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    I'm trying to do question 3b on this paper and have looked at the mark scheme but I can't understand what is supposed to happen to the 256 in the question as the answer seems to ignore it.
    http://www.mathsgeeks.co.uk/allpaper...AQAJan10QP.pdf
    here is the mark scheme
    http://www.mathsgeeks.co.uk/allpaper...AQAJan10MS.pdf
    i have found part a fine
    i would appreciate it if anyone could help me get a better understanding of this, thanks
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    (Original post by sw95)
    I'm trying to do question 3b on this paper and have looked at the mark scheme but I can't understand what is supposed to happen to the 256 in the question as the answer seems to ignore it.
    http://www.mathsgeeks.co.uk/allpaper...AQAJan10QP.pdf
    here is the mark scheme
    http://www.mathsgeeks.co.uk/allpaper...AQAJan10MS.pdf
    i have found part a fine
    i would appreciate it if anyone could help me get a better understanding of this, thanks
    Part b follows on from part a, in terms of content. So, we want to get our expression into a form similar to that of \displaystyle\left(1+\frac{3}{4}  x\right)^{-\frac{1}{3}}

    If we divide the top and bottom of our fraction under the cube root, we have:

    \displaystyle\sqrt[3]{\frac{256}{4+3x}}=\sqrt[3]{\frac{64}{1+\frac{3}{4}x}}=64^{  \frac{1}{3}}\left(1+\frac{3}{4}x  \right)^{-\frac{1}{3}}

    Giving us the "4" in front of the brackets in the MS.
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    (Original post by sw95)
    I'm trying to do question 3b on this paper and have looked at the mark scheme but I can't understand what is supposed to happen to the 256 in the question as the answer seems to ignore it.
    http://www.mathsgeeks.co.uk/allpaper...AQAJan10QP.pdf
    here is the mark scheme
    http://www.mathsgeeks.co.uk/allpaper...AQAJan10MS.pdf
    i have found part a fine
    i would appreciate it if anyone could help me get a better understanding of this, thanks
    When I was doing A level maths I used this website all the time:
    http://www.examsolutions.net/
    It is sooo good, really good videos explaining how to figure out all of the answers etc. it has past papers on there too.
    I managed to get an A in C4! All thanks to that website


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    (Original post by ghostwalker)
    Part b follows on from part a, in terms of content. So, we want to get our expression into a form similar to that of \displaystyle\left(1+\frac{3}{4}  x\right)^{-\frac{1}{3}}

    If we divide the top and bottom of our fraction under the cube root, we have:

    \displaystyle\sqrt[3]{\frac{256}{4+3x}}=\sqrt[3]{\frac{64}{1+\frac{3}{4}x}}=64^{  \frac{1}{3}}\left(1+\frac{3}{4}x  \right)^{-\frac{1}{3}}

    Giving us the "4" in front of the brackets in the MS.
    thanks
 
 
 
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