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    (Original post by RiahDawson)
    Name:  image.jpg
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    Firstly do you know what to do when you have a minus sign in the exponent?
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    (Original post by gagafacea1)
    Firstly do you know what to do when you have a minus sign in the exponent?
    Well I know you usually it's

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    (Original post by RiahDawson)
    Well I know you usually it's

    But having a fraction confuses me Name:  image.jpg
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    Great! Now isn't a the same as \frac{a}{1}? Meaning that to remove the minus sign, you turn the fraction upside down (ie turning it to its reciprocal). So if you do that to \frac{3}{c}, it becomes ...?

    Also remember that exponentiation (raising to a number) is distributive over multiplication, and therefore division. In other words, just like how you did part 19 by putting the 2 above the p and the q, you do the same for fractions, you put it above both numbers.
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    (Original post by gagafacea1)
    Great! Now isn't a the same as \frac{a}{1}? Meaning that to remove the minus sign, you turn the fraction upside down (ie turning it to its reciprocal). So if you do that to \frac{2}{c}, it becomes ...?
    Name:  image.jpg
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Size:  381.7 KB Ooooh now I'm starting to understand thank you ^_^" I just thought the 1 magically appears outa nowhere cause I learnt it as a rule.
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    (Original post by RiahDawson)
    Name:  image.jpg
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Size:  381.7 KB Ooooh now I'm starting to understand thank you ^_^" I just thought the 1 magically appears outa nowhere cause I learnt it as a rule.
    Yes exactly! And now you have \displaystyle \left(\frac{c}{3}\right)^2. Which btw I edited my post above to give you an idea of how to do.
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    (Original post by gagafacea1)
    Yes exactly! And now you have \displaystyle \left(\frac{c}{2}\right)^2. Which btw I edited my post above to give you an idea of how to do.
    Thank you so much
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    (Original post by gagafacea1)
    ...
    The question was \left(\frac{3}{c}\right)^{-2} not \left(\frac{2}{c}\right)^{-2} by the way.
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    (Original post by notnek)
    The question was \left(\frac{3}{c}\right)^{-2} not \left(\frac{2}{c}\right)^{-2} by the way.
    woops. Thanks !
 
 
 
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