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??

* Let H be a group, and let G be the direct product H × H. Then the subgroups

Let w < sub > j </ sub > be the ' price ' ( the rental ) of a certain factor j, let MP < sub > j1 </ sub > and MP < sub > j2 </ sub > be its marginal product in the production of goods 1 and 2, and let p < sub > 1 </ sub > and p < sub > 2 </ sub > be these goods ' prices.

Let X and Y be objects of a category D. The product of X and Y is an object X × Y together with two morphisms

Let u, v be arbitrary vectors in a vector space V over F with an inner product, where F is the field of real or complex numbers.

Let be a list of n linearly independent vectors of some complex vector space with an inner product.

Let K be the direct product

Let K be a field, and let A be a vector space over K equipped with an additional binary operation from A × A to A, denoted here by · ( i. e. if x and y are any two elements of A, x · y is the product of x and y ).

Let M × M be the Cartesian product of M with itself.

Let be the usual scalar product on ℝ < sup > 3 </ sup >.

Let V be a finite-dimensional Euclidean vector space, with the standard Euclidean inner product denoted by.

Let ( a, L ) and ( b, M ) be two Poincaré transformations, and let us denote their group product by ( a, L ).

Let W be a vector space, with an inner product.

Let us denote the expansion of our product by:

Let be a fixed number, and let be the set of pairs of numbers whose product is at least.

Let V be a vector space over a field K. For any nonnegative integer k, we define the k < sup > th </ sup > tensor power of V to be the tensor product of V with itself k times:

Let R be the ring of integers of an algebraic number field K and P a prime ideal of R. For each extension field L of K we can consider the integral closure S of R in L and the ideal PS of S. This may or may not be prime, but assuming is finite it is a product of prime ideals

Let us attempt to find a solution which is not identically zero satisfying the boundary conditions but with the following property: u is a product in which the dependence of u on x, t is separated, that is:

Let Q denote the set of rational numbers, and let d be a square-free integer ( i. e., a product of distinct primes ) other than 1.

Let S a multiplicatively closed subset of R, i. e. for any s and t ∈ S, the product st is also in S. Then the localization of M with respect to S, denoted S < sup >− 1 </ sup > M, is defined to be the following module: as a set, it consists of equivalence classes of pairs ( m, s ), where m ∈ M and s ∈ S. Two such pairs ( m, s ) and ( n, t ) are considered equivalent if there is a third element u of S such that

Let us assume we can express this infinite series as a ( normalized ) product of linear factors given by its roots, just as we do for finite polynomials:

Let be the product of with itself, let be an open tubular neighbourhood of the diagonal in.

Let H be the completion of H < sub > 0 </ sub > with respect to this inner product.

Let a be a root of P, and Q < sub > a, t </ sub > the product of P by the principal part of the Laurent series of f at a.

Suppose that the Fermat equation with exponent &# 8467 ; ≥ 3 had a solution in non-zero integers a, b, c. Let us form the corresponding Frey curve E. It is an elliptic curve and one can show that its discriminant Δ is equal to 16 ( abc )< sup > 2 &# 8467 ;</ sup > and its conductor N is the radical of abc, i. e. the product of all distinct primes dividing abc.