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МатематикаГДЗ по геометрии 7 класс Атанасян ФГОС Задание 1101
Задание 1101
\[\boxed{\mathbf{1101.ОК\ ГДЗ - домашка\ на}\ 5}\]
\[\mathbf{Формула}\mathbf{:\ \ \ }\]
\[C = 2\pi R \Longrightarrow \ R = \frac{C}{2\pi};\]
\[где\ \ \ \pi = 3,14.\]
| \[C\] | \[25,12\] | \[18,84\] | \[82\] | \[18\pi\] | \[4,4\] | \[6,28\] | \[637,42\] | \[14,65\] | \[2\sqrt{2}\] |
|---|---|---|---|---|---|---|---|---|---|
| \[R\] | \[4\] | \[3\] | \[13,06\] | \[9\] | \[0,7\] | \[1\] | \[101,5\] | \[2\frac{1}{3}\] | \[0,45\] |
\[1)\ C = 2 \bullet 3,14 \bullet 4 = 25,12;\]
\[2)\ C = 2 \bullet 3,14 \bullet 3 = 18,84;\]
\[3)\ R = \frac{82}{2 \bullet 3,14} = 13,06;\]
\[4)\ R = \frac{18\pi}{2\pi} = 9;\]
\[5)\ C = 2 \bullet 3,14 \bullet 0,7 = 4,4;\]
\[6)\ R = \frac{6,28}{2 \bullet 3,14} = 1;\]
\[7)\ C = 2 \bullet 3,14 \bullet 101,5 = 637,42;\]
\[8)\ C = 2 \bullet 3,14 \bullet 2\frac{1}{3} = 14,65;\]
\[9)\ R = \frac{2\sqrt{2}}{2 \bullet 3,14} = 0,45.\]
\[\boxed{\mathbf{1101.еуроки - ответы\ на\ пятёрку}}\]
\[\textbf{а)}\ sin\ \alpha = \frac{\sqrt{3}}{2}:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - \frac{3}{4} = \frac{1}{4}\]
\[\cos\alpha = \pm \frac{1}{2}.\]
\[\textbf{б)}\ sin\ \alpha = \frac{1}{4}:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - \frac{1}{16} =\]
\[= \frac{15}{16}\]
\[\cos\alpha = \pm \frac{\sqrt{15}}{4}.\]
\[\textbf{в)}\ sin\ \alpha = 0:\]
\[\text{si}n^{2}\alpha + cos^{2}\alpha = 1\]
\[\cos^{2}\alpha = 1 - \sin^{2}\alpha = 1 - 0 = 1\]
\[\cos\alpha = \pm 1.\]
