suprspi
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| Joined: 20 Oct 2014 |
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| 02 Nov 2014 10:07 PM |
I'm so stressed out. Haven't slept more then 4 hours in the last 3 days. My life is a mess.
Please.
The number N of bacteria in a culture is given by the model
N = 100e^kt
where t is the time (in hours). If N = 300 when t = 5, estimate the time required for the population to double in size.
WHAT I'M DOING
300 = 100e^k(5) ln3 = 5k 1.0986 / 5 = k = .2197
100e^(.2197)(5) = 299.96 hours, which is obviously wrong.
Please help... |
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suprspi
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| 02 Nov 2014 10:08 PM |
Answer is 3.15 hours, but I don't know how to get it. We have to include our work, and I can't do that if the work is wrong. |
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suprspi
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suprspi
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| 02 Nov 2014 10:12 PM |
300 = 100e^(k5) 300/100 = e^(k5) ln(3) = k5 k = ln(3)/5 = 0.219722458
So...
200 = 100e^0.219722458t 2 = e^0.219722458t ln(2) = 0.219722458t t = ln(2)/0.219722458 = 3.15464876 hours
rounded t = 3.15 hours. |
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| 02 Nov 2014 10:13 PM |
wolframalpha.com shows work solves problem
ur welcome |
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suprspi
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| 02 Nov 2014 10:14 PM |
omg thank you coolkid now i can get through like half my lesson knowing what to do |
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Mixernxer
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suprspi
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| 02 Nov 2014 10:17 PM |
mixxx can you do decays stuck on those too :( |
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suprspi
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Mixernxer
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suprspi
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| 02 Nov 2014 10:25 PM |
o yay okay heres da problem
The half-life of radioactive radium (226RA) is 1620 years. What percent of a present amount of radioactive radium will remain after 100 years?
would you still use the y = ce^kt equation or no |
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Mixernxer
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| 02 Nov 2014 10:30 PM |
It won't let me post the equation so Uhm ill try to make it different
S is equal to 100 times 1/2 to the power of (100 devided by total number of years) |
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suprspi
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Mixernxer
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| 02 Nov 2014 10:35 PM |
| so basically 100 is the original amount and whatever you get is the "percentage" of what remains. s = 100 * (.5)^100/total time |
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suprspi
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| 02 Nov 2014 10:45 PM |
agh I've been staring at that and still... :L |
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Mixernxer
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| 02 Nov 2014 10:49 PM |
uhm what is confusing you?
┬──┬ ︵(╯。□。)╯ |
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suprspi
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| 02 Nov 2014 10:52 PM |
when you try to explain it to me I like to see the actual equation
this is what i'm doing
y = ce^kt
50 = 100e^k100 ln(1/2) = 100k k = -.00693
now wat |
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Mixernxer
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| 02 Nov 2014 10:53 PM |
uhm why did you set it to 50? thats saying 50% will remain, but the equation doesnt work becuase 50 percent wont
┬──┬ ︵(╯。□。)╯ |
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