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Q4\A\ Find the value of the integral fet. dt using two and three points of Gaussian-Quadrature Method. B\ Given the following points:- X Y 1.1 1.7 3 10.6 13.2 20.3 Find the value of f(x) when x-2 then do the inverse using Lagrange's Inverse formula. حيث: موازي × 309.3 = cos(20.5°) × 309.3 = = 0.9397 290.36 ن عمودي = 309.3 × °20.5)sin = 309.3 × 105.76 0.3420 = ن ثم نحسب المكون الثقلي للقوة على طول المنحدر: F الجاذبية sin(0) x g × m = حيث و هو تسارع الجاذبية (9.8 م /ث ): F الجاذبية = 37.7 × 9.8 × (°20.5)sin 129.98 0.3420 × 9.8 × 37.7 x = ن =

Q4\A\ Find the value of the integral fet. dt using two and three points of Gaussian-Quadrature Method. B\ Given the following points:- X Y 1.1 1.7 3 10.6 13.2 20.3 Find the value of f(x) when x-2 then do the inverse using Lagrange's Inverse formula. حيث: موازي × 309.3 = cos(20.5°) × 309.3 = = 0.9397 290.36 ن عمودي = 309.3 × °20.5)sin = 309.3 × 105.76 0.3420 = ن ثم نحسب المكون الثقلي للقوة على طول المنحدر: F الجاذبية sin(0) x g × m = حيث و هو تسارع الجاذبية (9.8 م /ث ): F الجاذبية = 37.7 × 9.8 × (°20.5)sin 129.98 0.3420 × 9.8 × 37.7 x = ن =

Structural Analysis
Structural Analysis
6th Edition
ISBN:9781337630931
Author:KASSIMALI, Aslam.
Publisher:KASSIMALI, Aslam.
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
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Q4\A\ Find the value of the integral fet. dt using two and three points of Gaussian-Quadrature Method. B\ Given the following points:- X Y 1.1 1.7 3 10.6 13.2 20.3 Find the value of f(x) when x-2 then do the inverse using Lagrange's Inverse formula.
Transcribed Image Text:Q4\A\ Find the value of the integral fet. dt using two and three points of Gaussian-Quadrature Method. B\ Given the following points:- X Y 1.1 1.7 3 10.6 13.2 20.3 Find the value of f(x) when x-2 then do the inverse using Lagrange's Inverse formula.
حيث: موازي × 309.3 = cos(20.5°) × 309.3 = = 0.9397 290.36 ن عمودي = 309.3 × °20.5)sin = 309.3 × 105.76 0.3420 = ن ثم نحسب المكون الثقلي للقوة على طول المنحدر: F الجاذبية sin(0) x g × m = حيث و هو تسارع الجاذبية (9.8 م /ث ): F الجاذبية = 37.7 × 9.8 × (°20.5)sin 129.98 0.3420 × 9.8 × 37.7 x = ن =
Transcribed Image Text:حيث: موازي × 309.3 = cos(20.5°) × 309.3 = = 0.9397 290.36 ن عمودي = 309.3 × °20.5)sin = 309.3 × 105.76 0.3420 = ن ثم نحسب المكون الثقلي للقوة على طول المنحدر: F الجاذبية sin(0) x g × m = حيث و هو تسارع الجاذبية (9.8 م /ث ): F الجاذبية = 37.7 × 9.8 × (°20.5)sin 129.98 0.3420 × 9.8 × 37.7 x = ن =
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