## 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.

1. amistre64

using math notation for staters

2. amistre64

$(-2y^{-1})^{-2}$

3. amistre64

this?

4. amistre64

$-2^{-2} y^2$ $-\frac{y^2}{4}$ maybe?

Is it still a minus (if it was inside the paren)?

6. anonymous

Yes, you set it up right, amistre.

I suspect the answer may be $y ^{2}\over 4$

Maybe amisstrte64 will come back and verify his answer.

9. anonymous

I think it is 4y^2

10. anonymous

...because it looks like this is the expression... (-2y^-1)^-2 Multiplying exponents (-2y)^2 Expanding brackets 4y^2

I 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$y ^{2}\over 4$

12. anonymous

$(-2y^{-1})^{-2}=(-2)^{-2}y^2=\frac{y^2}{(-2)^2}=\frac{y^2}{4}$\]

Help me out here satellite73

14. anonymous

if that was the problem to beginwith

15. anonymous

we get $\frac{y^2}{4}$ yes?

16. anonymous

hello radar! here to celebrate amistre's 1000 medal

that is what i got, amistre got a -y^2/4

18. anonymous

hope it is soon cause i got to run

before you run look at gianfranco solution above, what is wrong with that approach beside getting a different answer???

20. anonymous

well if the question is as he wrote it, you have to raise (-2) to the power of -2, not the same as $-2^{-2}$

21. anonymous

you should get a 4 in the denominator

22. anonymous

gianfraco was acting as if it was $((-2y)^{-1})^{-2}$

23. anonymous

but 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

24. anonymous

i read it the way you did

I can see that that would make a difference.

26. anonymous

of course. like the difference between $(2\times 3)^2$ and $2\times 3^2$

Understand. Thanks for clearing up a few things.

28. anonymous

O never mind everybody..I think I've figured it out.