A bacteria population is 6000 at time t = 0 and its rate of growth is 1000 Â· 9t bacteria per hour after t hours. what is the population after one hour? (round your answer to the nearest whole number.)

dP/dt = 1000 * 9^t is the bacteria population's growth rate.where:P(t) is the population at time t, andP(0) = 6000.
The population after one hour would be:
P(1) = P(0) + <Integral of dP/dt from 0 to 1>
P(1) = P(0) + [ 1000 * 9^1 / ln(9) - 1000 * 9^0 / ln(9) ]
P(1) = 6000 + 3641 = 9461 would be the answer

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lublana lublana · Answer 2 · 2019-12-23T05:07:16+01:00

Answer:

14384

Step-by-step explanation:

We are given that

P(0)=6000

[tex]\frac{dP}{dt}=1000\cdot 9^t[/tex]

We have to find the population after 1 hour.

[tex]dP=1000(9^t)dt[/tex]

Taking integration on both sides

[tex]P=1000\int 9^t dt[/tex]

[tex]P(t)=\frac{1000}{log 9}9^t[/tex]+C

Using formula:[tex]\int a^t=\frac{1}{log a}a^t+C[/tex]

Substitute t=0 and P(0)=6000

[tex]6000=\frac{1000}{log9}+C[/tex]

[tex]C=6000-\frac{1000}{log 9}=6000-\frac{1000}{0.954}[/tex]

[tex]C=4952[/tex]

Substitute the values then we get

[tex]P(t)=1047.95(9^t)+4952[/tex]

Substitute t=1

Then, we get

[tex]P(1)=1047.95(9)+4952[/tex]

[tex]P(1)=14384[/tex]

Hence, the population after one hour=14384