What is an octet rule?
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You asked another question about  valence electrons, and I'm guessing this question is related.  An octet rule implies that an atom "wants" to have 8 electrons in its outermost shell, and its chemistry is driven by this.  One or two extra electrons implies that the atom wants to give them away and is a metal (or is hydrogen), one or two less than the desired 8 means that the atom wants to pick up one or two and will "accept" electrons from a donor.  So, for example, sodium has 1 electron in its outermost shell, chlorine has 7, so sodium donates the one to chlorine.  Sodium now has a positive charge, and chlorine a negative one, so they stick together and form sodium chloride, or table salt.  \n\nIt's really a wild oversimplification, but works as a first approximation.
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I don't think the concept you are talking about goes by the name of "Simpson's rule", not sure what its called but it should be (MO + ML*4 + MP)/6 or with your example (2+4*4+6)/6=4\n\nSimpson's rule on the other hand has to do with approximating integrals (the area under a curve).  Which unfortunately its been a few years and I haven't retained that information.\n\nAnd maybe I'm wrong and there are 2 Simpson's rules, it's possible...
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The rule is:\n\noutput = input - 3
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I assume you mean, Big-O notation.  It is a way to describe the execution time of an algorithm, given the problem size n, which is usually the number of items.\n\nBig-O basically represents worst-case.  Big-Omega is best-case. Big-Theta is average.\n\nAs a few examples:\n\nBig-O of a nested for loop, is O(n^2) since for each n elements in the outer loop, n elements in the inner loop must be executed.\n\nBig-O of a binary search is O( log_2(n) ).
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wHAT ARE THR RULES...?
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There is a concept called "Rule of ten", which you can find explained in:\n1) your textbook\n2) this same site, by using the Search Box\n3) many other places
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i think the false statement is C. as the degree of any number is always 1. i dont understand the statement b tho :S
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There's a very explicit algorithm which explains\nhow to handle number, operators (including left\nor right precedence), functions, etc. on the \nWikipedia site for RPN (see reference #1).\n\nMostly, it's very straight-forward.  If it's an operand\n(ie. a number), then push it on the stack.  If it's an\noperator, then pop off as many operands are needed and\nevaluate.\n\nBut do look at the full algorithm for the details.\n\nOh yeah, for a full-blown implementation, see the source\ncode to the Unix "dc" (desk calculator) program.  The\nSolaris source is at reference #2.
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Because the las energy lawyer is complete. The octet rule says that the last energy lawyer must be complete or have 8 electrons to be inreactive.
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I think you can do this.  Did you know that octahedron means 8-sided figure?  and the fact that it is regular means that all faces are the same size (area).  So the answer would be 32 / 8 = ?
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EAN Rule states that in forming a complex, ligands are added until the total number of e- in the central metal atom/ion becomes the same as the number of e- in the next inert gas.\n\ni.e. at.no. of the central metal atom+no. of e- lost in forming the complex+no. of e- gained by co-ordination should be equal to the no. of e- in the next inert gas.\n\nCr(CO)6\nat. no. of Cr+no. of e- lost in forming the ion+no. of e- gained (bcoz of CO).\n24+0+6*2=36 \nwhich is the at.no. of the next noble gas(Kr).\n(here the no. of e- lost in forming the ion is 0 bcoz Cr(CO)6 is a neutral complex & CO is also neutral in charge.Each CO group donates 2 e- , making a total of 12e- for 6 CO groups.\nHence it obeys EAN Rule.\n\nsimilarly you can work out for others also.\n\nwell the answer for your ques is Mn(CO)6 & V(CO)6.
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C  H   O\n.6  12   6\n\nWITHOUT DECIMAL POINT
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It depends on the octet rule. If the element fulfills its octet, or has 8 valence electrons(such as the noble gases) then it will be very stable. The farther they are from fulfilling the octet rule, the less stable the element is.
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Yes.  NH3 obeys the octet rule because nitrogen shares two electrons in each of three covalent bonds to hydrogen, and has a free electron pair. 2*3+2=8.
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three principles that govern this process: \n1) the Aufbau (build-up) principle, \n2) the Pauli exclusion principle, \n3) Hund's rule\n\nExclusion principle \nNo more than one electron can have all four quantum numbers the same. What this translates to in terms of our pictures of orbitals is that each orbital can only hold two electrons, one "spin up" and one "spin down". \n\nBuild-up principle \nYou may consider an atom as being "built up" from a naked nucleus by gradually adding to it one electron after another, until all the electrons it will hold have been added. Much as one fills up a container with liquid from the bottom up, so also are the orbitals of an atom filled from the lowest energy orbitals to the highest energy orbitals. \n\nHowever, the three p orbitals of a given shell all occur at the same energy level. So, how are they filled up? Is one of them filled full with the two electrons it can hold first? Or do each of the three orbitals receive one electron apiece before the any single orbital is double occupied. As it turns out, the latter situation occurs. \n\nHund's rule \nThis rule is applicable only for those elements that have d electrons, and so is less important in organic chemistry (though it is important in organometallic chemistry). It says that filled and half-filled shells tend to have additional stability. In some instances, then, for example, the 4s orbitals will be filled before the 3d orbitals.
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Three flops - A, B and C - A is the least significant bit.\nNew values for flops (inputs) are A', B' and C'.\nUp/down input is U - set to 1 for up\n\nA' = not A\nB' = B xor ((U and A) or (not U and not A))\nC' = C xor ((U and A and B) or (not U and not A and not B))
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I believe that the integral of 1/x is equal to the natural log of x + C.  (ln x + C).  Usually you take the antiderivative of the function but there are exceptions to the rule, like 1/x.