Water is pumped from a tank at constant rate, and no more water enters the tank. If the tank contains 10452 L at 3:56PM and 8092 L at 4:12PM the same day, at what time will the tank contain only 6617 L?

35 minutes from 4:12 pm at 4:47 pm the tank will contain only 6617 liters of water.

Step-by-step explanation:

3:56-4:12 = -00:56 minutes passed

10452-8092=2360 liters of water lost

8092-6617=1475 liters

meaning: 56 mins lost 2360L

                x mins lost 1475L

(the 2360 liters lost)/(the 56 minutes during which they were lost) is equal to the rate at which (the 1475 liters would be lost)/(the unknown amount of time)

2360/56 = 1475/x                               (cross multiply)

2360x = 1475(56)                                            

2360x = 82600                                   (simplify)

x = 82600/2360                                  (divide to isolate x)

x = 35                                                    (the unknown amount of time during                             which the water is lost)

4:12 pm plus 35 minutes is 4:47 pm.

f-1(x)= cube root of 19x squared/x

step-by-step explanation:

idk

step-by-step explanation: