Решебник по математике 6 класс Мерзляк ФГОС Задание 211

 

\[\boxed{\mathbf{211\ }\left( \mathbf{211} \right)\mathbf{.}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

Пояснение.

Решение.

\[1)\ \frac{3}{12} = \frac{3\ :3}{12\ :3} = \frac{1}{4}\]

\[2)\ \frac{4}{12} = \frac{4\ :4}{12\ :4} = \frac{1}{3}\]

\[3)\ \frac{6}{54} = \frac{6\ :6}{54\ :6} = \frac{1}{9}\]

\[4)\ \frac{25}{70} = \frac{25\ :5}{70\ :5} = \frac{5}{14}\]

\[5)\ \frac{26}{65} = \frac{26\ :13}{65\ :13} = \frac{2}{5}\]

\[6)\ \frac{12}{60} = \frac{12\ :12}{60\ :12} = \frac{1}{5}\]

\[7)\ \frac{36}{48} = \frac{36\ :12}{48\ :12} = \frac{3}{4}\]

\[8)\ \frac{35}{105} = \frac{35\ :35}{105\ :35} = \frac{1}{3}\]

\[9)\ \frac{480}{720} = \frac{480\ :10}{720\ :10} = \frac{48}{72} =\]

\[= \frac{48\ :24}{72\ :24} = \frac{2}{3}\]

\[10)\ \frac{204}{306} = \frac{204\ :102}{306\ :102} = \frac{2}{3}\]

\[\boxed{\mathbf{211}\mathbf{.}}\]

\[\angle BOC = x;\]

\[\angle AOC = (x + 38{^\circ});\]

\[\angle AOB = 180{^\circ}.\]

\[x + x + 38 = 180\]

\[2x = 180 - 38\]

\[2x = 142\]

\[x = 142\ :2 = 71{^\circ} - \angle BOC.\]

\[\angle AOC = x + 38 = 71 + 38 =\]

\[= 109{^\circ}.\]

\[Ответ:71{^\circ}\ и\ 109{^\circ}.\]