## Precalculus (6th Edition)

$2$
To solve the item mentally, note that: $\sqrt{0.1}\cdot\sqrt{40} = \sqrt{0.1(40)}=\sqrt{4}$ (because of the product rule for radicals) Since $4=2^2$. then $\sqrt{4} = \sqrt{2^2} = 2$ Thus. $\sqrt{0.1}\cdot\sqrt{40}=2$