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dybincka [34]
2 years ago
6

I NEED HELP SOLVING THIS MATH PROBLEM.

Mathematics
1 answer:
olasank [31]2 years ago
3 0

Answer:

19s

Step-by-step explanation:

s = 1s

1s+8s+10s=(1+8+10)s=19s

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Add them together to get the total which = 5
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1:100 because 20 divide by 20=1 and 2000 divide by 20 =100
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dolphi86 [110]

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3 years ago
A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium d
nikitadnepr [17]

Answer:

a) k=2.08 1/hour

b) The exponential growth model can be written as:

P(t)=Ce^{kt}

c) 977,435,644 cells

d) 2.033 billions cells per hour.

e) 2.81 hours.

Step-by-step explanation:

We have a model of exponential growth.

We know that the population duplicates every 20 minutes (t=0.33).

The initial population is P(t=0)=58.

The exponential growth model can be written as:

P(t)=Ce^{kt}

For t=0, we have:

P(0)=Ce^0=C=58

If we use the duplication time, we have:

P(t+0.33)=2P(t)\\\\58e^{k(t+0.33)}=2\cdot58e^{kt}\\\\e^{0.33k}=2\\\\0.33k=ln(2)\\\\k=ln(2)/0.33=2.08

Then, we have the model as:

P(t)=58e^{2.08t}

The relative growth rate (RGR) is defined, if P is the population and t the time, as:

RGR=\dfrac{1}{P}\dfrac{dP}{dt}=k

In this case, the RGR is k=2.08 1/h.

After 8 hours, we will have:

P(8)=58e^{2.08\cdot8}=58e^{16.64}=58\cdot 16,852,338= 977,435,644

The rate of growth can be calculated as dP/dt and is:

dP/dt=58[2.08\cdot e^{2.08t}]=120.64e^2.08t=2.08P(t)

For t=8, the rate of growth is:

dP/dt(8)=2.08P(8)=2.08\cdot 977,435,644 = 2,033,066,140

(2.033 billions cells per hour).

We can calculate when the population will reach 20,000 cells as:

P(t)=20,000\\\\58e^{2.08t}=20,000\\\\e^{2.08t}=20,000/58\approx344.827\\\\2.08t=ln(344.827)\approx5.843\\\\t=5.843/2.08\approx2.81

3 0
3 years ago
Please help quickly i really need it
wariber [46]

Answer:

(x - 6)² = 18

Step-by-step explanation:

(x - 6)² = 18

x² - 12x + 36 = 18

x² - 12x + 18

6 0
2 years ago
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