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Solve (x+2y-1)dx + (2x+y+1)dy = 0 using %3D methods of substitution. A. (y+1)+ 4(x+1) (y+1) + (x+1)? = o В. (у-1)2 + 2(х+1) (у-1) + (x+1)? %3Dо С.(у-1)* + 4(х+1) (у-1) + 2(х+1)2 %3D0 D. (y-1)? + 4(x+1) (y-1) + (x+1)² = 0 E. None of the above: = 0

Solve (x+2y-1)dx + (2x+y+1)dy = 0 using %3D methods of substitution. A. (y+1)+ 4(x+1) (y+1) + (x+1)? = o В. (у-1)2 + 2(х+1) (у-1) + (x+1)? %3Dо С.(у-1)* + 4(х+1) (у-1) + 2(х+1)2 %3D0 D. (y-1)? + 4(x+1) (y-1) + (x+1)² = 0 E. None of the above: = 0

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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solve using methods of substitution

Solve (x+2y-1)dx + (2x+y+1)dy = 0 using methods of substitution. A. (y+1)+ 4(x+1) (y+1) + (x+1)? = 0 B. (y-1)? + 2(x+1) (y-1) + (x+1)² = 0 C. (y-1)? + 4(x+1) (y-1) + 2(x+1)² = 0 D. (y-1)? + 4(x+1) (y-1) + (x+1)² = 0 E. None of the above: = 0
Transcribed Image Text:Solve (x+2y-1)dx + (2x+y+1)dy = 0 using methods of substitution. A. (y+1)+ 4(x+1) (y+1) + (x+1)? = 0 B. (y-1)? + 2(x+1) (y-1) + (x+1)² = 0 C. (y-1)? + 4(x+1) (y-1) + 2(x+1)² = 0 D. (y-1)? + 4(x+1) (y-1) + (x+1)² = 0 E. None of the above: = 0
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