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HI!!

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ready?
yes
first the \(-2\) outside means a) flip everything, then b) square it
Okay
so first \[\huge\frac{1}{\left(-3u^2v^3\right)^2}\]
then we need so square everything that means square \(-3\) to get \(9\) and also double each exponent
Okay but where did the 2 come from in the parenthesis
\[\huge\frac{1}{\left(-3u^3v^3\right)^2}=\frac{1}{9u^6v^6}\]
oh oops that was a typo
Oh okay :) i'm following
should have been \[\huge\frac{1}{\left(-3u^3v^3\right)^2}\]
then square all answer above is right though, looks like your choice A
thank you :) okay there is another question like it... can i post it and tell you how i would do it and then you tell me if i am doing it right or not??
ok sure \[\color\magenta\heartsuit\]
So i would flip it first right
no not here
the reason we flipped before was because there was a \(-2\) outside the parentheses
Oh okay so then i wouldn't flip it i would leave it the same
there is only one choice here that makes sense take a look you have \[\huge\frac{x^4}{x^{-5}}\] right?
the \(-5\) in the deominator means bring it up to the numerator as \(+5\) so \[\frac{x^4}{x^{-5}}=x^4\times x^5=x^{4+5}=x^9\]
only one choice as \(x^9\) in it so we dont really need to do the rest
so i take the two common ones and add them
yeah you want to do it all?
Yes please cause i have a lot more questions like this and i want to make sure i understand them
ok lets take it slow
first off, unlike the last one there is no parentheses anywheres, so it is somewhat easier
you have a minus sign out front that stays there
you also have \[\frac{2}{4}\] which is the same as \(\frac{1}{2}\) so there will be a 2 in the denominator
Okay
as for the x terms, you have \[\frac{x^4}{x^{-5}}\] the \(-5\) has a minus sign, so that comes upstairs as \(x^5\) which is why you get \[\frac{x^4}{x^{-5}}=x^4\times x^5=x^9\]
Alright
and for \[\frac{y^2}{y^5}\] the 5 is bigger than the 2, so subtract 2 from 5 in the denominator \[\frac{y^2}{y^5}=\frac{1}{y^{5-2}}=\frac{1}{y^3}\]
so in total, a) there is a - sign out front b) there is a 2 in the denominator c) there is a \(x^9\) in the numerator and d) a \(y^3\) in the denominator
Okay so which ever number is bigger you use that to decide whether or not you add or subtract and whether or not it says on the top or the bottom, am i right?
yes more or less
if the exponent is negative a) if it is up bring it down b) if it is down bring it up
if both terms have positive exponents, subtract the smaller one from the bigger one
here is an example \[\frac{x^7}{x^{-3}}=x^{10}\] wheras \[\frac{x^4}{x^{10}}=\frac{1}{x^6}\]
Okay that makes a little more since now
whew
Thanks

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