Afilter has been designed to have a run length of 48 hr while operating at a filtration rate of 12 m/h. the design backwash rate is 40 m/h, backwash duration is 15 min, and filter-to-waste duration is 10 min. the plant operators decided to clean the filters more thoroughly and they backwash at a rate of 55 m/h for 25 min. as a result, ripening takes longer and the filter-to-waste duration is 50 min. calculate the ufrv, ubwv, ufwv, and recovery (a) as designed and (b) as operated. what is the percent increase in the volume of treated water lost as waste backwash water and filter-to-waste water?

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."

option (b) is correct

step-by-step explanation:

to find the solution of this system, we need to convert it into reduced row-echelon form. we have,

after applying row operations r2 -> r2/2 and r3 -> r3/3, we get,

the correct answer is $7.38

step-by-step explanation: