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Nov 4th, 2002, 07:25 PM
#1
Thread Starter
Stuck in the 80s
[Resolved] Easy Question, The Sequel
Okay, I'm presented with this problem:
The half-life of element X is 9 hours. If 80mg are present now, how much will exist after 2 days?
So I used the formula A(t) = A0 * e ^ kt
Code:
.5 = e ^ 9k
ln .5 = 9k
k = ln (.5)/9
k = -0.0077
So now I have A(t) = A0 * e ^ -.0077t
So I do:
Code:
A = 80e ^ -.0077(48)
A = 3810.55
Which obviously isn't correct. What am I doing wrong?
Last edited by The Hobo; Nov 5th, 2002 at 12:18 AM.
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Nov 4th, 2002, 07:31 PM
#2
Thread Starter
Stuck in the 80s
Nevermind.
I forgot to put these parenthesis:
Code:
A = 80e ^ (-.0077(48))
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Nov 4th, 2002, 07:33 PM
#3
Thread Starter
Stuck in the 80s
hmm...that gives me 55.28, which still doesn't seem right. Any ideas?
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Nov 4th, 2002, 09:47 PM
#4
transcendental analytic
Use  
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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Nov 4th, 2002, 09:49 PM
#5
Thread Starter
Stuck in the 80s
Originally posted by kedaman
-0.0770
Bah. Thanks kedaman 
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Nov 4th, 2002, 09:52 PM
#6
transcendental analytic
'twas an optical illusion  ..waaaah.. liquidgeneration sabotage!
Use
writing software in C++ is like driving rivets into steel beam with a toothpick.
writing haskell makes your life easier:
reverse (p (6*9)) where p x|x==0=""|True=chr (48+z): p y where (y,z)=divMod x 13
To throw away OOP for low level languages is myopia, to keep OOP is hyperopia. To throw away OOP for a high level language is insight.
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