Question

(help me pleaseee. I'll give you a fan and a medal)

What is the value of 2x^-2y^-2 for x = 3 and y=-2

a. 1/18
b. 72
c. 2(-6)^-4
d. 1/648

Answer 1

@DanJS

Answer 2

\[\frac{ 2 }{ x^2*y^2 }\]

Answer 3

the negative exponents, mean put them on the other side of the fraction and change the sign on the exponent
\[\frac{ a^{-2} }{ 1 } = \frac{ 1 }{ a^2 }\]

Answer 4

oki O_O wait what Im confused?

Answer 5

@DanJS

Answer 6

*confused again*

Answer 7

\[2*x^{-2}*y^{-2}\]

those are negative exponents, you need to move them to the denominator of the fraction , then the exponents will become positive

Answer 8

\[\frac{ 2 }{ x^2*y^2 }\]

Answer 9

so all I gotta do is put it in like that ^

Answer 10

right, then use the values they give x = 3 and y=-2 , in there and evaluate it

Answer 11

(-2)^2 = (-2)*(-2) = +4

Answer 12

\[\frac{ 2 }{ 3^2*(-2)^2 }\]

Answer 13

I got 0.8888889

Answer 14

\[\frac{ 2 }{ 3^2*(-2)^2 } = \frac{ 2 }{ 9 * 4 } = \frac{ 2 }{ 36 } = \frac{ 1 }{ 18 }\]

Answer 15

ohh okay. I see now

Answer 16

that look ok?

Answer 17

yes it does :)