ГДЗ по алгебре 11 класс Никольский Параграф 6. Первообразная и интеграл Задание 93

\[\boxed{\mathbf{93}.}\]

\[T_{0} = 100{^\circ};\ \ T_{1} = 28{^\circ}:\]

\[\frac{\text{dT}}{\text{dt}} = - k(T - 28)\]

\[\frac{d(T - 28)}{\text{dt}} = - k(T - 28).\]

\[\int_{}^{}\frac{d(T - 28)}{\text{dt}} = \int_{}^{}{- k(T - 28)}\]

\[\int_{}^{}\frac{d(T - 28)}{\text{dt}} = - k\int_{}^{}{dt + C}\]

\[\ln|T - 28| = - kt + C\]

\[T - 28 = e^{- kt + C}\]

\[T = 28 + Ae^{- kt}.\]

\[T(0) = 28 + Ae^{- ok} = 100;\]

\[T(12) = 28 + Ae^{- 12k} = 52.\]

\[28 + Ae^{- ok} = 100\]

\[A = 72.\]

\[Подставим:\]

\[28 + Ae^{- 12k} = 52\]

\[28 + 72 \cdot e^{- 12k} = 52\]

\[72e^{- 12k} = 24\]

\[e^{- 12k} = \frac{1}{3}\]

\[e^{- k} = \left( \frac{1}{3} \right)^{\frac{1}{12}}.\]

\[Температура\ через\ 24\ мин:\]

\[T(24) = 28 + Ae^{- 24k} =\]

\[= 28 + 72 \cdot \left( e^{- 12k} \right)^{2} =\]

\[= 28 + 72 \cdot \left( \frac{1}{3} \right)^{2} = 36{^\circ}.\]

\[Ответ:36{^\circ}.\]