ГДЗ по алгебре и начала математического анализа 10 класс Колягин Проверочная работа №1 (страница 78)

\[\mathbf{Проверочная\ работа\ №1\ (стр.78)}\]

\[\boxed{\mathbf{1}.}\]

\[\frac{\left( a^{3} \right)^{5} \cdot a^{0} \cdot a^{2}}{a^{- 2}} =\]

\[= a^{15} \cdot 1 \cdot a^{2} \cdot a^{2} = a^{19}.\]

\[\boxed{\mathbf{2}.}\]

\[0,00038 = 3,8 \cdot 10^{- 4}.\]

\[\boxed{\mathbf{3}.}\]

\[\left\{ \begin{matrix} 2x - 3y = 7\ \ | \cdot 4 \\ 3x + 4y = 2\ \ | \cdot 3 \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} 8x - 12y = 28 \\ 9x + 12y = 6\ \ \ \\ \end{matrix} \right.\ ( + )\]

\[17x = 34\]

\[x = 2.\]

\[2x - 3y = 7\]

\[3y = 2x - 7\]

\[3y = 2 \cdot 2 - 7 = - 3\]

\[y = - 1.\]

\[\left\{ \begin{matrix} x = 2\ \ \ \\ y = - 1 \\ \end{matrix} \right.\ \]

\[Ответ:(2; - 1).\]

\[\boxed{\mathbf{4}.}\]

\[\left\{ \begin{matrix} 5x + 3 > 0 \\ \frac{1}{2}x - 4 < 0 \\ \end{matrix} \right.\ \text{\ \ \ \ }\left\{ \begin{matrix} 5x > - 3 \\ \frac{1}{2}x < 4\ \ \\ \end{matrix} \right.\ \text{\ \ \ \ \ \ }\]

\[\left\{ \begin{matrix} x > - 0,6 \\ x < 8\ \ \ \ \ \ \\ \end{matrix} \right.\ \]

\[Ответ:\ - 0,6 < x < 8.\]

\[\boxed{\mathbf{5}.}\]

\[x < 0;\ \ y < 0:\]

\[\sqrt{9x^{3}y^{5}} = - 3xy^{2}\sqrt{\text{xy}}.\]

\[\boxed{\mathbf{6}.}\]

\[5x + 3 - 2x^{2} = 0\]

\[2x^{2} - 5x - 3 = 0\]

\[D = 25 + 24 = 49\]

\[x_{1} = \frac{5 + 7}{4} = 3;\ \]

\[\ x_{2} = \frac{5 - 7}{4} = - 0,5.\]

\[Ответ:x = - 0,5;\ \ x = 3.\]

\[\boxed{\mathbf{7}.}\]

\[y = x^{2} - 5x + 6\]

\[\boxed{\mathbf{8}.}\]

\[a_{n + 1} = - a_{n}^{2} + 1;\ \ \ a_{1} = 2:\]

\[a_{2} = - 2^{2} + 1 = - 3;\]

\[a_{3} = - ( - 3)^{2} + 1 = - 8;\]

\[a_{4} = - ( - 8)^{2} + 1 = - 63;\]

\[a_{5} = - ( - 63)^{2} + 1 = - 3968.\]

\[\boxed{\mathbf{9}.}\]

\[Упорядочим\ ряд\ данных:\]

\[3,\ 4,\ 5,\ 5,\ 7,\ 8.\]

\[Мода = 5.\]

\[Медиана = \frac{5 + 5}{2} = 5.\]

\[Среднее\ значение:\]

\[\frac{3 + 4 + 5 \cdot 2 + 7 + 8}{6} = \frac{32}{6} = 5\frac{1}{3}.\]