## anonymous 5 years ago the number of bacteria in a culture is growing at a rate of 1500e^(3t/4) per unit of time t. at t=0, the number of bacteria present was 2000. find the number present at t=4.

$P=\int\limits_{}^{}1500e^{\frac{3}{4}t}dt=1500\frac{4}{3}e^{\frac{3}{4}t}+C$ at t=0 P=2000 2000=2000e^0+C C=0 P(t)=2000e^{3t/4} P(4)=2000e^3 P(4)=40171.1