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    I am doing the paper June 2013(R)
    Stuck on 4(b) where it says: using your expansion to estimate an approximate value of (7100)^1/3
    For my expansion I got 2-(3/4)x-(9/32)x^2-(45/256)x^3
    The Original function is (8-9x)^1/3
    If anyone could help it would be good as I don't understand atand what the mark scheme is doing
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    (Original post by Dnaap)
    I am doing the paper June 2013(R)
    Stuck on 4(b) where it says: using your expansion to estimate an approximate value of (7100)^1/3
    For my expansion I got 2-(3/4)x-(9/32)x^2-(45/256)x^3
    The Original function is (8-9x)^1/3
    If anyone could help it would be good as I don't understand atand what the mark scheme is doing
    This question:
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    ~ Realised I misread one part of the question which made my post wrong ~
    :getmecoat:
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    If you make x=1/10 what do you get inside the cube root?

    you get 71/10 so we have  \displaystyle \left (\frac{71}{10} \right )^{1/3} \approx \text{your expansion when x=0.1} .
    Now you need to multiply this by some number to get  \sqrt[3]{7100} \approx \text{your expansion}\times \text{the number you multiplied by on the LHS} .
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    (Original post by Dnaap)
    I am doing the paper June 2013(R)
    Stuck on 4(b) where it says: using your expansion to estimate an approximate value of (7100)^1/3
    For my expansion I got 2-(3/4)x-(9/32)x^2-(45/256)x^3
    The Original function is (8-9x)^1/3
    If anyone could help it would be good as I don't understand atand what the mark scheme is doing
    7100 = 71 \times 1000 so:

    \displaystyle

\begin{equation*}\sqrt[3]{7100} = \sqrt[3]{71 \times 1000} = \sqrt[3]{71} \times \sqrt[3]{1000} = 10 \times \sqrt[3]{71}\end{equation*}.

    Then, if you let x = \frac{1}{10} you get \displaystyle (8 - 9x)^{1/3} = \left(8 - \frac{9}{10}\right)^{1/3} = \left(\frac{71}{10}\right)^{1/3}

    So plugging x = \frac{1}{10} into your expansion gets you an approximation for \left(\frac{71}{10}\right)^{1/3} = 7.1^{1/3}

    But 7.1^{1/3} = \frac{\sqrt[3]{7100}}{10} \Rightarrow \sqrt[3]{7100} = 10 \times 7.1^{1/3}

    So once you put in x = 0.1 into your expansion, you approximate 7.1^{1/3} - you need to multiply it by 10 to get \sqrt[3]{7100}.
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    (Original post by B_9710)
    If you make x=1/10 what do you get inside the cube root?

    you get 71/10 so we have  \displaystyle \left (\frac{71}{10} \right )^{1/3} \approx \text{your expansion when x=0.1} .
    Now you need to multiply this by some number to get  \sqrt[3]{7100} \approx \text{your expansion}\times \text{the number you multiplied by on the LHS} .
    Why x=1/10 though? How do you know to put 1/10 in
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    (Original post by Zacken)
    7100 = 71 \times 1000 so:

    \displaystyle

\begin{equation*}\sqrt[3]{7100} = \sqrt[3]{71 \times 1000} = \sqrt[3]{71} \times \sqrt[3]{1000} = 10 \times \sqrt[3]{71}\end{equation*}.

    Then, if you let x = \frac{1}{10} you get \displaystyle (8 - 9x)^{1/3} = \left(8 - \frac{9}{10}\right)^{1/3} = \left(\frac{71}{10}\right)^{1/3}

    So plugging x = \frac{1}{10} into your expansion gets you an approximation for \left(\frac{71}{10}\right)^{1/3} = 7.1^{1/3}

    But 7.1^{1/3} = \frac{\sqrt[3]{7100}}{10} \Rightarrow \sqrt[3]{7100} = 10 \times 7.1^{1/3}

    So once you put in x = 0.1 into your expansion, you approximate 7.1^{1/3} - you need to multiply it by 10 to get \sqrt[3]{7100}.
    Thanks very thorough I'm just not sure why you chose to put 1/10 in
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    (Original post by Zacken)
    7100 = 71 \times 1000 so:

    \displaystyle

\begin{equation*}\sqrt[3]{7100} = \sqrt[3]{71 \times 1000} = \sqrt[3]{71} \times \sqrt[3]{1000} = 10 \times \sqrt[3]{71}\end{equation*}.

    Then, if you let x = \frac{1}{10} you get \displaystyle (8 - 9x)^{1/3} = \left(8 - \frac{9}{10}\right)^{1/3} = \left(\frac{71}{10}\right)^{1/3}

    So plugging x = \frac{1}{10} into your expansion gets you an approximation for \left(\frac{71}{10}\right)^{1/3} = 7.1^{1/3}

    But 7.1^{1/3} = \frac{\sqrt[3]{7100}}{10} \Rightarrow \sqrt[3]{7100} = 10 \times 7.1^{1/3}

    So once you put in x = 0.1 into your expansion, you approximate 7.1^{1/3} - you need to multiply it by 10 to get \sqrt[3]{7100}.
    Never mind I got it reading over your working. Thank you a lot!
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    (Original post by Dnaap)
    Never mind I got it reading over your working. Thank you a lot!
    Glad it helped!
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    (Original post by Zacken)
    7100 = 71 \times 1000 so:

    \displaystyle

\begin{equation*}\sqrt[3]{7100} = \sqrt[3]{71 \times 1000} = \sqrt[3]{71} \times \sqrt[3]{1000} = 10 \times \sqrt[3]{71}\end{equation*}.

    Then, if you let x = \frac{1}{10} you get \displaystyle (8 - 9x)^{1/3} = \left(8 - \frac{9}{10}\right)^{1/3} = \left(\frac{71}{10}\right)^{1/3}

    So plugging x = \frac{1}{10} into your expansion gets you an approximation for \left(\frac{71}{10}\right)^{1/3} = 7.1^{1/3}

    But 7.1^{1/3} = \frac{\sqrt[3]{7100}}{10} \Rightarrow \sqrt[3]{7100} = 10 \times 7.1^{1/3}

    So once you put in x = 0.1 into your expansion, you approximate 7.1^{1/3} - you need to multiply it by 10 to get \sqrt[3]{7100}.
    How come I get different answers when I put both the LHS and RHS into my calculator? According to what you wrote they should be equal right?
    \displaystyle\begin{equation*} \sqrt[3]{7100} =  10 \times \sqrt[3]{71}\end{equation*}.
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    (Original post by Glavien)
    How come I get different answers when I put both the LHS and RHS into my calculator? According to what you wrote they should be equal right?
    \displaystyle\begin{equation*} \sqrt[3]{7100} =  10 \times \sqrt[3]{71}\end{equation*}.
    Sorry, yes. The first line of my answer is a fail. (the rest is fine) It should be \sqrt[3]{7100} = 10 \times \sqrt[3]{7.1}.
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    (Original post by Zacken)
    Sorry, yes. The first line of my answer is a fail. (the rest is fine) It should be \sqrt[3]{7100} = 10 \times \sqrt[3]{7.1}.
    Ahh thanks, that must mean the first line of the mark scheme is wrong too.
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    (Original post by Glavien)
    Ahh thanks, that must mean the first line of the mark scheme is wrong too.
    Sounds like it.
 
 
 
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