Question

Calcule:

a)
$\sqrt{6 {}^{2} \: + 8 {}^{2} }$
b)
$\sqrt[3]{8} + \sqrt{100}$
c)
$\sqrt{9 {}^{2} - 4 \times 1 \times 20}$

Answer 1

$a) \\ \sqrt{6^2+8^2} = \sqrt{36+64} = \sqrt{100} =10 \\ \\ b) \\ \sqrt[3]{8} + \sqrt{100} = \sqrt[3]{2^3} +10=2+10=12 \\ \\ c) \\ \sqrt{9^2-4\times1\times20} = \sqrt{81-80} = \sqrt{1} =1$