Решебник по математике 6 класс Виленкин ФГОС Часть 1, 2 Часть 1. П. 3. Отношения и пропорции Задание 120

\[\boxed{\mathbf{3.120}}\]

Пояснение.

Решение.

\[\textbf{а)}\ \frac{2\frac{7}{12}}{6\frac{1}{5}} = \frac{y}{2,2}\]

\[6\frac{1}{5}y = 2\frac{7}{12} \cdot 2,2\]

\[\frac{31}{5}y = \frac{31}{12} \cdot \frac{22}{10}\]

\[y = \frac{31}{12} \cdot \frac{11}{5} \cdot \frac{5}{31}\]

\[y = \frac{11}{12}.\]

\[\textbf{б)}\ \frac{2\frac{3}{4}}{4\frac{1}{8}} = \frac{1,6}{t}\]

\[2\frac{3}{4}t = 1,6 \cdot 4\frac{1}{8}\]

\[\frac{11}{4}t = \frac{16}{10} \cdot \frac{33}{8}\]

\[t = \frac{8}{5} \cdot \frac{33}{8} \cdot \frac{4}{11}\]

\[t = \frac{3 \cdot 4}{5} = \frac{12}{5}\]

\[t = 2,4.\]