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>> No.11479020 [View]
File: 168 KB, 900x650, __remilia_scarlet_and_flandre_scarlet_touhou_drawn_by_space_jin__61b4892201886c218a6d311f36a81f80.jpg [View same] [iqdb] [saucenao] [google]
11479020

>>11479001
The "group elements" form a basis for the group ring as a free module (take ten seconds to check the definition), so yes.

>> No.11432332 [View]
File: 168 KB, 900x650, __remilia_scarlet_and_flandre_scarlet_touhou_drawn_by_space_jin__61b4892201886c218a6d311f36a81f80.jpg [View same] [iqdb] [saucenao] [google]
11432332

>>11432252
Seconding the other anon who said it's nonsense.

>> No.10204998 [View]
File: 168 KB, 900x650, 1540861668393.jpg [View same] [iqdb] [saucenao] [google]
10204998

>>10204917
You can't post Touhous on reddit.
Also can't [math]\mathcal{do~this}[/math].

>> No.10126654 [View]
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10126654

>>10126492
Yes.
Draw a circle and lay a diameter line on it from points A to B. Mark the center O of the circle. Chance upon a point C of the triangle, and connect the triangle ABC. Also connect OC.
Since a circle is a set of points at the same distance of the center, OA, OB and OC have the same measure. Since the two triangles are isosceles, OAC and OCA have the same angle, and similarly do OBC and OCB. But BOC and AOC sum two right angles, and the whole of a triangle also sums two right angles, so the remaining angles of the two triangles, OAC, OCA, OBC and OCB, sum two right angles. Since OAC is equal to OCA, and OBC is equal to OCB, OCA and OCB sum a right angle.

Man I love Euclidposting.
>>10126495
>analytic geometry
No, friend.
>>10126533
What the fuck even is this.

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