Solve the initial value problem. = 20t(5t2 - 1)3, s(1) = -4

A. s = (5t2 - 1)4 - 260
B. s = (5t2 - 1)4 - 4
C. s = (5t2 - 1)4
D. s = (5t2 - 1)4 - 132

Answer: D

Mathematics

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