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The solution set for the equations x − y = −1 and 3x + y = 9 is the single point (2, 3). A solution of a linear system is an assignment of values to the variables ,, …, such that each of the equations is satisfied. The set of all possible solutions is called the solution set. [5]
where f (2k−1) is the (2k − 1)th derivative of f and B 2k is the (2k)th Bernoulli number: B 2 = 1 / 6 , B 4 = − + 1 / 30 , and so on. Setting f ( x ) = x , the first derivative of f is 1, and every other term vanishes, so [ 15 ]
Half of the leading zero's neighbor is 2. 93×5=465 Half of 3's neighbor is 0, plus 5 because 3 is odd, is 5. ... Write 4, carry the 1. 3 + half of 5 (2) + 5 (since 3 ...
where 1 Z and 1 S are the units of the tensor product operations (which vary depending on which category of -adic torsion sheaves is under consideration). If S is regular and g : X → S, and if K is an invertible object in the derived category on S with respect to ⊗ L, then define D X to be the functor RHom(—, g! K).
The idea becomes clearer by considering the general series 1 − 2x + 3x 2 − 4x 3 + 5x 4 − 6x 5 + &c. that arises while expanding the expression 1 ⁄ (1+x) 2, which this series is indeed equal to after we set x = 1.
This is the minimum number of characters needed to encode a 32 bit number into 5 printable characters in a process similar to MIME-64 encoding, since 85 5 is only slightly bigger than 2 32. Such method is 6.7% more efficient than MIME-64 which encodes a 24 bit number into 4 printable characters.
The regular skew polyhedron {4,5| 4} can be realized within the 5-cube, with its 32 vertices, 80 edges, and 40 square faces, and the other 40 square faces of the 5-cube become square holes. This polytope is one of 31 uniform 5-polytopes generated from the regular 5-cube or 5-orthoplex .
The pile remaining after each division must contain an integral number of coconuts. If there were only one such division, then it is readily apparent that 5 · 1+1=6 is a solution. In fact any multiple of five plus one is a solution, so a possible general formula is 5 · k – 4, since a multiple of 5 plus 1 is also a multiple of 5 minus 4. So ...