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A rare isotope of a nuclear material is very unstable, decaying at a rate of 15% each second. Find how much isotope remains 1...Question
A rare isotope of a nuclear material is very unstable, decaying at a rate of 15% each second. Find how much isotope remains 8 seconds after 5 grams of the isotope is created. Use y equals 5 left parenthesis 2.7 right parenthesis Superscript negative 0.15 t, where t represents time in seconds.
Question content area bottom
Part 1
There would be
enter your response here grams of the isotope.
(Round to the nearest tenth as needed.)
Question content area bottom
Part 1
There would be
enter your response here grams of the isotope.
(Round to the nearest tenth as needed.)
Answers
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Answered by
GPT-4o mini
To find how much of the isotope remains after 8 seconds, we'll use the given formula:
\[
y = 5 \cdot (2.7)^{-0.15t}
\]
In this formula, \( y \) represents the remaining amount of isotope, \( t \) is the time in seconds, and the initial amount of the isotope is 5 grams.
### Step 1: Plug in the value of \( t \)
We need to find \( y \) when \( t = 8 \):
\[
y = 5 \cdot (2.7)^{-0.15 \cdot 8}
\]
### Step 2: Calculate the exponent
First, calculate \( -0.15 \cdot 8 \):
\[
-0.15 \cdot 8 = -1.2
\]
### Step 3: Calculate \( (2.7)^{-1.2} \)
Next, calculate \( (2.7)^{-1.2} \):
Using a calculator:
\[
(2.7)^{-1.2} \approx \frac{1}{(2.7)^{1.2}} \approx \frac{1}{4.054} \approx 0.246
\]
### Step 4: Calculate \( y \)
Now, plug this value back into the equation:
\[
y = 5 \cdot 0.246 \approx 1.230
\]
### Step 5: Round to the nearest tenth
Finally, rounding to the nearest tenth, we get:
\[
y \approx 1.2 \text{ grams}
\]
So the final answer is:
There would be **1.2 grams** of the isotope remaining after 8 seconds.
\[
y = 5 \cdot (2.7)^{-0.15t}
\]
In this formula, \( y \) represents the remaining amount of isotope, \( t \) is the time in seconds, and the initial amount of the isotope is 5 grams.
### Step 1: Plug in the value of \( t \)
We need to find \( y \) when \( t = 8 \):
\[
y = 5 \cdot (2.7)^{-0.15 \cdot 8}
\]
### Step 2: Calculate the exponent
First, calculate \( -0.15 \cdot 8 \):
\[
-0.15 \cdot 8 = -1.2
\]
### Step 3: Calculate \( (2.7)^{-1.2} \)
Next, calculate \( (2.7)^{-1.2} \):
Using a calculator:
\[
(2.7)^{-1.2} \approx \frac{1}{(2.7)^{1.2}} \approx \frac{1}{4.054} \approx 0.246
\]
### Step 4: Calculate \( y \)
Now, plug this value back into the equation:
\[
y = 5 \cdot 0.246 \approx 1.230
\]
### Step 5: Round to the nearest tenth
Finally, rounding to the nearest tenth, we get:
\[
y \approx 1.2 \text{ grams}
\]
So the final answer is:
There would be **1.2 grams** of the isotope remaining after 8 seconds.
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