College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.4 - Linear Functions - 2.4 Exercises - Page 215: 81

Answer

$2\sqrt{10}$

Work Step by Step

We find the distance between the second and fourth points, $(1,-3)$ and $(3,3)$ by using the distance formula: $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2)}$ $=\sqrt{(3-1)^{2}+(3-(-3))^{2}}$ $=\sqrt{2^{2}+6^{2}}$ $=\sqrt{4+36}$ $=\sqrt{40}$ $=\sqrt{4*10}$ $=2\sqrt{10}$
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