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The following is a list of centroids of various two-dimensional and three-dimensional objects. The centroid of an object in -dimensional space is the intersection of all hyperplanes that divide into two parts of equal moment about the hyperplane.
Cuboid – , where , , and are the sides' length; Cylinder – π r 2 h {\textstyle \pi r^{2}h} , where r {\textstyle r} is the base's radius and h {\textstyle h} is the cone's height; Ellipsoid – 4 3 π a b c {\textstyle {\frac {4}{3}}\pi abc} , where a {\textstyle a} , b {\textstyle b} , and c {\textstyle c} are the semi-major and semi ...
Some superquadrics. In mathematics, the superquadrics or super-quadrics (also superquadratics) are a family of geometric shapes defined by formulas that resemble those of ellipsoids and other quadrics, except that the squaring operations are replaced by arbitrary powers.
A Pythagorean quadruple is called primitive if the greatest common divisor of its entries is 1. Every Pythagorean quadruple is an integer multiple of a primitive quadruple. The set of primitive Pythagorean quadruples for which a is odd can be generated by the formulas = +, = (+), = (), = + + +, where m, n, p, q are non-negative integers with greatest common divisor 1 such that m + n + p + q is o
In mathematics, a quadric or quadric surface is a generalization of conic sections (ellipses, parabolas, and hyperbolas).In three-dimensional space, quadrics include ellipsoids, paraboloids, and hyperboloids.
The rhombic dodecahedron can be seen as a degenerate limiting case of a pyritohedron, with permutation of coordinates (±1, ±1, ±1) and (0, 1 + h, 1 − h 2) with parameter h = 1. These coordinates illustrate that a rhombic dodecahedron can be seen as a cube with six square pyramids attached to each face, allowing them to fit together into a ...
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Because of the factorization (2n + 1)(n 2 + n + 1), it is impossible for a centered cube number to be a prime number. [3] The only centered cube numbers which are also the square numbers are 1 and 9, [4] [5] which can be shown by solving x 2 = y 3 + 3y, the only integer solutions being (x,y) from {(0,0), (1,2), (3,6), (12,42)}, By substituting a=(x-1)/2 and b=y/2, we obtain x^2=2y^3+3y^2+3y+1.