into how many parts will five random planes divide space?
919905
26.  The pattern has to start 1, 2, 4, 8, 15.  I see the differences are 1, 2, 4, 7, and the differences of the differences are 1, 2, 3.  The next difference of the difference has to be 4, so, working my way back up, I figure the next difference has to be 11, making the next number 26.
137083
usually rectangular-4 0f each suite
590008
They're all wrong.  72 x 48 would be 1/100th of the screen.  You've got to remember we're dealing with AREA here - 2 dimensions.  Look at it this way:  The total area of the screen (number of pixels) is:\n\n720 x 480 = 345,600\n\nSo, each section should be 1/10 of that total, or 34,560 pixels.\n\nThere are a couple easy ways to divide it into 10 sections:\n\n72 x 480\n720 x 48\nThese options would result in a 10x1 grid or a 1x10 grid.\n\nThere are a variety of other ways as well.  If you want a 2x5 grid:\n\n144 x 240\n360 x 96
568866
is there the same probability it will go into each bin?\n\nthen 350/9=38.8888 so about 39 in each
856650
This isn't the answer, but it might spark something in your mind for an answer.  If you needed to make 5 lines, with 5 trees, using a total of 19 trees, you could make a square with a diaganal line running through it.  I'll keep trying to think of another way.  May we inquire where the question came from?\n\n.1...2...3....4...5\n.6..............7...8\n.9........10......11\n12..13...........14\n15..16.17.18.19
233027
There are 8! different arrangements (permutations) of the 8 friends.\n\nThere are 4! different arrangements of the 4 boys.\n\nThere are 5 different locations where the leftmost boy can sit when the 4 boys sit together.\n\nTherefore, there are 5*4! (or 5!) arrangements of the boys where all the boys sit together.  But there are also 4! arrangements of the 4 girls in the remaining slots.  So there are 5!*4! arrangements where the 4 boys sit together.\n\nSince there are 8! arrangements overall, the probability of all boys sitting together is (5!*4!)/8!.\n\nSimilarly, if the boys sit to the left and the girls sit to the right, then the probability is (4!*4!)/8!.\n\nInterestingly, this last answer is also the same probability as boy-girl-boy-girl-boy-girl-boy-girl and also the probability of girl-boy-girl-boy-girl-boy-girl-boy.  Therefore, the probability of alternating seats is (2*4!*4!)/8!.\n\nRecall that n! (read n-factorial) is n*(n-1)*(n-2)*...*3*2*1.
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)
737814
see in a ||gm we need two || lines in a opposite side so we chose 2 out of 6 \ni.e, 6C2\n\nand two more || lines for other side then we have to chose 2 out of 5 \ni.e, 5C2\n\ntotal cases 6C2 * 5C2\n\nwhich will gives you \n(6*5*4*3*2*1/2*1*4*3*2*1) * (5*4*3*2*1/2*1*3*2*1)\n(15)*(10)\n150
747450
Sounds like Combinatorics.  Combinations, permutations, a bit of graph theory, discrete math, I usually describe it as "Advanced Counting".
953741
Just break the pentagon up into 5 triangles.  You can then add together the area of the 5 triangles to get the total area.
1022630
None of those are "planes".\n\nA plane is drawn as a square simply because of the limitations of a sheet of paper.\n\nA plane is infinite in two directions.  It has no start or end point on the xy axis.  A sheet of paper of course does have start end end points.  Usually 8x11\n\nNone of rectangels, squares, or triangles are planes.  They are all subsets of them.  And darned small ones at that.\n\n-Dio
792270
Hi I assume you mean an unsymmetrical item. I also assume you want a best fit. I have done similar things on a computer but not by hand. The computer finds three point most distant from the centroid and uses these points as pivots for three orthogonal planes. Not sure of the rest.
35955
like this???            \n              \n                x x x x x \n                x x x x x \n                x x x x x \n                x x x x x \n               (4x5/5x4)
341037
-5 x 2 x 2 x 2 x 2 x 2 (or, I guess you could say -1 x 5 x 2 x 2 x 2 x 2 x 2)
223867
It depends on the object.  You find the area of all the sides and then add them together.
747559
Simply draw a chart that has each of the 6 variables at right angles to each other, draw lines between the linked variables such coke-greenbike (#3), then use process of elimination.\n\nYeah, right... draw a 6-dimensional hypercube....\n\n Jeez, I've spent about 3 hours on this, and so far it looks like Carla, who was on the yellow bike which was 1st in line, was Bigbird and brought rootbeer and jelly beans.  Either Emily, who was dressed as either dracula or the clown, or Daniel dressed as the other, brought the candycorn.  But at the moment I have a contradiction, which means that somewhere I made a logical error.  I'll try to get back to this sometime later - good luck
1111648
200 x 200\n\nor\n\n40, 000 colors