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3 years ago

HIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIIII

Mathematics

1 answer:

FrozenT [24]3 years ago

7 0

Answer:

The x-intercept is a point on the x-axis where the line crosses the x-axis and you can easily find it at point (-7, 0) because the line crosses the x-axis there.

Similarly, the y-intercept is a point on the y-axis where the line crosses the y-axis and you can easily find it at point (0, 2) because the line crosses the y-axis there.

You might be interested in

Help me please <br><br> -4.5+4.4+_____=0

s2008m [1.1K]

Answer:

+0.1

Step-by-step explanation:

Note that -4.5 + 4.4 = -0.1.

Adding 0.1 to -0.1 results in 0.

The unknown was +0.1.

4 0

3 years ago

What's a statement that is true for any number or variable<br>

KATRIN_1 [288]

Digit 3 has a value of 3*10000

5 0

4 years ago

Read 2 more answers

A triangle has sides xcm, (x + 4) cm and 11cm, where x is a whole number of cm. if the perimeter of the triangle is less than 32

NARA [144]

9514 1404 393

Answer:

{4, 5, 6, 7, 8}

Step-by-step explanation:

The inequality related to the perimeter is ...

x +(x +4) +11 < 32

2x < 17 . . . . . . . . . . . . subtract 15

x < 8.5

The inequality related to the triangle inequality is ...

(x) +(x +4) > 11

2x > 7

x > 3.5

Then possible whole-number values of x are ...

{4, 5, 6, 7, 8}

6 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

4 years ago

Which of the below descriptions shows a possible HAMILTON PATH?

Vadim26 [7]

Option B: FECBAD is the Hamilton path

Explanation:

Given that ABCDEF is a graph with vertices A,B,C,D,E and F

We need to determine the Hamilton path.

Since, we know that a Hamilton path is a path that touches every vertex in a graph exactly once.

Option A: EFADECBA

The path touches the vertices E and A twice.

Hence, the path EFADECBA is not a Hamilton path.

Therefore, Option A is not the correct answer.

Option B: FECBAD

The path touches every vertex in the graph exactly once.

Hence, the path FECBAD is a Hamilton path.

Therefore, Option B is the correct answer.

Option C: ADEFBD

The path touches the vertex A twice because when the path travels from F to B the only way to travel through A.

Hence, the path ADEFBD is not a Hamilton path.

Therefore, Option C is not the correct answer.

Option D: ADECBAFE

The path touches the vertices E and A twice.

Hence, the path ADECBAFE is not a Hamilton path.

Therefore, Option D is not the correct answer.

8 0

4 years ago