a and b values for transformations?

    a=-5
    a=3
    B=-2/3
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    (Original post by Sznsnsn)
    All you do on that one is e^3
    http://pastpapers.download.wjec.co.u...ematics-ms.pdf
    Like look in part b , thats exactly what i did tbh
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    (Original post by Sogolhosseini)
    a=-5
    a=3
    B=-2/3
    I got a=5 & a=-3
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    (Original post by Ayaz789)
    http://pastpapers.download.wjec.co.uk/s12-0975-01.pdf
    I silll am quite sure im right :/

    http://pastpapers.download.wjec.co.u...matics1-ms.pdf

    Check out the answers
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    (Original post by Ayaz789)
    http://pastpapers.download.wjec.co.u...ematics-ms.pdf
    Like look in part b , thats exactly what i did tbh
    That's because that one is just on its own not a power of an exponential
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    (Original post by 98matt)
    What about the parametric differentiation q where you had to give the second derivative in terms of y?

    Posted from TSR Mobile
    -1/y^3
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    (Original post by Sznsnsn)
    That's because that one is just on its own not a power of an exponential
    Well i should still get an A* hopefully considering i messed that and 5a up!
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    (Original post by Sogolhosseini)
    http://pastpapers.download.wjec.co.u...matics1-ms.pdf

    Check out the answers
    ^^^^^
    It's just times by e^3
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    (Original post by FusionNetworks)
    -1/y^3
    YES ! I had that too I think

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    (Original post by Ayaz789)
    I got a=5 & a=-3
    Agree
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    WJEC need to sort out this gardening obsession of theirs


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    (Original post by Ayaz789)
    I got a=5 & a=-3

    Yea yea you are right. The I forgot the the shape for a sec
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    what did you get for d2y/d2x in terms of t
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    So for hh(x), was h-1(x) = h(x)?
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    (Original post by FusionNetworks)
    So for hh(x), was h-1(x) = h(x)?
    Yeah and then 1/9 when subbed the value in?
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    (Original post by FusionNetworks)
    So for hh(x), was h-1(x) = h(x)?
    Yeah I did that, and for b i got 1/9
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    (Original post by Mango88)
    what did you get for d2y/d2x in terms of t
    If I remember correctly: -1/8cos^3(3t)
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    What did you guys do for 2 part b?
 
 
 
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