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3 years ago

Consider the graph below:

Mathematics

2 answers:

zalisa [80]3 years ago

8 0

Answer:

Option C

Step-by-step explanation:

Given is a graph of the function and we have to find the equation

The graph has x intercepts as -1,2 and 3

Hence the graph will have these terms as factors

(x+1), (x-2) and (x-3)

So function would be of the form

$y=a(x+1)(x-2)(x-3)$

To find a:

Use the fact that y intercept is 6

When x=0, y=6

i.e. y =a(1)(6)

a=1

Hence funciton is

$y=(x+1)(x-2)(x-3)$

Option C is right

Julli [10]3 years ago

6 0

It would be C) y = ( x + 1 ) ( x - 3 ) ( x - 2 ) because when you graph the points it falls to the left and rises to the right.

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Lewis has 13 cats and 2 dogs. How many animals does he have in all?

Stels [109]

Answer:15

Step-by-step explanation:13 +2 = 15

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3 years ago

Is 2x and 5x^2 like terms?

ki77a [65]

No, they are not like terms because they cannot be added together. 5x^2 is squared, whereas 2x is not. Though they both have x variables attached to them, they are not like terms because one is squared and the other is not. Examples of like terms include 5 and 8, 4y and 7y, 8z^2 and 10z^2.

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2 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.

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3 years ago

Please help and explain how to solve these! Thanks

spin [16.1K]

2. C
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3. D
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Slope of graph d = 81/27 = 3

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