Answer
$\sqrt{5}$
Work Step by Step
The magnitude of a vector $v=ai+bj$ is: $||v||=\sqrt{a^2+b^2}$.
Hence,
$||v||=\sqrt{(-2)^2+1^2}\\
||v||=\sqrt{4+1}\\
||v||=\sqrt{5}$
Precalculus (10th Edition)
by Sullivan, Michael
Published by
Pearson
ISBN 10:
0-32197-907-9
ISBN 13:
978-0-32197-907-0
Chapter 9 - Polar Coordinates; Vectors - Chapter Review - Review Exercises - Page 635: 33
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