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46 simplify $2sqrt{3} cdot sqrt{27}$

Question

46 simplify $2sqrt{3} cdot sqrt{27}$

Explanation:

Step1: Simplify √27

We know that \( \sqrt{27}=\sqrt{9\times3} \), and since \( \sqrt{ab}=\sqrt{a}\times\sqrt{b} \) (for \( a\geq0,b\geq0 \)), we have \( \sqrt{9\times3}=\sqrt{9}\times\sqrt{3}=3\sqrt{3} \).

Step2: Multiply the two terms

Now we have \( 2\sqrt{3}\times3\sqrt{3} \). Using the property \( \sqrt{a}\times\sqrt{b}=\sqrt{ab} \) and \( a\sqrt{b}\times c\sqrt{d}=ac\sqrt{bd} \), we get \( (2\times3)\times(\sqrt{3}\times\sqrt{3}) = 6\times3 = 18 \).

Answer:

\( 18 \)