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ladessa [460]
3 years ago
15

True or False - Explain the graphs of two different direct variations can be parallel

Mathematics
2 answers:
Montano1993 [528]3 years ago
6 0
False.because u have to do the vlt
Kruka [31]3 years ago
3 0
False because you in comepare and coaiser
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Completely simplified the expression for x plus 10 plus 6 x plus 2 x by combining like terms into your answer in the box.
siniylev [52]
We can first turn the text into a proper formula:

x plus 10 plus 6 x plus 2 x

x+10+6x+2x

The like terms are the numbers with

We can then put the like terms together to continue the calculation:

x+6x+2x+10

=9x + 10

Therefore, the answer is 9x + 10.

Hope it helps!
7 0
3 years ago
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Which of the following could be the equation of the function below? On a coordinate plane, a curve crosses the y-axis at y = 2.
Neko [114]

Answer:

I believe according to the information provided that the answer would be would be B

Step-by-step explanation:

3 0
3 years ago
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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,
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
4 years ago
How do i answer this?
Vedmedyk [2.9K]
Somesomeone answer this please !!!
4 0
3 years ago
Can you please send help
love history [14]

Answer:

yes ill send the fbi

Step-by-step explanation:

7 0
3 years ago
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