Let exactly 1'' '' of `` A '' extend beyond `` B '' and use a square to check your angle to exactly 90 degrees.

: Theorem on projections: Let the function f: X → B be such that a ~ b → f ( a )

Gloria Gaynor ( born September 7, 1949 ) is an American singer, best known for the disco era hits ; " I Will Survive " ( Hot 100 number 1, 1979 ), " Never Can Say Goodbye " ( Hot 100 number 9, 1974 ), " Let Me Know ( I Have a Right )" ( Hot 100 number 42, 1980 ) and " I Am What I Am " ( R & B number 82, 1983 ).

Let a trapezoid have vertices A, B, C, and D in sequence and have parallel sides AB and CD.

Let us call the particles A and B.

Let the input power to a device be a force F < sub > A </ sub > acting on a point that moves with velocity v < sub > A </ sub > and the output power be a force F < sub > B </ sub > acts on a point that moves with velocity v < sub > B </ sub >.

Let us take a hypothetical single scan line, with B representing a black pixel and W representing white:

Let A, B, C and D be the hexes that make up a rhombus, with A and C being the non-touching pair.

# Let e1 be an edge that is in A but not in B.

Let us call them A, B and C.

Other chart hits by White included " Never, Never Gonna Give Ya Up " (# 2 R & B, # 7 Pop in 1973 ), " Can't Get Enough of Your Love, Babe " (# 1 Pop and R & B in 1974 ), " You're the First, the Last, My Everything " (# 1 R & B, # 2 Pop in 1974 ), " What Am I Gonna Do with You " (# 1 R & B, # 8 Pop in 1975 ), " Let the Music Play " (# 4 R & B in 1976 ), " It's Ecstasy When You Lay Down Next to Me " (# 1 R & B, # 4 Pop in 1977 ) and " Your Sweetness is My Weakness " (# 2 R & B in 1978 ).

# Let B < sub > 1 </ sub >, B < sub > 2 </ sub > be base elements and let I be their intersection.

* Let B be a base for X and let Y be a subspace of X.

Let M be the intersection of all subgroups of the free Burnside group B ( m, n ) which have finite index, then M is a normal subgroup of B ( m, n ) ( otherwise, there exists a subgroup g < sup >-1 </ sup > Mg with finite index containing elements not in M ).

Let the two horses be horse A and horse B.

Let B contains all the sentences of A except ¬ φ.

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 Q be a nonsingular quadric surface bearing reguli Af and Af, and let **zg be a Af curve of order K on Q.

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??

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

' Let Me Go, Love '... McDonald's entrancing vocal presence ... so overshadows Larson's that she seems to be playing second fiddle rather than sharing the lead.

Let G be a group with identity element e, N a normal subgroup of G ( i. e., N ◁ G ) and H a subgroup of G. The following statements are equivalent:

Let M be a monoid with identity element e and let A be the set of all subsets of M. For two such subsets S and T, let S + T be the union of S and T and set ST =

Let G be a finite simple undirected graph with edge set E. The power set of E becomes a Z < sub > 2 </ sub >- vector space if we take the symmetric difference as addition, identity function as negation, and empty set as zero.

Let be a Lie group and be its Lie algebra ( thought of as the tangent space to the identity element of ).

* Let be a Lie group homomorphism and let be its derivative at the identity.

Let G be a Lie group and let be its Lie algebra ( which we identify with T < sub > e </ sub > G, the tangent space to the identity element in G ).

Let, where is the identity matrix.

Let be the identity morphism on and set.

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 M be an algebra of bounded operators on a Hilbert space H, containing the identity operator and closed under taking adjoints.

Let A be a superalgebra over a commutative ring K. The submodule A < sub > 0 </ sub >, consisting of all even elements, is closed under multiplication and contains the identity of A and therefore forms a subalgebra of A, naturally called the even subalgebra.

" Following the success of his first two hits, he found it hard to maintain an identity, as most of his songs were later covered by other Motown artists, most prominently " Everybody Needs Love ," a hit when covered by Gladys Knight & The Pips, " Maria ( You Were The Only One )," a hit for Michael Jackson and " If You Let Me ," a minor hit for Eddie Kendricks.

Let S be a countable semigroup ( in which we denote the operation by juxtaposition ) with identity e and with an involution *

Let f be the map from S to itself which is the identity outside of A and inside A we have

Let x be a normal element of a C *- algebra A with an identity element e ; then there is a unique mapping π: f → f ( x ) defined for f a continuous function on the spectrum Sp ( x ) of x such that π is a unit-preserving morphism of C *- algebras such that π ( 1 )

This identity is derived from the divergence theorem applied to the vector field: Let φ and ψ be scalar functions defined on some region U in R < sup > 3 </ sup >, and suppose that φ is twice continuously differentiable, and ψ is once continuously differentiable.

Let Diff ( M ) be the orientation-preserving diffeomorphism group of M ( only the identity component of mappings homotopic to the identity diffeomorphism if you wish ) and Diff < sub > x </ sub >< sup > 1 </ sup >( M ) the stabilizer of x.

* Let K < sup > a </ sup > be an algebraic closure of K containing L. Every embedding σ of L in K < sup > a </ sup > which restricts to the identity on K, satisfies σ ( L ) = L. In other words, σ is an automorphism of L over K.

Let be the tangent space of a Lie group at the identity ( its Lie algebra ).

The idea that all things were created “ by mine Only Begotten ” ( i. e., Jesus Christ, in his premortal state ) is made clear, as is the Son ’ s identity as the co-creator at the time when God said “ Let us make man .” Otherwise, the structure and basic premises of the Genesis account of the Creation were left intact.

Let denote the identity element of.

Let G be a covering group of H. The kernel K of the covering homomorphism is just the fiber over the identity in H and is a discrete normal subgroup of G. The kernel K is closed in G if and only if G is Hausdorff ( and if and only if H is Hausdorff ).

Let H be a topological group and let G be a covering space of H. If G and H are both path-connected and locally path-connected, then for any choice of element e * in the fiber over e ∈ H, there exists a unique topological group structure on G, with e * as the identity, for which the covering map p: G → H is a homomorphism.

Let PH be the path group of H. That is, PH is the space of paths in H based at the identity together with the compact-open topology.