Решебник по математике 4 класс Петерсон рабочая тетрадь Часть 1, 2, 3 Часть 3 | Страница 12

\[\boxed{\mathbf{Страница\ 13.}еуроки - ответы\ на\ пятёрку}\]

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

\[\angle\text{AOM} = 155{^\circ}\]

\[\angle\text{BOM} = 130{^\circ}\]

\[\angle\text{COM} = 90{^\circ}\]

\[\angle\text{COD} = 75{^\circ}\]

\[\angle\text{EOM} = 30{^\circ}\]

\[\angle\text{COD} = 15{^\circ}\]

\[\angle BOE = 100{^\circ}\]

\[\angle AOE = 125{^\circ}\]

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

\[\angle A = 100{^\circ};\]

\[\angle B = 30{^\circ};\]

\[\angle C = 20{^\circ};\]

\[\angle D = 135{^\circ}.\]

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

\[\textbf{а)}\ 180\ :9 \cdot 7 = 20 \cdot 7 = 140{^\circ}\]

\[\textbf{б)}\ 90\ :2 \cdot 3 = 45 \cdot 3 = 135{^\circ}\]

\[\textbf{в)}\ 60\ :4 \cdot 5 = 15 \cdot 5 = 75{^\circ}\]

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

\[\textbf{а)}\ x + 3\frac{5}{9}\text{.\ }\]

\[x = \frac{4}{9};\ \ 1\frac{2}{9};\ \ 6\frac{7}{9}:\]

\[\frac{4}{9} + 3\frac{5}{9} = 3\frac{9}{9} = 4.\]

\[1\frac{2}{9} + 3\frac{5}{9} = 4\frac{7}{9}.\]

\[6\frac{7}{9} + 3\frac{5}{9} = 9\frac{12}{9} = 10\frac{3}{9}.\]

\[\textbf{б)}\ y - 1\frac{6}{14}.\]

\[y = 2;3\frac{11}{14};\ \ 7\frac{1}{14}:\]

\[2 - 1\frac{6}{14} = 1\frac{14}{14} - 1\frac{6}{14} = \frac{8}{14}.\]

\[3\frac{11}{14} - 1\frac{6}{14} = 2\frac{5}{14}.\]

\[7\frac{1}{14} - 1\frac{6}{14} = 6\frac{15}{14} - 1\frac{6}{14} = 5\frac{9}{14}.\]

\[\boxed{\mathbf{7.}}\]

\[1\ яблоко - 1\ часть\]

\[2\ яблока - 3\ части\]

\[600\ :4 = 150\ (г) - 1\ яблоко.\]

\[Ответ:150\ грамм.\]