## Equating Sentence Examples

- But what was matter of immanent assumption with Erigena is in them an
**equating**of two things which have been dealt with on the hypothesis that they are separate, and which, therefore, still retain that external relation to one another. - He first divides by the factor x -x', reducing it to the degree m - I in both x and x' where m>n; he then forms m equations by
**equating**to zero the coefficients of the various powers of x'; these equations involve the m powers xo, x, - of x, and regarding these as the unknowns of a system of linear equations the resultant is reached in the form of a determinant of order m. - = 0, we find that, eliminating x, the resultant is a homogeneous function of y and z of degree mn;
**equating**this to zero and solving for the ratio of y to z we obtain mn solutions; if values of y and z, given by any solution, be substituted in each of the two equations, they will possess a common factor which gives a value of x which, corn bined with the chosen values of y and z, yields a system of values which satisfies both equations. - X (1 +PD1+12D2+...+ï¿½8D8+...) fm, and now expanding and
**equating**coefficients of like powers of /t D 1 f - Z(Difi)f2f3. - This can be verified by
**equating**to zero the five coefficients of the Hessian (ab) 2 axb2. - For instance, by
**equating**coefficients of or in the expansions of (I +x) m+n and of (I dx) m . - (iv.) On the other hand, the method of
**equating**coefficients often applies without the assumption of these laws. - The equations of motion can be established in a similar way by considering the rate of increase of momentum in a fixed direction of the fluid inside the surface, and
**equating**it to the momentum generated by the force acting throughout the space 5, and by the pressure acting over the surface S. - If other vortices are present, any one may be supposed to move with the velocity due to the others, the resultant stream function being = gy m log r =log IIrm; (9) the path of a vortex is obtained by
**equating**the value of 1P at the vortex to a constant, omitting the rm of the vortex itself. - The moment of inertia of the body about the axis, denoted by But if is the moment of inertia of the body about a mean diameter, and w the angular velocity about it generated by an impluse couple M, and M' is the couple required to set the surrounding medium in motion, supposed of effective radius of gyration k', If the shot is spinning about its axis with angular velocity p, and is precessing steadily at a rate about a line parallel to the resultant momentum F at an angle 0, the velocity of the vector of angular momentum, as in the case of a top, is C i pµ sin 0- C2µ 2 sin 0 cos 0; (4) and
**equating**this to the impressed couple (multiplied by g), that is, to gN = (c 1 -c 2)c2u 2 tan 0, (5) and dividing out sin 0, which equated to zero would imply perfect centring, we obtain C21 2 cos 0- (c 2 -c 1)c2u 2 sec 0 =o. **Equating**(I) and (2) u/U = wÃ† (3) which gives the particle velocity in terms of the pressure excess.**Equating**this to the momentum entering at B and subtracting P' from each X+W+p(U - u)2 =poU 2.- Differentiating and
**equating**to zero, the cost is least when dC _ LP +La =o, dl = l2 P=ale=G; that is, when the cost of one pier is equal to the cost erected of the main girders of one span. **Equating**the muzzle-energy and the work in foot-tons (2) E= w V 2 _XM.E.P.- With this purpose in view, he not only notes carefully the length of the reign of each king in both kingdoms, but also (as long as the northern kingdom existed) brings the history of the two kingdoms into relation with one another by
**equating**the commencement of each reign in either kingdom with the year of the reign of the contemporary king in the other kingdom. **Equating**The Above Two Values Of X, We Have 15 N' 3 =19 M' 16; Whence N' =M' 4'N1 13.**Equating**the expression (30) to ir/m, we find that f(a)=C/a, where n=3mO.**Equating**this to Mgh sin 0 ~t, and dividing out by sin 0, we obtain A cos 0 ~Cn~,i+Mgh=o, (3) as the condition in question.**Equating**this to the rate of increase of the angular momentum about OB, investigated as above, we find (C+Ma2+A~ cos e)~~=Mg ~- cot 0, (4)**Equating**this force with the resultant of the tension 7rpgr 2 h = 21rrT cos a, or h = 2T cos a/pgr.**Equating**the forces pghla =2/T cos a, whence It= 2T cos a/pga.- Hence
**equating**the forces which act on the portion included between ACB and PRQ, ry2p-27ryT cos a= - F (9). - Then a is the resultant horizontal pressure with an over turning moment of (I)
**Equating**the moment of resistance (2) to the overturning moment (I), we have pxzd =d3 3 6 and x =?2 p ..