Can somebody explain left hand sums and right hand sums?
826403
hmmm taking me way back here.  when you break the area under a curve into rectangles of differential width,  either the left top corner will lie on the curve or the right top corner will.  \n\nas you decrease the differential width the area will become more exact until bada bing you are doing calculus.  i think.
943561
remember the order of operations. for this particular problem, division and multiplication come first.  since they are both present in the problem, and neither one is supposed to be performed before or after the other in the order of operations, you should start from left to right, the same way you read.  \n\ntimes the 80 and the 2 together, and then divide it by 5. \n\n now do the addition and subtraction. again, neither subtraction nor addidtion is supposed to be performed before or after each other in the order of operations, so you start from the left to the right. \n\nsimply complete the steps.
740929
Without a little more context I can't say for sure but, most likely, it's just another way of asking for the vector sum. That's the one vector that is equivalent to adding up the directions and magnitudes of all other vectors in the problem.
823906
not quite sure what exactly you mean, but i would guess\n\n2+2+2+2 or 4+4 or 1+1+1+1+1+1+1+1
22859
Subract the two terms on the left giving:\n-2t < 16\n\nAdd 2t to both sides giving\n0 < 16 + 2t\n\nSubtract 16 from both sides\n\n-16 < 2t\n\nDivide both sides by 2\n\n-8 < t
440005
Pseudocode, I'm not up on my Java syntax:\n\nint sum=0;\nfor(int i=0;i<array.size;i++)\n  sum+=array[i];
951662
well the number to the utmost left gets added by 11 and in every row 1 gets added more than the previous row so if u see the difference between any two consecutive row is always 12 !!\nthts wat i thnk
548405
please dont hate math\nshow interest\nthen you do well\n\nFOIL\nFirst-Out-Inner-Last\n\nThis method is used for multiplying two binomials.\nsay (a+b)(x+y)\n\n( a + b ) (x + y)\n\nFirst------first terms in the two binomials. they are to be multiplied. so ax\nthen Outer a and y. so ay\nInner-----bx\nLast-----last terms in the binomials. so 'by'\nthen add all the terms.\nax+ay+bx+by\n\nyou may consult me
724685
r-s = t\nadd s to both sides\nr = t+s
740336
1. 101\n2.  There are 2 ones.  The first and last entries of the row.\n3.  Each entry is the sum of the 2 entries above and to     each side of it.\n4.  If the 2'nd entry in a row is a prime, it divides each of the other entries in the row (except for the last entry in the row which is always a 1)\n\nCheck out http://ptri1.tripod.com/ for all manner of kewl sh|t about Pascal's Triangle.  Also Fibonacci Series and other fun things ☺\n\n\nDoug\n\nEDIT:  Give Peter D the points.  He needs 'em worse than I do ☺
1102516
by any chance is this a question from your maths homework? lol :D\n\nI cant work it out myself because to put it simple, im crap at maths.
414270
Add the two tops and add two bottoms.Then the sum is\n(Top)[1+0 , 3+0] and (Bottom)[2+0 , 6+0] = E\nE is any 2x2 matrix and O is the 2x2 zeromatrix.\nAlways is E+O = O+E = E.\nThis is also true for nxn matrices.
972642
(A^2+2ab+b^2)/(a+b) + (a^2-b^2)/(a-b) + (a-b)/(a+b)/(a^2-b^2)/a+b)\n(a+b)^2/(a+b)+(a+b)(a-b)/(a-b)+ (a-b)/(a+b)*(a+b)/(a^2-b^2)\na+b+a+b+(a-b)/((a+b)(a-b))\n2a+2b+1/(a-b)\n\nif I got all the ( ) in the right places.
245176
u*(u-2)+v*(u-2)\n(u+v)*(u-2)
706696
EDIT:  Left_Field is right, n = no of periods.  Corrections below\n\n\nIt is done by calculus.  Continuous compounding is\n\nA = lim(n->∞) P*(1+r/n)^nt\n\n= lim(n->∞) P*[(1+r/n)^n]^t\n\nlim(n->∞) (1+r/n)^n = e^r, e is the base of the natural logs\n\nA = P*[e^r]t = P*e^rt
201475
Do the math or at least show us the factors to get the sum and then we can decide whether or not it's true.  You can't get a sum without addends right?
1067729
The sum of (-1)^n from 0 to infinite...1-1+1-1+1-1....\nGrouping this sum like this first: (1-1)+(1-1)+(1-1)...=0+0+0+...=0\nOR Use the following grouping: 1+(-1+1)+(-1+1)+...=1+0+0+0...=1\n\nThis is a fun problem, and shows that when using summation, be careful.