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Question 1 Let A be the value of the triple integral SSS₂ (x + 22) = 1 pts dV where D is the region in 0, y = 2, y = 2x, z = 0, and the first octant bounded by the planes x z = 1 + 2x + y. Then the value of cos(A/4) is -0.411 0.709 0.067 -0.841 0.578 -0.913 -0.908 -0.120

Question 1 Let A be the value of the triple integral SSS₂ (x + 22) = 1 pts dV where D is the region in 0, y = 2, y = 2x, z = 0, and the first octant bounded by the planes x z = 1 + 2x + y. Then the value of cos(A/4) is -0.411 0.709 0.067 -0.841 0.578 -0.913 -0.908 -0.120

Elements Of Modern Algebra
Elements Of Modern Algebra
8th Edition
ISBN:9781285463230
Author:Gilbert, Linda, Jimmie
Publisher:Gilbert, Linda, Jimmie
Chapter5: Rings, Integral Domains, And Fields
Section5.4: Ordered Integral Domains
Problem 5E: 5. Prove that the equation has no solution in an ordered integral domain.
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Question 1 Let A be the value of the triple integral SSS₂ (x + 22) = 1 pts dV where D is the region in 0, y = 2, y = 2x, z = 0, and the first octant bounded by the planes x z = 1 + 2x + y. Then the value of cos(A/4) is -0.411 0.709 0.067 -0.841 0.578 -0.913 -0.908 -0.120
Transcribed Image Text:Question 1 Let A be the value of the triple integral SSS₂ (x + 22) = 1 pts dV where D is the region in 0, y = 2, y = 2x, z = 0, and the first octant bounded by the planes x z = 1 + 2x + y. Then the value of cos(A/4) is -0.411 0.709 0.067 -0.841 0.578 -0.913 -0.908 -0.120
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