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Type: Posts; User: jemidiah
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I'm not sure what advice you're hoping to get. You say you want a mathematical "pattern", but that's really vague. You might want an algorithm to choose a subset of a block of seats which satisfies...
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One approach is to compute Z*OP/|OP| = Z*(P.x - O.x, P.y - O.y)/sqrt((P.x - O.x)^2 + (P.y - O.y)^2) and add it to P. Here OP/|OP| is the unit vector in the direction you wish to move. In your...
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Huh, I apparently missed that "l" repeatedly, sorry about that. You didn't give the important details so I suppose I latched onto a perceived typo instead.
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Well, the volume of a cylinder is pi*r^2*h where r is the radius and h is the height. Your "V = pir^2" is at leaving off the h.
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I think I understand what you want from post #4, which is a function f(x, y) = (u, v) which sends two fixed pairs (x1, y1), (x2, y2) to two fixed pairs (u1, v1), (u2, v2) while scaling distance at a...
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Posting a full example calculation will probably provide enough details for us to describe your situation rigorously and work out the conversion formula. Otherwise it's too vague to really say. As...
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Well, my Z is just whatever you're multiplying the dimensions of the image by. So, if the image started at 100x100, doubled to 200x200, and then cut down to 50x50, you would use Z=2 in the first...
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Suppose your image has upper left corner (P.X, P.Y). I'll assume the rectangular viewing area has upper left corner (0, 0) and height/width V.height/V.width. Similarly suppose you wish the part of...
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I'm not familiar with JavaScript, but apparently variable names are case-sensitive, so you should use .Y in each case instead of .y. Even then, your radiiDifference is 15-15 = 0, so when running "var...
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Just glancing at it, you wrote "Math.pow((Pn1.Point.Y - Pn.Point.y) ^ 2)" instead of "Math.pow((Pn1.Point.Y - Pn.Point.y), 2)".
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https://en.wikipedia.org/wiki/Binomial
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To find the height of the four isosceles triangles, draw a right triangle from the vertex to the center of the base to the midpoint of one of the sides. If the sides have length A and the overall...
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Why do you VB* people do everything the hard way? Weakly typed languages for the win. :)
(Though to be fair, the .NET IDE is really quite nice.)
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Ah, that's far too complicated to hope for an analytic solution. There's also a problem of uniqueness--there are surely quite a few values of your three variables which "on average" with purely...
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More details are needed to give an exact solution; you'll probably end up wanting something like the inverse of a cumulative distribution function.
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I find the code in post #4 very difficult to interpret--a lot of it doesn't seem to make much sense. The code in post #5 looks pretty close, but I think it should instead be...
'Assume the Sides...
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They're not symbol-for-symbol the same, but yes, they're equal, since 3-c is -(c-3), and 12c-6a is -(6a-12c). I had checked to make sure. You can also have Wolfram Alpha check for symbolic equality,...
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You're correct, and Wolfram Alpha agrees.
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Minor mistake: "mx = (x2 - x1) / circ2Rad" should be pulled out of the If statement and put right before it. mx is not getting properly reset in the x1=x2 case, which only shows up in the a=0 (or...
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Note quite. Again the issue is the SSA case, which comes down to the non-invertibility of the sine function on the range 0 to 180. In particular, given a, A, and B, you can compute sin(b), but sin is...
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You can't do much with just 1 side and its opposite angle. There are certainly many solutions give just those two pieces of information. You should be able to pick a second angle arbitrarily (subject...
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This is getting at the standard similarity/congruence tests for triangles. Here they are:
AAA: three angles determine the corresponding sides up to an overall scale factor. Since the angles add up...
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This isn't really a math question. It should be in one of the language-specific forums. There are also a zillion code examples of the zillion and one sorting algorithms out there, though they may not...
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Minor thing: I noticed a typo in my second graph's equation from post #8. I had written y = −x² + 21x + 11 instead of y = −x² + 21x + 22. This didn't affect the default Wolfram Alpha view of the...
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Well, y is 0 at the x-intercepts, since the x-intercepts are on the x-axis, and the x-axis is precisely the line consisting of points whose y-coordinate is 0.
Perhaps this will help? Consider the...
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I think I'd have to see the question to know what you're trying to do. You again said you needed to "solve for x", and I've told you how to solve for x in terms of y. If you mean you need to solve...
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You wrote an expression, not an equation, so what do you mean by it "seems insoluble without a value for y"?
To take a simpler example, start with y=x^2. Doing the same operation, i.e. saying x^2...
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Well, if you want to solve your equation for x in terms of y, you'll have to use the quadratic formula, and there are two "branches" of solutions in general (the function is not invertible in...
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Oh, I agree my solution is easy to implement with a for loop, it's just cleaner and (potentially) faster to use a linked list/queue style data structure, at least if there are often many unused spots...
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Yeah, I'm not a big fan of the two methods you suggested either. The first would make earlier entities *much* more likely to be chosen than later ones, while the second would make the last entity...
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That article is nice and brief, and it gives some notion of what the parts of Heron's formula individually "mean". It does something quite similar to what my proof does, but with the incircle rather...
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@boops boops: for what it's worth, I made an effort to avoid Heron's formula since it seemed like its use just pushes the problem farther down the line--why is Heron's formula true? On the other hand...
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First, here's why the circumcenter exists. Take two sides of a triangle and consider their perpendicular bisectors. Since the sides are not parallel, neither are the perpendicular bisectors, so they...
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Just to be clear, my comment about precision was directed at how precisely you wanted to draw "the maximum sized scale triangle within the bounding rectangle [subject to some given angles]", rather...
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The complexity of what you want depends a lot on how precise you want it all to be and what you're willing to say is "good enough". For instance, your "acute" case in post #1 allows non-acute...
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@dday9: cool beans.
Totally understandable, though in this case there was a very easy check besides looking at the original source code. The matrix form of the operation needed to have...
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I see one big mistake still in #3, namely the "- x" should be "+ x". That would produce extremely erratic results.
I disagree with passel in that I think the integer-based coordinates for your...
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You seem to be referring to the fact that the rows of Pascal's triangle, when plotted after normalization, look very much like a normal distribution, especially as you take longer and longer rows....
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Ah, my mistake. That makes the computation easier. Since 0 < N < 100, we may evidently assume every such sequence terminates, since a non-terminating sequence occurs with 0 probability. Each sequence...
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Unless I'm really, really misunderstanding your question, it's not (N/100)M times. For instance, at N=100, this gives M, but as you say it should obviously give infinity. As I understand it, you...
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