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Newton's law of cooling is given by $\theta=\theta_0^{\circ}e^{-kt},$ where the excess of temperature at zero time is $0^{\circ}_{\circ}C $ and at time $t$ seconds is $\theta ^{\circ}C $. Determine the rate of change of temperature after $40 s,$ given that $\theta_{\cdot}=16^{\circ}C $ and $k=-0.03.\qquad (e^{1.2}=3.3201)$

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