-- 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.
So WX is 7y +4 and XY is 21, and when you add those lines you get the whole thing, which is 12y so write that like an equation
12y = 7y+4+21
12y = 7y+25
5y = 25
y = 5
Answer:
-8100
Step-by-step explanation:
(2/5)^-2(-3)^4
Given data
and second term
The first term= (2/5)^-2 and
Second term=(-3)^4
Simplify the first term
(2/5)^-2= 2^-2/ 5^-2= 1/2^2 * 1/5^2
=1/4/ 1/25
=1/4*25/1
=100
Simplify the second term
=(-3)^4
= -81
Hence, 100*81
=-8100
Answer:
x
=
5
±
i
√
11
/6
Step-by-step explanation:
The roots (zeros) are the x values where the graph intersects the x-axis. To find the roots (zeros), replace y with 0 and solve for x
.
