Let $p ( t ) = \frac { 1,000 \left( e ^ { 0.1 t } + 3 \right) } { e ^ { 0.1 t } + 9 }$ be the population of a bacteria colony at time t hours. Find the growth rate of the bacteria after 10 hours.

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The Answer of Let $p ( t ) = \frac { 1,000 \left( e ^ { 0.1 t } + 3 \right) } { e ^ { 0.1 t } + 9 }$ be the population of a bacteria colony at time t hours. Find the growth rate of the bacteria after 10 hours.

Let p(t)=1,000(e0.1t+3)e0.1t+9p ( t ) = \frac { 1,000 \left( e ^ { 0.1 t } + 3 \right) } { e ^ { 0.1 t } + 9 }p(t)=e0.1t+91,000(e0.1t+3)​

Let $p ( t ) = \frac { 1,000 \left( e ^ { 0.1 t } + 3 \right) } { e ^ { 0.1 t } + 9 }$