Решебник по алгебре 7 класс Макарычев ФГОС Задание 299

 

\[\boxed{\text{299.}\text{\ }еуроки - ответы\ на\ пятёрку}\]

\[Подставим\ в\ формулу\ \]

\[значения\ \text{x\ }и\ найдем\ \text{y.}\]

\[если\ x = - 9:\ \ \]

\[\ y = - \frac{1}{6} \cdot ( - 9) = \frac{9}{6} = \frac{3}{2} = 1,5;\]

\[если\ x = 0:\ \]

\[y = - \frac{1}{6} \cdot 0 = 0;\]

\[если\ x = 1:\text{\ \ }\]

\[y = - \frac{1}{6} \cdot 1 = - \frac{1}{6};\ \]

\[если\ x = 4:\]

\[y = - \frac{1}{6} \cdot 4 = - \frac{4}{6} = - \frac{2}{3}.\]

\[\boxed{\text{299.}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[если\ x = - 9:\ \ \]

\[\ y = - \frac{1}{6} \cdot ( - 9) = \frac{9}{6} = \frac{3}{2} = 1,5;\]

\[если\ x = 0:\ \]

\[y = - \frac{1}{6} \cdot 0 = 0;\]

\[если\ x = 1:\text{\ \ }\]

\[y = - \frac{1}{6} \cdot 1 = - \frac{1}{6};\ \]

\[если\ x = 4:\]

\[y = - \frac{1}{6} \cdot 4 = - \frac{4}{6} = - \frac{2}{3}.\]