Rewrite the irrational square root 48−−√ as the product of an integer and another irrational square root.(1 point) Responses 212−−√ 2 start root 12 end root 43–√ 4 start root 3 end root 412−−√ 4 start root 12 end root 224−−√

Answer 1

To rewrite \(\sqrt{48}\) as the product of an integer and another irrational square root, we can factor \(48\) as follows:

\[ 48 = 16 \times 3 \]

Now, we can take the square root:

\[ \sqrt{48} = \sqrt{16 \times 3} = \sqrt{16} \times \sqrt{3} = 4\sqrt{3} \]

Thus, \(\sqrt{48}\) can be rewritten as \(4\sqrt{3}\).

The correct response is:

\(4\sqrt{3}\).