## University Calculus: Early Transcendentals (3rd Edition)

$\lt 4,-10\gt$ and $2 \sqrt {29}$
The formula to find the magnitude of a vector is: $|n|=\sqrt{n_1^2+n_2^2}$ Here, $-2v=-2\lt -2,5 \gt =\lt 4,-10\gt$ and $|\lt 4,-10\gt|=\sqrt{4^2+(-10)^2}=\sqrt {116}=2 \sqrt {29}$ Hence, our final answers are: $\lt 4,-10\gt$ and $2 \sqrt {29}$