# Pq Sentence Examples

pq
• The number of partitions of a biweight pq into exactly i biparts is given (after Euler) by the coefficient of a, z xPy Q in the expansion of the generating function 1 - ax.

• From the law of angular motion of the latter its radius vector will run ahead of PQ near A, PQ will overtake and pass it at apocentre, and the two will again coincide at pericentre when the revolution is completed.

• Recalling the formulae above which connect s P4 and a m, we see that dP4 and Dp q are in co-relation with these quantities respectively, and may be said to be operations which correspond to the partitions (pq), (10 P 01 4) respectively.

• Since dp4+(-)P+T1(p +q qi 1)!dd4, the solutions of the partial differential equation d P4 =o are the single bipart forms, omitting s P4, and we have seen that the solutions of p4 = o are those monomial functions in which the part pq is absent.

• B is the point at which the effect is required, distant a+b from 0, so that AB= b, AP=s, PQ ds.

• Generally if S denotes any closed surface, fixed in the fluid, M the mass of the fluid inside it at any time t, and 0 the angle which the outward-drawn normal makes with the velocity q at that point, dM/dt = rate of increase of fluid inside the surface, (I) =flux across the surface into the interior _ - f f pq cos OdS, the integral equation of continuity.

• Let PQ be the ordinate of the point P FIG.

• Then the locus of the intersection of PQ and OM is the quadratrix of Dinostratus.

• Let ON(= OP cos 0) be the perpendicular on PQ.

• Since the conditions in the region PQ remain always the same, the momentum perpendicular to AB entering the region at Q is equal to the momentum perpendicular to AB leaving the region at P. But, since the motion at Q is along AB, there is no momentum there perpendicular to AB.

• Again if PQ be any segment of the beam which is free from load, Q lying to the right of P, we find FPFQ, MPMQ=--F.PQ; (12)

• If PQ be a short segment containing an isolated load W, we have FeFi.=W, MQ=MP; (3) hence F is discontinuous at a p concentrated load, diminishing by an amount equal to the load as we pass the loaded point to the right, whilst M is continuous.

• The plane in question is called the null-plane of P. If the null-plane of P pass through Q, the null-plane of will pass through P, since PQ is a null-line.

• It is often convenient to resolve the forces on an element PQ

• In the case of a chain hanging freely under gravity it is usually convenient to formulate the conditions of equilibrium of a finite portion PQ.