Water is being pumped into a 10-foot-tall cylindrical tank at a constant rate. the depth of the water increases linearly

please help soon I have been stuck on this problem for a really long time.

at 1:30 pm the water depth was 2.4ft

it is now 4:00 and the water depth is 3.9 ft

what will the water depth be at 5:00?

The depth of water at 5:00 is 4.5 ft.

Given that,  

Now  

Depth increased in 2.5 hours is

= 3.9 - 2.4

= 1.5 ft

Depth increased in 1 hour = = 0.6 ft

Depth at 5:00 PM = Depth at 4:00 PM + Depth increased in 1 hour

= 3.9 + 0.6

= 4.5 ft

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The depth of water at 5:00 is 4.5 ft.

Step-by-step explanation:

Depth of water at 1:30 PM = 2.4 ft

Depth of water at 4:00 PM = 3.9 ft

Change in time = 4 - 1:30 = 2.5 hours

Depth increased in 2.5 hours = 3.9 - 2.4 = 1.5 ft

Now, we are given that the depth of water is increasing linearly, so after 4:00 PM as well, the water depth will keep on increasing with the same rate as that of previous time intervals.

We have to find out the depth at 5:00 PM i.e. we need to find out the increase from 4:00 PM to 5:00 PM (1 hour) and then add it to depth at 4:00 PM.

Depth increased in 2.5 hours = 1.5 ft

Depth increased in 1 hour = 1.5 / 2.5 = 0.6 ft

Depth at 5:00 PM = Depth at 4:00 PM + Depth increased in 1 hour

3.9 + 0.6 = 4.5 ft

Depth increased in 2.5 hours = 1.5 ft

So, The depth of water at 5:00 is 4.5 ft.

the balance is found at the link

step-by-step explanation:

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine