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/lit/ - Literature

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>> No.15607772 [View]
File: 28 KB, 410x461, 67a86f04a706bf46f6d60b0fa89e566e.jpg [View same] [iqdb] [saucenao] [google]
15607772

stop glamorizing mental illness
stop glamorizing mental illness
stop glamorizing mental illness
stop glamorizing mental illness
stop glamorizing mental illness

>> No.15471577 [View]
File: 28 KB, 410x461, 67a86f04a706bf46f6d60b0fa89e566e.jpg [View same] [iqdb] [saucenao] [google]
15471577

>>15471566
Then why are you claiming things you know nothing about? Just shut the fuck up kiddo and let the adults talk.

>> No.15374230 [View]
File: 28 KB, 410x461, 1587995446455.jpg [View same] [iqdb] [saucenao] [google]
15374230

>>15371316
Mine too

>> No.15225268 [View]
File: 28 KB, 410x461, 1587995446455.jpg [View same] [iqdb] [saucenao] [google]
15225268

>>15225242
It like that

>> No.15209443 [View]
File: 28 KB, 410x461, 67a86f04a706bf46f6d60b0fa89e566e.jpg [View same] [iqdb] [saucenao] [google]
15209443

>>15209390
Bound the area of the excess above with rectangles of width 2rsin(t) and height r(1-cos(t)) where t is the angle of one of the triangles (i.e. 360/n degrees). We have n such triangles so the total excess is 2 n r^2 sin(t)(1-cos(t)) = 2r^2 * 360 sin(t)(1-cos(t))/t.
Now you can prove that sin(t)/t tends to a constant because (in radians) sin(t)<t for t small enough, and (1-cos(t)) tends to 0 so the total excess area tends to 0 and so the are of the regular polygon does indeed tend to the area.
Simple.

>> No.15047282 [View]
File: 28 KB, 410x461, 67a86f04a706bf46f6d60b0fa89e566e.jpg [View same] [iqdb] [saucenao] [google]
15047282

>>15047127
My post hit too close to home?

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