University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Practice Exercises - Page 638: 4

Answer

$\lt 10,-25 \gt$ and $5 \sqrt {29}$

Work Step by Step

The formula to find the magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $5v=5\lt 2,-5 \gt =\lt 10,-25 \gt$ and $|\lt 10,-25\gt|=\sqrt{(10)^2+(-25)^2}=\sqrt {725}=5 \sqrt {29}$ Hence, our final answers are: $\lt 10,-25 \gt$ and $5 \sqrt {29}$
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