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 Blake Webmaster2082 Posts11 Ineo 5:1 - 14.43.1960.829 days ago r = 5.6666666666666666666666666666667 Call the intersection of the two line segments A Call the point where the line segment of length 5 intersects the circle B Call the center of the circle C Call the radius r r = 3 + MYSTERYNUMBER Now look at the triangle ABC AB = 5 AC = MYSTERYNUMBER BC = r = 3 + MYSTERYNUMBER Go go gadget Pythagorean theorem: BC2 = AB2 + AC2 (3 + M)2 = 52 + M2 9 + 6M + M2 = 25 + M2 6M = 16 M = 16/6 = 2.6666666666666666666667 ... r = M + 3 r = 2.666666666666666666667 + 3 = 5.666666666666666666667 [Quote] [Link]
 Oh! Thank you very much. :D [Quote] [Link]
 Just something fun: Find the tangent line of circle (x-a)2+(y-b)2=r2 at point (c,d). [Quote] [Link]
 Blake Webmaster2082 Posts11 Ineo 5:2 - 13.6.16959.898 days ago Draw a line from c, d (the point on the circle) to a, b (the center of the circle). The slope of this line would be perpendicular to the tangent line. Now you have the slope. Slope of c,d-a,b = (b - d) / (a - c) Perpendicular slope = (c - a) / (b - d) That tangent line goes through c, d. Now you have a point. (y - d) = slope * (x - c) y - d = (c - a) / (b - d) * (x - c) (y - d) * (b - d) = (c - a) * (x - c) [Quote] [Link]
 ywkbme said:This thread makes me feel like an idiot... This. [Quote] [Link]
 And nobody's even done any integration yet... [Quote] [Link]
 I was thinking more along the lines of ∫xxdx, but whatever. [Quote] [Link]
 Hydrogen777 Administrator5988 Posts11 Ineo 6:0 - 15.35.9955.783 days ago Similarly difficult problems, no? [Quote] [Link]
 ∫e^x^2dx = ∫Σ((x^2k)/k!) = Σ[(x^(2k+1))/((2k+1)k!)] + C uh...I think...and k goes from 0 to ∞ oh and according to wolfram alpha, ∫x^xdx has no indefinite integral that can be represented with standard functions. [Quote] [Link]
 Hydrogen777 Administrator5988 Posts11 Ineo 6:1 - 5.15.60955.293 days ago Even infinite series? That's intense. [Quote] [Link]
 JSan4122 Posts11 Ineo 6:2 - 9.61.92954.07 days ago Here's something I find a little funny. Sine, Cosine, and tangent are all use to describe angles. But when graphed, the look nothing like they would be involved with angles. Does anyone else find that funny? [Quote] [Link]
 I think they look like they're involved with angles... Then again, nothing math related really looks all that much like what it really is. [Quote] [Link]