Question

MIKY93 TE ROOOGGGG!!!

Answer 1

$0,(3) \to \frac{3}{9} \to \frac{1}{3} \\\\ 0,0(3) \to \frac{3}{90} \to \frac{1}{30} \\\\ 0,00(3) \to \frac{3}{900} \to \frac{1}{300} \\\\\\ a;b;c \ \ \ \ d.p \\\\ \frac{1}{3};\frac{1}{30};\frac{1}{300} \\\\\\ \frac{a}{\frac{1}{3}}=\frac{b}{\frac{1}{30}}=\frac{c}{\frac{1}{300}}=k \\\\ 3a=30b=300c=k \\\\ a=\frac{k}{3} \\\\ b=\frac{k}{30} \\\\ c=\frac{k}{300} \\\\\\\ \frac{2a+5b}{3b+10c}= \frac{2* \frac{k}{3}+5* \frac{k}{30}}{3*\frac{k}{30}+10*\frac{k}{300}} =$

$= \frac{\frac{2k}{3}^{(2}+ \frac{k}{6}}{\frac{k}{10}^{(3}+\frac{k}{30}}= \\\\\\ = \frac{\frac{4k+k}{6}}{\frac{3k+k}{30}}= \\\\ = \frac{5k}{\not 6}* \frac{\not 30}{4k}= \\\\ =\frac{5\not k*5}{4\not k}= \\\\ =\boxed{\frac{25}{4}}$