Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.4 - Adding, Subtracting, and Multiplying Radical Expressions - Exercise Set - Page 440: 88b

Answer

$10$

Work Step by Step

Using the properties of radicals, the given expression, $ 2\sqrt{5}\cdot\sqrt{5} ,$ simplifies to \begin{array}{l}\require{cancel} 2\sqrt{5}\cdot1\sqrt{5} \\\\= 2(1)\sqrt{5(5)} \\\\= 2\sqrt{(5)^2} \\\\= 2(5) \\\\= 10 .\end{array}
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