Advanced Placement Calculus 2016 Graphical Numerical Algebraic Fifth    Edition Student Edition + Mathxl 1-year License
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60! 5!.7!.15!.33!
• • Let > be a potential for the vector field F = (−2 y³, −6 xy² − 4 z³, −12 yz² + 4 2). Then the value of sin((-1.63, 2.06, 0.57) – (0,0,0)) is - 0.336 -0.931 -0.587 0.440 0.902 0.607 -0.609 0.146
The value of cos(4M) where M is the magnitude of the vector field with potential ƒ = e² sin(лy) cos(π²) at x = 1, y = 1/4, z = 1/3 is 0.602 -0.323 0.712 -0.816 0.781 0.102 0.075 0.013
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