MHB Summation Problems: Solve Delta ti * T(Delta ti) for Age in Days

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The discussion revolves around solving the equation involving Delta ti, which represents time in days at specific temperatures, and T(Delta ti), the temperature during that time. The user is struggling to obtain meaningful results, as their calculations yield values below one day, which seems illogical. Another participant provides an example calculation using a temperature of 27°C and a Delta ti of 10 days, resulting in approximately 13.8 days. The conversation seeks clarification on the original equation and its application to achieve correct age in days.
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Hello,

Some of you may know this equation and I need help solving it.
Delta ti is the time in days at a certain temperature (0 - 80).
T(Delta ti) is the temperature during Delta ti.

The answer is supposed to be an age in days but my tries have given me answers that are below 1, which doesn't make any sense. Can someone please try to open this up for me?
 

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Cer said:
Hello,

Some of you may know this equation and I need help solving it.
Delta ti is the time in days at a certain temperature (0 - 80).
T(Delta ti) is the temperature during Delta ti.

The answer is supposed to be an age in days but my tries have given me answers that are below 1, which doesn't make any sense. Can someone please try to open this up for me?

Hi Cer! Welcome to MHB! :)

Mmm... let's try $T(\Delta t_i)=27^\circ \text C$ with $\Delta t_i=10 \text{ days}$.

Then we get:
$$e^{-(4000/300-13.65)} \cdot \Delta t_i= e^{0.32} \cdot 10 \approx 13.8 \text{ days}$$

What is the problem? :confused:
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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