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bagirrra123 [75]
2 years ago
10

Mrs. Farad goes to an office supply store to buy 6 markers and a poster board for each of the 20 girls in her troop. Let m repre

sent the cost of 1 marker and b represent the cost of 1 poster board. Write two different expressions that represent the total amount Mrs. Farad paid for her purchase.
Mathematics
1 answer:
Elina [12.6K]2 years ago
8 0

Answer:

ok so if m is maker and b is poster then

6m+b since she buys 6 markers and 1 poster or

b+6m if she bought it in a different order

Hope This Helps!!!

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The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,
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Step-by-step explanation:

Assuming that the differential equation is

\frac{dP}{dt} = 0.04P\left(1-\frac{P}{500}\right).

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First of all, this is an example of the logistic equation, which has the general form

\frac{dP}{dt} = kP\left(1-\frac{P}{K}\right).

In order to make the calculation easier we are going to solve the general equation, and later substitute the values of the constants, notice that k=0.04 and K=500 and the initial condition P(0)=100.

Notice that this equation is separable, then

\frac{dP}{P(1-P/K)} = kdt.

Now, intagrating in both sides of the equation

\int\frac{dP}{P(1-P/K)} = \int kdt = kt +C.

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\frac{1}{P(1-P/K)} = \frac{1}{P} - \frac{1}{K-P}.

So,

\int\frac{dP}{P(1-P/K)} = \ln|P| - \ln|K-P| = \ln\left| \frac{P}{K-P} \right| = -\ln\left| \frac{K-P}{P} \right|.

We have obtained that:

-\ln\left| \frac{K-P}{P}\right| = kt +C

which is equivalent to

\ln\left| \frac{K-P}{P}\right|= -kt -C

Taking exponentials in both hands:

\left| \frac{K-P}{P}\right| = e^{-kt -C}

Hence,

\frac{K-P(t)}{P(t)} = Ae^{-kt}.

The next step is to substitute the given values in the statement of the problem:

\frac{500-P(t)}{P(t)} = Ae^{-0.04t}.

We calculate the value of A using the initial condition P(0)=100, substituting t=0:

\frac{500-100}{100} = A} and A=4.

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