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| Автор | Сообщение |
|---|---|
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Alex1985
#
30 ноя 2011 |
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|
o_a
#
30 ноя 2011 |
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Alex1985
#
30 ноя 2011 |
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|
o_a
#
30 ноя 2011 |
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Alex1985
#
30 ноя 2011 |
|
|
o_a
#
30 ноя 2011 |
Чтобы написать сообщение, необходимо войти или зарегистрироваться
![$poly := [[0, 1], [0, 2], [.5, 2.75], [1.25, 3], [2, 2.75], [2.5, 2.25], [1.75, 1.5], [2.5, .75], [2, .25], [1.25, 0], [.5, .25]];$ $poly := [[0, 1], [0, 2], [.5, 2.75], [1.25, 3], [2, 2.75], [2.5, 2.25], [1.75, 1.5], [2.5, .75], [2, .25], [1.25, 0], [.5, .25]];$](http://teacode.com/service/latex/latex.png?latex=poly+%3A%3D+%5B%5B0%2C+1%5D%2C+%5B0%2C+2%5D%2C+%5B.5%2C+2.75%5D%2C+%5B1.25%2C+3%5D%2C+%5B2%2C+2.75%5D%2C+%5B2.5%2C+2.25%5D%2C+%5B1.75%2C+1.5%5D%2C+%5B2.5%2C+.75%5D%2C+%5B2%2C+.25%5D%2C+%5B1.25%2C+0%5D%2C+%5B.5%2C+.25%5D%5D%3B&fontsize=21)
2) можно указать функциональную зависимость, используя команду ![$pos := [seq([cos(2*Pi*i/n), sin(2*Pi*i/n)], i = 1 .. n)]; $ $pos := [seq([cos(2*Pi*i/n), sin(2*Pi*i/n)], i = 1 .. n)]; $](http://teacode.com/service/latex/latex.png?latex=pos+%3A%3D+%5Bseq%28%5Bcos%282*Pi*i%2Fn%29%2C+sin%282*Pi*i%2Fn%29%5D%2C+i+%3D+1+..+n%29%5D%3B+&fontsize=21)