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 =
Chapter2: Loads On Structures
Section: Chapter Questions
Problem 1P
Related questions
Question
![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.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2915ef6c-7c04-4657-be8b-7fcf329bf76f%2Ffdb18449-beb8-42e1-82ce-5ed7b827ceec%2Frvbaip_processed.jpeg&w=3840&q=75)
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
=](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2915ef6c-7c04-4657-be8b-7fcf329bf76f%2Ffdb18449-beb8-42e1-82ce-5ed7b827ceec%2Fb0ju7lb_processed.jpeg&w=3840&q=75)
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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