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The degree diameter problem seeks tight relations between the diameter, number of vertices, and degree of a graph. One way of formulating it is to ask for the largest graph with given bounds on its degree and diameter. For any fixed degree, this maximum size is exponential in the diameter, with the base of the exponent depending on the degree. [1]
A 10,000 point Monte Carlo simulation of the distribution of the sample mean of a circular uniform distribution for N = 3 Probability densities (¯) for small values of . Densities for N > 3 {\displaystyle N>3} are normalised to the maximum density, those for N = 1 {\displaystyle N=1} and 2 {\displaystyle 2} are scaled to aid visibility.
Since the diameter is twice the radius, the "missing" part of the diameter is (2r − x) in length. Using the fact that one part of one chord times the other part is equal to the same product taken along a chord intersecting the first chord, we find that (2r − x)x = (y / 2) 2. Solving for r, we find the required result.
Let A′ be the point opposite A on the circle, so that A′A is a diameter, and A′AB is an inscribed triangle on a diameter. By Thales' theorem , this is a right triangle with right angle at B. Let the length of A′B be c n , which we call the complement of s n ; thus c n 2 + s n 2 = (2 r ) 2 .
In applied sciences, the equivalent radius (or mean radius) is the radius of a circle or sphere with the same perimeter, area, or volume of a non-circular or non-spherical object. The equivalent diameter (or mean diameter ) ( D {\displaystyle D} ) is twice the equivalent radius.
In this context, a diameter is any chord which passes through the conic's centre. A diameter of an ellipse is any line passing through the centre of the ellipse. [7] Half of any such diameter may be called a semidiameter, although this term is most often a synonym for the radius of a circle or sphere. [8] The longest diameter is called the ...
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