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Then and element
Then our choice function can choose the least element of every set under our unusual ordering.
Then is a group whose identity element is The group inverse of an arbitrary group element is the function inverse
Then an element e of S is called a left identity if e * a = a for all a in S, and a right identity if a * e = a for all a in S. If e is both a left identity and a right identity, then it is called a two-sided identity, or simply an identity.
* Let the index set I of an inverse system ( X < sub > i </ sub >, f < sub > ij </ sub >) have a greatest element m. Then the natural projection π < sub > m </ sub >: X → X < sub > m </ sub > is an isomorphism.
Then, after the pump energy stored in the laser medium has approached the maximum possible level, the introduced loss mechanism ( often an electro-or acousto-optical element ) is rapidly removed ( or that occurs by itself in a passive device ), allowing lasing to begin which rapidly obtains the stored energy in the gain medium.
For f a real polynomial in x, and for any a in such an algebra define f ( a ) to be the element of the algebra resulting from the obvious substitution of a into f. Then for any two such polynomials f and g, we have that ( fg ) ( a )
Then, for any non-zero element x of M, the cyclic submodule xR must equal M. Fix such an x.
Then, in 1846, the element ilmenium was claimed to have been discovered, but later was determined to be impure niobium.
Then in 1908, the Japanese chemist Masataka Ogawa found evidence in the mineral thorianite, which he thought indicated the presence of element 43.
Then each element in G is represented in some way by a product
Suppose a partially ordered set P has the property that every chain ( i. e. totally ordered subset ) has an upper bound in P. Then the set P contains at least one maximal element.
Suppose a non-empty partially ordered set P has the property that every non-empty chain has an upper bound in P. Then the set P contains at least one maximal element.
Then p + q = ( p < sub > 1 </ sub >+ q < sub > 1 </ sub >, p < sub > 2 </ sub >+ q < sub > 2 </ sub >); since p < sub > 1 </ sub > = p < sub > 2 </ sub > and q < sub > 1 </ sub > = q < sub > 2 </ sub >, then p < sub > 1 </ sub > + q < sub > 1 </ sub > = p < sub > 2 </ sub > + q < sub > 2 </ sub >, so p + q is an element of W.
# Let p = ( p < sub > 1 </ sub >, p < sub > 2 </ sub >) be an element of W, that is, a point in the plane such that p < sub > 1 </ sub > = p < sub > 2 </ sub >, and let c be a scalar in R. Then cp = ( cp < sub > 1 </ sub >, cp < sub > 2 </ sub >); since p < sub > 1 </ sub > = p < sub > 2 </ sub >, then cp < sub > 1 </ sub > = cp < sub > 2 </ sub >, so cp is an element of W.
Then for each x in I, there is a base element B < sub > 3 </ sub > containing x and contained in I.
Then, for example, (−∞, 1 ) and ( 0, ∞) would be in the topology generated by S, being unions of a single base element, and so their intersection ( 0, 1 ) would be as well.
Then if we intersect each element of B with Y, the resulting collection of sets is a base for the subspace Y.
" Then with her right hand she would sprinkle water at the head of the child and say, " Behold this element without whose assistance no mortal being can survive.
Then the union ( written as +) and the concatenation ( written as ·) of two elements of A again belong to A, and so does the Kleene star operation applied to any element of A.
Let B be a complex Banach algebra containing a unit e. Then we define the spectrum σ ( x ) ( or more explicitly σ < sub > B </ sub >( x )) of an element x of B to be the set of those complex numbers λ for which λe − x is not invertible in B.
Then the film was rewound, and everything except the foreground element matted out so that the foreground element would now photograph in the previously blacked out area.

Then and c
which is the length of chain whose weight is equal in magnitude to the tension at c. Then
Let a, b, and c be elements of G. Then:
Let t and s ( t > s ) be the sides of the two inscribed squares in a right triangle with hypotenuse c. Then s < sup > 2 </ sup > equals half the harmonic mean of c < sup > 2 </ sup > and t < sup > 2 </ sup >.
It is frequently stated in the following equivalent form: Suppose that is continuous and that u is a real number satisfying or Then for some cb, f ( c ) = u.
Consider some set P and a binary relation ≤ on P. Thenis a preorder, or quasiorder, if it is reflexive and transitive, i. e., for all a, b and c in P, we have that:
Then Fischer and Rabin ( 1974 ) proved that any decision algorithm for Presburger arithmetic has a worst-case runtime of at least, for some constant c > 0.
Then, they were ranked to the east, when they were buried in the 5th and later to the beginning of the 6th c. We can notice a strong Anglo-Saxon influence in the middle of the period, that disappears later.
Then the n-th series coefficient c < sub > n </ sub > is given by:
Then the Alexandroff extension of X is a certain compact space X * together with an open embedding c: X → X * such that the complement of X in X * consists of a single point, typically denoted ∞.
Then there is an exact sequence relating the kernels and cokernels of a, b, and c:
Let be the mean of the values in associated with class c, and let be the variance of the values in associated with class c. Then, the probability of some value given a class,, can be computed by plugging into the equation for a Normal distribution parameterized by and.
Then the area of the parallelogram with vertices at a, b and c is equivalent to the absolute value of the determinant of a matrix built using a, b and c as rows with the last column padded using ones as follows:
* Then t ( n ( c )) =
Abstractly, we can say that D is a linear transformation from some vector space V to another one, W. We know that D ( c ) = 0 for any constant function c. We can by general theory ( mean value theorem ) identify the subspace C of V, consisting of all constant functions as the whole kernel of D. Then by linear algebra we can establish that D < sup >− 1 </ sup > is a well-defined linear transformation that is bijective on Im D and takes values in V / C.
Denote the orthocenter of triangle ABC as H, denote the sidelengths as a, b, and c, and denote the circumradius of the triangle as R. Then
Suppose a vertex joins three units with spin numbers a, b, and c. Then, these requirements are stated as:
Then c # = 2 3 5 = 30.
Then the brothers were in The Stompers later called The Heartbeats formed in c 1957 with Geoff on rhythm guitar ; Dave on lead guitar ; Denny Driscoll on lead vocals ; Johnny Stark on drums and Ton Edwards on bass.
Then 0 < sup >#</ sup > is defined to be the set of Gödel numbers of the true sentences about the constructible universe, with c < sub > i </ sub > interpreted as the uncountable cardinal ℵ < sub > i </ sub >.

Then and =
Then we are told that P1 is not sped up, so S1 = 1, while P2 is sped up 5 ×, P3 is sped up 20 ×, and P4 is sped up 1. 6 ×.
Then I < sub > x </ sub > and I < sub > x </ sub >< sup > 2 </ sup > are real vector spaces and the cotangent space is defined as the quotient space T < sub > x </ sub >< sup >*</ sup > M = I < sub > x </ sub > / I < sub > x </ sub >< sup > 2 </ sup >.
Then f = u + iv is complex-differentiable at that point if and only if the partial derivatives of u and v satisfy the Cauchy – Riemann equations ( 1a ) and ( 1b ) at that point.
Indeed, following, suppose ƒ is a complex function defined in an open set Ω ⊂ C. Then, writing for every z ∈ Ω, one can also regard Ω as an open subset of R < sup > 2 </ sup >, and ƒ as a function of two real variables x and y, which maps Ω ⊂ R < sup > 2 </ sup > to C. We consider the Cauchy – Riemann equations at z = 0 assuming ƒ ( z ) = 0, just for notational simplicity – the proof is identical in general case.
Then, considering all three colour channels, and assuming that the colour channels are expressed in a γ = 1 colour space ( that is to say, the measured values are proportional to light intensity ), we have:
Then G is a group under composition, meaning that ∀ x ∈ A ∀ g ∈ G ( = ), because G satisfies the following four conditions:
Then setting x =
Then letting y < sub > k </ sub > =
Then Φ ( x ) = φ.
Then by definition, torque τ = r × F.
Then the map T admits one and only one fixed-point x < sup >*</ sup > in X ( this means T ( x < sup >*</ sup >) = x < sup >*</ sup >).
Then a fuzzy subset s: S of a set S is recursively enumerable if a recursive map h: S × N Ü exists such that, for every x in S, the function h ( x, n ) is increasing with respect to n and s ( x ) = lim h ( x, n ).
Then u + v = ( u < sub > 1 </ sub >+ v < sub > 1 </ sub >, u < sub > 2 </ sub >+ v < sub > 2 </ sub >, 0 + 0 ) = ( u < sub > 1 </ sub >+ v < sub > 1 </ sub >, u < sub > 2 </ sub >+ v < sub > 2 </ sub >, 0 ).

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