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Precalculus 17 Online
OpenStudy (anonymous):

a chemical storage tank is to be filled by three pipes. the first two transmit 930 L/h of liquid each and the third transmits 560 L/h. if the third pipe starts working 90 minutes after the first two begin and the tank has a capacity of 4000 L, how long (in hours) will it take after the first two pipes begin working to fill the tank?

OpenStudy (anonymous):

we have two different rates first period: 2x 930L/h = 1860 L/h second period: 1860 L/h + 560 L/h = 2420 L/h

OpenStudy (anonymous):

also, we know the first period lasted 90m which is 1,5 hours tank so far: 1,5 x 1860 = 2790 L

OpenStudy (anonymous):

tell me if everything is clear so far

OpenStudy (anonymous):

yes I understand

OpenStudy (anonymous):

theres an error

OpenStudy (anonymous):

where?

OpenStudy (anonymous):

I should not have divided by the time it takes, but multiplied

OpenStudy (anonymous):

we are still 4000L - 2790 L = 1210 L from the tank being filled the remaining 1210 L are filled at the second rate of 2420 L/h the time it takes: \[1210L = \frac{ 2420 L }{ 1h } \times x h\]

OpenStudy (anonymous):

if we input at 2420L/h for one hour, we get 2420L if we input at 2420L/h for half an our, we get... 1210 L

OpenStudy (anonymous):

the equation is satisfied with "additional hours": x = 0.5 \[1210L = \frac{ 2420 L \times 0.5 h}{ 1h } =\frac{ 1210 L }{ 1 }\]

OpenStudy (anonymous):

so 1210 L

OpenStudy (anonymous):

wait i need hours though ...how do i find that?

OpenStudy (anonymous):

"how long (in hours) will it take after the first two pipes begin working to fill the tank?" it will take 90 m + 30 m to fill the tank

OpenStudy (anonymous):

oh ok

OpenStudy (anonymous):

because the x value we have chosen is 0.5 that is half an hour additional to 90m already passed

OpenStudy (anonymous):

go it

OpenStudy (anonymous):

in total it takes 1.5 + 0.5 = 2 hours

OpenStudy (anonymous):

thanks

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