Question

resolva as expressões com as quatro óperaçoes,potencia e raízes
$\sqrt{49}-\sqrt{36}.{3}^{3}-{2}^{3} +\sqrt{16}.{3}^{0} =$

Answer 1

$\sqrt{49} - \sqrt{36} \times {3}^{3} - {2}^{3 } + \sqrt{16} \times 3 {}^{0} = \\ 7 - 6 \times 27 - 8 + 4 \times 1 = \\ 7 - 162 - 8 + 4 = \\ - 162 - 1 + 4 = \\ - 162 + 3 = \\ - 159$