MA = 8x – 2, mB = 2x – 8, and mC = 94 – 4x. List the sides of ABC in order from shortest to longest.

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

A line passes through (3, 7) and (6, 9). Which equation best represents the line?

sp2606 [1]

The name is the second option y= (2/3)x +5. Hope this helps

8 0

3 years ago

Read 2 more answers

How many solutions does the equation |x + 6| − 4 = c have if c = 5? If c = −10? Complete the explanation

mixer [17]

Answer:

<h2>For c = 5 → two solutions</h2><h2>For c = -10 → no solutions</h2>

Step-by-step explanation:

We know

$|a|\geq0$

for any real value of a.

|a| = b > 0 - two solutions: a = b or a = -b

|a| = 0 - one solution: a = 0

|a| = b < 0 - no solution

|x + 6| - 4 = c

for c = 5:

|x + 6| - 4 = 5 add 4 to both sides

|x + 6| = 9 > 0 TWO SOLUTIONS

for c = -10

|x + 6| - 4 = -10 add 4 to both sides

|x + 6| = -6 < 0 NO SOLUTIONS

Calculate the solutions for c = 5:

|x + 6| = 9 ⇔ x + 6 = 9 or x + 6 = -9 subtract 6 from both sides

x = 3 or x = -15

5 0

3 years ago

Amelia randomly selects two cards, with replacement, from a normal deck of cards. Calculate the probability that: the first card

Lena [83]

Answer:

1 / 2704

Step-by-step explanation:

Number of cards in a deck = 52

9 of clubs in a deck = 1

10 of clubs in a deck = 1

Probability = required outcome / Total possible outcomes

P(9 of club) = 1 / 52

With replacement ;

Then ;

P(10 of club) = 1 / 52

Hence,

P(9 of club, then 10 of club) = 1/52 * 1/52 = 1 / 2704

3 0

3 years ago

Calculate the discriminant and use it to determine how many real-number roots the equation has.

Vladimir79 [104]

In a quadratic equation with the general formula of:

ax^2 + bx + c = 0

The discriminant is equal to b^2 - 4(a)(c). If the answer is a perfect square, then there are two real numbers. If not, then there are no real number root.

The discriminant for this equation is

(-6)^2 - 4(3)(1) = 24

Since 24 is not a perfect square, there are no real number roots.

4 0

3 years ago