At noon, ship A is 180 km west of ship B. Ship A is sailing east at 35 km/h and ship B is sailing north at 25 km/h. How fast is the distance between the ships changing at 4:00 PM?

Step-by-step explanation:

Gotta love these motion  problems!!! This one  is especially tricky!  At noon, ship A is 180 km west of ship B.  If ship A is traveling east, it is closing its distance to ship B (if ship B didn't move).  But ship B is moving north at the same time.  We need to find the distance each can travel in 4 hours using the d = rt formula.  For ship A.  We know it's moving at 35 km/h, so in 4 hours it can move d = 35(4) which is 140 km.  Remember that is closing the distance to ship B.  So it is 180 - 140 directly south of ship B.  This problem will turn out to be a right triangle problem of sorts.  The distance of 40 km serves as the base of this right triangle.  

With ship B moving north at 25 km/hr, it can travel d = 25(4) which is 100 km.  This serves as the height of the triangle.  We are asked to find the rate at which the distance is changing 4 hours from their starting points.  We know how far each can travel in those 4 hours, so in order to find the rate of change (which is the same thing as the slope!), we take the height and divide it by the base because that is the same thing as taking the rise over the run.  100/40 is 10/4 which is a rate of change of 2.5 km/hr

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.