Question

1. \\(sqrt{-18} cdot sqrt{-10}\\)

Explanation:

Step1: Rewrite using imaginary unit

Recall that \(\sqrt{-a}=i\sqrt{a}\) for \(a>0\). So, \(\sqrt{-18}=i\sqrt{18}\) and \(\sqrt{-10}=i\sqrt{10}\).
\[
\sqrt{-18}\cdot\sqrt{-10}=i\sqrt{18}\cdot i\sqrt{10}
\]

Step2: Multiply the imaginary units and the radicals

Multiply \(i\) with \(i\) to get \(i^{2}\), and multiply \(\sqrt{18}\) with \(\sqrt{10}\) using the property \(\sqrt{a}\cdot\sqrt{b}=\sqrt{ab}\).
\[
i\sqrt{18}\cdot i\sqrt{10}=i^{2}\sqrt{18\times10}
\]

Step3: Simplify \(i^{2}\) and the radical

We know that \(i^{2}=-1\), and \(18\times10 = 180\). Also, simplify \(\sqrt{180}\): \(\sqrt{180}=\sqrt{36\times5}=6\sqrt{5}\).
\[
i^{2}\sqrt{180}=-1\times6\sqrt{5}=-6\sqrt{5}
\]

Answer:

\(-6\sqrt{5}\)