-- The filler pipe can fill 1/6 of the pool every hour.
-- The drainer pipe can drain 1/10 of the pool every hour.
-- When they're filling and draining at the same time, the filler pipe
will win eventually, because it finishes more of the pool in an hour
than what the drain pipe can finish in an hour.
-- When they're filling and draining at the same time, then every hour,
1/6 of the pool fills and 1/10 of it empties. The difference is (1/6) - (1/10).
To do that subtraction, we need a common denominator.
The smallest denominator that works is 30.
1/6 = 5/30
1/10 = 3/30 .
So in every hour, 5/30 of the pool fills, and 3/30 of the pool empties.
The result of both at the same time is that 2/30 = 1/15 fills each hour.
If nobody notices what's going on and closes the drain pipe, it will take
<em><u>15 hours</u></em> to fill the pool.
If the drain pipe had <em><u>not</u></em> been open, the filler pipe alone could have filled
the pool <em><u>2-1/2 times</u></em> in that same 15 hours. With both pipes open,
1-1/2 pool's worth of water went straight down the drain during that time,
and it was wasted.
I would say that the school should take the cost of 1-1/2 poolsworth out
of Ms. Charles' pay at the rate of $5 a week. I would, but that would
guarantee her more job security than she deserves after pulling a stunt
like that.
I hope this did not take place in California.
tysm luv this #FleePointsForFlee thing!!!!! :))))
If you would like to order the following values in descending order, you can do this using the following steps:
- 3/4 = - 0.75
1/8 = 0.125
- 0.735
7/8 = 0.875
1/80 = 0.0125
0.056
The result is: 7/8, 1/8, 0.056, 1/80, - 0.735, -0.75.
108 = ( x)(48-x)
108 = 48x - x^2
48x -x^2 -108 = 0
I'm assuming you're doing solve the square right now, so you should be able to do the rest.
Answer:
{x}\lt\frac{{13}}{{2}}
Step-by-step explanation:
