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Question
simplify. $sqrt{3}(sqrt{6}+sqrt{15})$ o $3sqrt{7}$ o $2sqrt{3}+3sqrt{5}$ o $3sqrt{2}+9sqrt{5}$ o $3sqrt{2}+3sqrt{5}$
Step1: Apply distributive property
$\sqrt{3}(\sqrt{6}+\sqrt{15})=\sqrt{3}\times\sqrt{6}+\sqrt{3}\times\sqrt{15}$
Step2: Use square - root multiplication rule $\sqrt{a}\times\sqrt{b}=\sqrt{ab}$
$\sqrt{3}\times\sqrt{6}=\sqrt{3\times6}=\sqrt{18}=3\sqrt{2}$ and $\sqrt{3}\times\sqrt{15}=\sqrt{3\times15}=\sqrt{45}=3\sqrt{5}$
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$3\sqrt{2}+3\sqrt{5}$