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What is cutting edge maths? - Mathematics Stack Exchange To come back to your question, the cutting edge is often in the refinement and well considered combination of equations, 'paragraphs' in this metaphor Where the metaphor differs is that the english language allows for endless break down of the rules, such that hundreds of paragraphs can be written quickly, whereas a single mathematical
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arithmetic - Can a piece of A4 paper be folded so that its thick . . . Say that during the cutting process, the cellulose fibers unravel somewhat, leaving only two layers of fiber Since the fibers are 2-20 nm in diameter , let's say that the two-fiber-layer sheets are about 10nm thick $2^{42} \times 10nm = 43,980 m$, which, according to Wolfram Alpha, is about five times the height of Mount Everest
geometry - Compass-and-straightedge construction of the square root of . . . For example, suppose you want to find the square root of 5 Construct a right triangle with side lengths 1 and 2 This can be done with straight edge and compass Then the hypotenuse has length $\sqrt{5} $ (times the unit) The procedure can get more complicated
Online tool for making graphs (vertices and edges)? Changing style of nodes and edges (color, shape, thickness of edge, line style, node size) Bending edges; Shortcuts support; Displaying the last action with possibility to undo; Copying, cutting, pasting of nodes and edges; Support for mobile and touch devices; The application is still in a development state – any suggestions and feedback are
Is it possible to draw this picture without lifting the pen? $\begingroup$ @JohannesD: No, because either the cross occurs at a vertex which divides the line (11-8) anyway; or you cross the same edges here as well chi is absolutely correct, and the suggestion does not introduce or remove any edge crossing that wasn't there already $\endgroup$ –
How can I find the points at which two circles intersect? $\begingroup$ There is only one plane in $\mathbb{R^2}$, and this is $\mathbb{R^2}$ What you do is the change of the coordinate plane or coordinate system $(\vec{a},\vec{b})$ do not define a coordinate plane you additionally need an origin which should be $\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right)$, I think
differential geometry - Poincaré hyperbolic geodesics in half-plane and . . . $\begingroup$ What a mess! It would be easier to use the pseudo-Euclidean hyperboloid model The "outside" of the Poincare disk is simply the lower sheet of the hyperboloid, while the inside is the upper sheet; and the "extended" geodesic is the intersection of a plane with both sheets (resulting in a hyperbola)