susyyy21
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One airplane is located 200 km north and 50 km east of an airport. A second plane at the same altitude is located 30 km north and 100 km north and 100 km west.
The distance between the planes is closest to:
A. 150 km
B. 200 km
C. 300 km
D. 350 km
E. 400 km

Respuesta :

facundo3141592

Answer: B

Step-by-step explanation:

We can define the North as our positive y-axis, and the East as the positive x-axis.

The position of the airport is the (0, 0)

Then the position of the first plane is: (200 km north and 50 km east)

(50km, 200km)

The position of the other plane is: (30 km north and 100 km west)

(-100km, 30km)

Now, if we have two points (a, b) and (c, d)

The distance between those points is:

D = √( (a - c)^2 + (b - d)^2)

Then the distance between the planes is:

D = √( (50km - (-100km))^2 + (200km - 30km)^2)

D = √( (150km)^2 + (170km)^2)

D = 226.7km

Then the distance is closest to 200km, the correct option is B.

MrRoyal

On a coordinate plane, we have:

  1. North represents the positive y-axis
  2. South represents the negative y-axis
  3. East represents the positive x-axis
  4. West represents the negative x-axis

The closest distance between the planes is 200km

The positions of the two planes is given as:

[tex]A = (50,200)[/tex] --- 50km east and 200km north

[tex]B = (-100,30)[/tex] --- 100km west and 30km north

The distance between the planes is calculated using the following distance formula:

[tex]d = \sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2}[/tex]

So, we have:

[tex]d = \sqrt{(50--100)^2 + (200-30)^2}[/tex]

[tex]d = \sqrt{150^2 + 170^2}[/tex]

[tex]d = \sqrt{51400}[/tex]

[tex]d = 226.72km[/tex]

From the list of given options, 200km is the closest to 226.72km

Hence, the closest distance between the planes is 200km

See attachment for the positions of both planes

Read more about distance at:

https://brainly.com/question/16310813

Ver imagen MrRoyal

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