Решебник по математике 6 класс Мерзляк ФГОС Задание 400

 

\[\boxed{\mathbf{400\ (400).}\mathbf{\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

Пояснение.

Решение.

\[\frac{5}{8} \cdot m \Longrightarrow выражение\ для\ \]

\[нахождения\ порций\ эскимо.\]

\[при\ m = 120:\]

\[\frac{5}{8} \cdot 120 = \frac{5 \cdot \mathbf{120}}{\mathbf{8}} =\]

\[= 75\ (порций) - эскимо\ \]

\[продали.\]

\[Ответ:75\ порций\ эскимо.\]

\[\boxed{\mathbf{400}\mathbf{.}}\]

\[1)\ \frac{11}{18} - \frac{14}{27}x = \frac{5}{12}\]

\[\frac{14}{27}x = \frac{11^{\backslash 2}}{18} - \frac{5^{\backslash 3}}{12}\]

\[\frac{14}{27}x = \frac{22}{36} - \frac{15}{36}\]

\[\frac{14}{27}x = \frac{7}{36}\]

\[x = \frac{7}{36}\ :\frac{14}{27}\]

\[x = \frac{7}{36} \cdot \frac{27}{14}\]

\[x = \frac{1 \cdot 3}{4 \cdot 2}\]

\[x = \frac{3}{8}.\]

\[2)\ \frac{1^{\backslash 20}}{3}x + \frac{1^{\backslash 15}}{4}x + \frac{1^{\backslash 12}}{5}x =\]

\[= 1\frac{19}{75}\]

\[\frac{20 + 15 + 12}{60}x = 1\frac{19}{75}\]

\[\frac{47}{60}x = \frac{94}{75}\]

\[x = \frac{94}{75}\ :\frac{47}{60}\]

\[x = \frac{94}{75} \cdot \frac{60}{47}\]

\[x = \frac{2 \cdot 4}{5 \cdot 1} = \frac{8}{5}\]

\[x = 1\frac{3}{5}.\]