Find the best answers to your questions at Westonci.ca, where experts and enthusiasts provide accurate, reliable information. Experience the ease of finding quick and accurate answers to your questions from professionals on our platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
To determine how long ago the Egyptian died, we can use the given exponential decay formula for carbon-14 in the mummy:
[tex]\[ y = y_0 e^{-0.0001216 t} \][/tex]
Here:
- [tex]\( y_0 \)[/tex] is the initial amount of carbon-14 in living human beings.
- [tex]\( y \)[/tex] is the amount of carbon-14 found in the mummy.
- [tex]\( t \)[/tex] is the time in years since the Egyptian died.
- The decay constant is [tex]\( 0.0001216 \)[/tex].
We know that the amount of carbon-14 found in the mummy is about two-fifths of the amount found in living beings, thus:
[tex]\[ y = \frac{2}{5} y_0 \][/tex]
Substituting this into the equation, we get:
[tex]\[ \frac{2}{5} y_0 = y_0 e^{-0.0001216 t} \][/tex]
Next, we can simplify by dividing both sides by [tex]\( y_0 \)[/tex]:
[tex]\[ \frac{2}{5} = e^{-0.0001216 t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides:
[tex]\[ \ln\left(\frac{2}{5}\right) = \ln\left(e^{-0.0001216 t}\right) \][/tex]
Since the natural logarithm and the exponential function are inverse functions, we can simplify further:
[tex]\[ \ln\left(\frac{2}{5}\right) = -0.0001216 t \][/tex]
Now, solve for [tex]\( t \)[/tex] by isolating it on one side:
[tex]\[ t = \frac{\ln\left(\frac{2}{5}\right)}{-0.0001216} \][/tex]
Evaluating this expression, we find:
[tex]\[ t \approx 7535.28562396509 \][/tex]
Rounding to the nearest integer gives us:
[tex]\[ t \approx 7535 \][/tex]
Therefore, the mummy had died approximately 7535 years ago.
About [tex]\( \boxed{7535} \)[/tex] years ago the Egyptian had died.
[tex]\[ y = y_0 e^{-0.0001216 t} \][/tex]
Here:
- [tex]\( y_0 \)[/tex] is the initial amount of carbon-14 in living human beings.
- [tex]\( y \)[/tex] is the amount of carbon-14 found in the mummy.
- [tex]\( t \)[/tex] is the time in years since the Egyptian died.
- The decay constant is [tex]\( 0.0001216 \)[/tex].
We know that the amount of carbon-14 found in the mummy is about two-fifths of the amount found in living beings, thus:
[tex]\[ y = \frac{2}{5} y_0 \][/tex]
Substituting this into the equation, we get:
[tex]\[ \frac{2}{5} y_0 = y_0 e^{-0.0001216 t} \][/tex]
Next, we can simplify by dividing both sides by [tex]\( y_0 \)[/tex]:
[tex]\[ \frac{2}{5} = e^{-0.0001216 t} \][/tex]
To solve for [tex]\( t \)[/tex], we take the natural logarithm of both sides:
[tex]\[ \ln\left(\frac{2}{5}\right) = \ln\left(e^{-0.0001216 t}\right) \][/tex]
Since the natural logarithm and the exponential function are inverse functions, we can simplify further:
[tex]\[ \ln\left(\frac{2}{5}\right) = -0.0001216 t \][/tex]
Now, solve for [tex]\( t \)[/tex] by isolating it on one side:
[tex]\[ t = \frac{\ln\left(\frac{2}{5}\right)}{-0.0001216} \][/tex]
Evaluating this expression, we find:
[tex]\[ t \approx 7535.28562396509 \][/tex]
Rounding to the nearest integer gives us:
[tex]\[ t \approx 7535 \][/tex]
Therefore, the mummy had died approximately 7535 years ago.
About [tex]\( \boxed{7535} \)[/tex] years ago the Egyptian had died.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. We're here to help at Westonci.ca. Keep visiting for the best answers to your questions.