Question #7bc10

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Question #7bc10

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Answer 1 · 2017-03-30T10:35:48

$\sqrt{3}$ $sqrt3$

Explanation:

$9 \sqrt{8} = 9 \sqrt{2^{3}} = 3^{2} \cdot 2^{\frac{3}{2}}$ $9sqrt8 = 9sqrt(2^3) = 3^2*2^(3/2)$
$6 \sqrt{6} = 3 \cdot 2 \cdot \sqrt{3 \cdot 2} = 3^{1} \cdot 2^{1} \cdot 3^{\frac{1}{2}} \cdot 2^{\frac{1}{2}} = 3^{\frac{3}{2}} \cdot 2^{\frac{3}{2}}$ $6sqrt6 = 3*2*sqrt(3*2) = 3^1*2^1*3^(1/2)*2^(1/2) = 3^(3/2)*2^(3/2)$

$therefore$ $(9sqrt8) / (6sqrt6) = (3^2*2^(3/2))/(3^(3/2)*2^(3/2))$

$= (3^(2 - 3/2)) * (2^(3/2 - 3/2))$
$= 3^(1/2)*1 = 3^(1/2) =$ $sqrt(3)$

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Question #7bc10

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