S dg = f dg cos 0 where, cos e = VR+xa GS Rd0 x J (VR² + x²)²VR² + x² dg GS Rx de (VR2 + x²)7 GS Rx de (R2 + x?)% GS Rx (R2 + x2)7; G S Rx [2л — 0] 3 (R2 + x2)Z 2n G S Rx 9 = 3 (R2 + x2)?

Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter2: Equations And Inequalities
Section2.7: More On Inequalities
Problem 44E
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Kindly explain the process of this solution:

S dg = f dg cos 0
where, cos 0
VR+x
GS Rde x
(VR2 + x?)?VR² + x²
2n
GS Rx de
3
° (VR2 + x?)7
GS Rx
de
3
(R2 + x2)7%
GS Rx
3
(R2 + x2)Z%
GS Rx
3 [2n – 0]
3
(R2 + x2)Z
2n G S Rx
3
(R2 + x?)7
Transcribed Image Text:S dg = f dg cos 0 where, cos 0 VR+x GS Rde x (VR2 + x?)?VR² + x² 2n GS Rx de 3 ° (VR2 + x?)7 GS Rx de 3 (R2 + x2)7% GS Rx 3 (R2 + x2)Z% GS Rx 3 [2n – 0] 3 (R2 + x2)Z 2n G S Rx 3 (R2 + x?)7
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