the amount of a radioactive tracer remaining after t days is givin by A=Ao e^-0.18t, where Ao is the starting?

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amount at the beginning of the time period. How much should be aquired now to have 40 grams remaining after 3 days?

2007-02-08 19:49:30 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics

Ao = A*e^0.18t where Ao is the starting amount and A the remainder

Ao =40 e^0.54 = =68.64 grams

2007-02-08 23:09:50 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋

Solve 40 = Ao exp(-0.54). A scientific calculator will do it easily.

2007-02-09 03:57:52 · answer #2 · answered by Anonymous · 0⤊ 0⤋

40 = Ao e^(-0.18t)
40 = Ao e^(-0.54)
40 e^(0.54) = Ao
Ao = 68.6
Should start with 68.6 g.

2007-02-09 05:16:08 · answer #3 · answered by Como 7 · 0⤊ 0⤋

A=Ao e^-0.18t
40=Ao e^-0.18(3)
40 = Ao (e^-0.54)
40 = Ao (0.5827)
Ao = 68.64g

2007-02-09 03:59:50 · answer #4 · answered by seah 7 · 0⤊ 0⤋