Awater tank can be emptied by using one pump for 4 hours. a second, smaller pump can empty the tank in 11 hours. if the larger pump is started at 1: 00 p. m., how much time should pass until the smaller pump is started so that the tank will be emptied at 4: 00 p. m.?

The second pump should be started at 13:15 hs.

Step-by-step explanation:

Pump number can empty the tank in 4 hours.

If it started at  1:00 PM then, at 5:00 PM would empty the tank.

Pump number 2, can empty the tank in 11 hours.

Pump 1, drains 1/4 of the tank each hour.

Pump 2, drains 1/11 of the tank each hour.

I need to complete the drainage in 3 hours.

The equation is. 3 hours using the pump one at a rate of 1/4 per hour plus unknown time using the second pump at a rate of 1/11 per hour is equal to the whole tank. (whole tank is 1 )

3*1/4 + X * 1/11 = 1

We are going to clear the equation and leave X.

So it is X = 11/4 . That is the time needed to have the large pump active.

That is 2 hours and 45 min.

So the second pump should be started at 4 pm - 2.45hs = 13:15 hs.

answer: domain and range mean all possible values of x and y that could be substituted to the inequality. a solution means all possible values that make the inequality statement true. so while the range is still all real numbers, the solution is the shaded area, for which y will always be greater than the line of the function.

answer: what do you mean