Решебник по алгебре 9 класс Макарычев Задание 109

 

\[\boxed{\text{109\ (109).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ y = 10x^{2} + 5\]

\[В\ І\ и\ ІІ\ четвертях.\]

\[\textbf{б)}\ y = - 7x^{2} - 3\]

\[В\ ІІІ\ и\ \text{IV}\ четвертях.\]

\[\textbf{в)}\ y = - 6x^{2} + 8\]

\[Во\ всех\ четвертях.\]

\[\textbf{г)}\ y = (x - 4)^{2}\]

\[В\ І\ и\ ІІ\ четвертях.\]

\[\textbf{д)}\ y = - (x - 8)^{2}\]

\[В\ ІІІ\ и\ \text{IV}\ четвертях.\]

\[\textbf{е)}\ y = - 3 \cdot (x + 5)^{2}\]

\[В\ ІІІ\ и\ \text{IV}\ четвертях.\]

\[\boxed{\text{109.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ y = 0,4x + 3;\ \ y = 5 - 0,6x\]

\[0,4x + 3 = 5 - 0,6x\]

\[0,4x + 0,6x = 5 - 3\]

\[x = 2.\]

\[y = 0,4 \cdot 2 + 3 = 3,8.\]

\[Ответ:(2;3,8).\]

\[\textbf{б)}\ y = \frac{1}{3}x + 11;\ \ y = - \frac{2}{9}x + 4\]

\[\frac{1}{3}x + 11 = - \frac{2}{9}x + 4\]

\[\frac{3}{9}x + \frac{2}{9}x = 4 - 11\]

\[\frac{5}{9}x = - 7\]

\[x = - 7 \cdot \frac{9}{5} = - \frac{63}{5} = - \frac{126}{10}\]

\[x = - 12,6.\]

\[y = \frac{1}{3} \cdot \left( - \frac{63}{5} \right) + 11 =\]

\[= - \frac{21}{5} + 11 = - 4,2 + 11 =\]

\[= 6,8.\]

\[Ответ:( - 12,6;6,8).\]