Question

Perform the indicated operation and simplify as much as possible: (2xy^4)^-2 divided by 3(x^4y)^-1

Answer 1

I'll get on this one in a sec

Answer 2

thanks!!

Answer 3

oh wonderful, division of negative exponents, so really, we're dividing the second by the first....

Answer 4

okay

Answer 5

\[\frac{ 3x^43y }{(2xy^4)^2} \]

Answer 6

\[\frac{3x^43y}{ 4x^2y^8 }\]

Answer 7

then common factors

Answer 8

\[\frac{ 3 }{ 4 } * \frac{ x^2 }{ y^4 }\] if I did my math right.

Answer 9

wouldn't the 3 stay in the denominator\[3(x^4y)^{-1}={3 \over (x^4y)}\]

Answer 10

so when you divide by this it is \[x^4y \over 3\]

Answer 11

you sure? maybe I read the problem wrong?

Answer 12

x^2 divided by 12y^7 is the answer but idk how to get it

Answer 13

Is it (3x^4y)^-1 or 3(x^4y)^-1

Answer 14

then chme is right

Answer 15

\[(2xy^4)^{-2} \over 3(x^4y)^{-1}\] is this the question?

Answer 16

I did the work right, just messed up the 3

Answer 17

\[\huge (x^a)^b=x^{ab}\]\[\huge \frac{ x^a }{ x^b }=x^{a-b}\]Things to memorize

Answer 18

^yes.... that's what I used right there

Answer 19

chme yes that is the correct problem

Answer 20

ChmE is right

Answer 21

yuppers, got that

Answer 22

chme come back!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 23

(x^4*3y)/3(2x*y^4)^2

Answer 24

\[(2xy^4)^{-2} \over 3(x^4y)^{-1}\]\[\frac{ x^4y }{ 3(2xy^4)^2 }\]\[\frac{ x^4y }{ 3(4x^2y^8) }\]\[\frac{ x^4y }{ 12x^2y^8 }\]\[\frac{ x^2 }{ 12y^7 }\]^Final Answer^

Answer 25

correct

Answer 26

medal

Answer 27

yes thank you you are the best!!

Answer 28

oh stop it