Let-Q-be-a-nonsingular-quadric-surface-bearing-reg

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Let and 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.

Let Q ( x ) denote the number of square-free ( quadratfrei ) integers between 1 and x.

Let Q be P's right child.

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 Q and R be the points of intersection of these two circles.

Let C be a non-singular algebraic curve of genus g over Q.

Let K be the rational number field Q and

Let Q ( H ) be the expected number of values we have to choose before finding the first collision.

Let h 0 be the hour angle when Q becomes positive.

Let K be a field lying between Q and its p-adic completion Q p with respect to the usual non-Archimedean p-adic norm

|| x || p on Q for some prime p. Let R be the subring of K defined by

Let A =( Q A , Σ, Δ A , I A , F A ) and B =( Q B , Σ, Δ B , I B , F B ) be Büchi automata and C =( Q C , Σ, Δ C , I C , F C ) be a finite automaton.

Let and be

Let the open enemy to it be regarded as a Pandora with her box opened ; ;

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 him be now ''!!

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 T be a linear operator on the finite-dimensional vector space V over the field F.

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 Af be the null space of Af.

Let N be a linear operator on the vector space V.

Let T be a linear operator on the finite-dimensional vector space V over the field F.

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 this be denoted by Af.

Let it be granted then that the theological differences in this area between Protestants and Roman Catholics appear to be irreconcilable.

Let not your heart be troubled, neither let it be afraid ''.

The same God who called this world into being when He said: `` Let there be light ''!!

For those who put their trust in Him He still says every day again: `` Let there be light ''!!

Let us therefore put first things first, and make sure of preserving the human race at whatever the temporary price may be ''.

Let her out, let her out -- that would be the solution, wouldn't it??

Let and nonsingular

: Let X be a projective nonsingular variety of dimension n over C and an ample line bundle.

Let be a real vector space equipped with a nonsingular real antisymmetric bilinear form ( i. e. a symplectic vector space ).

Let be a real symplectic vector space with nonsingular symplectic form.

Let be a real-graded vector space equipped with a nonsingular antisymmetric bilinear superform ( i. e. ) such that is real if either or is an even element and imaginary if both of them are odd.

Let and quadric

This is the affine cone over the projective quadric S. Let N + be the future part of the null cone ( with the origin deleted ).

Let and surface

Let the potential energy difference across the surface due to effective surface dipole be.

Let, where denote a point on Boy's surface.

Let M be a smooth manifold of dimension n ; for instance a surface ( in the case n = 2 ) or hypersurface in the Cartesian space R n + 1 .

Let the surface of the sphere be S. The volume of the cone with base area S and height r is, which must equal the volume of the sphere:.

Let us introduce the factor f j that describes how the actual charge density differs from the average and itself on a position on the surface of the j-th conductor:

Let X be a Riemann surface.

Let X ( u, v ) be a parametric surface.

Let us note that the surface integral of this 2-form is the same as the surface integral of the vector field which has as components, and.

Let us notice that we defined the surface integral by using a parametrization of the surface S. We know that a given surface might have several parametrizations.

Let be a quartic surface, and let p be a singular point of this surface.

As he approaches the surface of the sun, he shouts, " Let there be light!

Let be a point on the surface.

Suppose that c is a simple closed curve in a closed, orientable surface S. Let A be a tubular neighborhood of c. Then A is an annulus and so is homeomorphic to the Cartesian product of

Let X be a connected non-compact Riemann surface.

Let X be the infinite cyclic cover of the knot complement of K. This covering can be obtained by cutting the knot complement along a Seifert surface of K and gluing together infinitely many copies of the resulting manifold with boundary in a cyclic manner.

: Let be a compact, closed surface ( not necessarily connected ).

Let us look at the surface of the earth: here the ground is flat ; there it is hilly and mountainous ; in other places it is sandy ; in others it is barren ; and elsewhere it is productive.

Let S be a Riemann surface with a metric whose curvature is bounded above by.

Let us consider a simple example of a body of mass, M, having an acceleration, a, and surface area, A, of the surface upon which the accelerating force is acting.

1.416 seconds.