To find (L(t))^-1, we replace t by [L(t)]^-1 and L(t) by t in the expression L(t) = 1.25t + 73
With this, we got: t = 1.25*(L(t))^-1 + 73, which implies [L(t)]^-1 = (t - 73)/1.25
B) If x = L(t), we got [L(x)]^-1 = (x - 73)/1.25 = (L(t) - 73)/1.25 = [(1.25t + 73) - 73]/1.25 = t
A) Therefore, [L(x)]^-1 is the the amount of time (in minutes) it takes to have x liter os liquid
C) And finally, if t = 125 we got: [L(125)]^-1 = (125 - 73)/1.25 = 41.6
