Maths year 11

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    (Original post by z_o_e)
    Oh alright x



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    Oh I think I see where you're going wrong. It's just the way you write it.

    You wrote 27^{\frac{1}{3}} as 3\sqrt{27} when you really meant it as \sqrt[3]{27}. Place the number in that little gap of the root as opposed to next to it, it's where I got confused.

    Other than that, you got C and D wrong.
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    (Original post by RDKGames)
    Oh I think I see where you're going wrong. It's just the way you write it.

    You wrote 27^{\frac{1}{3}} as 3\sqrt{27} when you really meant it as \sqrt[3]{27}. Place the number in that little gap of the root as opposed to next to it, it's where I got confused.

    Other than that, you got C and D wrong.
    How do I do c and D ?

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    (Original post by z_o_e)
    How do I do c and D ?

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    For C, you were right at \frac{1}{36^{1/2}} but you are wrong to say that this is the same as \sqrt{36}.
    \displaystyle \frac{1}{36^{1/2}}=\frac{1}{\sqrt{36}}

    For D, it's a similar problem. \displaystyle \frac{1}{32^{1/5}}=\frac{1}{\sqrt[5]{32}}. The fifth root of 32 is 2.
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    (Original post by RDKGames)
    For C, you were right at \frac{1}{36^{1/2}} but you are wrong to say that this is the same as \sqrt{36}.
    \displaystyle \frac{1}{36^{1/2}}=\frac{1}{\sqrt{36}}

    For D, it's a similar problem. \displaystyle \frac{1}{32^{1/5}}=\frac{1}{\sqrt[5]{32}}. The fifth root of 32 is 2.
    How do I do b?


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    (Original post by z_o_e)
    How do I do b?


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    You are correct at \displaystyle \frac{1}{\frac{9}{25}}. Now multiply this fraction by \frac{25}{25}. Also your C is wrong on that too.
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    (Original post by RDKGames)
    You are correct at \displaystyle \frac{1}{\frac{9}{25}}. Now multiply this fraction by \frac{25}{25}. Also your C is wrong on that too.
    I got 225/25

    So do I divide 25 by 225 and see what whole number goes into it?


    What's wrong with C?


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    (Original post by RDKGames)
    You are correct at \displaystyle \frac{1}{\frac{9}{25}}. Now multiply this fraction by \frac{25}{25}. Also your C is wrong on that too.


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    C is now fine.

    For B:
    \displaystyle \frac{1}{\frac{9}{25}} \cdot \frac{25}{25} = \frac{1 \cdot 25}{\frac{9}{25} \cdot 25}=\frac{25}{9}
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    (Original post by RDKGames)
    C is now fine.

    For B:
    \displaystyle \frac{1}{\frac{9}{25}} \cdot \frac{25}{25} = \frac{1 \cdot 25}{\frac{9}{25} \cdot 25}=\frac{25}{9}
    How will I do these type of questions


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    (Original post by z_o_e)
    How will I do these type of questions


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    I'll do the first example for you.

    \displaystyle (\frac{8}{125})^{-4/3}=(\frac{\sqrt[3]8}{\sqrt[3]{125}})^{-4}=(\frac{2}{5})^{-4}=(\frac{2^4}{5^4})^{-1}=(\frac{16}{625})^{-1}=\frac{625}{16}
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    (Original post by RDKGames)
    I'll do the first example for you.

    \displaystyle (\frac{8}{125})^{-4/3}=(\frac{\sqrt[3]8}{\sqrt[3]{125}})^{-4}=(\frac{2}{5})^{-4}=(\frac{2^4}{5^4})^{-1}=(\frac{16}{625})^{-1}=\frac{625}{16}
    I gave it a go ;//


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    (Original post by z_o_e)
    I gave it a go ;//


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    Absolutely correct!
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    (Original post by RDKGames)
    Absolutely correct!
    Yuss♥

    Wb this?
    Do I flip them over in the end?

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    (Original post by z_o_e)
    Yuss♥

    Wb this?
    Do I flip them over in the end?

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    That's correct.

    When you have a fraction raised to -1, the "flipping over" procedure is called taking the reciprocal of the fraction. It is easy to remember this, and well worth it.

    The way it works is this;

    Take  \displaystyle (\frac{2}{3})^{-1}

    This is the same as \displaystyle \frac{1}{\frac{2}{3}}

    We can multiply the denominator by 3, which forces us to multiply the numerator by 3 as well:

    \displaystyle \frac{1}{\frac{2}{3}} \cdot \frac{3}{3} = \frac{1\cdot 3}{\frac{2}{3} \cdot 3} = \frac{3}{2} which is "flipped"
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    (Original post by RDKGames)
    That's correct.

    When you have a fraction raised to -1, the "flipping over" procedure is called taking the reciprocal of the fraction. It is easy to remember this, and well worth it.

    The way it works is this;

    Take  \displaystyle (\frac{2}{3})^{-1}

    This is the same as \displaystyle \frac{1}{\frac{2}{3}}

    We can multiply the denominator by 3, which forces us to multiply the numerator by 3 as well:

    \displaystyle \frac{1}{\frac{2}{3}} \cdot \frac{3}{3} = \frac{1\cdot 3}{\frac{2}{3} \cdot 3} = \frac{3}{2} which is "flipped"


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    Correct.
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    (Original post by RDKGames)
    Correct.
    How do I do this?


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    (Original post by z_o_e)
    How do I do this?


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    You can treat the 32 as a fraction of 32/1 if you wish.

    Otherwise:

    (32)^{-4/5}=(\sqrt[5]{32})^{-4}=(2)^{-4}=(2^4)^{-1}=(16)^{-1}=\frac{1}{16}
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    (Original post by RDKGames)
    You can treat the 32 as a fraction of 32/1 if you wish.

    Otherwise:

    (32)^{-4/5}=(\sqrt[5]{32})^{-4}=(2)^{-4}=(2^4)^{-1}=(16)^{-1}=\frac{1}{16}
    So for this... idk this is Confusing



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    (Original post by z_o_e)
    So for this... idk this is Confusing



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    Yeah that's right. The cube root of 27 is....
 
 
 
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