Решебник по математике 6 класс Виленкин Часть 1, 2 Часть 2. Параграф 4. Действия с рациональными числами Задание 315

\[\boxed{\text{4.\ 315}}\]

\[\textbf{а)} - z \cdot 5 = - 150\]

\[- z = - 150\ :5\]

\[- z = - 30\]

\[z = 30.\]

\[\textbf{б)}\ 4 \cdot ( - z) = - 32\]

\[- z = - 32\ :4\]

\[- z = - 8\]

\[z = 8.\]

\[\textbf{в)} - 0,4y = 44\]

\[y = 44\ :( - 0,4)\]

\[y = - 110.\]

\[\textbf{г)}\ \frac{1}{7}z = - 1\]

\[z\ :7 = - 1\]

\[z = - 1 \cdot 7\]

\[z = - 7.\]

\[\textbf{д)}\ \frac{5}{7}t = - \frac{25}{28}\]

\[t = - \frac{25}{28}\ :\frac{5}{7}\]

\[t = - \frac{25}{28} \cdot \frac{7}{5}\]

\[t = - \frac{5}{4} = - 1,25.\]

\[\textbf{е)} - \frac{4}{9}z = \frac{16}{27}\]

\[z = \frac{16}{27}\ :\left( - \frac{4}{9} \right)\]

\[z = - \frac{16}{27} \cdot \frac{9}{4}\]

\[z = - \frac{4}{3} = - 1\frac{1}{3}.\]

\[\textbf{ж)}\ - \frac{8}{11}t = - 1\frac{7}{33}\]

\[t = - \frac{40}{33}\ :\left( - \frac{8}{11} \right)\]

\[t = \frac{40}{33} \cdot \frac{11}{8}\]

\[t = \frac{5}{3} = 1\frac{2}{3}.\]

\[\textbf{з)} - \frac{7}{8}z + 7 = 2\frac{5}{8}\]

\[- \frac{7}{8}z = 2\frac{5}{8} - 7\]

\[- \frac{7}{8}z = - 4\frac{3}{8}\]

\[z = - \frac{35}{8}\ :\left( - \frac{7}{8} \right)\]

\[z = \frac{35}{8} \cdot \frac{8}{7}\]

\[z = 5.\ \]