Question

solve (2b^2/-y^5)^-2

Answer 1

i got y^10/4b^4 is this right

Answer 2

solve?

Answer 3

simplify sorry

Answer 4

\[(\frac{2b^2}{-y^5})^{-2}=\frac{(2b^2)^{-2}}{(-y^5)^{-2}}=\frac{(-y^5)^2}{(2b^2)^2}=\frac{-y^{10}}{4b^4}\]

Answer 5

yeah that shuold be (-y)^10 and that = y^10

Answer 6

oh i forgot the negative thanksihaveon e more i'm not sure about

Answer 7

\[(-y^5)^2=(-1*y^5)^2=(-1)^2(y^{10})=y^2\]

Answer 8

err y ^10...

Answer 9

you are correct:)

Answer 10

okay

Answer 11

thanks

Answer 12

np

Answer 13

i have one more to show

Answer 14

x-y^-1/x^-1-y ig ot xy-x-y^2/y is this correct?

Answer 15

\[\frac{x-y^{-1}}{x^{-1}-y}?\]

Answer 16

yes

Answer 17

\[\frac{x-\frac{1}{y}}{\frac{1}{x}-y}=\frac{\frac{xy-1}{y}}{\frac{1-xy}{x}}=\frac{xy-1}{y}*\frac{x}{-(xy-1)}=-\frac{x}{y }\]

Answer 18

sorry I got a call...

Answer 19

this is true for\[xy\ne1,x\ne0\]

Answer 20

okay