Answer:
The liters that the tank will contain at 5:11 PM that day are:
Step-by-step explanation:
Firstly, you must identify the outlet flow of the water pumped from the tank, for this, you must subtract the last volume given from the first volume:
- 19,140 L - 8,097 L = 11,043 L
And the minutes that passed from the first volume until the last volume given (18 minutes from 4:47 PM to 5:05 PM), so, you must divide that two values to obtain the outlet flow:
- Outlet flow =
- Outlet flow =
- Outlet flow = 613.5
Now, you must see the next hour given (5:11 PM), if you see, from 5:05 PM to 5:11 PM has passed 6 minutes, taking into account this, you replace the equation of outlet flow to clear the volume:
- Outlet flow =
- Volume = Outlet flow * time
And replace the values to obtain the new volume pumped:
- Volume = 613.5 * 6 min
- Volume = 3681 L.
At last, you must subtract these liters from the last volume identified in the tank:
- New Volume in the tank = 8097 L - 3681 L
- New Volume in the tank = 4416 L
The volume in the tank at 5:11 PM is <u>4416 Liters</u>.
X y
0 0
1 1
2 8
3 27
The answer is y=x^3
Ans
It 1750
Step-by-step explanation:
You have to multiple
Answer: 6.4
Step-by-step explanation:
you take help of ratio in this case.
8/x = 2.5/2
or, (2.5).(x) = (8).(2)
or, x = 16/2.5
so, x = 6.4
Answer:
4
Step-by-step explanation:
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