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 -24****t +60**

**Check the solution at 1,2 and 3 minutes to confirm**