## Intermediate Algebra (6th Edition)

Published by Pearson

# Chapter 7 - Review - Page 470: 130

#### Answer

$-\sqrt{10}$

#### Work Step by Step

Using $i=\sqrt{-1}$ and the properties of radicals, the expression $\sqrt{-2}\cdot\sqrt{-5}$ simplifies to \begin{array}{l} \sqrt{2}\cdot\sqrt{-1}\cdot\sqrt{5}\cdot\sqrt{-1} \\= (i\sqrt{2})(i\sqrt{5}) \\= i^2\sqrt{10} \\= (-1)\sqrt{10} \\= -\sqrt{10} .\end{array}

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