Решебник по математике 4 класс Петерсон рабочая тетрадь Часть 1, 2, 3 Часть 2 | Страница 10

\[\boxed{Страница\ 11.}\]

\[\boxed{\mathbf{1.}}\]

\[AB = \frac{3}{5}CD \rightarrow правильная\ часть.\]

\[CD = \frac{5}{3}AB \rightarrow неправильная\ \]

\[часть.\]

\[\boxed{\mathbf{2.}}\]

\[Правильные\ части\ отрезков\ \]

\[(подчеркните\ их\ в\ тетради):\]

\[\frac{4}{5}EF;\ \ \frac{2}{9}MK;\ \ \frac{5}{8}\text{XY.}\]

\[\boxed{\mathbf{3.}}\]

\[AB = 2\ части;\ \ \ CD = 5\ частей;\ \ \text{\ \ }\]

\[EF = 7\ частей.\]

\[AB = \frac{2}{5}\text{CD}\]

\[CD = \frac{5}{2}\text{AB}\]

\[AB = \frac{2}{7}\text{EF}\]

\[EF = \frac{7}{2}\text{AB}\]

\[CD = \frac{5}{7}\text{EF}\]

\[EF = \frac{7}{5}\text{CD}\]

\[\boxed{\mathbf{4.}}\]

\[\frac{20}{4} = 5\]

\[\frac{56}{7} = 8\]

\[\frac{72}{8} = 9\]

\[\frac{42}{6} = 7\]

\[\frac{30}{5} = 6\]

\[\boxed{\mathbf{5.}}\]

\[x + \frac{5}{9} = \frac{7}{9}\]

\[x = \frac{7}{9} - \frac{5}{9}\]

\[x = \frac{2}{9}.\]

\[y - \frac{4}{11} = \frac{6}{11}\]

\[y = \frac{6}{11} + \frac{4}{11}\]

\[y = \frac{10}{11}.\]

\[\frac{12}{10} - n = \frac{8}{10}\]

\[n = \frac{12}{10} - \frac{8}{10}\]

\[n = \frac{4}{10}.\]

\[\boxed{\mathbf{6.}}\]

\[56 = 7 \cdot 8\]

\[48 = 8 \cdot 6\]

\[Число\ 786.\]

\[Два\ числа\ превратились\ в\ \]

\[5648:\ \ \ 56\ и\ 48.\]

\[Ответ:C.\ 2.\]