Xa sentence example

xa
  • X (xa) ki (xb) k2 (xc) k3...axibx2cx3...xx = (AB) hi (AC) h2 (BC) h3...A11 4 13 A1,14131 A B I ?C"' B C "' X (XA) ki (XB) k2 (XC) k3...AXB122cCk...X If this be of order e and appertain to an nie L Eke-/1+2m =e, h i+h2+���+221+ji+j2+���+kl+li =n, hi+h3+..�+222+ji+j3+���+k2+12 = n, h2+h3+���+223+j2+%3+�.�+k3+13 =n; viz., the symbols a, b, c,...
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  • 4), and issues in a jet between two edges A and A'; the wall xA being bent at a corner B, with the external angle (3= 2Wr/n.
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  • The stream lines xBAJ, xA'J' are given by = 0, m; so that if c denotes the ultimate breadth JJ' of the jet, where the velocity may be supposed uniform and equal to the skin velocity Q, m=Qc, c=m/Q.
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  • If there are more B corners than one, either on xA or x'A', the expression for i is the product of corresponding factors, such as in (5) Restricting the attention to a single corner B, a = n(cos no +i sin 110) _ (b-a'.0-a) +1!
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  • The line XZ consists of a series of lengths, as XA, AB ...
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  • The various forms in areal co-ordinates may be derived from the above by substituting Xa for 1, µb for m and vc for n, or directly by expressing the condition for tangency of the line x+y+z = o to the conic expressed in areal coordinates.
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  • 7) that "he emptied himself and took upon him the form of a servant" (EauTOv µop4 v OovXoD Xa(3c7.v).
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  • Hruschka's extractor, first brought to public xa cto s.X?
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  • X (xa) ki (xb) k2 (xc) k3...axibx2cx3...xx = (AB) hi (AC) h2 (BC) h3...A11 4 13 A1,14131 A B I ?C"' B C "' X (XA) ki (XB) k2 (XC) k3...AXB122cCk...X If this be of order e and appertain to an nie L Eke-/1+2m =e, h i+h2+���+221+ji+j2+���+kl+li =n, hi+h3+..�+222+ji+j3+���+k2+12 = n, h2+h3+���+223+j2+%3+�.�+k3+13 =n; viz., the symbols a, b, c,...
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  • For the quadratic aoxi +2a i x i x 2 +a 2 x, we have (i.) ax = 7/1x1+2aixix2-I-7/24, (ii.) xx=xi+xzi (ab) 2 =2(aoa2 - ai), a a = a o+712, _ (v.) (xa)ax= i'?- (a2 - ao)xix2 - aix2.
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  • Passing to the point Xefa we find two known forces, the load XA acting downwards, and a push from the strut XE, which, being in compression, must push at both ends, as indicated by the arrow, fig.
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  • We then have the polygon of forces Exaf, the reciprocal figure of the lines meeting at that point in the frame, and representing the forces at the point Exaf; the direction of the forces on EH and XA being known determines the direction of the forces due to the elastic reaction of the members AF and EF,, showing AF to push as a strut, while EF is a tie.
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