Introductory Algebra for College Students (7th Edition)

Published by Pearson
ISBN 10: 0-13417-805-X
ISBN 13: 978-0-13417-805-9

Chapter 9 - Section 9.1 - Solving Quadratic Equations by the Square Root Property - Exercise Set: 58

Answer

$d=9 \text{ units} $

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 $( -\sqrt{3},4\sqrt{6} )$ and $( 2\sqrt{3},\sqrt{6} )$ is \begin{array}{l}\require{cancel} d=\sqrt{(-\sqrt{3}-2\sqrt{3})^2+(4\sqrt{6}-\sqrt{6})^2} \\\\ d=\sqrt{(-3\sqrt{3})^2+(3\sqrt{6})^2} \\\\ d=\sqrt{9\cdot3+9\cdot6} \\\\ d=\sqrt{27+54} \\\\ d=\sqrt{81} \\\\ d=\sqrt{(9)^2} \\\\ d=9 \text{ units} .\end{array}
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.