Ask question

Ask question

All categories

3 years ago

7

Is this a function or no (-3,2)(-1,0)(1,-2)(-3,4)(3,-2)?

1 answer:

Arisa [49]3 years ago

3 0

Answer:

-2-2-28

-2-2-2'&&&---6223,357%35372273753857270%

You might be interested in

58% of a $148 selling price goes toward paying expenses. How many dollars of the selling price is profit?

LuckyWell [14K]

Answer:

62.16

Step-by-step explanation:

6 0

2 years ago

Please do attached question

kondaur [170]

Answer:

ummmmmmmmmmmmmmmmm

Step-by-step explanation:

hmmmmmmmmmmmm

6 0

3 years ago

If a 20 ft tree casts a 15ft shadow, how long a shadow is cast

Irina-Kira [14]

Answer:

25 ft

Step-by-step explanation:

c = √a^2 + b^2 = √20^2 + 15^2 = 25ft

Hooke me up with a 5 star and a thanks :)

4 0

2 years ago

Simplify -4/5 divided by 3/-2

mixer [17]

Answer:

$\frac{8}{15}$

Step-by-step explanation:

8 0

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.

6 0

3 years ago