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

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\[\textbf{а)}\ \angle KMN = 55{^\circ} + 42{^\circ} = 97{^\circ}.\]

\[\textbf{б)}\ \angle KMN = 180{^\circ} - 64{^\circ} = 116{^\circ}.\]

\[\textbf{в)}\ \angle KMN = 90{^\circ} - 28{^\circ} = 62{^\circ}.\]

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\[\frac{2}{10} < \frac{7}{10}\]

\[\frac{5}{6} > \frac{5}{8}\]

\[\frac{4}{3} > \frac{3}{4}\]

\[\frac{9}{16} < \frac{16}{9}\]

\[1\frac{5}{7} < 6\frac{5}{12}\]

\[2\frac{4}{5} < 3\frac{1}{5}\]

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

\[\frac{17}{100} = 17\%\]

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\[1)\ 3\frac{5}{11} - x = 1\frac{9}{11}\]

\[x = 3\frac{5}{11} - 1\frac{9}{11}\]

\[x = 2\frac{16}{11} - 1\frac{9}{11}\]

\[x = 1\frac{7}{11}.\]

\[2)\ y - 5\frac{4}{6} = 3\frac{2}{6}\]

\[y = 3\frac{2}{6} + 5\frac{4}{6}\]

\[y = 8\ \frac{6}{6}\]

\[y = 9.\]

\[3)\ n + \frac{7}{8} = 5\frac{1}{8}\]

\[n = 4\frac{9}{8} - \frac{7}{8}\]

\[n = 4\frac{2}{8}.\]

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\[Острых\ углов:\ \ 12.\]

\[Тупых\ углов:4.\]

\[Прямых\ углов:0.\]

\[Смежных\ углов:\ 4\ пары.\]