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Mathematics 6 Online
OpenStudy (anonymous):

Put the following radical expression into simplified form. Assume all variables represent positive numbers.

OpenStudy (anonymous):

\[\sqrt{\frac{ 48x^2y^3 }{ 5 }}\]

OpenStudy (anonymous):

@iambatman

OpenStudy (anonymous):

\[\sqrt{48}=4\sqrt{3}\] x is now outside of the radical & so is on y & y^2 is inside the radical? & the 5 just syas there?

OpenStudy (anonymous):

*one

OpenStudy (anonymous):

\[\frac{ 4xy \sqrt{3y} }{ 5 }?\]

OpenStudy (anonymous):

@mathslover

mathslover (mathslover):

Well done @cookiibabii93

OpenStudy (anonymous):

Thank you ! :)

OpenStudy (anonymous):

@Jack1 , this one too...Sorry to work you, but this is for a quiz lol i need all these points

OpenStudy (jack1):

\[\large \sqrt{\frac {48 x^2 y^2} 5}\] \[\large \sqrt{\frac {48 x^2 y^2} 5}\] \[\large \sqrt{\frac {48}{5} {x^2 y^2}}\] \[\large \sqrt{\frac {48}{5}} \times \sqrt{x^2} \times\sqrt{ y^2}\] \[\large \sqrt{\frac {48}{5}} \times x \times y\]

OpenStudy (anonymous):

So, I dont factor the 48?

OpenStudy (jack1):

wait, is that a y^3 or y^2 ...?

OpenStudy (anonymous):

y^3

OpenStudy (jack1):

gotcha, then \[\huge 4 \sqrt \frac 35 \times x \times y^{\frac 32}\]

OpenStudy (anonymous):

that's the answer?! lol

OpenStudy (jack1):

yeahs, just an extra step and i had to change the power on the y

OpenStudy (anonymous):

Is it all under the radical?

OpenStudy (jack1):

nah, only the fraction is x becomes 2/2 so to the power of 1 y becomes 3/2, so you can see that's still there 4 is extracted out

OpenStudy (anonymous):

Oooooh ok, thanks...one more lol

OpenStudy (jack1):

k but i gotta go soon

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