Shade a model for 1.4 divided by 0.7

Which quadratic equation defines the function that has zeros at − 1/12 and 1/4 ?

Cerrena [4.2K]

$\bf \begin{cases}
x=-\frac{1}{2}\implies &x+\frac{1}{2}=0\\\\
x=\frac{1}{4}\implies &x-\frac{1}{4}=0
\end{cases}
\\\\\\
\left( x+\frac{1}{2} \right)\left( x-\frac{1}{4} \right)=\stackrel{y}{0}\implies \stackrel{FOIL}{x^2+\frac{1}{4}x-\frac{1}{8}}=y$

7 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

3 years ago

What is the exact length of AB?<br> 4 m<br> 8 mm<br> m<br> 2 TT m

dalvyx [7]

Answer:

The answer is

$\pi$

if I am not wrong

Step-by-step explanation:

First you need to convert 45 degress to radians,then you use the arc length formula which is = thr radius * the angle to get the answer

5 0

2 years ago

In kite PQRS, m∠TQP=63°. Identify m∠QPT.

mario62 [17]

Assuming m<PTQ is a right angle (measuring 90°) and angle TQP is already marked as 63° while knowing there are 180° in a triangle, you would take 180°- 63°- 90°= 27

8 0

3 years ago

Read 2 more answers

The measurements 45.367 cm and 43.43 cm are made in the same unit of measure. Which measurement is less precise? Explain your an

Kaylis [27]

Answer:

45.367 cm

Step-by-step explanation:

Approximating to a more number of decimal places gives a better estimation of the exact value.

For 45.367 cm, we approximated to three decimal places or five significant figures.

For 43.43 cm, the approximation is done to four decimal places.

Therefore the measurement 45.367cm is more precise because it has a relatively smaller error.

7 0

3 years ago