Решебник по математике 5 класс Мерзляк ФГОС Задание 1059

 

\[\boxed{\mathbf{1059\ (1059)}.\ Еуроки\ - \ ДЗ\ без\ мороки}\]

\[100\% - 73\% =\]

\[= 27\% - составляет\ сельское\ \]

\[население.\]

\[Ответ:27\%.\]

\[\boxed{\mathbf{1059}\mathbf{.}\mathbf{\ }}\]

\[2)\ \frac{81}{88} \cdot \left( 6 - 1\frac{13}{15} \cdot 1\frac{19}{21} \right) =\]

\[= \frac{81}{88} \cdot \left( 6 - \frac{28}{15} \cdot \frac{40}{21} \right) =\]

\[= \frac{81}{88} \cdot \left( 6 - \frac{28 \cdot 40}{15 \cdot 21} \right) =\]

\[= \frac{81}{88} \cdot \left( 6 - \frac{4 \cdot 8}{3 \cdot 3} \right) =\]

\[= \frac{81}{88} \cdot \left( 6 - \frac{32}{9} \right) =\]

\[= \frac{81}{88} \cdot \left( 5\frac{9}{9} - 3\frac{5}{9} \right) = \frac{81}{88} \cdot 2\frac{4}{9} =\]

\[= \frac{81}{88} \cdot \frac{22}{9} = \frac{81 \cdot 22}{88 \cdot 9} =\]

\[= \frac{9 \cdot 2}{8 \cdot 1} = \frac{9}{4} = 2\frac{1}{4}\]

\[3)\ \left( 5\frac{1}{16} - 1\frac{1^{\backslash 2}}{8} \right)\left( \frac{5^{\backslash 7}}{6} + \frac{3^{\backslash 3}}{14} \right) =\]

\[= \left( 4\frac{17}{16} - 1\frac{2}{16} \right)\left( \frac{35}{42} + \frac{9}{42} \right) =\]

\[= 3\frac{15}{16} \cdot \frac{44}{42} =\]

\[4)\ 5\frac{1}{16} - 1\frac{1}{8} \cdot \left( \frac{5^{\backslash 7}}{6} + \frac{3^{\backslash 3}}{14} \right) =\]

\[= 5\frac{1}{16} - \frac{9}{8} \cdot \left( \frac{35}{42} + \frac{9}{42} \right) =\]

\[= 5\frac{1}{16} - \frac{9}{8} \cdot \frac{44}{42} = 5\frac{1}{16} - \frac{9}{8} \cdot \frac{22}{21} =\]

\[= 5\frac{1}{16} - \frac{9 \cdot 22}{8 \cdot 21} = 5\frac{1}{16} - \frac{3 \cdot 11}{4 \cdot 7} =\]

\[= 5\frac{1}{16} - \frac{33}{28} = 5\frac{1^{\backslash 7}}{16} - 1\frac{5^{\backslash 4}}{28} =\]

\[= 5\frac{7}{112} - 1\frac{20}{112} =\]

\[= 4\frac{119}{112} - 1\frac{20}{112} = 3\frac{99}{112}\]