Решебник по алгебре 9 класс Макарычев Задание 785

 

\[\boxed{\text{785\ (785).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[y^{2} + y - 240 = 0\]

\[D = 1 + 4 \cdot 240 = 961\]

\[x_{1,2} = \frac{- 1 \pm 31}{2} = 15;\ - 16;\]

\[1)\ \left\{ \begin{matrix} y_{1} = 15 \\ x_{1} = 16 \\ \end{matrix} \right.\ ,\ \ 2)\ \left\{ \begin{matrix} y_{2} = - 16 \\ x_{2} = - 15. \\ \end{matrix} \right.\ \]

\[Ответ:(16;15)\ и\ ( - 15;\ - 16).\]

\[x^{2} - 12x + 32 = 0\]

\[D = 6^{2} - 32 = 4,\]

\[x_{1,2} = 6 \pm 2 = 8;\ 4;\]

\[1)\ \left\{ \begin{matrix} x_{1} = 4\ \ \ \\ y_{1} = - 7 \\ \end{matrix} \right.\ ,\ \ 2)\ \left\{ \begin{matrix} x_{2} = 8 \\ y_{2} = 1. \\ \end{matrix} \right.\ \ \]

\[Ответ:(4;\ - 7)\ и\ \ (8;1).\]

\[\boxed{\text{785.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[a_{1} = 28,\ \ S_{25} = 925,\]

\[\ \ d = ?,\ \ a_{13} = ?\]

\[S_{25} = \frac{2a_{1} + 24d}{2} \cdot 25 = 925\]

\[(28 + 12d) \cdot 25 = 925\]

\[12d = 37 - 28\]

\[12d = 9\]

\[d = - \frac{9}{12} = 0,75.\]

\[a_{30} = a_{1} + 29d = 28 + 29 \cdot \frac{3}{4} =\]

\[= 49,75.\]