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Help with maths question

image.jpgHow do you do.. Number 20
Original post by RiahDawson
image.jpgHow do you do.. Number 20


Firstly do you know what to do when you have a minus sign in the exponent?
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Avatar for RiahDawson
RiahDawson
OP
17
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

But having a fraction confuses me image.jpg
Original post by RiahDawson
Well I know you usually it's

But having a fraction confuses me image.jpg

Great! Now isn't aa the same as a1\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 3c\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.
(edited 8 years ago)
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RiahDawson
OP
17
Original post by gagafacea1
Great! Now isn't aa the same as a1\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 2c\frac{2}{c}, it becomes ...?


image.jpg 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.
Original post by RiahDawson
image.jpg 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 (c3)2\displaystyle \left(\frac{c}{3}\right)^2. Which btw I edited my post above to give you an idea of how to do.
(edited 8 years ago)
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RiahDawson
OP
17
Original post by gagafacea1
Yes exactly! And now you have (c2)2\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
Reply 7
Avatar for Notnek
Notnek
21
Original post by gagafacea1
...

The question was (3c)−2\left(\frac{3}{c}\right)^{-2} not (2c)−2\left(\frac{2}{c}\right)^{-2} by the way.
Original post by notnek
The question was (3c)−2\left(\frac{3}{c}\right)^{-2} not (2c)−2\left(\frac{2}{c}\right)^{-2} by the way.


woops. Thanks !

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