<PRE><B>Question 65741</B>
The amount of water remaining after t days is given by A=A0e^-0.18 where A0 is the starting amount at the beginning of the time period. How much should be acquired now to have 40 grams remaining after 3 days?
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The formula A = Ao*e^-.18 seems to be missing the time factor t:
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I assume it is A =  Ao * e^-.18t
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Substitute for A and t:
40 = Ao * e^(-.18*3)
40 = Ao * e^-.54
:
Find e^-.54 on a good calc:
40 = Ao * .582748
:
Ao = 40/.582748
Ao = 68.64 grams is the initial amt



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