Two hoses (x and y) are filling a pool. working alone at their individual constant rates, hose x could fill the pool in 6 hours and hose y could do it in 4 hours. if the two hoses work together to fill half the pool, and then hose y finishes alone, how long will it take to fill the pool?

Total time to fill pool is 3 hours 12 minutes

Step-by-step explanation:

Time taken for hose X to fill alone = 6 hours

Time taken for hose y to fill alone = 4 hours

Let the capacity of the pool = A

Ø  Work rate when X works alone = A/6

Ø  Work rate when Y works alone = A/4

Combined workrate when hoses X & Y work in conjunction will be

A/6 + A/4 = (4A + 6A)/24 = 10A/24

Therefore, combined rate = 5A/12

Notice that workrate = capacity/time

Ø  Time = capacity/workrate

Time taken when both hoses work = capacity/ combined workrate = A/(5A/12) = A * (12/5A) = 12/5

Time taken is 2.4 hours = 2 hours 24 minutes

Since both hoses work together to half the pool

Time taken to fill half the pool = 2 hours 24 minutes/2 = 1 hour 12 minutes

The remaining half is filled by Y alone

Total time to fill pool is 1 hour 12 minutes plus  4/2 hours (Y is filling up half the pool by itself) = 3 hours 12 minutes

13

step-by-step explanation:

distance formula is d=√(x2-x1)^2   +   (y2-y1)^2

i dont know sorry

step-by-step explanation: