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3 years ago

Y’all I need help fast

Mathematics

1 answer:

vitfil [10]3 years ago

7 0

Answer:

All Real Numbers, this is because it has infinitely MANY solutions.

Step-by-step explanation:

You might be interested in

What's 3,897.003 in expanded form.

Art [367]

3,897.003 in expanded form is:

3,000 + 800 + 90 + 7 + 0.003

3 0

3 years ago

leaks can write a 500-work essay in a hour. if she writes a essay in 10 minutes, how many words will it contain?

gavmur [86]

Answer:

about 83 words

Step-by-step explanation:

60/6=10

500/6 about 83

7 0

3 years ago

At a school carnival, 33 out of 55 tickets sold were early-admission tickets. What percentage of the tickets were early-admissio

7nadin3 [17]

Answer:

60%

Step-by-step explanation:

33/55 = .6

multiple by 100 to convert the decimal into a percentage

.6 × 100 = 60%

8 0

3 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

4 years ago

jan, mya, and sara ran a total of 64 miles last week. jan and mya ran the same number of miles. sara ran 8 less miles than jan a

kherson [118]

$x-\ Jan's\ destination\\\\y-\ Mya's \ destination \\\\ z-\ Sara's\ destination\\\\\
x+y+z=64\\
x=y\\
z-8=2x\\\\
x+x+z=64\\
z=2x+8\\\\
2x+2x+8=64\\\\
4x+8=64\\\\
4x=64-8\\\\
4x=56\ \ |:4\\\\x=14\\\\z=2x+8=2*14+8=28+8=36\\\\ Sara\ ran\ 36\ miles.$

5 0

3 years ago