Intermediate Algebra (6th Edition)

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

Chapter 7 - Test - Page 471: 34

Answer

$d=\sqrt{95}$

Work Step by Step

Using $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$ (or the Distance Formula), then the distance, $d$, between the given points $ (-2\sqrt{5},\sqrt{10}) \text{ and } (-\sqrt{5},4\sqrt{10}) ,$ is \begin{array}{l}\require{cancel} d=\sqrt{(-2\sqrt{5}-(-\sqrt{5}))^2+(\sqrt{10}-4\sqrt{10})^2} \\\\ d=\sqrt{(-2\sqrt{5}+\sqrt{5})^2+(\sqrt{10}-4\sqrt{10})^2} \\\\ d=\sqrt{(-\sqrt{5})^2+(-3\sqrt{10})^2} \\\\ d=\sqrt{5+9\cdot10} \\\\ d=\sqrt{5+90} \\\\ d=\sqrt{95} .\end{array}
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