Question

5x^-2y^10 over 2x^-1(-3x^-3y^-1)^-2 very hard for me

Answer 1

i just dont know i got this answer (45/2)x^-7 y^8

Answer 2

are there any options

Answer 3

no

Answer 4

gr8 .. :(

Answer 5

its super hard

Answer 6

not super hard it is super complex

Answer 7

\[\frac{5x^{-2}y^{10}}{2x^{-1}(-3x^{-3}y^{-1})^{-2}}\]\[=\frac{5x^{-2}y^{10}}{2x^{-1}((-3)^{(-2)}x^{(-3)(-2)}y^{(-1)(-2)})}\]\[=\frac{5x^{-2}y^{10}}{2x^{-1}((9)^{(-1)}x^{6}y^{2}}\]\[=\frac{5x^{-2}y^{10}}{2(9)^{(-1)}x^{5}y^{2}}\]\[=\frac{5(9)y^{10-2}}{2x^{5+2}}\]\[=\frac{45y^{8}}{2x^{7}}\]\[=\frac{45}{2}x^{-7}y^{8}\]To be frank, it's not that difficult...

Answer 8

gr8 @Callisto

Answer 9

was i right

Answer 10

ya u were right gr8 work @demarco1322

Answer 11

Wait... should you leave the answer in positive indices?

Answer 12

thanks the both of you

Answer 13

If you need, then the answer is the second last step; if not, the your answer is okay :)

Answer 14

thanks

Answer 15

Welcome!~

Answer 16

\[\frac{(5x{}^{\wedge}-2y{}^{\wedge}10)}{ (2x{}^{\wedge}-1(-3x{}^{\wedge}-3y{}^{\wedge}-1){}^{\wedge}-2)}=\frac{\frac{5 y^{10}}{x^2}}{ \left(\frac{2}{x}\right)\left(\frac{x^6 y^2}{9}\right)} =\frac{\frac{5 y^{10}}{x^2}}{ \frac{2 x^5 y^2}{9}}=\frac{45 y^8}{2 x^7} \]

Answer 17

thanks

Answer 18

Your are welcome.