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angelika.me | ||
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www.jeremykun.com
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| | | | | Problem: $ \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots = 1$ Solution: Problem: $ \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots = \frac{1}{2}$ Solution: Problem: $ \frac{1}{4} + \frac{1}{16} + \frac{1}{64} + \dots = \frac{1}{3}$ Solution: Problem: $ 1 + r + r^2 + \dots = \frac{1}{1-r}$ if $ r < 1$. Solution: This last one follows from similarity of the subsequent trapezoids: the right edge of the teal(ish) trapezoid has length $ r$, and so the right edge of the neighboring trapezoid, $ x$, is found by $ \frac{r}{1} = \frac{x}{r}$, and we see that it has length $ r^2$. | |
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lucas.art
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| | | | | I thought this was impossible, I thought that you'd need to lock the camera position and save it, so you could then go back to a different angle that was working for you. It turns out you can easily undo (and redo) camera movements by using the square brackets | |
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webrocker.de
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| | | | | Gruß von der #TagesLichtRunde... | |
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orjpap.github.io
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| | | Random morning thoughts: how many people actually visit this blog? Where do they come from? What if I could just spend a little time to build my own analytics? | ||
