what is the maximum number of regions that a circular region can be divided into by 100 segments?
830196
After careful counting, I come up with 5,051.  :)
939021
I am only using visual techniques and not math theorems here:\n\nThe first plane will divide space into 2 parts.\n\nIf totally random the second plane will probably not be parallel with the first.  This will divide the 2 parts into 4.  If it did happen to be parallel with the first, then the space is divided into 3 parts.\n\nThe third plane will most likely not be parallel with either of the other two.  It would divide the space into 8ths.  If it was parallel to either of the other two then the minimum parts is 4 and the maximum is 6 parts.\n\nThe fourth plane will most likely not be parallel with any of the first 3, and can divide the space into maximum of 15 parts.  The fifth will make that 26.  Parallel planes will reduce this number.\n\nThe space will be divided into a minimum of 6 and a maximum of 26 parts with 5 planes.\n\nI just found a great description at the site below:
1092713
triskaidecagon\n\nor a 13-sided polygon, haha.\n\nCheck the source for polygon names from 1-20 sides  :)
1165642
One way to reason is this.  Let n be any even integer greater than or equal to 4.  Divide the circle into n equal wedges (by drawing n/2 diameter lines with angles between successive lines equal to 360/n degrees).  Now, cut the wedges out and make a "rectangle" with them: n/2 wedges with the points downward and the arcs on top of adjacent pairs of wedges touching each other, then the other n/2 wedges with the points upward and adjacent arcs touching, filling the spaces of the first n/2 wedges.  What you end up with is almost a rectangle with top and bottom each made of n/2 360/n-degree arcs, and the lateral sides slanting slightly, made from the long edges of the wedges on each end.\n\nYou can estimate the area of this figure (which is the same as the area of the original circle of radius r) by pretending it IS a rectangle.  The length is the circumference of the circle, or 2*pi*r/2 (since half the circumference is the top, half is the bottom) (by definition of pi).  The width is the length of a wedge, or r.  So the area is about pi*r^2.  \n\nTo turn the estimate into an exact value, consider what happens as n goes to infinity, that is, it increases without bound.\n\nAs n goes to infinity, the slanted edges aproach being verticle and the number of arcs making up the top and bottom approaches infinity, and the arcs themselves approach straight horizontal lines.  Thus, the figure approaches a rectangle of exactly area pi*r^2.  However, for each n, you have the same circle of the same area, so the area of that circle is pi*r^2.
103215
With this given data, it is not possible\n\nalthough the area is totalling to be 70 for both cases (7*10 and two pieces of 8*8 + 1*6) but with one division only, this is not possible\n\nPLEASE DO LET ME KNOW IF THERE IS SOME SOLUTION (i found it very interesting)
822332
Since you have to divide the area into five rectangles, you will need rectangles from:\n\n0 to 1, 1 to 2, 2 to 3, 3 to 4, 4 to 5\n\nSince you are using the right endpoints, you need to be using the second value of each division; that is, in the "x to y" examples, you'll be using the y endpoints 1,2,3,4, and 5.\n\nThe area of each rectangle will then be 1*f(x), since the width of each rectangle is 1.  Therefore, the total sum formula should be as follows (with * representing multiplication).\n\n1 * f(1) + 1*f(2) + 1*f(3) + 1*f(4) + 1*f(5)\n\nOr, quite simply, f(1) + f(2) + f(3) + f(4) + f(5) to solve using the right endpoints.  Calculating this, we get\n\n(25 - 1^2) + (25 - 2^2) + (25 - 3^2) + (25 - 4^2) + (25 - 5^2)\n(24) + (21) + (16) + (9) + 0 = 70\n\nTo calculate using the left endpoints, recall that our intervals for the rectangles were:  0 to 1, 1 to 2, 2 to 3, 3 to 4, and 4 to 5.  In this case, we use the LEFT endpoints and take the FIRST value of each interval; that is, 0,1,2,3, and 4.  The width of the rectangle doesn't change at all (the width is 1), so our left endpoint summation formula because\n\n1 * f(0) + 1 * f(1) + 1 * f(2) + 1 * f(3) + 1 * f(4)\n\nor, quite simply\n\nf(0) + f(1) + f(2) + f(3) + f(4).  We calculated f(1) through f(4) already above, so we'll copy those values.  But f(0) = (25 - 0^2) = 25, so our total now becomes:\n\n25 + 24 + 21 + 16 + 9 = 95.\n\nArea using right endpoints = 70\nArea using left endpoints = 95
589996
Assuming that A, B, C are all on the same line.\n\nLarger circle:\nRadius = AC = 6\nArea = PI * r^2 = 36*PI\n\nSmaller circle:\nRadius = AC - AB = 6 - 2 = 4\nArea = PI * r^2 = 16*PI\n\nShaded area:\nArea = (Large A) - (Small A)\n     =  36*PI - 16*PI = 20*PI = 62.8 units^2
956315
The simplest way is to cut it into five slices, each with an arc of 72 degrees (or 2*pi/5 radians). You could probably also figure out a way to cut it into strips so that each strip has equal area, but that would be significantly more complicated.
991890
Try the 2-2 pattern. it's the best way. The 3 is done with that criss-cross pattern, but the 2-2 allows you to continue the circle, just make it have a big radius
515192
If it's a square, (60 x 60) then probably 8.\nIf it's a circle (r=60) then probably 10.\nIf it's a cirlce (d=60) then probably 6
587075
Would be happy to help, but some information is missing.  I am assuming that you are asking about the region bounded by the following four equations:\n{2x + y < 25}  {x + y < ?}  {x < 12}  {x > 0}\n\nNeed to know what ? is in order to give an exact answer.
967530
May be a Sphere/Circle of defined polygons or Trees.... with infinite edges and finite branching points....
1175861
Area of Semicircle = (PI * R * R)/2\n\nAns. = 2.262 sq. cm
149999
A regular n-gon has the centers of its inscribed and circumscribed circles the same point. If you draw segments from this center to each of the vertex, you will get n number of radii and n number of triangles, which are all congruent. Using derivation you will arrive at the conclusion that\nA = .5aP or A = .5ans where\nA is the area, a is the apothem, s is the side and P is the perimeter\na =2A/ns\nThe radius is acquired by\nsin [180degrees(n - 2)/n] = a/r\nr = a/sin[180degrees(n - 2)/n]
191858
If there are no other constraints, the roles can be assigned in \n15*14*13*12*11*10 ways. The first role can be filled in 15 ways and the second in 14 ways and so on. This is also called a permutation. See http://mathworld.wolfram.com/Permutation.html
901557
lol...I'm learning about this in school...  75divided 3=25, 1/3=25 soo...50 parts sand  is your answer
68097
do you mean things as far as max and mins, asymptotes and things like that?\n\nif you are a little more specific i can help you better, or e-mail me, it would be a little easier  matttlocke@yahoo.com