Question

help...

Answer 1

Helper.

Answer 2

hamburger helper

Answer 3

loool.

Answer 4

i know there is a question coming somewhere, i can just feel it...

Answer 5

i can see it coming.

Answer 6

\[^{} \sqrt[4]{162xy}\]

Answer 7

x is to the 4th y is to the 6th

Answer 8

I'm slow...

Answer 9

ok first off
\[162=2\times 81\] so
\[\sqrt{162}=\sqrt{81\times 2}=\sqrt{81}\times \sqrt{2}=9\sqrt{2}\]

Answer 10

\[3xy \sqrt[4]{2y^2}\]

Answer 11

that is a forthroot at sat.

Answer 12

basically you want to know what comes outside the radical, and what stays in

Answer 13

ooh
\[\sqrt[4]{162x^4y^6}\]

Answer 14

yes...I am jealous ur brilliance...everyone

Answer 15

in that case
\[162=2\times 3^4\] and so
\[\sqrt[4]{162}=\sqrt[4]{2\times 3^4}=\sqrt[4]{3^4}\times \sqrt[4]{2}=3\sqrt[4]{2}\]

Answer 16

likewise
\[\sqrt[4]{x^4}=x\]

Answer 17

and
\[\sqrt[4]{y^6}=\sqrt[4]{y^4y^2}=\sqrt[4]{y^4}\times \sqrt[4]{y^2}=y\sqrt[4]{y^2}\]

Answer 18

giving saifoo's answer of
\[3xy \sqrt[4]{2y^2}\]

Answer 19

haha!

Answer 20

if you are a stickler for detail it is also true that
\[\sqrt[4]{y^2}=\sqrt{y}\]

Answer 21

please dont be jealous.

Answer 22

@makenan

Answer 23

i am pouting now

Answer 24

i did a bit mistake for sure.

Answer 25

thank you greatly