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anonymous
 5 years ago
the rate at which a radioactive isotope disntegrates is porportional to the amount present. if 30g sample will contain only 20g after 10 minutes, what is the half life of the isotope, correct to the nearest 10th. y=yoe^kt. :)
anonymous
 5 years ago
the rate at which a radioactive isotope disntegrates is porportional to the amount present. if 30g sample will contain only 20g after 10 minutes, what is the half life of the isotope, correct to the nearest 10th. y=yoe^kt. :)

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0your formula is \[y=y_0e^{kt}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[y_0=30\] and you know that when t=10, y = 20

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so set \[20=30e^{10k}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and solve for k: \[20=30e^{10k}\] \[\frac{2}{3}=e^{10k}\] \[ln(\frac{2}{3})=10k\] \[k=\frac{ln(\frac{2}{3})}{10}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0thank you thank youi thank you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so now you know your formula is \[y=30e^{4.055t}\]and you want to know when you have half of what you started with. you started with 30 and half of that is 15 so if you want you can write \[15=30e^{4.055t}\] or just cut to the chase and write \[\frac{1}{2}=e^{4.055t}\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and now solve for t: \[ln(.5)=4.055t\] \[t=\frac{ln(.5)}{4.055}=\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0big mistake on my part

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ln(\frac{2}{3}}{10}=.04055\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so wrong from there on in. it is \[\frac{ln(.5)}{.04055}=15.32\]\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0very sorry i was off by two decimal places

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hope you are still here to see my correction
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