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stira [4]
3 years ago
9

I am so confused right now plz help me

Mathematics
1 answer:
Maksim231197 [3]3 years ago
3 0
You have to remember P.E.M.D.A.S. when doing problems like this.
Drew used PEMDAS so we know he isn't wrong. However, Sam didn't use PEMDAS. She multiplied 2&5 instead of raising 5 to the second. 
The correct answer for Sams' problem would be 30.
Hope this helps! :)
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About 5 percent of tourists who go to Egypt visit the pyramids of Dahshur. Let's say that a group of 480 tourists goes to Egypt.
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A. 24

Step-by-step explanation:

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2 years ago
The Department of Agriculture is monitoring the spread of mice by placing 100 mice at the start of the project. The population,
uranmaximum [27]

Answer:

Step-by-step explanation:

Assuming that the differential equation is

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

We need to solve it and obtain an expression for P(t) in order to complete the exercise.

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.

In order to calculate the integral in the left hand side we make a partial fraction decomposition:

\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.

So,

\frac{500-P(t)}{P(t)} = 4e^{-0.04t}.

Finally, as we want the value of t such that P(t)=200, we substitute this last value into the above equation. Thus,

\frac{500-200}{200} = 4e^{-0.04t}.

This is equivalent to \frac{3}{8} = e^{-0.04t}. Taking logarithms we get \ln\frac{3}{8} = -0.04t. Then,

t = \frac{\ln\frac{3}{8}}{-0.04} \approx 24.520731325.

So, the population of rats will be 200 after 25 months.

6 0
3 years ago
Help 6th grade math i will give brainliest
andriy [413]

Answer:

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Step-by-step explanation:

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Answer:

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Answer:

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