At noon, ship A is 40 miles due west of ship B. Ship A is sailing west at 24 mph and ship B is sailing north at 22 mph. How fast (in mph) is the distance between the ships changing at 4 PM? (Note: 1 knot is a speed of 1 nautical mile per hour.)

Step-by-step explanation:

Distance traveled by A in 4 hours =

Distance traveled by B in 4 hours =

Total distance between A and the initial point of B is

Distance between A and B 4 hours later

From Pythagoras theorem we have

Differentiating with respect to time we get

So, the the distance between the ships at 4 PM is changing at

a=r

d=q

c=p

b=s

hope this !

the quotient is 4x^2 - 27x + 167 with a remainder of -996

step-by-step explanation:

in synthetic division, use -6 as the divisor and use the coefficients {4, -3, 5, 6}:

-6     )   4     -3     5     6

                -24   162 -1002

          4   -27   167   -996

thus, the quotient is 4x^2 - 27x + 167 with a remainder of -996.

note:   your "4x°-3x+5x+6" is ambiguous / incomplete.   i had to assume that you meant '4x^3 - 3x^2 + 5x + 6."