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    Really struggling with these questions would really appreciate some help,
    answers need to be given with positive indices.

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    (Original post by John Robbo)
    Really struggling with these questions would really appreciate some help,
    answers need to be given with positive indices.

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    Use similar application from these examples, for
    q1, consider this x^n \times x^m=x^{m+n} note that the only different part about this on the first part is that you need to also multiply the coefficients of each part together 9 and 2, and add a slightly more complicated power together.

    q2, a very similar idea to the first here, note that x=x^{1} use the above rule to solve this problem by multiplying each part of the bracket by the outside.

    q3, another rule here,  (x^{m})^n=x^{mn}, this is a very fast solve tbh
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    (Original post by Robbie242)
    Use similar application from these examples, for
    q1, consider this x^n \times x^m=x^{m+n} note that the only different part about this on the first part is that you need to also multiply the coefficients of each part together 9 and 2, and add a slightly more complicated power together.

    q2, a very similar idea to the first here, note that x=x^{1} use the above rule to solve this problem by multiplying each part of the bracket by the outside.

    q3, another rule here,  (x^{m})^n=x^{mn}, this is a very fast solve tbh
    Thanks for your help is there any chance you could give me answers too each question too help me revise them.
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    (Original post by John Robbo)
    Thanks for your help is there any chance you could give me answers too each question too help me revise them.
    Okay, although it isn't the standard policy I've put the answers in the Spoiler

    Spoiler:
    Show

    q1
    9a^{\frac{1}{7}} \times 2a^{\frac{-5}{7}}=18a^{\frac{-4}{7}}
    Note that the minus would look better on the middle, just don't know how to do it :P

    q2

    3x^{-3}(x-4x^{2})=3x^{-2}-12x^{-1}

    q3

    (9x^8)^{\frac{-1}{2}}=\frac{1}{3}x^{-4}

    I hope you understand where these answers come from and glad to help!

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    (Original post by John Robbo)
    Thanks for your help is there any chance you could give me answers too each question too help me revise them.
    If you tell us what answers you get, we can tell you if they are correct, and if necessary help you get to the correct answer.
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    (Original post by Robbie242)
    Okay, although it isn't the standard policy I've put the answers in the Spoiler

    Spoiler:
    Show

    q1
    9a^{\frac{1}{7}} \times 2a^{\frac{-5}{7}}=18a^{\frac{-4}{7}}
    Note that the minus would look better on the middle, just don't know how to do it :P

    q2

    3x^{-3}(x-4x^{2})=3x^{-2}-12x^{-1}

    q3

    (9x^8)^{\frac{-1}{2}}=9x^{-4}

    I hope you understand where these answers come from and glad to help!

    On the third one, you forgot to apply the power to the 9 as well . It should have been:

    Spoiler:
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    \frac{1}{3}x^{-4}
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    (Original post by brittanna)
    If you tell us what answers you get, we can tell you if they are correct, and if necessary help you get to the correct answer.
    Dumb question but look at my spoiler please, is my 3rd answer wrong? I was wondering if I do 9^-1/2 as well? Aha nevermind, that was a silly mistake!
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    (Original post by Robbie242)
    Okay, although it isn't the standard policy I've put the answers in the Spoiler

    Spoiler:
    Show

    q1
    9a^{\frac{1}{7}} \times 2a^{\frac{-5}{7}}=18a^{\frac{-4}{7}}
    Note that the minus would look better on the middle, just don't know how to do it :P

    q2

    3x^{-3}(x-4x^{2})=3x^{-2}-12x^{-1}

    q3

    (9x^8)^{\frac{-1}{2}}=9x^{-4}

    I hope you understand where these answers come from and glad to help!

    so with positive indices would the answers be :

    1. 18/a^4/7

    2. 3/x^2 - 12/x^1

    3. 9/x^4
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    (Original post by John Robbo)
    so with positive indices would the answers be :

    1. 18/a^4/7

    2. 3/x^2 - 12/x^1

    3. 9/x^4
    3rd one I did wrong, it was \frac{1}{3}x^{-4} since 9^-1/2=1/3 And yup other two are right
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    how do work out the 1/3x bit?
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    (Original post by John Robbo)
    how do work out the 1/3x bit?
    (9x^8)^{-\frac{1}{2}}=9^{-\frac{1}{2}}*(x^8)^{-\frac{1}{2}}

    =\dfrac{1}{\sqrt{9}} * \dfrac{1}{\sqrt{x^8}}

    =\dfrac{1}{3} * \dfrac{1}{x^4}

    =\dfrac{1}{3x^4} = \boxed{\dfrac{1}{3}x^{-4}}
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    (Original post by John Robbo)
    how do work out the 1/3x bit?
    Given you have (9x^8)^{\frac{-1}{2}} you need to use your algebra rules on the power of 8 as well as 9, as both are intricate parts of the bracket affected by this negative power
    notice that 9^\frac{-1}{2}=\frac{1}{\sqrt 9} which is 1/3
 
 
 
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