Odpowiedź:
[tex]2\frac{1}{2}\cdot\sqrt{0,04}+3\frac{1}{3}\cdot\sqrt{0,09}-6\frac{1}{6}\cdot\sqrt{0,36}=\frac{5}{2}\cdot0,2+\frac{10}{3}\cdot0,3-\frac{37}{6}\cdot0,6=\\\\=\frac{\not5^1}{\not2_{1}}\cdot\frac{\not2^1}{\not10_{2}}+\frac{\not10^1}{\not3_{1}}\cdot\frac{\not3^1}{\not10_{1}}-\frac{37}{\not6_{1}}\cdot\frac{\not6^1}{10}=\frac{1}{2}+1-\frac{37}{10}=1\frac{1}{2}-\frac{37}{10}=\frac{3}{2}-\frac{37}{10}=\frac{15}{10}-\frac{37}{10}=\\\\=-\frac{22}{10}=-2\frac{2}{10}=-2\frac{1}{5}[/tex]
