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AngelaB97
 one year ago
can someone please explain this to me thoroughly because i keep getting confused on how o simplify this expression
AngelaB97
 one year ago
can someone please explain this to me thoroughly because i keep getting confused on how o simplify this expression

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AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1\[(x^2/y^2)^2 (2y^3/x^2)^3\]

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1dw:1441669920633:dw

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1it is supposed to be 8/x^10y^13 bu i keep getting 8/x^10y^5

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\large \left(\frac{x^2}{y^{2}}\right)^{2}\left(\frac{2y^{3}}{x^2}\right)^3\]Change the variables inside the parenthesis so they are all positive. \[\large ( x^2y^2)^{2}\left(\frac{2}{x^2y^3}\right)^3\]Now that we've got all the variables INSIDE the parenthesis with positive powers, now we take care of the outer powers. Take the first one and put it over 1. \[\frac{1}{(x^2y^2)^2}\cdot \left(\frac{2}{x^2y^3}\right)^3\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now let's distribute the powers through all variables inside the parenthesis. \[\frac{1}{(x^2)^2(y^2)^2} \cdot \frac{2^3}{(x^2)^3(y^3)^3}\]

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1so since the exponent outside of the parentheses is negative aren't you supposed to flip it so it becomes y^2/x^2 ?

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1r it that where i am making the mistake?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now let's distribute the powers through all variables inside the parenthesis. \[\frac{1}{(x^2)^2(y^2)^2} \cdot \frac{2^3}{(x^2)^3(y^3)^3}\]

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1so since the exponent outside of the parentheses is negative aren't you supposed to flip it so it becomes y^2/x^2 ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hmm... not exactly. We take care of each step individually.

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1see, its (x^2/y^2) isn't it supposed to be (y^2/x^2)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So we can write out \(\left(\dfrac{x^2}{y^{2}}\right)^{2}\) as \(\dfrac{1}{\left( \dfrac{x^2}{\dfrac{1}{y^2}}\right)^2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Do you see whats happening here?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't think it's proper to say \(\left(\dfrac{x^2}{y^{2}}\right)^{2} \iff \left(\dfrac{y^2}{x^2}\right)^2\)

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1can you also help me with another problem please?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Are you finished with this one?

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1okay so it's (2y^1z/z^2)^1 (y/3z^2)^2

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a new problem?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\left(\frac{2y^{1}z}{z^2}\right)^{1}\cdot \left(\frac{y}{3z^2}\right)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So like our previous problem, what will you want to do first?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No, what do we do to all the negative variables within the parenthesis?

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1we witch them dont we?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0By switch you mean... make them positive? :)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You are correct. Therefore we will have \[\left(\frac{2z}{yz^2}\right)^{1}\cdot \left(\frac{y}{3z^2}\right)^2\]

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1yes so then it becomes (yz^2/2z)^1 ?

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1multiplied by (y/3z^2)^2

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1so first you make the numbers inside the parentheses positive, then you flip it whenever there's a negative exponent over the parentheses?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\color{#0cbb34}{\text{Originally Posted by}}\) @AngelaB97 so first you make the numbers inside the parentheses positive, `then you flip it whenever there's a negative exponent over the parentheses?` \(\color{#0cbb34}{\text{End of Quote}}\) Im not understanding what you mean by that highlighted portion.

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1as in whenever you have a negative exponent both outside and inside the parentheses, you have to deal with the numbers inside the parentheses?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, you deal with the numbers inside the ( ) first, making them + , then we take care of the outside ( ) by putting over 1. [i.e 1/ ( ) ]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Does that make sense?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Cool! \(\checkmark\) Glad you're following along.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now we have: \[\left(\frac{2z}{yz^2}\right)^{1}\cdot \left(\frac{y}{3z^2}\right)^2\] We're going to put the first one over 1 to make it +. \[\frac{1}{\left(\dfrac{2z}{yz^2}\right)} \cdot \left(\frac{y}{3z^2}\right)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0SOrry, bear with me, Im lagging a lot right now.

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1don't worry you're helping me alot

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This becomes: \[\frac{yz^2}{2z} \cdot \left(\frac{y}{3z^2}\right)^2\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Now we can reduce our fraction. Remember, \(\dfrac{x^m}{x^n} = x^{mn} \therefore \dfrac{z^2}{z} = z^{21} = z\) Now we can rewrite this as: \[\frac{yz}{2} \cdot \left(\frac{y}{3z^2}\right)^2\]

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1then yz/2 * y^2/9z^4

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0We get \[=\frac{y^3z}{18z^4}\] \[\frac{z}{z^4} =z^{14} =z^{3}\] \[=\frac{y^3}{18z^{3}} \iff \boxed{\frac{y^3z^3}{18}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Whoop whoop wait a minute.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0we got \(z^{3}\) therefore it would be: \[\frac{y^3z^{3}}{18} \iff \boxed{\frac{y^3}{18z^3}} \]

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1gotcha, thanks sosososo much!!! :) <3

AngelaB97
 one year ago
Best ResponseYou've already chosen the best response.1you helped me out so much today

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0No problem :) I've got to head off now. SO good luck on the rest!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im glad you participated instead of blatantly asking for the answer.
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