suprspi
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| Joined: 20 Oct 2014 |
| Total Posts: 480 |
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| 02 Nov 2014 11:12 PM |
All I have.
PRECAL QUESTION:
The half-life of radioactive radium (226Ra) is 1620 years. What percent of a present amount of radioactive radium will remain after 100 years?
WHAT I HAVE DONE SO FAR y = ce^kt
50 = 100e^k620
ln(1/2) = k1620 k = -0.000428
y = ce^-0.000428(t)
Answer in the end SHOULD BE 95.8% but i have no idea how to get to it :( |
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suprspi
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| Joined: 20 Oct 2014 |
| Total Posts: 480 |
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suprspi
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| Joined: 20 Oct 2014 |
| Total Posts: 480 |
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suprspi
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| Joined: 20 Oct 2014 |
| Total Posts: 480 |
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suprspi
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| Joined: 20 Oct 2014 |
| Total Posts: 480 |
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| 02 Nov 2014 11:20 PM |
ermg l2k why must roblox block ur replies :( |
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suprspi
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| 02 Nov 2014 11:22 PM |
| ROBLOX STAWP BLOCKING HIS POSTS DIS BOY NEEDS HELP |
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suprspi
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QuitBye
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| Joined: 03 Apr 2008 |
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| 02 Nov 2014 11:24 PM |
Its not an e^k one Its a half life, which would need (0.5)^(t/1620)
So the setup would be y = s(0.5)^(100/1620)
Because you were given the half life, you know it'll decrease by exactly 0.5 -- 0.5^1 -- in the 1620th year. That gives the 0.5^(t/1620). Then you're given a time of 100 years. So t = 100.
You take the original value - 100% - and plug that into the formula for the value after 100 years:
y = (100)(0.5)^(100/1620) = (100)(0.95811) (approximately) = 95.811 (approximately)
So approx 96% will be left after the 100 years
WOOHOO. https://www.youtube.com/watch?v=cBVeNQ-mtcY |
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| 02 Nov 2014 11:25 PM |
| I have percale and I'm at function right now xd |
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suprspi
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| Joined: 20 Oct 2014 |
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| 02 Nov 2014 11:27 PM |
omg tyyyy l2k make dat t-shirt pls omg ty I'm struggling so much in precal this year skipped algebra 2 so I'm freaking out over this stuff |
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| 02 Nov 2014 11:41 PM |
Javascript is crazy useful xd Had to split everything with invisible characters!
But yeye I loved my calc course lolol |
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