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stepladder [879]
4 years ago
8

A colony of bacteria has 10,000 bacteria and is growing exponentially. The inequality 10,000(4x) ≥ 320,000 represents the hours,

x, for which the colony will have at least 320,000 bacteria.
Which interval represents the hours when the colony will have at least 320,000 bacteria?


[2.5, ∞)

[3, ∞)

[32, ∞)

[0, 2.5]

I have no clue what it is
Mathematics
1 answer:
Tresset [83]4 years ago
8 0

Answer:

The correct option is;

[2.5, ∞)

Step-by-step explanation:

The given exponential equation can be expressed as follows;

10,000 × 4ˣ ≥ 320,000

Therefore we have;

4ˣ ≥ 320,000/10,000

4ˣ ≥ 32

㏒(4ˣ) ≥ ㏒(32)

x·㏒4 ≥ ㏒(32)

x ≥ ㏒(32)/㏒4

∴ x ≥ 2.5

Therefore, the interval the colony will have at least 320,000 bacteria is where x ≥ 2.5 or 2.5 ≤ x < ∞ which gives, the interval  [2.5, ∞)

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The solution of the differential equation is B=5+25e^{-4t+4}

Step-by-step explanation:

The differential equation \frac{dB}{dt}+4B=20 is a first order separable ordinary differential equation (ODE). We know this because a separable first-order ODE has the form:

y'(t)=g(t)\cdot h(y)

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We can rewrite our differential equation in the form of a first-order separable ODE in this way:

\frac{dB}{dt}+4B=20\\\frac{dB}{dt}=20-4B\\\frac{dB}{dt}=4(5-B)\\\frac{1}{5-B}\frac{dB}{dt}=4

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\frac{1}{5-B}\frac{dB}{dt}=4\\\frac{1}{5-B}\cdot dB=4\cdot dt\\\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {4} \, dt

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\int\limits {\frac{1}{5-B}} \, dB\\\mathrm{Apply\:u-substitution:}\:u=5-B\\\int\limits {\frac{1}{5-B}} \, dB=\int\limits {\frac{1}{u}} \, dB\\\mathrm{du=-dB}\\-\int\limits {\frac{1}{u}} \,du\\\mathrm{Use\:the\:common\:integral}:\quad \int \frac{1}{u}du=\ln \left(\left|u\right|\right)\\-\int\limits {\frac{1}{u}} \,du =-\ln \left|u\right|\\\mathrm{Substitute\:back}\:u=5-B\\-\ln \left|5-B\right|\\\mathrm{Add\:a\:constant\:to\:the\:solution}\\-\ln \left|5-B\right|+C

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-\ln \left|5-B\right|+C=4t + C\\-\ln \left|5-B\right|=4t + D

If we want to find the explicit general solution of the differential equation

We isolate B

-\ln \left|5-B\right|=4t + D\\\ln \left|5-B\right|=-4t+D\\\left|5-B\right|=e^{-4t+D}

Recall the definition of |x|

|x|=\left \{ {{x, \:if \>x\geq \>0 } \atop {-x, \:if \>x0}} \right.

So

\left|5-B\right|=e^{-4t+D}\\5-B= \pm \:e^{-4t+D}\\B=5 \pm \:e^{-4t+D}\\B=5\pm \:e^{-4t}\cdot e^{D}\\B=5+Ae^{-4t}

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B=5+Ae^{-4t}\\30=5+Ae^{-4}\\30-5=Ae^{-4}\\25e^{4}=A

And the solution is

B=5+(25e^{4})e^{-4t}\\B=5+25e^{-4t+4}

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