Suppose you know that the series
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Chapter 11 Solutions
CALCULUS: EARLY TRANSCENDENTALS
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- Given that the outward flux of a vector field through the sphere of radius r centered at the origin is 5(1 cos(2r)) sin(r), and D is the value of the divergence of the vector field at the origin, the value of sin (2D) is -0.998 0.616 0.963 0.486 0.835 -0.070 -0.668 -0.129arrow_forward10 The hypotenuse of a right triangle has one end at the origin and one end on the curve y = Express the area of the triangle as a function of x. A(x) =arrow_forwardIn Problems 17-26, solve the initial value problem. 17. dy = (1+ y²) tan x, y(0) = √√3arrow_forward
- could you explain this as well as disproving each wrong optionarrow_forwardcould you please show the computation of this by wiresarrow_forward4 Consider f(x) periodic function with period 2, coinciding with (x) = -x on the interval [,0) and being the null function on the interval [0,7). The Fourier series of f: (A) does not converge in quadratic norm to f(x) on [−π,π] (B) is pointwise convergent to f(x) for every x = R П (C) is in the form - 4 ∞ +Σ ak cos(kx) + bk sin(kx), ak ‡0, bk ‡0 k=1 (D) is in the form ak cos(kx) + bk sin(kx), ak 0, bk 0 k=1arrow_forward