If one considers a version of QCD with N < sub > f </ sub > flavors of massless quarks, then there is a global ( chiral ) flavor symmetry group SU < sub > L </ sub >( N < sub > f </ sub >) × SU < sub > R </ sub >( N < sub > f </ sub >) × U < sub > B </ sub >( 1 ) × U < sub > A </ sub >( 1 ).

If the electrophile is chiral, it typically maintains its chirality, though the S < sub > N </ sub > 2 product's configuration is flipped as compared to that of the original electrophile.

If the substrate under nucleophilic attack is chiral, this can lead, although not necessarily, to an inversion of stereochemistry called a Walden inversion ( the nucleophile attacks the electrophilic carbon center, inverting the tetrahedron, much like an umbrella turning inside out in the wind ).

If so, there will be more ( spontaneously broken ) chiral symmetries and therefore more Goldstone bosons than are eaten by the Higgs mechanism.

If f is any superfield, is always a chiral superfield.

If the organic molecule is chiral, it may react preferentially with a particular enantiomer or diastereomer of substrate.

If all quarks have equal mass, then this chiral symmetry is broken to the vector symmetry of the " diagonal flavour group " which applies the same transformation to both helicities of the quarks.

If a chiral condensate forms, then the chiral symmetry is spontaneously broken into a diagonal subgroup SU ( N ) since the condensate leads to a pairing of the left-handed and the right-handed flavors.

If the migrating carbon is chiral, the stereochemistry is retained.

If we take N = 1 chiral super QCD with N < sub > c </ sub > colors and N < sub > f </ sub > flavors with, then by the Seiberg duality, this theory is dual to a nonabelian gauge theory which is trivial ( i. e. free ) in the infrared limit.

If the mass m is nonzero, the model is massive classically, otherwise it enjoys a chiral symmetry.

If the teacher said that the learner clearly wants to stop, the experimenter replied, " Whether the learner likes it or not, you must go on until he has learned all the word pairs correctly, so please go on ".

If S contains two elements that are not pairwise orthogonal ( in particular, the set of all quantum states includes such pairs ) then an argument like that given above shows that the answer is no.

If a player has two small pairs, and he believes that it will be necessary for him to make a full house to win, then he has four outs: the two remaining cards of each rank that he holds.

If a hand contained one of the tiles on the left and one of the tiles on the right, these would not form a pair at all, since the tiles that make pairs are defined by tradition.

* If a player has no pairs, straights or flushes, he should set the second-and third-highest cards in his two-card hand.

If a player's side cards are small or his larger pair is large, he should split the pairs.

If a player has no side cards higher than a jack, he should always split pairs, even 2s and 3s ( most house ways split if there's a pair of 6s or higher, and split small pairs if there's no ace for the low hand ).

If all three pairs of opposite edges of a tetrahedron are perpendicular, then it is called an orthocentric tetrahedron.

If the position was found to be r < sub > 0 </ sub > then in an interpretation satisfying CFD, the statistical population describing position and momentum would contain all pairs ( r < sub > 0 </ sub >, p ) for every possible momentum value p, whereas an interpretation that rejects counterfactual values completely would only have the pair ( r < sub > 0 </ sub >,⊥) where ⊥ denotes an undefined value.

* SAS ( Side-Angle-Side ): If two pairs of sides of two triangles are equal in length, and the included angles are equal in measurement, then the triangles are congruent.

* SSS ( Side-Side-Side ): If three pairs of sides of two triangles are equal in length, then the triangles are congruent.

* ASA ( Angle-Side-Angle ): If two pairs of angles of two triangles are equal in measurement, and the included sides are equal in length, then the triangles are congruent.

* AAS ( Angle-Angle-Side ): If two pairs of angles of two triangles are equal in measurement, and a pair of corresponding non-included sides are equal in length, then the triangles are congruent.

If the chromosome pairs fail to separate properly during cell division, the egg or sperm may end up with a second copy of one of the chromosomes.

If a binary star happens to orbit in a plane along our line of sight, its components will eclipse and transit each other ; these pairs are called eclipsing binaries, or, as they are detected by their changes in brightness during eclipses and transits, photometric binaries.

* If we fix a homomorphism of abelian groups, we can consider the category consisting of all pairs where is an abelian group and is a group homomorphism with.

If the biproduct A ⊕ B exists for all pairs of objects A and B in the category C, then all finite biproducts exist.

If the two pairs of wings are considered as interchangeable, homologous structures, this may be described as a parallel reduction in the number of wings, but otherwise the two changes are each divergent changes in one pair of wings.

* If every voter's preference between X and Y remains unchanged, then the group's preference between X and Y will also remain unchanged ( even if voters ' preferences between other pairs like X and Z, Y and Z, or Z and W change ).

If the leader starts with a pair, only pairs may be played on top of it.

If two texts in this language are laid side by side, then the following pairs can be expected:

If a Wing falls to fewer than 12 healthy dragon / rider pairs, the Wing is disbanded and absorbed by other Wings.

If an upper bound can be placed on the velocity of the physical bodies in a scene, then pairs of objects can be pruned based on their initial distance and the size of the time step.

If D denotes the differentiation operator and P is the polynomial Af then V is the null space of the operator p (, ), because Af simply says Af.

If F and G are ( covariant ) functors between the categories C and D, then a natural transformation η from F to G associates to every object X in C a morphism in D such that for every morphism in C, we have ; this means that the following diagram is commutative:

If φ is C < sup > k </ sup >, then the inhomogeneous equation is explicitly solvable in any bounded domain D, provided φ is continuous on the closure of D. Indeed, by the Cauchy integral formula,

If this output is then used as the clock signal for a similarly arranged D flip-flop ( remembering to invert the output to the input ), you will get another 1 bit counter that counts half as fast.

If v < sub > n </ sub > is the number of vertices of degree n and D is the maximum degree of any vertex,

If you work as a No. 2 in a New York City D. A.

If one chooses a critical value of the test statistic D < sub > α </ sub > such that P ( D < sub > n </ sub > > D < sub > α </ sub >)

In 2001, David Wiley criticized learning object theory in his paper, The Reusability Paradox which is summarized by D ' Arcy Norman as, If a learning object is useful in a particular context, by definition it is not reusable in a different context.

The " diamond problem " ( sometimes referred to as the " deadly diamond of death ") is an ambiguity that arises when two classes B and C inherit from A, and class D inherits from both B and C. If D calls a method defined in A ( and does not override the method ), and B and C have overridden that method differently, then from which class does it inherit: B, or C?

If ( D ) is the wheel diameter, the torque on the runner is: T = F ( D / 2 ) = ρQD ( V < sub > i </ sub > − u ).

* Matter, William D. If It Takes All Summer: The Battle of Spotsylvania ( 1988 )

If this type of contact utilizes a " make before break " functionality, then it is called a Form D contact.

If D is a derivation at x, then D ( ƒ )

If every object X < sub > i </ sub > of C admits a initial morphism to U, then the assignment and defines a functor V from C to D. The maps φ < sub > i </ sub > then define a natural transformation from 1 < sub > C </ sub > ( the identity functor on C ) to UV.

Similar statements apply to the dual situation of terminal morphisms from U. If such morphisms exist for every X in C one obtains a functor V: C → D which is right-adjoint to U ( so U is left-adjoint to V ).

If the ratio of the two resistances in the known leg is equal to the ratio of the two in the unknown leg, then the voltage between the two midpoints ( B and D ) will be zero and no current will flow through the galvanometer.

If the opponent places a stone at B or D, the remaining hex can be filled to join the original two stones into a single group.