At noon, ship A is 130 km west of ship B. Ship A is sailing east at 25 km/h and ship B is sailing north at 20 km/h. How fast is the distance between the ships changing at 4:00 PM?

answer = 12.87 km/h

Step-by-step explanation:

Given

Ship A is sailing east at 25 km/h =

ship B is sailing north at 20 km/h =

here x and y are the  sailing at t = 4 : 00 pm for ship A and B respectively

so we get x = 4 ×25 =100 km/h

                 y = 4× 20 = 80 km/h

let z is the distance between the ships, we need to find at t = 4 hr

At noon, ship A is 130 km west of ship B (12:00 pm)

so equation will be

derivative first equation w . r. to t we get

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)

1. x> -5

2. x< 8

step-by-step explanation:

1. 2x> -10

divide by 2 from both sides of equation.

2x/2> -10/2

simplify, to find the answer.

-10/2=-5

x> -5 is the correct answer.

2. 3x< 24

divide by 3 from both sides of equation.

3x/3< 24/3

simplify, to find the answer.

24/3=8

x< 8 is the correct answer.

i hope this you, and have a wonderful day!