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For many practical problems, the detailed Bode plots can be approximated with straight-line segments that are asymptotes of the precise response. The effect of each of the terms of a multiple element transfer function can be approximated by a set of straight lines on a Bode plot. This allows a graphical solution of the overall frequency ...
The "nine dots" puzzle. The puzzle asks to link all nine dots using four straight lines or fewer, without lifting the pen. The nine dots puzzle is a mathematical puzzle whose task is to connect nine squarely arranged points with a pen by four (or fewer) straight lines without lifting the pen or retracing any lines.
The number of points (n), chords (c) and regions (r G) for first 6 terms of Moser's circle problem. In geometry, the problem of dividing a circle into areas by means of an inscribed polygon with n sides in such a way as to maximise the number of areas created by the edges and diagonals, sometimes called Moser's circle problem (named after Leo Moser), has a solution by an inductive method.
G (w,n) = 1 / (sqrt (1 + w ** (2 * n))) dB (x) = 20 * log10 (abs (x)) # Phase is for first order P (w) =-atan (w) * 180 / pi # Gridlines set grid # Set x axis to logarithmic scale set logscale x 10 # No need for a key set no key #0.1,-25 # Frequency response's line plotting style set style line 1 lt 1 lw 2 # Asymptote lines and slope lines are ...
This work has been released into the public domain by its author, Mik81.This applies worldwide. In some countries this may not be legally possible; if so: Mik81 grants anyone the right to use this work for any purpose, without any conditions, unless such conditions are required by law.
For each pair of lines, there can be only one cell where the two lines meet at the bottom vertex, so the number of downward-bounded cells is at most the number of pairs of lines, () /. Adding the unbounded and bounded cells, the total number of cells in an arrangement can be at most n ( n + 1 ) / 2 + 1 {\displaystyle n(n+1)/2+1} . [ 5 ]
It was really epic," Bode said, noting she was "laughing a lot" while "everyone else around me looked so concerned. "When anyone falls, it's the most terrifying thing," Slater, 32, chimed in.
Proof without words: One house is temporarily deleted. The lines connecting the remaining houses with the utilities divide the plane into three regions. Whichever region the deleted house is placed into, the similarly shaded utility is outside the region. By the Jordan curve theorem, a line connecting them must intersect one of the existing lines.