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Elenna [48]
3 years ago
6

Find the function p(x) = f(x) × g(x). f(x)=3, g(x)=6x-4

Mathematics
1 answer:
trasher [3.6K]3 years ago
6 0

Answer:

p(x) = 18x - 12

Step-by-step explanation:

given p(x) = f(x) × g(x), then

p(x) = 3(6x - 4 ) ← distribute by 3

p(x) = 18x - 12


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Cory is hiking up a hill with a constant slope. The double number line shows that after Cory hikes 3 \text{ km}3 km3, start text
Andrews [41]

Answer:

The answer is "3 km ".

Step-by-step explanation:

Distance in (km): 3  \to 1

Elevation value in (m):  120 \to 40

\to \frac{3}{1} = 3 \ km \\\\\to \frac{120}{40} =\frac{12}{4} = 3 \ km

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3 years ago
In 10 words or fewer, what other numbers do you think are in the domain of this function?​
Maru [420]

Answer:

Numbers greater than or equal to 0.

Step-by-step explanation:

The domain of this function is {x∈R | x≥0}, meaning that x can be anything greater than or equal to 0.

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3 years ago
How do i solve this?
storchak [24]
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6 0
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Read 2 more answers
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
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