Решебник по алгебре 9 класс Макарычев Задание 118

 

\[\boxed{\text{118\ (118).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 5x - 0,7 < 3x + 5,1\]

\[5x - 3x < 5,1 + 0,7\]

\[2x < 5,8\]

\[x < 5,8\ :2\]

\[x < 2,9.\]

\[Ответ:x < 2,9.\]

\[\textbf{б)}\ 0,8x + 4,5 \geq 5 - 1,2x\]

\[0,8x + 1,2x \geq 5 - 4,5\]

\[2x \geq 0,5\]

\[x \geq 0,5\ :2\]

\[x \geq 0,25.\]

\[Ответ:x \geq 0,25.\]

\[\textbf{в)}\ 2x + 4,2 \leq 4x + 7,8\]

\[2x - 4x \leq 7,8 - 4,2\]

\[- 2x \leq 3,6\]

\[2x \geq - 3,6\]

\[x \geq - 3,6\ :2\]

\[x \geq - 1,8\]

\[Ответ:x \geq - 1,8.\]

\[\textbf{г)}\ 3x - 2,6 > 5,5x - 3,1\]

\[3x - 5,5x > - 3,1 + 2,6\]

\[- 2,5x > - 0,5\]

\[2,5x < 0,5\]

\[x < 0,5\ :2,5\]

\[x < 0,2.\]

\[Ответ:x < 0,2.\]

\[\boxed{\text{118.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[y = \frac{1}{4}x²\]

\[\textbf{а)}\ y( - 2,5) = \frac{1}{4} \cdot ( - 2,5)^{2} =\]

\[= \frac{{2,5}^{2}}{4} = 1,5625;\]

\[y( - 1,5) = \frac{1}{4} \cdot ( - 1,5)^{2} = 0,5625;\]

\[y(3,5) = \frac{1}{4} \cdot {3,5}^{2} = 3,0625.\]

\[\textbf{б)}\ y = 5:\]

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

\[x^{2} = 20\ \ \]

\[x = \pm \sqrt{20}\text{\ \ }\]

\[x = \pm 2\sqrt{5}.\]

\[y = 3:\]

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

\[x^{2} = 12\]

\[x = \pm \sqrt{12}\text{\ \ }\]

\[x = \pm 2\sqrt{3}.\]

\[y = 2:\]

\[2 = \frac{1}{4}x^{2}\ \]

\[x^{2} = 8\ \ \]

\[x = \pm 2\sqrt{2}.\]

\[\textbf{в)}\ Функция\ убывает\ на\ \]

\[промежутке\ ( - \infty;0\rbrack\ и\ \]

\[возрастает\]

\[на\ промежутке\ \lbrack 0;\ + \infty).\]