## Elementary Algebra

$\approx 5.2$
Using $\sqrt{3}=1.73$ and the properties of radicals, the value of the given expression, $\sqrt{27} ,$ to the nearest tenth, is \begin{array}{l}\require{cancel} \sqrt{9\cdot3} \\\\= \sqrt{(3)^2\cdot3} \\\\= 3\sqrt{3} \\\\= 3(1.73) \\\\ \approx 5.2 .\end{array}