Simplectic Manefolds?
594436
In mathematics, a symplectic manifold is a smooth manifold M equipped with a closed, nondegenerate, 2-form ω called the symplectic form. The study of symplectic manifolds is called symplectic topology. Symplectic manifolds arise naturally in abstract formulations of classical mechanics and analytical mechanics as the cotangent bundles of manifolds, e.g. in the Hamiltonian formulation of classical mechanics, which provides one of the major motivations for the field: The set of all possible configurations of a system is modelled as a manifold, and this manifold's cotangent bundle describes the phase space of the system.\n\nAny real-valued differentiable function H on a symplectic manifold can serve as an energy function or Hamiltonian. Associated to any Hamiltonian is a Hamiltonian vector field; the integral curves of the Hamiltonian vector field are solutions to the Hamilton-Jacobi equations. The Hamiltonian vector field defines a flow on the symplectic manifold, called a Hamiltonian flow or symplectomorphism. By Liouville's theorem, Hamiltonian flows preserve the volume form on the phase space.
189752
sphere, cube, pyramid, cilinder
1802
I haven't tried it, but check out MatPy (linked below).  Another possibility is Numerical Python (also linked below).
24869
List is a bit long, so heres the link:\nhttp://en.wikipedia.org/wiki/List_of_geometric_shapes
387286
Karnaugh map:\n\nCheck out the following link which will help.
180412
Sure, try kinkos. But, it's relatively easy to do it through ms word (tools->envelopes and labels). Just type in the dimensions of the envelope.
577894
Plyometrics with weights.
932841
This might help --> http://www.uwm.edu/~ericskey/TANOTES/Geometry/node15.html
209284
It is not easy to draw when Yahoo compresses out the spaces, so try:\n\n-------E-------\n---X-------N---\n-A---U---------\nM-F------------\n\nor\n\n(E (X (A (M) (F)) (U)) (N))\n\nor\n\nE\n-X\n--A\n---M (leaf)\n---F (leaf)\n--U (leaf)\n-N (leaf)\n\nE to X and N\nX to A and U\nA to M and F
207831
yahoo answers\nmyspace\nmyspace groups
630158
Since you don't want to subdivide (subdivision with chi-square is perfectly valid BTW) you could try a Tukey's HSD. However if you are doing this by hand it will probably be faster to subdivide.
628839
No difficulties encountered...\n\nA more oriented question would help us all. That forum has full of people who can perform embedding with eyes closed and cut virtually anything, from animal and vegetable to rubbers and cheese.\n\nWe will help you!
759812
Ok, so the relative co-efficient of x is 1, so the unrelative co-hedrent homeomorphic must obviously be 2 to the power of Y-as Y is homeomorphic to X's relative undescribed co-efficient.\nUnderstand my friend. I hope I helped.
648070
http://www.anagramgenius.com/archive/algebr.html\n\nedit: sorry didn't answer your real question. I don't know of a program that does what you want, but it shouldn't be too hard to write it. \n\nEdit: Contact me at algebra@envisagecapital.com if you would like the solution. I wrote a quick program to do what you wanted.
801416
sphere none of the list\ncone one vertex and one face\ncylinder two faces-two circular edges
691515
algorithms, data structures, computer organization
220375
Try Microsoft Office templates.