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May 15

tons of money

I'm trying to figure out what a ton of some American dollar bills and American coins would be worth. Many resources give me conflicting information. But it's the exact sums are what I'm trying to seek. If anyone out there, could help me, that would be wonderful. Thank you so very much.2604:2000:7113:9D00:18B9:8A5:B682:3FBE (talk) 16:09, 15 May 2017 (UTC)[reply]

It depends on the denominations. The one that is probably worth the least is pennies. Penny (United States coin) says that one is 2.5 grams. One ton is 910 kg=910,000 grams so it would take 364000 pennies or $3640 to make a tone. This is actually more than the price of of Zinc, which is mostly what pennies are made of. The difference isn't all that big though, so I imagine it's possible what I've heard that it costs more to make pennies than they're worth. --RDBury (talk) 23:20, 15 May 2017 (UTC)[reply]

According to the people who make them, "the approximate weight of a note" (e.g. a dollar bill) is 1 gram. So one ton of $1 bills would be worth about $910,000. "The exact sum" is likely not published and if you wanted it down to the exact dollar then it would surely depend on the condition of the money. --76.71.6.254 (talk) 07:56, 17 May 2017 (UTC)[reply]

Number of diagonal intersections in a polygon

The article Diagonal says

If no three diagonals of a convex polygon are concurrent at a point in the interior, the number of interior intersections of diagonals is given by ${\binom {n}{4}}$ .^{[citation needed]}

This looks right to me for n up to 6. Can someone find a reference? Thanks. Loraof (talk) 19:07, 15 May 2017 (UTC)[reply]

Poonen, Bjorn; Rubinstein, Michael. The number of intersection points made by the diagonals of a regular polygon. SIAM J. Discrete Math. 11 (1998), no. 1, 135–156 link to a version on Poonen's website link to journal page. has it in the third sentence. --JBL (talk) 20:28, 15 May 2017 (UTC)[reply]

You might be interested in the YouTube [1] which contains a proof as part of another problem. --RDBury (talk) 23:53, 15 May 2017 (UTC)[reply]

{\binom {n}{4}}

makese sense. Each interior point of intersection lies on exactly two diagonals (since no three diagonals intersect) and the end points of these diagonals consist of four distinct vertices. Conversely, each set of four vertices defines a convex quadrilateral (because the whole polygon is convex) and the diagonals that connect opposite corners of this quadrilateral intersect at an interior point. Some of the sides of the quadrilateral may represent diagonals as well, but these only intersect at vertices, not at interior points. So there is a 1-1 correspondence between interior intersection points of diagonals and the

{\binom {n}{4}}

sets of four different vertices. Gandalf61 (talk) 10:51, 17 May 2017 (UTC)[reply]

Number of diagonals in a polyhedron

Given a convex polyhedron that has the same number of edges on all faces, is there a formula for the number of space diagonals in terms of the number F of faces, the number E of edges, and the number V of vertices (any one of which is of course redundant with respect to the other two)? I'd like to put this into both articles, with a source. Loraof (talk) 19:33, 15 May 2017 (UTC)[reply]

The number of space diagonals of any convex polyhedron is

{\binom {V}{2}}+E-\sum _{f{\text{ a face}}}{\binom {|f|}{2}}

. And of course this can be written lots of other ways, for the reason you mention. In particular, if all the faces have size k then this becomes

{\binom {Fk/2-F+2}{2}}+{\frac {Fk}{2}}-F{\binom {k}{2}}

. This is straightforward, but I have no idea where it might be written down. --JBL (talk) 20:35, 15 May 2017 (UTC)[reply]