Решебник по алгебре 7 класс Мерзляк ФГОС Задание 994

 

\[\boxed{\text{994\ (994).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ A\ ( - 4;0);\ \ B\ (0;2):\]

\[y = kx + b\]

\[0 = - 4k + b\]

\[4k = b\]

\[2 = 0 + b \Longrightarrow \ \ b = 2\]

\[Тогда:\ \ \]

\[4k = 2 \Longrightarrow \ \ k = 0,5\]

\[2)\ C\ (0;\ - 3);\ \ D\ (5;0):\]

\[y = kx + b\]

\[- 3 = 0 + b \Longrightarrow \ \ b = - 3\]

\[0 = 5k - 3\ \ \]

\[5k = 3 \Longrightarrow \ \ k = \frac{3}{5} = 0,6\]

\[\boxed{\text{994.\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[g(x) = \frac{20}{x} - 3\ \ \ и\ \ \ \ \ \]

\[h(x) = 8 - 3x\]

\[1)\ g(1) = \frac{20}{1} - 3 = 17;\ \ \ \ \ \]

\[\text{\ h}(1) = 8 - 3 \cdot 1 = 5\]

\[g(1) > h(1).\]

\[2)\ g(5) = \frac{20}{5} - 3 = 4 - 3 = 1;\ \ \]

\[\text{\ h}(2) = 8 - 3 \cdot 2 = 8 - 6 = 2\]

\[g(5) < h(2).\]

\[3)\ g( - 2) = \frac{20}{- 2} - 3 =\]

\[= - 10 - 3 = - 13;\ \ \ \]

\[h(6) = 8 - 3 \cdot 6 = 8 - 18 =\]

\[= - 10\]

\[g( - 2) < h(6).\]