Then, to conclude on an indeterminate note: `` Nevertheless, if fallout increased substantially, or remained high for a long time, it would become far more important as a potential health hazard in this country and throughout the world ''.

Then, too, the utmost clinical flexibility is necessary in judiciously combining carefully timed family-oriented home visits, single and group office interviews, and appropriate telephone follow-up calls, if the worker is to be genuinely accessible and if the predicted unhealthy outcome is to be actually averted in accordance with the principles of preventive intervention.

Then if you change jobs you won't necessarily have to sell the farm.

Then, if the middle number is activated to its greatest potential in terms of this square, through multiplying it by the highest number, 9 ( which is the square of the base number ), the result is 45 ; ;

Then it will be a `` fearful thing to fall into the hands of the living God '' if you have abused Him in your hands.

Then he fled, not waiting to see if she minded him or took notice of his cry.

Then, if Myra does nothing about fetching her, I'll pack her right back to her mother -- if I have to take her myself ''!!

Then, a little later, Shilkret discovered there was no one to play the brief celesta solo during the slow section, so he hastily asked Gershwin if he might play the solo ; Gershwin said he could and so he briefly participated in the actual recording.

Then another priesthood holder joins in, if available, and pronounces a " sealing " of the anointing and other words of blessing, as he feels inspired.

Then he proves the antithesis, that time has no beginning, by showing that if time had a beginning, then there must have been " empty time " out of which time arose.

Then X is reflexive if and only if each X < sub > j </ sub > is reflexive.

Then X is separable if and only if X ′ is separable.

Then g ( E ) is called the density of states if the energy spectrum is continuous.

Then X is compact if and only if X is a complete lattice ( i. e. all subsets have suprema and infima ).

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.

Then Goursat's theorem asserts that ƒ is analytic in an open complex domain Ω if and only if it satisfies the Cauchy – Riemann equation in the domain.

Then, if these " digital modules " were able to build a self-sustaining business, the company would be free to use them to develop a complete computer in their Phase II.

Then the Cartesian product set D < sub > 1 </ sub > D < sub > 2 </ sub > can be made into a directed set by defining ( n < sub > 1 </ sub >, n < sub > 2 </ sub >) ≤ ( m < sub > 1 </ sub >, m < sub > 2 </ sub >) if and only if n < sub > 1 </ sub > ≤ m < sub > 1 </ sub > and n < sub > 2 </ sub > ≤ m < sub > 2 </ sub >.

Then the overall runtime is O ( n < sup > 2 </ sup >).

Then the periodic Bernoulli functions P < sub > n </ sub > are defined as

Then, in terms of P < sub > n </ sub >( x ), the remainder

Then, using the periodic Bernoulli function P < sub > n </ sub > defined above and repeating the argument on the interval, one can obtain an expression of ƒ ( 1 ).

For instance, suppose that each input is an integer z in the range 0 to N − 1, and the output must be an integer h in the range 0 to n − 1, where N is much larger than n. Then the hash function could be h

The algorithm for deciding this is conceptually simple: it constructs ( the description of ) a new program t taking an argument n which ( 1 ) first executes program a on input i ( both a and i being hard-coded into the definition of t ), and ( 2 ) then returns the square of n. If a ( i ) runs forever, then t will never get to step ( 2 ), regardless of n. Then clearly, t is a function for computing squares if and only if step ( 1 ) terminates.

Then, we can show that if the game starts with n spots, it will end in no more than 3n − 1 moves and no fewer than 2n moves.

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 the statistician must analyze the properties of and, which are viewed as random vectors since a randomly different selection of n cases to observe would have resulted in different values for them.

Then for arbitrary ε > 0 there is an embedding ( or immersion ) ƒ < sub > ε </ sub >: M < sup > m </ sup > → R < sup > n </ sup > which is

Then the n-th series coefficient c < sub > n </ sub > is given by:

Let K be a closed subset of a compact set T in R < sup > n </ sup > and let C < sub > K </ sub > be an open cover of K. Then

Then, using a simple greedy algorithm, the easy knapsack can be solved using O ( n ) arithmetic operations, which decrypts the message.

Then n is palindromic if and only if a < sub > i </ sub > = a < sub > k − i </ sub > for all i. Zero is written 0 in any base and is also palindromic by definition.

* Then t ( n ( c )) =

Then Y = u ( X < sub > 1 </ sub >, X < sub > 2 </ sub >, ..., X < sub > n </ sub >) is a sufficient statistic for θ if and only if, for some function H,

Then N < sub > x </ sub > is a directed set, where the direction is given by reverse inclusion, so that S ≥ T if and only if S is contained in T. For S in N < sub > x </ sub >, let x < sub > S </ sub > be a point in S. Then ( x < sub > S </ sub >) is a net.

Then p < sub > k </ sub > converges monotonically to π ; with p < sub > k </ sub >-π ≈ 10 < sup >− 2 < sup > k + 1 </ sup ></ sup > for k ≥ 2. s

For every r ≥ 0, let n ( r, f ) be the number of poles, counting multiplicity, of the meromorphic function f in the disc | z | ≤ r. Then define the Nevanlinna counting function by

Then the map Tor < sub > i </ sub >< sup > A </ sup >( W, R ) → Tor < sub > i </ sub >< sup > A </ sup >( W, S ) is zero for all i ≥ 1.

In symbols, let A be a Noetherian local ring with maximal ideal m, and suppose a < sub > 1 </ sub >, ..., a < sub > n </ sub > is a minimal set of generators of m. Then in general n ≥ dim A, and A is defined to be regular if n = dim A.

Then there is a constant C depending only on b such that ( g < sub > b </ sub >( n ))< sub > n ≥ 1 </ sub > satisfies

Then for i ≥ 0 define

Examples of non-closed subgroups are plentiful ; for example take G to be a torus of dimension ≥ 2, and let H be a one-parameter subgroup of irrational slope, i. e. one that winds around in G. Then there is a Lie group homomorphism φ: R → G with H as its image.

Then the Birkhoff interpolation problem with k = 2 has a unique solution if and only if S < sub > m </ sub > ≥ m for all m. showed that this is a necessary condition for all values of k.

* Let R be a local ring and M a finitely generated module over R. Then the projective dimension of M over R is equal to the length of every minimal free resolution of M. Moreover, the projective dimension is equal to the global dimension of M, which is by definition the smallest integer i ≥ 0 such that

Then there exists an integer p ≥ 1 depending only on L such that every string w in L of length at least p ( p is called the " pumping length ") can be written as w

Then for n ≥ 1

Then by the triangle inequality the polygon perimeter > AB + AX + XB = AB + AC + CX + XB ≥ AB + AC + BC.