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• <center>$\frac{d_{n}(fg)}{dx^{n}}=\sum _{k}[n:k]\frac{d^{n-k}f}{dx^{n-k}}\frac{d^{k}g}{dx^{k}}</mat 501 bytes (83 words) - 17:01, 17 April 2018 • .../b> <b>time</b>=''CG''/''V''<sub>1</sub>&#x2013;''CD''/''V''<sub>2</sub>=''FG''/''V''<sub>1</sub>. Source delay time+geophone delay time=head-wave interc 386 bytes (64 words) - 15:36, 20 December 2011 • The recording cable length is ''FG'' and the line length is ''AD''. The number of dots along the offset axis ( 6 KB (961 words) - 13:14, 17 September 2014 • ...receiver location ''G'' is given by ''SR'' + ''RG'' = ''FR'' + ''RG'' = ''FG'' = ''vt'', where point ''F'' is the mirror image of the source location '' To compute the traveltime ''t'' associated with the distance ''FG'', we shall need to compute the coordinates of ''F'' : (''x<sub>F</sub>'', 69 KB (11,212 words) - 09:58, 19 September 2014 • ...her factors include rock properties and anisotropy. The gradient is called FG or fracture gradient. 9 KB (1,341 words) - 13:10, 4 January 2018 • <center>[itex]\frac{d_{n}(fg)}{dx^{n}}=\sum _{k}[n:k]\frac{d^{n-k}f}{dx^{n-k}}\frac{d^{k}g}{dx^{k}}</mat 430 bytes (73 words) - 15:27, 3 September 2019 • <center>(''d''<sup>''n''</sup>(''fg''))/(''dx''<sup>''n''</sup>)= &#x2211; <sub>''k''</sub>&#x005B;''n'':''k''& 666 bytes (104 words) - 16:39, 21 December 2017 • ...= SB/ V1 - AB/V2 = SE/V1; tiempo de retardo del geófono = CG/V1 - CD/V2 = FG/V1 . Tiempo de retardo de la fuente + tempo de retardo del geófono = tiemp 2 KB (266 words) - 12:01, 29 November 2020 • ...H'' and ''BDA'' and thus coincides with [itex]AC$. Clearly $(EF+FG+GH)=BD+DA$. $DA=h/\cos \theta$, [itex]BD=DA \cos2\theta</m
6 KB (926 words) - 14:08, 30 July 2019