At noon, ship A is 120 km west of ship B. Ship A is sailing east at 20 km/h and ship B is sailing north at 15 km/h. How fast is the distance between the ships changing at 4:00 PM?

  1.39 km/h

Step-by-step explanation:

Let the initial position of ship B represent the origin of our coordinate system. Then the position of ship A as a function of time t is ...

  A = -120 +20t . . . (east of the origin)

and the position of B is ...

  B = 15t . . . (north of the origin)

Then the distance between them is ...

  d = √(A² +B²) = √((-120 +20t)² +(15t)²) = √(625t² -4800t +1440)

And the rate of change is ...

  d' = (625t -2400)/√(625t² -4800t +14400)

At t = 4, the rate of change is ...

  d' = (625·4 -2400)/√(625·16 -4800·4 +14400) = 100/√5200 = 1.39 . . . km/h

The distance between the ships is increasing at about 1.39 km/h.

step-by-step explanation:

6.48/4 4 goes into 6.48 1.62 times

2160ft3

step-by-step explanation: