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how to solve equations → 162= z 3(8A+4C+2D)+zY-6A+8B-C-D) +z(A-6B+4C+2D)+(B-c-D) Equating the coefficients, we aet 8A + 4C +2D =0 -6A +8B - C - D=16 A- 6B +4C+2D =0 B-C- D %3D 32 112 •: Solving, we det A=- 96. 85 C= %3D

how to solve equations → 162= z 3(8A+4C+2D)+zY-6A+8B-C-D) +z(A-6B+4C+2D)+(B-c-D) Equating the coefficients, we aet 8A + 4C +2D =0 -6A +8B - C - D=16 A- 6B +4C+2D =0 B-C- D %3D 32 112 •: Solving, we det A=- 96. 85 C= %3D

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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how to solve equations → 162²= z3(8A+4C+2D)+zY-6A+8B-C-D) +=(A-6B+4C+2)+(B-c-D) Equating the coefficients, we get 8A + 4C +2D =0 -6A +8B - C - D=16 A- 6B +4C+2D =0 B-C- D = 0 32 •: Solving. we get 112 B = A=- %3D 85
Transcribed Image Text:how to solve equations → 162²= z3(8A+4C+2D)+zY-6A+8B-C-D) +=(A-6B+4C+2)+(B-c-D) Equating the coefficients, we get 8A + 4C +2D =0 -6A +8B - C - D=16 A- 6B +4C+2D =0 B-C- D = 0 32 •: Solving. we get 112 B = A=- %3D 85
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