how many derees is in quadrangel?
914112
It's usually called a quadrilateral, and the total of the internal angles of a convex quadrilateral is 360º.\n\nTo figure the sum of the internal angles of any convex polygon, the formula is (n-2)*180º, so a pentagon (five sides) has 540º, a hexagon (six sides) has 720º and so forth.
24421
NONE.Well... how many did u get?
837543
If I have interpreted you correctly, each of the 49 packages must contain either BBB, BBS, BSS, or SSS.\nThen there are:\n22,100 combinations.\n\nThis is the same as the number of solutions of the equation:\nx1 + x2 + x3 + x4 = 49\nin non-negative integers.\n(i.e., x1 is the number of BBB packages, x2 is the number of BBS, x3 the number of BSS, and x4 the number of SSS)\n\nThis is C(49+4-1, 4-1) = C(52,3) = 52*51*50/(3*2*1) = 22,100.\n\n*************************\nADDITIONAL DETAILS:\nOk, I saw that you reposted this question.  However, you are still missing something or have the numbers or the question incorrect.\n\n285 = 3*5*19 .  This number is not a result formable by the combination function: C(a,b) = a!/[(a-b)!b!]\n\nThe closest would be 286 = C(13,3).  This would be the number of ways that 3 items could be chosen from 13 items.\n\nAlso, as you have it written on the other post, my answer above is still correct.  There are 22,100 ways that 4 different types of items could be distributed into 49 packages.  This assumes that the packages are indistinguishable from each other.  If they weren't, i.e., if BBB in package #1 and BBS in package #2 is different from BBS in #1 and BBB in #2, then the number of combinations is MUCH larger.  (It would be 4^49 ~ 3.2 * 10^29)
797156
go to http://www.webspawner.com/users/qrangerx
1017768
the dimansions don't stop at 4 or 5.  according to some there are at least 11 or 13.  Check out the elegant uiverse by Brian Greene.  He'll tell you all about it.
971669
1. 1s2 2s2 2p6 3s1\n2. 1s2 2s2 2p6 \n3. 1s2 2s2 2p4\n4. 1s2 2s2 2p6\n5. 1s2 2s2 2p6 3s2 3p6 3d7 4s2
203508
Two.\n\nYou can rewrite it as x^2 = 15 or x = sqrt(15)\n\nx solves to (rounded) 3.87 or -3.87
807574
what is the space between the 14 and the 105?\n\nOh well, I guess it doesn't matter.  I added the numbers and came up with 153.  306 if you want me to add them twice.
911275
Quin 5\nSex 6\nSept 7\nOct 8\nNon 9\nDec 10
793308
1. sqrt 91a\n2. asqrt75\n3. sqrt15 - 3\n4.-4sqrt2-19
859764
I'll do the first one.  The others are the same\n\nSQRT(250*y^6) = SQRT(250) * SQRT(y^6)\n= SQRT(25)*SQRT(10)*y3\n= 5*SQRT(10)*y^3
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)
931594
1)15,49,55.87,105.\n2)57,59,61,67,69,71,73\n3)6 gps of 3,4 gps of 8,3gps of 10\n4)2*3*5*11\n5)7&4\n6)36
900057
a Q1 since it is between 0 and 67\nb Q2 since it is between 67 and 88\nc Q3 since it is between 88 and 105
720929
Let me just work through the steps to solve the problem:\n\nsqrt(2x^2-1)=x   <---Original Equation\n2x^2-1=x         <---Square both sides\n2x^2-x-1=0       <---Subtract x\n\nNow, apply the quadradic equation:\nx=(-b+/-(sqrt(b^2-4ac))/(2a)\n\nFor this equation, the equation is:\nx=(-(-1)+/-(sqrt((-1^2)-4*2*(-1))/(2*(2))\n\nSimplified it is:\nx=(1+/-sqrt(1+8))/(4)\nx=(1+3)/4 or x=(1-3)/4\n\nSo, x=1 or x=-1/2
236101
permutation?\n\ni'm not sure what you mean by "see how many it is."
509467
1: x<12\n2: x>257\n3: x>42\n4: x>84\n5: x>25\n6: x>179\n7: x<8\n8: x>577\n9: x<113