Question

Como Resolver:
2² + 1 + $\frac{1}{3}$ + √0,36

Answer 1

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=4 + 1 + \frac{1}{3}+\sqrt{\frac{36}{100}}$

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=\frac{5*3}{3} + \frac{1}{3}+\frac{\sqrt{36}}{\sqrt{100}}$

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=\frac{15}{3} + \frac{1}{3}+\frac{6}{10}$

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=\frac{15+1}{3}+\frac{6}{10}$

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=\frac{16}{3}+\frac{6}{10}$

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=\frac{16*10+6*3}{3*10}$

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=\frac{160+18}{30}$

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=\frac{178}{30}$

$2^{2} + 1 + \frac{1}{3} + \sqrt{0,36}=\frac{89}{15}$