## Intermediate Algebra (6th Edition)

The given expression, $\sqrt[]{8} ,$ is closest to $3$ since $3^2=9.$ The given expression, $\sqrt[]{20} ,$ is closest to $4$ since $4^2=16.$ Hence, the length of the bent wire, $\sqrt{8}+\sqrt{20} ,$ is closest to \begin{array}{l}\require{cancel} 3+4 \\\\= 7 .\end{array} Hence, $\text{ Choice C .}$