Since the remaining water changes quadractically with time,
litres remaining can be written as ` L = at^2 + bt + c` where `t` is time since you pulled the plug.
At `t=1` `at^2 + bt + c = a + b + c= 38.4`
At `t=2` `at^2 + bt...
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Since the remaining water changes quadractically with time,
litres remaining can be written as ` L = at^2 + bt + c` where `t` is time since you pulled the plug.
At `t=1` `at^2 + bt + c = a + b + c= 38.4`
At `t=2` `at^2 + bt + c = 4a + 2b + c = 21.6`
At `t=3` `at^2 + bt + c = 9a + 3b + c = 9.6`
Reduce to Echelon form by subtracting row 1) from rows 2) and 3)
1) 1 1 1 38.4 1) 1 1 1 38.4
2) 4 2 1 21.6 2)-4*1) 0 -2 -3 -132
3) 9 3 1 9.6 3)-9*1) 0 -6 -8 -336
Then multiplying row 2) by -1/2
1) 1 1 1 38.4
2) 0 1 3/2 66
3) 0 -6 -8 -336
Then adding row 2) to row 3)
1) 1 1 1 38.4
2) 0 1 3/2 66
3)+6*2) 0 0 1 60
` `
From equation 3) `c = 60`
substituting into equation 2) `b + 3/2(60) = 66 implies b = -24`
substituting into equation 1) `a -24 + 60 = 38.4 implies a = 2.4`
Therefore the equation for remaining litres after t minutes is given by
L = 2.4^2 -24t +60
Check the solution at 1,2 and 3 minutes to confirm