At 1:59 PM a car leaves an intersection heading East at a constant rate of 40 ft/sec. One minute later, at 2 PM, a second car leaves the same intersection heading North at the constant rate of 100 ft/sec. Find the rate at which the distance between the two cars is changing at 2:01 PM.

128.1 ft/s

Step-by-step explanation:

The first car at 1:59 PM moving at a rate of 40 ft/s heading East moves a distance of 40 ft/s × 1 s = 40 ft from 1:59 PM to 2:00 PM.

At the instant of 2:00 PM, the second car crosses the same intersection at a constant rate of 100 ft/s. From 2:00 PM to 2:01 PM which is 1 s, the first car moves a distance of 40 ft/s × 1s = 40 ft

The second car also moves a distance of 100 ft/s × 1 s = 100 ft.

So, the total distance moved by the first car at 2:01 PM is 40 ft + 40 ft = 80 ft and that moved by the second car is 100 ft.

Since their directions are perpendicular, we apply Pythagoras' theorem to find the distance between them at 2:01 PM.

So d = √(100² + 80²)

= √(10000 + 6400)

= √16400

= 128.06 ft

≅ 128.1 ft

So the rate at which the distance between the two cars is increasing at 2:01 PM is their distance at 2:01 PM divided by the time duration from 2:00 PM to 2:01 PM which their distance apart changed which is 1 s.

rate = d/t = 128.1 ft/1s = 128.1 ft/s

domain is all the x values included in the

0 ≤ x ≤ 2

range is all the y values of the graph

-2 ≤ y ≤ 2

***keep in mind that there are different ways to show domain and range

hope this !

~just a girl in love with shawn mendes

(0,,,0)

step-by-step explanation:

the table of x and y values is shown in the attached pic(part of question)

we have to decide according that graph

given points are (0, 0) (–1, 0), (2, 0)

when we graph these points on graph they show an increasing line as shown in the attached pic