ГДЗ по алгебре 7 класс Макарычев ФГОС Задание 76

\[\boxed{\text{76.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

Пояснение.

Решение.

\[\textbf{а)}\ 8,6 < 8,63 < 8,7\]

\[\textbf{б)}\ \frac{1}{8} < \frac{22}{168} < \frac{1}{7}\]

\[\frac{1^{\backslash 24}}{7} = \frac{24}{168};\ \ \ \ \ \ \ \ \ \ \frac{1^{\backslash 21}}{8} = \frac{21}{168}\ \]

\[\textbf{в)} - 3,7 < - 3,67 < - 3,6\]

\[\textbf{г)}\ \frac{3}{4} < \frac{19}{24} < \frac{5}{6}\]

\[\frac{3^{\backslash 6}}{4} = \frac{18}{24};\ \ \ \ \ \ \frac{5^{\backslash 4}}{6} = \frac{20}{24}\]

\[\boxed{\text{76\ (76).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 3\frac{1}{8} \cdot 5 = \left( 3 + \frac{1}{8} \right) \cdot 5 =\]

\[= 3 \cdot 5 + \frac{1}{8} \cdot 5 = 15 + \frac{5}{8} = 15\frac{5}{8}\]

\[\textbf{б)}\ 7 \cdot 2\frac{3}{7} = 7 \cdot \left( 2 + \frac{3}{7} \right) =\]

\[= 7 \cdot 2 + 7 \cdot \frac{3}{7} = 14 + 3 = 17\]

\[\textbf{в)}\ 2\frac{2}{5} \cdot 10 = \left( 2 + \frac{2}{5} \right) \cdot 10 =\]

\[= 2 \cdot 10 + \frac{2}{5} \cdot 10 = 20 + 4 = 24\]

\[\textbf{г)}\ 6 \cdot 4\frac{5}{12} = 6 \cdot \left( 4 + \frac{5}{12} \right) =\]

\[= 6 \cdot 4 + 6 \cdot \frac{5}{12} = 24 + \frac{5}{2} =\]

\[= 24 + 2\frac{1}{2} = 26\frac{1}{2} = 26,5\]