Question

1. \\(sqrt{-15} cdot sqrt{-20}\\)

Explanation:

Step1: Rewrite using imaginary unit

Recall that \(\sqrt{-a}=i\sqrt{a}\) for \(a>0\). So, \(\sqrt{-15}=i\sqrt{15}\) and \(\sqrt{-20}=i\sqrt{20}\).
The expression becomes \(i\sqrt{15}\cdot i\sqrt{20}\).

Step2: Multiply the imaginary units and the radicals

First, multiply the imaginary units: \(i\cdot i = i^{2}\).
Then, multiply the radicals: \(\sqrt{15}\cdot\sqrt{20}=\sqrt{15\times20}=\sqrt{300}\).
Simplify \(\sqrt{300}\): \(\sqrt{300}=\sqrt{100\times3}=10\sqrt{3}\).
And we know that \(i^{2}=-1\).

Step3: Combine the results

Now, combine \(i^{2}\) and \(\sqrt{300}\): \(i^{2}\cdot\sqrt{300}=-1\times10\sqrt{3}=-10\sqrt{3}\).

Answer:

\(-10\sqrt{3}\)