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Svetllana [295]
3 years ago
9

The size of a slow-growing bacteria strain increases continuously over time. This equation describes the predicted size of the b

acteria strain, in square millimeters, after t minutes: y = 112(1.58)t. According to the equation, what will the approximate size of the bacteria strain be after 5 minutes?
Mathematics
1 answer:
Artemon [7]3 years ago
6 0

Answer:

  1102.8 mm^2

Step-by-step explanation:

We presume your equation is intended to be ...

  y = 112(1.58)^t

Since t is in minutes, fill in the given value and do the arithmetic:

  y = 112(1.58)^5 = 112(9.84658) ≈ 1102.8

The approximate size of the bacteria strain will be 1102.8 mm^2 after 5 minutes.

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Answer: x = 2

Step-by-step explanation:

5x + 11 - 2x = 17

Combine like terms

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Subtract 11 from each side

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Find the general solution to each of the following ODEs. Then, decide whether or not the set of solutions form a vector space. E
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Answer:

(A) y=ke^{2t} with k\in\mathbb{R}.

(B) y=ke^{2t}/2-1/2 with k\in\mathbb{R}

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Step-by-step explanation

(A) We can see this as separation of variables or just a linear ODE of first grade, then 0=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y \Rightarrow  \frac{1}{2y}dy=dt \ \Rightarrow \int \frac{1}{2y}dy=\int dt \Rightarrow \ln |y|^{1/2}=t+C \Rightarrow |y|^{1/2}=e^{\ln |y|^{1/2}}=e^{t+C}=e^{C}e^t} \Rightarrow y=ke^{2t}. With this answer we see that the set of solutions of the ODE form a vector space over, where vectors are of the form e^{2t} with t real.

(B) Proceeding and the previous item, we obtain 1=y'-2y=\frac{dy}{dt}-2y\Rightarrow \frac{dy}{dt}=2y+1 \Rightarrow  \frac{1}{2y+1}dy=dt \ \Rightarrow \int \frac{1}{2y+1}dy=\int dt \Rightarrow 1/2\ln |2y+1|=t+C \Rightarrow |2y+1|^{1/2}=e^{\ln |2y+1|^{1/2}}=e^{t+C}=e^{C}e^t \Rightarrow y=ke^{2t}/2-1/2. Which is not a vector space with the usual operations (this is because -1/2), in other words, if you sum two solutions you don't obtain a solution.

(C) This is a linear ODE of second grade, then if we set y=e^{mt} \Rightarrow y''=m^2e^{mt} and we obtain the characteristic equation 0=y''-4y=m^2e^{mt}-4e^{mt}=(m^2-4)e^{mt}\Rightarrow m^{2}-4=0\Rightarrow m=\pm 2 and then the general solution is y=k_1e^{2t}+k_2e^{-2t} with k_1,k_2\in\mathbb{R}, and as in the first items the set of solutions form a vector space.

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