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sasho [114]
3 years ago
5

Please help me with this question 25pts

Mathematics
2 answers:
Lisa [10]3 years ago
7 0

C, the third option.

  • -5/3 ≅ -1.6
  • √2 ≅ 1.4
  • √5 ≅ 2.2
ss7ja [257]3 years ago
4 0
The last one, the second one, the first one, the third one. I think that’s it but I could be wrong
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Step-by-step explanation:

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8 0
3 years ago
X - Y = 2. Wouldn't this be changed to Y= -x + 2 ???
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X - y = 2
Rewriting in terms of y,
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4 0
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A culture of the bacterium Salmonella enteritidis initially contains 50 cells. When introduced into a nutrient broth, the cultur
photoshop1234 [79]

Answer:

The expression for the number of bacteria after t hours is P(t) = 50e^{0.2922t}

Step-by-step explanation:

When introduced into a nutrient broth, the culture grows at a rate proportional to its size.

This means that the size of the population, after t hours, is modeled by the following differential equation:

\frac{dP}{dt} = rP

In which R is the growth rate.

The solution of this differential equation is:

P(t) = P(0)e^{rt}

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A culture of the bacterium Salmonella enteritidis initially contains 50 cells.

This means that P(0) = 50, and so:

P(t) = P(0)e^{rt}

P(t) = 50e^{rt}

After 1.5 hours, the population has increased to 775.

This means that P(1.5) = 775. We use this to find r. So

P(t) = P(0)e^{rt}

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e^{1.5r} = \frac{775}{500}

\ln{e^{1.5r}} = \ln{\frac{775}{500}}

1.5r = \ln{\frac{775}{500}}

r = \frac{\ln{\frac{775}{500}}}{1.5}

r = 0.2922

The expression is:

P(t) = 50e^{rt}

P(t) = 50e^{0.2922t}

The expression for the number of bacteria after t hours is P(t) = 50e^{0.2922t}

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3 years ago
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