How do you divide binomials by binomials?
439000
Since you cannot factor any binomial to form a monomial, the only factoring you can do is by "common monomial factoring". After that, cancel like factors, but if there are no common factors, that means you cannot factor it anymore.\nExample:\n\n(3x - 3)/(2x - 2)\nyou can get common monomial factors for each:\n3(x - 1)/2(x - 1)\ncancel the common factors\n3/2\n\n(4x² - 6x)/(2x + 10)\n= 2x(2x - 3)/2(x + 5)\n= x(2x - 3)/(x + 5)\n\n(x + 3)/(x + 5)\n---> this cannot be simplified anymore\n\n^_^
579225
Division by zero is undefined, even if you divide zero into zero.\n\nhttp://en.wikipedia.org/wiki/Division_by_zero
943885
if you take the numbers one-by-one: 25! / ( (25-1)!*1! )\n\ncombination from n numbers taken by k:  n! / ( (n-k)!*k! )\n\nfor example:\n    -25 numbers, taken by 7: 25! / (18!*7!)\n    -25 numbers, taken by 4: 25!/ (21!*4!)
789066
Firstly, it has to start with either a 1 or a 2, and end with a 2 or a 6. So we just take each case one at a time.\n1..2: that leaves 23556, so there are 5!/2! ways, ie 60 ways.\n1..6: that leaves 22355, so there are 5!/(2!2!) ways, ie 30 ways.\n2..2: that leaves 13556, so there are 5!/2! ways, ie 60 ways.\n2..6: that leaves 12355, so there are 5!/2! ways, ie 60 ways.\n\nNote that in each case there are 5! ways of arranging the other 5 numbers, but we need to divide by 2 if there is a pair of repeated numbers.\nThat gives a total of 210.\n\n(note that both of the other answers given so far are wrong - its not 360, because that person has forgotten they have to be even; and its not 120. They counted 2x2 possibilities for the first and last digits which is correct, but one of those cases already uses both 2s, so you *don't* want to divide by 2 in the end for that particular case (you're not counting them twice are you!)\nTheres really no other way of doing it than splitting them into cases like I did.
978661
Have fun:\nhttp://www.math.nyu.edu/~griffith/teaching/06numanalysis/labs/bisection/
1025267
Here is how:\n\n6+6+3\n6+5+4\n5+5+5\n\nnow change the first two around until you have all the possiblities.
1139003
(n/(n - 3)) + (5/(2n - 6))\n(n/(n - 3)) + (5/(2(n - 3)))\nMultiply everything by 2(n - 3)\n(2n + 5)/(2n - 6)
906623
Sure.  To multiply 3/4 x 1/2, for example, you multiply the numerators (3 x 1) and the denominators (4 x 2) and end up with 3/8.  To divide by 3/4 by 1/2, you invert the multiplier, making it 2/1, and then multiply the numerators (3 x 2) and denominators (4 x 1) and end up with 6/4 or 1 1/2.
791300
Use the FOIL method; First, Outer, Inner, Last.\n\nFirst, n * n\n\nOuter, n * 5\n\nInner, n * 7\n\nLast 7 * 5.\n\nThen add them all together.\n\n22 NOV 06, 0420 hrs, GMT.
258044
divide the exponent by two
5300
(n+k-1) over (k) = (n+k-1) / (k)\n\n=> (n+k-1) / (k) = n/k +1 -1/k
741529
Divide the top # by the bottom #.  Then multiply by 100
554599
4.2\n\nfor a binomial distribution the mean is the number * the probability
395828
haha, but this does not work. you cannot mathematically divide by (a-b) because this quantity is equal to 0. (1-1)=0. Dividing by zero yields an indeterminate answer.
1094108
Use a calculator and raise it to the 1/3 power.\nDon't try to do it by hand...
496936
the first can be shown by induction.\n\nthe 2nd is.. pretty obvious\n\nI couldn't do the 2nd one with Binomial Theorem... btw, what is Binomial theorem? but it is very easy to see 99^50+100^50<101^50\n\nu can write 99^50+100^50<101^50 as 0.99^50+1<1.01^50 and use a calculator, or\n\nTo prove this is true, show 101^50-(99^50)>100^50\n\nWith Binomial expansion\n101^50=(100+1)^50=100^50+50(100^49) +a +b +c +d +...\n99^50=(100-1)^50=100^50-50(100^49) +a -b +c -d +....\n\nwith a,b,c,d,e,f... being some positive real numbers.\n\nu can see that the difference is (50+50)100^49 +2b +2d +2f +... =(100)100^49 +2b +2d +2f +... =100^50 +2b +2d +2f +...> 100^50\n\nI'm not sure if this has anything to do with the Binomial theorem u mentioned.. but.. it has Binomial expansion
795924
Binomial distribution, (a) p=0.5, n=12; (b) p=0.5, n=13.  You can do it!