Question

(4u^3v^2) ^ -2 / (-2u^2v^3)^-3

Answer 1

rewite the negative power to a positive power. It gets easier then

Answer 2

okay after I got, -8u^6v^9/16u^6v^4 what do I do now?

Answer 3

the answer is -1/128v^2 I dont know how they got that

Answer 4

\[for example: 6^{-2} = 1/6^2\]

Answer 5

there has to be some value of u is given or else you will not be able to cancel u from the give equation.

Answer 6

If the equation is correct than answer as to be -1/2*U^6V

Answer 7

mmm, I too get a different answer. Let me try to write it out again

Answer 8

Is this the equation we're talking about?

\[\frac{(4u^3v^2)^{-2}}{(-2u^22v^3)^{-3}}\]

Answer 9

weird... @sheetalvee is supposed to be right...

Answer 10

yes @pauledenburg

Answer 11

are you sure you're viewing at the right question that has an answer as what you gave?

Answer 12

-8u^6v^9/16u^6v^4
\[\Large \frac{-8u^6v^9}{16u^6v^4} \implies \frac{-8}{16} \times \frac{u^6}{u^6} \times \frac{v^9}{v^4} \implies -\frac 12 \times 1 \times v^5\]
this is very weird...

Answer 13

yesss

Answer 14

you're supposed to be right

Answer 15

maybe the question is wrong then , thank you anyways :)

Answer 16

welcome :D

Answer 17

Be careful, you forgot an 8 in front of v^9:

\[ \frac{-8u^{6}8v^9}{16u^6v^4}\]

Which gives me:
\[\frac{-64u^{6}v^9}{16u^6v^4} => \frac{-4}{1}\times\frac{u^6}{u^6}\times\frac{v^9}{v^4} => -4v^5\]