Question

radical question.

Answer 1

@DanJS

Answer 2

\[\sqrt[3]{81x^6y^9}\]

Answer 3

I can show you the process that I went through if you want.

Answer 4

that one is pretty straight forward

Answer 5

\[\sqrt[3]{81x^6y^9}=\sqrt[3]{81^\frac{ 1 }{ 3 }x^\frac{ 6 }{ 3 }y^\frac{ 9 }{ 3 }}\]

Answer 6

you did it right, and divided the powers by 3, but you left the root on it

Answer 7

\[\sqrt[3]{27x^2y^3}\]

Answer 8

how did you get 81 to become three?

Answer 9

\[\sqrt[3]{27x^2y^3} = \sqrt[3]{3^3} * \sqrt[3]{x^2}*\sqrt[3]{y^3} = 3*x^{2/3}*y\]

Answer 10

sorry the first one i messed it up

Answer 11

\[\sqrt[3]{81} = \sqrt[3]{3*3*3*3} = \sqrt[3]{3^3*3} = 3*\sqrt[3]{3}\]

Answer 12

okay. I saw it like this:
\[\sqrt[3]{81}=\sqrt[3]{27}=3\]

Answer 13

nah, i messed it up, the cubed root of 81 is 3 times the cubed root of 3

Answer 14

Okay. So we now have:
\[3\sqrt{x^2y^3}\]

Answer 15

sorry i meant:
\[3\sqrt[3]{x^2y^3}\]

Answer 16

lets start over here...
\[\sqrt[3]{81x^6y^9} = \sqrt[3]{81}*\sqrt[3]{x^6}*\sqrt[3]{y^9}\]

Answer 17

\[\sqrt[3]{81} = \sqrt[3]{3^3*3} = 3\sqrt[3]{3}\]

Answer 18

\[\sqrt[3]{x^6} = x^2~~~and~~~~\sqrt[3]{y^9} = y^3\]

Answer 19

\[\sqrt[3]{81x^6y^9} = 3*\sqrt[3]{3}*x^2*y^3\]

Answer 20

okay. So I didn't put the root 3. Sorry I see. So final answer: (written how my class shows it)
\[3x^2y^3\sqrt[3]{3}\]

Answer 21

ok, yeah that is the same as the post above, they just put the radical at the end

Answer 22

either way is right

Answer 23

yeah that's what i meant by the way my class does it. they always have the y and x variables on the left side.

Answer 24

they dont take points off if you put them in a different order do they? it is all multiplication

Answer 25

i am just used to writing the variables after numbers, it doesnt really matter though

Answer 26

No they don't. I was just told that on the final, which is multiple choice, they will have it written like that.

Answer 27

ahh, ok

if you want to do another i have some time

Answer 28

sweet! I'll start a new thread.