Решебник по алгебре 9 класс Макарычев Задание 120

\[\boxed{\text{120\ (120).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 1,5\ с;\ \ \ \]

\[\textbf{б)}\ 2,5\ с;\ \ \ \]

\[\textbf{в)}\ 31,25\ м;\ \ \ \ \]

\[\textbf{г)}\ 4\ с.\]

\[\boxed{\text{120.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[y_{1}(0,5) = {0,5}^{2} = 0,25\]

\[y_{2}(0,5) = 1,8 \cdot {0,5}^{2} =\]

\[= 1,8 \cdot 0,25 = 0,45\]

\[y_{3}(0,5) = \frac{1}{3} \cdot {0,5}^{2} = \frac{1}{3} \cdot 0,25 \approx\]

\[\approx 0,083 \Longrightarrow y_{3} < y_{1} < y_{2}.\]

\[y_{1}(1) = 1^{2} = 1\]

\[y_{2}(1) = 1,8 \cdot 1^{2} = 1,8\]

\[y_{3}(1) = \frac{1}{3} \cdot 1^{2} \approx 0,3 \Longrightarrow\]

\[\Longrightarrow y_{3} < y_{1} < y_{2}.\]

\[y_{1}(2) = 2^{2} = 4\]

\[y_{2}(2) = 1,8 \cdot 2^{2} = 1,8 \cdot 4 = 7,2\]

\[y_{3}(2) = \frac{1}{3} \cdot 2^{2} = \frac{4}{3} \approx 1,3\]

\[\Longrightarrow y_{3} < y_{1} < y_{2}.\]