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Let Q be P's right child.
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Let and Q
Let Q be a nonsingular quadric surface bearing reguli Af and Af, and let **zg be a Af curve of order K on Q.
* Let Q be a set enclosed between two step regions S and T. A step region is formed from a finite union of adjacent rectangles resting on a common base, i. e. S ⊆ Q ⊆ T. If there is a unique number c such that a ( S ) ≤ c ≤ a ( T ) for all such step regions S and T, then a ( Q )
Let now x ' and y ' be tuples of previously unused variables of the same length as x and y respectively, and let Q be a previously unused relation symbol which takes as many arguments as the sum of lengths of x and y ; we consider the formula
Let the line of symmetry intersect the parabola at point Q, and denote the focus as point F and its distance from point Q as f. Let the perpendicular to the line of symmetry, through the focus, intersect the parabola at a point T. Then ( 1 ) the distance from F to T is 2f, and ( 2 ) a tangent to the parabola at point T intersects the line of symmetry at a 45 ° angle.
Beginning with From Russia with Love in 1963, Llewelyn appeared as Q, the quartermaster of the MI6 gadget lab ( also known as Q branch ), in almost every Bond film until his death ( 17 ), only missing appearances in Live and Let Die in 1973, and Never Say Never Again, the latter of which is not part of the official James Bond film series.
Let K be a field lying between Q and its p-adic completion Q < sub > p </ sub > with respect to the usual non-Archimedean p-adic norm
Let A =( Q < sub > A </ sub >, Σ, Δ < sub > A </ sub >, I < sub > A </ sub >, F < sub > A </ sub >) and B =( Q < sub > B </ sub >, Σ, Δ < sub > B </ sub >, I < sub > B </ sub >, F < sub > B </ sub >) be Büchi automata and C =( Q < sub > C </ sub >, Σ, Δ < sub > C </ sub >, I < sub > C </ sub >, F < sub > C </ sub >) be a finite automaton.
Let and be
Let every policeman and park guard keep his eye on John and Jane Doe, lest one piece of bread be placed undetected and one bird survive.
Let us assume that it would be possible for an enemy to create an aerosol of the causative agent of epidemic typhus ( Rickettsia prowazwki ) over City A and that a large number of cases of typhus fever resulted therefrom.
Let p be the minimal polynomial for T, Af, where the Af, are distinct irreducible monic polynomials over F and the Af are positive integers.
Let V be a finite-dimensional vector space over an algebraically closed field F, e.g., the field of complex numbers.
Let N be a positive integer and let V be the space of all N times continuously differentiable functions F on the real line which satisfy the differential equation Af where Af are some fixed constants.
Let us take a set of circumstances in which I happen to be interested on the legislative side and in which I think every one of us might naturally make such a statement.
Let the state of the stream leaving stage R be denoted by a vector Af and the operating variables of stage R by Af.
Let it be granted then that the theological differences in this area between Protestants and Roman Catholics appear to be irreconcilable.
Let us therefore put first things first, and make sure of preserving the human race at whatever the temporary price may be ''.
Let and right
Let P be the root of the unbalanced subtree, with R and L denoting the right and left children of P respectively.
" Let there arise from you a group of people inviting to what is good, enjoining what is right, and forbidding what is wrong ; these are the ones who will be successful.
On this sign in Russian memorializing an anniversary of the city of Balakhna, the word Balakhna on the right is in the nominative case, while the word Balakhne is in the dative case in Balakhne 500 Let (' Balakhna is 500 years old ') on the front of the sign.
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 >.
Let us assume the bias is V and the barrier width is W. This probability, P, that an electron at z = 0 ( left edge of barrier ) can be found at z = W ( right edge of barrier ) is proportional to the wave function squared,
Let M and N be ( left or right ) modules over the same ring, and let f: M → N be a module homomorphism.
Let there be a right angle ABC, r a line parallel to BC passing by A and s a line parallel to AB passing by C. Let D be the point of intersection of lines r and s ( Note that it has not been proven that D lies on the circle )
Let us very roughly consider the line between the " leftmost " and " rightmost " two points of the k selected points ( for some arbitrary left / right axis: we can choose top and bottom for the exceptional vertical case ).
Let us see how he may free himself from this enslavement and achieve an harmonious inner integration, true Self-realization, and right relationships with others.
Let R < sub > h </ sub > denote the ( right ) action of h ∈ H on P. The derivative of this action defines a vertical vector field on P for each element ξ of: if h ( t ) is a 1-parameter subgroup with h ( 0 )= e ( the identity element ) and h '( 0 )= ξ, then the corresponding vertical vector field is
Let P be the underlying principal homogeneous space of G. A Klein geometry is the homogeneous space given by the quotient P / H of P by the right action of H. There is a right H-action on the fibres of the canonical projection
0.252 seconds.