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.

tickets for the harlem globetrotters

show at michigan state university in

2010 cost $16, $23, or, for vip seats, $40.

total revenue = 16x + 23y + 40z

if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, if 9 times as many $16 tickets were sold

as vip tickets,

x = 9z => x – 9z = 0, sales of all three kinds of tickets would

total $46,575.

16x + 23y + 40z = 46575, how many of each kind of ticket would

have been sold?

solve for x, y and z

x − 9z = 0

x − y − z = 55

16x + 23y + 40z = 46575

x = 1170

y = 985

z = 130

welcome!

the y will be in still but your answer will be y x 3

step-by-step explanation:

jk: )