<span>30 hours
For this problem, going to assume that the actual flow rate for both pipes is constant for the entire duration of either filling or emptying the pool. The pipe to fill the pool I'll consider to have a value of 1/12 while the drain that empties the pool will have a value of 1/20. With those values, the equation that expresses how many hour it will take to fill the pool while the drain is open becomes:
X(1/12 - 1/20) = 1
Now solve for X
X(5/60 - 3/60) = 1
X(2/60) = 1
X(1/30) = 1
X/30 = 1
X = 30
To check the answer, let's see how much water would have been added over 30 hours.
30/12 = 2.5
So 2 and a half pools worth of water would have been added. Now how much would be removed?
30/20 = 1.5
And 1 and half pools worth would have been removed. So the amount left in the pool is
2.5 - 1.5 = 1
And that's exactly the amount needed.</span>
Answer:
Step-by-step explanation:
To rid the denominators, multiply both sides by 
.
We get:
Simplifying, we have:
Solving, we get:
Therefore, the solution with the greatest value is 
Answer:
Water in container B will be at 6.own
Step-by-step explanation:
calculate the volume in cont. A
VoA=400×6
=2400cm cube
VoB=420ml
=420cm cube [1ml=1cm cube]
now add both volumes
V=2400+420
=2820cm cube
Now,
2820 cm cube in jar B so
2820=420×h
h=6.71
30 is the answer. You do 10X 3 which is 30- 2 which is 28 and 28 to the nearest 10th is 30
Answer:
The answer is 114 which is the first answer which is A
Step-by-step explanation:
The rest 99, 87, and 25 just dont
Hope this helps you
