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Re: Ugh, I need help my Precal so bad...

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suprspi is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
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 is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
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 is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
02 Nov 2014 10:10 PM
:(
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suprspi is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
02 Nov 2014 10:12 PM
Bump
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CoolKid9387 is not online. CoolKid9387
Joined: 24 Feb 2010
Total Posts: 538
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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Excitedness is not online. Excitedness
Joined: 16 Oct 2012
Total Posts: 1822
02 Nov 2014 10:13 PM
wolframalpha.com
shows work
solves problem

ur welcome
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suprspi is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
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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CoolKid9387 is not online. CoolKid9387
Joined: 24 Feb 2010
Total Posts: 538
02 Nov 2014 10:15 PM
ya
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Mixernxer is online. Mixernxer
Joined: 06 Sep 2008
Total Posts: 14112
02 Nov 2014 10:16 PM
HIII I KNOW PRECAL!!
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suprspi is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
02 Nov 2014 10:17 PM
mixxx
can you do decays
stuck on those too
:(
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suprspi is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
02 Nov 2014 10:21 PM
mix
u ther
:C
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Mixernxer is online. Mixernxer
Joined: 06 Sep 2008
Total Posts: 14112
02 Nov 2014 10:23 PM
N=No(1/2)^t/h
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suprspi is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
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 is online. Mixernxer
Joined: 06 Sep 2008
Total Posts: 14112
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 is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
02 Nov 2014 10:31 PM
erm
im confuzzled
:C
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Mixernxer is online. Mixernxer
Joined: 06 Sep 2008
Total Posts: 14112
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 is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
02 Nov 2014 10:45 PM
agh
I've been staring at that and still...
:L
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Mixernxer is online. Mixernxer
Joined: 06 Sep 2008
Total Posts: 14112
02 Nov 2014 10:49 PM
uhm what is confusing you?


┬──┬ ︵(╯。□。)╯
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suprspi is not online. suprspi
Joined: 20 Oct 2014
Total Posts: 480
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 is online. Mixernxer
Joined: 06 Sep 2008
Total Posts: 14112
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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