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3 years ago

3. Use the diagram below to answer the following questions.

Mathematics

2 answers:

Sauron [17]3 years ago

7 0

Answer:

a. Three points are C, D and G.

b. Four different line segments are FD, FG, DG and DC.

c. Two other names for (FG) are (FD+DG) or GF.

d. Two different rays are GF and FG.

Step-by-step explanation:

In this given figure we have four points F, D, C and G. We can choose any three from them.

Therefore the three points are C, D and G.

If a straight line connect two points, then the it is called a line segment.

The points F and D, F and G, D and G, D and C are connected by straight lines.

Therefore the four different line segments are FD, FG, DG and DC.

The line segments FG connects two points F and G. So it can be written as FG or GF.

Since point D lies between the line segments FG, therefore FG can be written as (FD+DG).

Therefore, the two other names for (FG) are (FD+DG) or GF.

Line segment has two end point but a ray has one end point and it continuing indefinitely one directions.

FG continuing indefinitely in right direction and GF continuing indefinitely in left direction.

Therefore the two different rays are GF and FG.

ANEK [815]3 years ago

4 0

A. C D and G
B. FD, DG, FG, and DC

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3 years ago

on the football team, two out of every seven players are seniors. if the team has 84 players, how many are not seniors.

zaharov [31]

There are 60 players on football team that are not seniors.

Step-by-step explanation:

<u>Method 1:</u>

Given

Total students = 84

Also given that

2 out of 7 are seniors

Which is 2/7

So in order to find the total number of seniors we will multiply the whole number of students to 2/7

$Senior\ students = \frac{2}{7} * 84\\= 24$

So there are 24 seniors on the football team.

Players that are not seniors = 84 - 24 = 60

There are 60 players who are not seniors.

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We can divide the total number by 7, to get how many sets of 7 players will be there

= 84/7 = 12

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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.

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3 years ago