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2 years ago

7

In triangle MPQ points N and R are located on sides MP and QP with NR drawn. which set of measurements below would justify that

triangle MPQ is similar to triangle NPR
number 9

1 answer:

NNADVOKAT [17]2 years ago

6 0

The answer is number 1. Because 12/10 =20/18 is a true statement. None of the others are true

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The vertices of a trapezoid are located at (1, 2), (3, 1), (3, 5), (1, 4). The trapezoid is translated 4 units to the right. Wha

padilas [110]

Answer:

Suppose that we have a given point (x, y)

If we translate this point N units to the right, then the new coordinates of the point are:

(x + N, y)

Ok, now if we know that the vertices of the trapezoid are:

(1, 2), (3, 1), (3, 5), (1, 4)

And we move the whole figure 4 units to the right, then all the vertices are moved 4 units to the right.

Then the new vertices of the figure will be:

(1 + 4, 2) = (5, 2)

(3 + 4, 1) = (7, 1)

(3 + 4, 5) = (7, 5)

(1 + 4, 4) = (5, 4)

Then the coordinates of the image of the trapezoid (of the new vertices) are:

(5, 2), (7, 1), (7, 5), (5, 4)

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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.

6 0

3 years ago

What is the area? Bit confused on this one.

Nutka1998 [239]

Answer:

A = 9x² + 12x

Step-by-step explanation:

The area (A) of the rectangle is calculated as

A = height × width

= 3(3x² + 4x) ← distribute parenthesis by 3

= 9x² + 12x

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3 years ago

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Tara used 8 centimeters of tape to wrap 4 presents. How many presents did Tara wrap if she used 10 centimeters of tape?

Akimi4 [234]

Answer:

She wrapped 5 presents

Step-by-step explanation:

She used 2cm to wrap each present

8cm = 4x

x is the amount of presents

divide by 4

x= 2 cm

Given 10 cm, divide by 2 to determine amount of x, or presents

7 0

3 years ago

Ingrid walked her dog and washed her car. The time she spent walking her dog was one-fourth the time it took her to wash her car

STALIN [3.7K]

1/4 of 14 is 7/2 which is 3 1/2 which is 3.5

4 0

3 years ago

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