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Hunter-Best [27]

3 years ago

10

The number of branches on a large tree after the year 2000 is represented by the following table:

2 answers:

Leya [2.2K]3 years ago

7 0

Answer:

B(t)= 16x (1.2)^2

Step-by-step explanation:

xenn [34]3 years ago

7 0

Answer:

B(t)= 16x (1.2)^2

Step-by-step explanation:

khan academy HIGH-key told me

but didn't at the same time

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Make v the subject of the formula

EleoNora [17]

$T =\dfrac 5{V+1}\\\\\\\implies VT+T = 5\\\\\implies VT = 5-T\\\\\implies V =\dfrac{5-T}T$

7 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

Y=4x+5 has a gradient of 4. What is the equation of a line parallel to it?

Setler79 [48]

It is -3,-2 is<span> Y=4x+5</span>

8 0

3 years ago

Elvin read 4/7 of the pages in his book. He has 226 pages left to read. How many pages has he read already?

Brut [27]

Answer:

he has read 129 pages

8 0

3 years ago

What is the formula to find the circumference of a circle using the radius?

Greeley [361]

Answer:

c = 2 *pie* r

Step-by-step explanation:

8 0

3 years ago