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3 years ago

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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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:

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3 years ago

How do i solve this?

storchak [24]

I hope it help you bye

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3 years ago

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.

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3 years ago