How do you write #(9sqrt8)/(12sqrt16)# in simplest radical form?

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How do you write #(9sqrt8)/(12sqrt16)# in simplest radical form?

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Answer 1 · 2015-04-09T22:18:33

#(9 sqrt8)/(12 sqrt16) # = # (3 . 3 . sqrt2 . sqrt4)/ (3 . 2. 2 . sqrt 16) # = #(cancel3 . 3 . sqrt2 . cancel 2)/ (cancel3 . cancel 2 . 2 . 4)# = #(3sqrt2)/8#
this is the simplest form i can write. I hope it helps.

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How do you write #(9sqrt8)/(12sqrt16)# in simplest radical form?

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