# PQ Sentence Examples

- 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. - 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. - For write (
**pq**) =sï¿½ and take logarithms of both sides of the fundamental relation; we obtain slox +soot' = + (3ly) S20x 2 +2siixy+s02y 2 = E(aix+(3 ly) 2, &C., and siox+SOly - (S 20 x2 + 2s ii x y+ s ooy 2) +... - From the above D p4 is an operator of order
**pq**, but it is convenient for some purposes to obtain its expression in the form of a number of terms, each of which denotes**pq**successive linear operations: to accomplish this write d ars and note the general result exp (mlodlo+moldol +... - 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. - The i t " power of that appertaining to a x and b x multiplied by the j t " power of that appertaining to a x and c x multiplied by &c. If any two of the linear forms, say p x, qx, be supposed identical, any symbolic expression involving the factor (
**pq**) is zero. - Calling the discriminate D, the solution of the quadratic as =o is given by the formula a: = o (a0+a12_x2 (a0x+aix2 If the form a 2 be written as the product of its linear factors p.a., the discriminant takes the form -2(
**pq**) 2. - Hence, from the identity ax (
**pq**) = px (aq) -qx (ap), we obtain (pet' = (aq) 5px - 5 (ap) (aq) 4 pxg x - (ap) 5 gi, the required canonical form. - We can replace pr or -pr by (+p)r or (- p)r, subject to the conditions that (+p) (+q) = (-p) (- q) = (+
**pq**), (+p) (-q) = (- p) (+ q) = (-**pq**), and that + (-s) means that s is to be subtracted. - In order to find the difference of optical distances between the courses QAQ', QPQ', we have to express QP-QA,
**PQ'-AQ**'. - B is the point at which the effect is required, distant a+b from 0, so that AB= b, AP=s,
**PQ**ds. - Taking as the standard phase that of the secondary wave from A, we may represent the effect of
**PQ**by cos 27r (_) .ds, where, l = BP - AP is the retardation at B of the wave from P relatively to that from A. - 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. - On the circle, and let M be another point on the circle so related to P that the ordinate
**PQ**moves from A to 0 in the same time as the vector OM describes a quadrant. - Then the locus of the intersection of
**PQ**and OM is the quadratrix of Dinostratus. - Newton m m _ _ there states that (p+
**pq**) - = p n -}- T naq ?- m 2n n bq -? - 5, the volume of the element between these planes, when h is very small, is approximately h X AB X arc
**PQ**= h. - Y where K-=4, X qth moment with regard to plane y =o, Lm yn X pth moment with regard to plane x =o, and R is the volume of a briquette whose ordinate at (x,.,y s) is found by multiplying by
**pq**x r P - 1 ys 4-1 the volume of that portion of the original briquette which lies between the planes x =xo, y =yo, y = ys. - 12 represent a horizontal section of the dome through the source P. Let OPA be the radius through P. Let
**PQ**represent a ray of sound making the angle B with the tangent at A. - 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. - Comparing this equation with ux 2 +vy 2 +w2 2 +22G'y2+2v'zx+2W'xy=0, we obtain as the condition for the general equation of the second degree to represent a circle :- (v+w-2u')Ia 2 = (w +u -2v')/b2 = (u+v-2w')lc2 In tangential q, r) co-ordinates the inscribed circle has for its equation(s - a)qr+ (s - b)rp+ (s - c)
**pq**= o, s being equal to 1(a +b +c); an alternative form is qr cot zA+rp cot ZB +**pq**cot2C =o; Tangential the centre is ap+bq+cr = o, or sinA +q sin B+rsinC =o. - D'Arzo Cr
**Pq**C Tr ' PgL. - 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. - The weight of
**PQ**and the tensions at P,Q. - A vector OU drawn parallel to
**PQ**, of length proportional to**PQ**/~I on any convenient scale, will represent the mean velocity in the interval 1t, i.e. - Experience the same resultant displacement
**PQ**in the same time. - For in time t the mutual action between two particles at P and Q produces equal and opposite momenta in the line
**PQ**, and these will have equal and opposite moments about the fixed axis. - The path of a point P in or attached to the rolling cone is a spherical epitrochoid traced on the surface of a sphere of the radius OP. From P draw
**PQ**perpendicular to the instantaneous axis. - Let
**PQ**bubble. - 1, curve a); the fixed line LM is termed the base, and the line
**PQ**which divides the curve symmetrically is the axis. - 1 AQP be the reflecting circle Caustics having C as centre, P the luminous point, and
**PQ**any incident ray, and we join CQ, it follows, by the law of the b efiection. - Conceive a perpendicular
**PQ**to be dropped from this point on the fundamental plane, meeting the latter in the point Q;**PQ**will then be parallel to OZ. - Are intersected by the lines RK, RM, RN, we have SA/AX = TP/
**PQ**= SP/**PQ**, since the angle PST = angle PTS.