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2004 P2 Q10(a) - HSN forum

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2004 P2 Q10(a)


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#1 sarmad

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Posted 16 May 2006 - 08:41 PM

hi i am having some problems with getting the answer, could somebody please show me how to this question, cheers. I got part (b) fine it was just part (a) that caused the problems.
sarmad

#2 aldo

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Posted 16 May 2006 - 09:26 PM

[attachmentid=189]

I am assuming this is the paper II question.

My solution is attached, hope this helps.

Attached Files



#3 ad absurdum

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Posted 16 May 2006 - 09:28 PM

QUOTE(sarmad @ May 16 2006, 09:41 PM) View Post

hi i am having some problems with getting the answer, could somebody please show me how to this question, cheers. I got part (b) fine it was just part (a) that caused the problems.
sarmad

It's after 1000 years, so t=1000, so:


\begin{align*}A_t = A_0e^{-0.002*1000} \\
A_t = A_0e^{-2} \\ \end{align*}

600 micrograms are left when t=1000, so:


\begin{align*} 600 = A_0e^{-2} \\
A_0 = \frac{600}{e^{-2}} \\
A_0 = 4433 \\ \end{align*}

Edit - Too slow tongue.gif
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#4 aldo

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Posted 16 May 2006 - 09:30 PM

Haha, luckily I had already done the solution for that some time ago, was nice and easy to post!!

#5 Steve

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Posted 17 May 2006 - 08:22 AM

Part (a) tells us that when t=1000, we have A_{1000}=600. We want to know A_0. Plugging this into the equation gives:



\begin{align*}600&=A_0e^{-2}\\
A_0&=600e^2\end{align*}

I hope this helps smile.gif
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#6 sarmad

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Posted 17 May 2006 - 08:29 AM

QUOTE(Steve @ May 17 2006, 09:22 AM) View Post

Part (a) tells us that when t=1000, we have A_{1000}=600. We want to know A_0. Plugging this into the equation gives:



\begin{align*}600&=A_0e^{-2}\\
A_0&=600e^2\end{align*}

I hope this helps smile.gif


cheers for that, i got that then for some reason i tried to get rid of the 600, then i confused myself. It was a lot easier than i thought at first!





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