A cylindrical tank with radius 3 meters is filled with a solution. The volume at t minutes is given by V = (-t^2 + 12t + 24)Ï€ m3. At 4 minutes, there are d meters of unused vertical space in the tank. At 6 minutes, there are only 1/3d meters of space remaining. What is the height of the tank?
Choices
A 6*(5/9)
B 6*(8/9)
C 7 *(1/3)
D 7* (5/9)
E 8
[spoiler]Ans: B[/spoiler]
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Cylindrical tank
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Neo Anderson
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dear friend, check the question you have typed, seems to be a typo, cause assuming -t2 in
Volume of a cylinder is pie r^2 h = 9 x pie x h
from V@4 = (-(4)^2 + 12x4 +24)pie = 56 pie or 9 x pie x (56/9) => h - 56/9 = d --- (1)
from V@6 = (-(6)^2 + 12x6 +24)pie = 60 pie or 9 x pie x (60/9) => h - 60/9 = d/3 --- (2)
from (1) and (2) you get d = 2/3 and hence h = 62/9
pls do check and confirm
is -(t)^2 and alsoV = (-t2 + 12t + 24)Ï€ m3
is d/3, i get the height as 62/9 m which is not the option. here goes the solution:-1/3d
Volume of a cylinder is pie r^2 h = 9 x pie x h
from V@4 = (-(4)^2 + 12x4 +24)pie = 56 pie or 9 x pie x (56/9) => h - 56/9 = d --- (1)
from V@6 = (-(6)^2 + 12x6 +24)pie = 60 pie or 9 x pie x (60/9) => h - 60/9 = d/3 --- (2)
from (1) and (2) you get d = 2/3 and hence h = 62/9
pls do check and confirm
