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N'(t) = 4e^(−0.05t) (0 ≤ t ≤ 20) intergrating we get N = 4[1/(- 0.05)] e^(−0.05t) + C ; or we can write N = - 80 e^(−0.05t) + C
at t = 0, N = 75 therefore 75 = - 80 e^[(- 0.05)(0)] + C
C = 155
hence N(t) = 155 - 80 e^(−0.05t) for (0 ≤ t ≤ 20)
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Jul 23rd, 2015
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