So ~ realizing the the fifty percent the quantity of 100mg would certainly be 50mg I put the 50 mg in my formula together the result. Thus,

\$50 mg=100e^k30\$

I divided both sides by 100:

\$frac50100=frac100e^k30100\$

\$ln 0.5=ln e^k30\$

\$fracln 0.530=k\$

k= -0.023

Answering part (b) should be easy after i find part (a)

(b) just how much of the sample remains after 100 years?

\$P(100)=100e^-0.023cdot 100\$

\$10.025 mg=100e^-0.023cdot 100\$

(c) After how long will just 1 mg remain?

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edited Aug 6 "15 at 13:15
sunny
inquiry Aug 6 "15 at 12:46

SunnySunny
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2
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Your answer come a) is correct.Now it is really easy to deal with b) and also c) becuase you recognize the degeneration equation: \$\$P(t)=P_0 e^-0.023 t\$\$

so, for b) friend have:\$\$P(100)=100 e^-2.3 approx 10 mbox mg\$\$and because that c) you have to solve:\$\$1=100 e^-0.023 t\$\$ and you find:\$\$t=-dfracln0.010.023 approx 199.3 mbox years\$\$

re-superstructure
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edited Aug 6 "15 in ~ 13:32
answered Aug 6 "15 at 13:26

Emilio NovatiEmilio Novati
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In a, you require to provide the systems of \$k\$, yet are otherwise correct. Because that b, the is no rounded correctly and also not offered as an answer, but OK. For c, usage the same equation together you have, \$1 mg=100 mg e^-0.23t\$ and also solve for \$t\$

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answer Aug 6 "15 in ~ 13:19

Ross MillikanRoss Millikan
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