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2 years ago

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The cost of a long-distance telephone call increases with the number of minutes of the call. Identify the independent and depend

ent quantity in the situation.

1 answer:

KonstantinChe [14]2 years ago

4 0

<u>The dependent and independent quality in the given situation:</u>

The given situation is that a long-distance telephone call cost increases with the number of minutes of the call. The two variables are dependent and independent.

The independent variable here in the given situation is the number of minutes of the call since it can be changed as much as we want

The dependent variable is the cost of the long-distance call as it relies on the number of minutes of the call. The cost increases when the number of minutes increases.

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¿Cuántos litros de agua se precisa para llenar una piscina inflable de 1/4 de metro cúbico?

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Answer:

250 litters

Step-by-step explanation:

To solve this question, we first need to know the conversion from cubic meter to liter.

1 cubic meter is equal to 1000 liters.

So, if we have 1/4 cubic meter, we can use a rule of three to find how many liters we will need:

1 m3 -> 1000 L

1/4 m3 -> x L

x * 1 = (1/4) * 1000

x = 1000 / 4 = 250 L

So we need 250 liters of water.

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2 years ago

Use a calculator to graph the functions y = –3|x – 2| and y = x3 – 3x – 3. How many points of intersection do the functions shar

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They have 3 intersections

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2 years ago

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The Florida Fish and Wildlife Conversation Commission (FWC) plans to hire hunters to kill the Burmese python, an invasive specie

andrew-mc [135]

Complete Question

The complete question is shown on the first uploaded image

Answer:

FWOC should hire 70,000 hunters to completely remove the python in the area

Step-by-step explanation:

From the question we are given that the Initial Pythons $P_o=$ 100,000

And the final python $P_f$ $=100,000+\frac{P}{10,000}(100,000-P)$

$P_f = 100,000 + 10P -\frac{P^2}{10,000}$

For Max Number of Python

=> $\frac{dP_f}{dP} = 0$

=> $\frac{dP_f}{dP} =$ $100,000 -2P = 0$

=> $P = 50,000$

Hence the maximum number of pythons $Max_P$ $=100,000 +\frac{50,000}{10,000} (50,000)$

$=350,000$

Using Mathematical relations

from the Question

5 python is killed by = 1 hunter

1 python is killed by = $\frac{1}{5} \ hunters$

350 , 000 pythons are killed by = $\frac{350,000}{5}$

$= 70,000 \ hunters$

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2 years ago

Phil bought 12 toy cars. Of these cars, 2 are red, 4 are white, 3 are yellow, and 3 are green. If Phil puts them all in a bag, w

insens350 [35]

The probability of Phil choosing a red to car is a 2 out of 12 chance. Whitten as 2:12. Hope this helps!

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2 years ago

A small radio transmitter broadcasts in a 30 mile radius. If you drive along a straight line from a city 33 miles north of the t

Paha777 [63]

Answer:

34.26miles from 49.58miles of the drive (69.11% of the drive).

Step-by-step explanation:

A problem of Analytic Geometry. This question can be solved using the resulting values from equaling the equations of the line (for the drive) and circle (with radius equals 30miles for radio transmitter broadcasting), and calculating the total distances from coincident points between circle and line, and total drive.

A graph is attached showing part of the circle and line with coincident points.

<h3>Line's Equation</h3>

Assuming transmitter is located (0, 0). From a city 33 miles north of the transmitter (0, 33) to a second city 37 miles east of the transmitter (37, 0).

Then, the equation for the line is (taking the two points from the start to the end):

$\\ y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}*(x-x_{1})$

$\\ y-33=\frac{0-33}{37-0}*(x-0)}$

$\\ y-33=\frac{-33}{37}*x}$

$\\ y=\frac{-33}{37}*x} + 33$ [1]

<h3>Circle's Equation</h3>

The equation for the circle whose center (0, 0) is:

$\\ (x-0)^2 + (y-0)^2 = 30^2$

$\\ x^2 + y^2 = 30^2$

$\\ y^2 = 30^2 - x^2$

$\\ y = \frac{+}{-}\sqrt{30^2- x^2}$ [2]

Equaling [1] and [2], to determine the points where starting to receive the radio transmitter signals to the point where finishing those signals:

$\\ \frac{-33}{37}*x + 33 = \sqrt{30^2- x^2}$

Solving this equation for <em>x</em>, we have two solutions for it (from <em>WolframAlpha</em>):

$\\ x_{1} = 40293/2458 - (111 \sqrt(80151))/2458 \approx 3.60775$

$\\ x_{2} = 40293/2458 + (111 \sqrt(80151))/2458 \approx 29.1774$

Then, using [1], the corresponding values for <em>y</em> are:

$\\ y_{1}=\frac{-33}{37}*(3.60775) \approx 29.78$

$\\ y_{2}=\frac{-33}{37}*(29.1774) \approx 6.97$

So,

$\\ (x_{1}=3.61, y_{1}=29.78)$

$\\ (x_{2}=29.18, y_{2}=6.97)$

Well, the distance of the drive is:

$\\ d = \sqrt{(x_{2}-x_{1})^2 + (y_{2}-y_{1})^2}$

$\\ d = \sqrt{(29.18-3.61)^2 + (6.97-29.78)^2}$

$\\ d \approx 34.2654$ miles.

The total distance traveled is:

$\\ d = \sqrt{(37-0)^2 + (0-33)^2}$

$\\ d \approx 49.5782$ miles

Thus, during the drive, the signal of the radio transmitter was picked up for 34.26miles from the total of the 49.58 traveled, that is, a fraction equivalent to $\\ \frac{34.2654}{49.5782} \approx 0.6911$ or 69.11% of the drive.

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2 years ago