Then, p < sup > 2 </ sup > is the fraction of the population homozygous for the first allele, 2pq is the fraction of heterozygotes, and q < sup > 2 </ sup > is the fraction homozygous for the alternative allele.

Then each point p of the line can be specified by its distance from O, taken with a + or − sign depending on which half-line contains p.

Then b < sub > 0 </ sub > is the value of p ( x < sub > 0 </ sub >).

#* Proof: suppose that p is composite, hence can be written with a and Then is prime, but and contradicting statement 1.

Then p < sub > 1 </ sub >= MU < sub > 1 </ sub >, p < sub > 2 </ sub >= MU < sub > 2 </ sub >.

Then the first case of Fermat's Last Theorem holds true for p.

Then the Zariski tangent space at a point p ∈ X is the collection of K-derivations D: O < sub > X, p </ sub >→ K, where K is the ground field and O < sub > X, p </ sub > is the stalk of O < sub > X </ sub > at p.

# 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, depending on whether a laptop or compact or extended keyboard type is used, the shortcut is ++ 5 or + 5 ( numeric keypad ) or ++ i ( laptop ).

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.

For about three decades immediately before 1902 it was negative, reaching − 6. 64 s. Then it increased to + 63. 83 s at 2000.

Then simply add 17. 5 + 18 which equals 35. 5 MPH ( 57KPH ).

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 ).

Then we have 3 − (− 2 ) = 3 + 2 = 5.

Then its negation ¬ φ, together with the field axioms and the infinite sequence of sentences 1 + 1 ≠ 0, 1 + 1 + 1 ≠ 0, …, is not satisfiable ( because there is no field of characteristic 0 in which ¬ φ holds, and the infinite sequence of sentences ensures any model would be a field of characteristic 0 ).

Then f + g is in L < sup > p </ sup >( S ), and we have the triangle inequality

Then with Black to move, both 3 ... Kxf7 and 3 ... Bxf2 + are not possible.

We can use this fact to prove part of a famous result: for any prime p such that p ≡ 1 ( mod 4 ) the number (− 1 ) is a square ( quadratic residue ) mod p. For suppose p = 4k + 1 for some integer k. Then we can take m = 2k above, and we conclude that

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

Then letting ψ < sub > β </ sub >( λ ) be the λ < sup > th </ sup > β-inaccessible cardinal, the fixed points of ψ < sub > β </ sub > are the ( β + 1 )- inaccessible cardinals ( the values ψ < sub > β + 1 </ sub >( λ )).

Then the term Ax < sup > α </ sup > y < sup > β </ sup > is approximately Dx < sup > α + βp / q </ sup >.

Then the tautological projection R < sup > n + 1, 1 </ sup > −

Then, for all coefficients λ + ( 1 − λ ) =

Then, if a and b are two periods of f such that < sup > a </ sup >⁄< sub > b </ sub > is not real, consider the parallelogram P whose vertices are 0, a, b and a + b. Then the image of f is equal to f ( P ).

Then, criticality occurs when ν · q = 1.

Then express it in lowest possible terms ( i. e., as a fully reduced fraction ) as < sup > m </ sup >⁄< sub > n </ sub > for natural numbers m and n, and let q be the largest integer no greater than √ k.

Then it is also a bounded operator from L < sup > r </ sup > to L < sup > r </ sup > for any r between p and q.

Then for any r between p and q we have that F is dense in, that Tƒ is in for any ƒ in F and that T is bounded in the norm.

Then T is also a bounded operator from to for any r between p and q.

Then the following interpolation inequality holds for all r between p and q and all f ∈ L < sup > r </ sup >:

Then locally, one may choose canonical coordinates ( q < sup > 1 </ sup >, ..., q < sup > n </ sup >, p < sub > 1 </ sub >, ..., p < sub > n </ sub >) on M, in which the symplectic form is expressed as

Then a curve γ ( t )=( q ( t ), p ( t )) is an integral curve of the Hamiltonian vector field X < sub > H </ sub > if and only if it is a solution of the Hamilton's equations:

Then a Cassini oval with foci q < sub > 1 </ sub > and q < sub > 2 </ sub > is defined to be the locus of points p so that the product of the distance from p to q < sub > 1 </ sub > and the distance from p to q < sub > 2 </ sub > is b < sup > 2 </ sup >.

Then in the above formula q refers to the number of automobiles produced, z < sub > 1 </ sub > refers to the number of tires used, and z < sub > 2 </ sub > refers to the number of steering wheels used.

Then, unless, then number q has a parent in the Stern – Brocot tree given by the continued fraction expression

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 >.

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 =

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 c ∈ b, f ( c ) = u.

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.

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 ).