Question

Water flows into a tank according to the rate

Answer 1

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Answer 2

repaste the question here

Answer 3

@dan815

Answer 4

Plug 10 into where you see t. Then subtract the answer for the second equation E(t) from the answer from the first equation F(t).

Answer 5

@MxuesFire can you help me finish on this problem?

Answer 6

Sure. But I might have to go soon.

Answer 7

First equation, (t+6)/(1+t). Plug in 10 and tell me what you get:)

Answer 8

F(t)=(t+6)/(1+t) and E(t)=(ln(t+2)/t+1)
F(t)=(10+6)/(1+10) and E(t)=(ln(10+2)/10+1)
F(t)=(16)/(11) and E(t)=(ln(12)/11)
F(t)-E(t)=(16-(ln(12))/(11)
=1.22864485

Answer 9

@MxuesFire

Answer 10

what do i plug in 10?

Answer 11

Thats your answer for the first part.

Answer 12

Now just add gallons to the end and you are done:)

Answer 13

@MxuesFire why did i plug 10 in the first place?

Answer 14

Because you wanted to find out how much water was in the tank after 10 minutes. t stood for time.

Answer 15

Hi Iam here :)

Answer 16

How far have you gotten?

Answer 17

F(t)=(t+6)/(1+t) and E(t)=(ln(t+2)/t+1)
F(t)=(10+6)/(1+10) and E(t)=(ln(10+2)/10+1)
F(t)=(16)/(11) and E(t)=(ln(12)/11)
F(t)-E(t)=(16-(ln(12))/(11)
=1.22864485

Answer 18

essentially this is what i have done yet it seems to be wrong to me for some reason @dan815

Answer 19

Okay let me read the question

Answer 20

Wait, i found something

Answer 21

okay so we know the dv/dt

Answer 22

You had F(t)=16/11. You forgot to divide them.

Answer 23

got to go.

Answer 24

F(t)-E(t)=16-(ln(12) / 11
=1.22864485

Answer 25

@dan815

Answer 26

umm wat are u guys doing -.-

Answer 27

idk thats y i was confused

Answer 28

What we are given is dv/dt , and we dont know the initial volume of the tank, lets just assume is some constant

Answer 29

so we see that the total change in Volume over time
dv/dt= Rate in - Rate out
=(t+6)/(1+t) - (ln(t+2)/t+1)

Answer 30

yes

Answer 31

Now integrate to find V(t)