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    ok so can anyone tell me how you could re-arange this equation (T^2=(4pie^2*A*B)/C*D)) into something that looks like y-mx+c using Logs Where T^2=y
    and B=x.
    Any help would be appreciated
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    Anyone?
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    (Original post by ryan051991)
    Anyone?
    This looks too similar to an AQA Physics ISA I did on simple harmonic motion... And it doesn't involve logs by the way.
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    (Original post by Freerider101)
    This looks too similar to an AQA Physics ISA I did on simple harmonic motion... And it doesn't involve logs by the way.
    ]


    yeh dude ive done it i was just so confused about it coz i thought you would have to do logs to get a (+c) because it did not go through origin
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    (Original post by ryan051991)
    ok so can anyone tell me how you could re-arange this equation (T^2=(4pie^2*A*B)/C*D)) into something that looks like y-mx+c using Logs Where T^2=y
    and B=x.
    Any help would be appreciated
    From that equation you simply want to write
    m=\frac{4\pi^2 A}{CD}

    c=0
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    Ah dude i am totally lost think i am just never going to get it LOL
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    (Original post by ryan051991)
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    yeh dude ive done it i was just so confused about it coz i thought you would have to do logs to get a (+c) because it did not go through origin
    There is a good reason for that.
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    (Original post by Freerider101)
    This looks too similar to an AQA Physics ISA I did on simple harmonic motion... And it doesn't involve logs by the way.
    Yeah you can Log both sides as it would give you m x and c (where aplicable) so would show relationship
 
 
 
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