Решебник по алгебре 9 класс Макарычев Задание 92

 

\[\boxed{\text{92\ (92).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

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

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

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

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

\[= \frac{1}{3} \cdot 0,25 \approx 0,083 \Longrightarrow\]

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

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

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

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

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

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

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

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

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

\[\boxed{\text{92.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\textbf{а)}\ \sqrt{5} + \sqrt{7} \approx 2,236 + 2,645 \approx\]

\[\approx 4,881 \approx 4,9 \approx 4,88.\]

\[\textbf{б)}\ \frac{4}{11} - \sqrt{8} \approx 0,3636 - 2\sqrt{2} \approx\]

\[\approx 0,3636 - 2,828 \approx\]

\[\approx - 2,4644 \approx - 2,5 \approx - 2,64.\]

\[\textbf{в)}\ \frac{11}{9} \cdot \left( - \sqrt{5} \right) \approx \frac{11}{9} \cdot ( - 2,236) \approx\]

\[\approx - 11 \cdot 0,248 \approx - 2,728 \approx\]

\[\approx - 2,7 \approx - 2,73.\]

\[\textbf{г)}\ \sqrt{2} - \frac{5}{8} \approx 1,414 - 0,625 \approx\]

\[\approx 0,789 \approx 0,8 \approx 0,79.\]

\[\textbf{д)}\ \frac{3}{16}\ :\sqrt{3} \approx \frac{3}{16}\ :1,732 \approx\]

\[\approx 0,1875\ :1,732 \approx 0,10826 \approx\]

\[\approx 0,1 \approx 0,11.\]