According to certain laws of mathematics \nif the numerator and the dinominator are the same \nthe answer obtained is always one\nso according to my question what will 0/0 be?
128411
Anything divided by 0 is actually undefined, not infinity or 0. When you think about how that works, it makes sense:\n\nDividing a numerator by smaller numbers makes your number larger. Divide 4 by 2, you get 2. Divide 4 by 1, you get 4. Divide 4 by 0.5, you get 8. And so on until you are dividing by a number so small it's pretty close to 0. The overall number gets larger and goes off into infinity.\n\nBut, if you divide a negative number in the same manner, your number gets smaller. -4 by 2 is -2, -4 by 1 is -4, -4 by 0.5 is -8, etc. As you approach 0, your numbers go off into negative infinity.\n\nYou'll have the same weird effect if you divide by negative numbers - a positive divided by an increasingly smaller negative will yield negative infinity, and a negative divided by an increasingly smaller negative will yield positive infinity.\n\nYou can't have it both ways - dividing by zero yields both a positive and negative infinity! So someone a long time ago decided that since a number can't really have two values, it will be undefined.
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We sometimes see that 0!=1. Why is that? It would seem that we cannot do the multiplications from one "up to" zero. And, if we go backwards (1x0), we get zero.\n\nExperimenting with factorials, we come up with n!=n(n-1)!. For example 17!=17x(16!):\n\n16!=1x2x...x16\n17!=(1x2x...x16)x17\n\nThat equation (n!=n(n-1)!) just dictated to us where to put the parentheses. By making n=1, we can find 0!:\n\n1!=1(0!)\n0!=1\n\nAnd, it turns out that 0!=1 works very well in many situations (in probability, for example).\n\n\n--------------------------------------------------------------------------------\n\nAddendum:\n\nThe above proof that 0!=1 is based upon n!=n(n-1)!, which is in turn based upon the definition of factorial. So, it would seem to be a valid proof. But, 0! cannot be defined directly from the definition of factorial. So, mathematicians like to define 0! as 1, without proving it. So, the proof just amounts to a demonstration that defining 0! as 1 is consistent with the definition of factorial.\n\nI received email saying that my proof that 0!=1 is invalid because 0! is a constant, and you cannot solve for a constant. Wrong. I can find 4! (in a number of ways), and 4! is a constant. I can solve for the square root of 7, probably using my calculator, and it is a constant, too. There are numerous ways of finding pi, and it too is a constant.\n\nn! is also the number of permutations (ways of arranging) exactly n things. It makes sense to say that there is one way to arrange zero things. So again, 0!=1.
420342
Can you factor the numerator?\n\nI guess not, but UMR math major ignores your denominator in the original problem.
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So is this sequence based purely on a mathematical function?\n\nOkay, so there's a 52 in there, could this sequence somehow relate to the calendar?\n\nWhat sort of scientist? I'm a computer scientist and nothing's glaringly obvious about those numbers.\n\n** GOT IT *** 68! Er or Erbium, the sequence spells out Brainteaser based on the atomic weights of B, Ra, In, Te, As, and Er.
483980
it shall terminate if denominator (when numerator and denominator are co-primes) is product of power or 2 and power of 5. The reason is the numerator has to be multiplied by 10^n when n zeroes are added.\n\n10 has factor 2 and 5.\n\nbasically we need to multiply and devide by 10^n.\n2^k5^l shall devide 10^n only when n >= k and n>= l\n\nother wise it shall not teminate\n1/11 = .099999.... no\n1/10 = .1 ( 10 = 2* 5)\n1/8 = .125 (8 = 2^3) so on\n\nthe reason that numerator and denominator are co-prime becase \n1/8 = 3/24 . As 3 devides the numerator so need not devide denominator\n1/25 terminates (1/5*2 = 100/5*2*100 = 4/100)\n1/30 does not as 3 does not divide any power of 10 so .033333...
154830
When V(t)=0 the particle stops.\nIt can have a(t)=0, but still moving (constant speed)
19662
I know it.\nIts one more than the Infinity.\nIf you know the value of infinity then substitute it and you'll get the answer.
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The sin of pi approaches 0 which makes the numerator 0.  One minus the cos of pi approaches 2.  Which makes the denominator 2.  This makes the limit 0/2 or 0 so none of the answers are correct so choice (e).
591523
the first one is 2, the last one is.. doesn't exist.\n\nLet Pn be the n-th prime number. For all interger N, (P1xP2xP3x....xPN)+1 is not divisible by Pn for all n from 1 to N, and hence is another prime number. So, for every discovered prime number, there is at least one other prime ((P1xP2xP3x..xPN)+1) larger than it.\n\n\nregarding your additional comment..\n\nprime number is defined to be is a positive integer that has no positive integer p>1 divisors other than 1 and p itself. So, from definition alone, it has excluded 0 and 1.\n\nActually, 0/0 is not 0. It is undefined. However, u can simply define the discontinuity at 0/0 to be 0 in this case. But even so, this has nothing to do with prime number, from the definition of prime.\n\nI know u find 0 fascinating. In fact, like infinity, it is. It always has a special place in mathematics. A lot of time, when defining mathematical objects, 0 is defined explicitly because it has a special place. It is the point of inflection of +ve and negative, it is a additive identity, A+0=A, and for any number is its multiplicative identity, A0=0...with the exception of infinity(infinity is not a number anyway). For example, the set {0} with its normal + - and x operations is a Ring. And if u define 0/0=0, and adding / operation, it is a Field. But, this is trivial and of not much use. The point is, 0 is indeed special although it is not a prime.
480196
You have probably figured out that no number will make this statement true. But don't forget about zero. \nYou will also find problems where even zero does not work. In those cases, "no solution" is the correct answer.
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Gary, that's a good one, but one mistake...\nyou said that a=b, thus you cannot divide by (a-b) b/c that equals zero, and you cant divide by zero.
283451
If you are approaching zero from the left, then -1 divided by 0 is positive infinity.\nIf you are approaching zero from the right, then -1 divided by 0 is negative infinity.\n-1 divided by 0 is undefined because you get different results when coming in from the right side of 0 and the left side of 0.\nI learned this in Calc 1 (possibly Calc 2).\nThere are different levels (degrees) of infinity.
595151
The trivial zeros are at (-2K+0i), where K is a positive integer. The nontrivial zeros are all the others. It has been shown that the nontrivial zeroes must have a real part between 0 and 1.
312427
The answer is 1 - you're raising a quantity to a power of 0, which always results in 1. The quantity within the parentheses contains no division by zero, so I know it's a real number, too.
195569
1=0 when reference to derivatives in calculus or when in reference to places numbers are from 0 - 9 \nthen they are repeated. etc 1_0 1_1 1_2 .. you get the idea. \n Any way first place is  = 0
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It's true.\nThe answer is indeed y = 0.\nYou can find this by dividing each side by -4.\n-4y = 0\n-4y/-4 = 0/-4\ny = 0\nRemember that zero divided by any nonzero number is always zero.
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n=3: that's any orbital on the third electron level, which is made up of s, p, and d sublevels.  \ns has one orbital\np has three orbitals\nd has five orbitals\nAdd them up-- 1+3+5=9 orbitals\n\nn=3, l=1: The l=1 means that we're talking about the p sublevel, which has three orbitals.  \n\nn=3, l=1, ml=-1: The ml value denotes which one of the three orbitals we're talking about, so this is one orbital. \n\nn=3, l=0, ml=-1: The l=0 means that we're talking about the s sublevel, which only has one orbital.  The ml=-1 tells us that we're talking about one of at least three orbitals, so there is no such orbital as the one described by the three numbers.  The answer is zero orbitals.  \nHint: ml is always greater than or equal to negative l and less than or equal to positive l. If l is 0, ml can only be 0.\n\nThe first person who answered this question doesn't know what he or she is talking about.  ms is what stands for spin, and that isn't in your question, so who knows what the first person is talking about?