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anonymous
 5 years ago
Can somebody explain how to simplify this expression? Negative 2y to the negative 1 power, all inside parentheses, with a negative 2 exponent outside the parentheses.
anonymous
 5 years ago
Can somebody explain how to simplify this expression? Negative 2y to the negative 1 power, all inside parentheses, with a negative 2 exponent outside the parentheses.

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amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0using math notation for staters

amistre64
 5 years ago
Best ResponseYou've already chosen the best response.0\[2^{2} y^2\] \[\frac{y^2}{4}\] maybe?

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Is it still a minus (if it was inside the paren)?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Yes, you set it up right, amistre.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I suspect the answer may be \[y ^{2}\over 4\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Maybe amisstrte64 will come back and verify his answer.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0...because it looks like this is the expression... (2y^1)^2 Multiplying exponents (2y)^2 Expanding brackets 4y^2

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I went this direction, but I do see your logic. I may have violated order of operations:\[(2y ^{1})^{2}\] to:\[1\over (2y ^{1})^{2}\]then to:\[1\over 4 y ^{2}\] finally\[y ^{2}\over 4\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[(2y^{1})^{2}=(2)^{2}y^2=\frac{y^2}{(2)^2}=\frac{y^2}{4}\]\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Help me out here satellite73

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if that was the problem to beginwith

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we get \[\frac{y^2}{4}\] yes?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hello radar! here to celebrate amistre's 1000 medal

radar
 5 years ago
Best ResponseYou've already chosen the best response.0that is what i got, amistre got a y^2/4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hope it is soon cause i got to run

radar
 5 years ago
Best ResponseYou've already chosen the best response.0before you run look at gianfranco solution above, what is wrong with that approach beside getting a different answer???

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0well if the question is as he wrote it, you have to raise (2) to the power of 2, not the same as \[2^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you should get a 4 in the denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0gianfraco was acting as if it was \[((2y)^{1})^{2}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but that is now how i read the problem. i read it as only the y being raised to the power of 1. if there were no parentheses that is what it means

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i read it the way you did

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I can see that that would make a difference.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0of course. like the difference between \[(2\times 3)^2\] and \[2\times 3^2\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Understand. Thanks for clearing up a few things.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0O never mind everybody..I think I've figured it out.
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