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 Apr5 comment Would like some pointers on this geometry problem Sorry, it's $AB \cdot BC = BD^2$, the blockquote messed it up. But the idea of assigning some random value to a side might be helpful. Apr5 revised Would like some pointers on this geometry problem deleted 4 characters in body Apr5 asked Would like some pointers on this geometry problem Mar26 revised A problem with a trigonometric equation latex and translated the trig functions' names Mar26 suggested approved edit on A problem with a trigonometric equation Mar22 comment Prove that the set of positive real numbers is not bounded from above @PeterT.off: Well, wouldn't you have to show that $\mathbb{N}$ is unbounded as well, using a similar proof? Mar21 revised Prove that the set of positive real numbers is not bounded from above added 213 characters in body Mar21 comment Prove that the set of positive real numbers is not bounded from above @Arturo: I didn't say it, but I can use the basic axioms of the real numbers. Mar21 asked Prove that the set of positive real numbers is not bounded from above Mar19 answered Using polar form to prove $|z| = 1 \implies \text{Re}\left(\frac{1-z}{1+z}\right) = 0$ Mar16 comment See the sign of the double derivative from just looking at the graph? In this case, it looks like it, but I don't think you can be completely sure without knowing the function's formula. Mar13 accepted Are there exact expressions for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$? Mar13 comment Are there exact expressions for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$? Well, that was simple :). Thanks. Mar13 asked Are there exact expressions for $\sin \frac{3\pi}{8}$ and $\cos \frac{3\pi}{8}$? Mar8 comment directional derivative unit vector If you have an angle $\theta$, the unit vector in that direction is $(\cos \theta, \sin \theta)$. Feb29 accepted How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$? Feb29 comment How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$? It wasn't my physics test, it was one some friends in my same year did. But never mind that, thank you for all your help. Feb29 comment How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$? I end up with $\int \frac{\mathrm{d}y}{\sqrt{\frac{2k}{y}+C}} = \int \mathrm{d}x$. According to WolframAlpha, the integral on the left hand side is quite complicated, and it seems impossible to solve for $y=y(x)$. Am I doing something wrong? Feb29 comment How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$? Well, I get $y' = \pm \sqrt{2k(\frac1{y}-\frac1{y(0)})}$ (I made it clearer in the title that the right hand side of the equation should be negative), but I don't know how to go from there. Feb29 revised How to solve $y'' = -\frac{k}{y^2}$, with $k > 0$? edited title