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This problem involves exponential decay, which follows the form:

\[
G(t) = G_0 \cdot e^{-kt}
\]

where:
- \(G(t)\) is the amount remaining at time \(t\),
- \(G_0\) is the initial amount,
- \(k\) is the decay constant,
- \(t\) is time in minutes.

Let’s go through the steps needed to solve the three parts of this problem.

---

### 1. **What is the half-life of the goo in minutes?**
We know:
- Initial amount: \(G_0 = 348 \, \text{grams}\),
- After 105 minutes: \(G(105) = 10.875 \, \text{grams}\).

The formula we need to use is:

\[
G(t) = G_0 \cdot e^{-kt}
\]

Plugging in the known values:

\[
10.875 = 348 \cdot e^{-105k}
\]

Solve for \(k\):

\[
\frac{10.875}{348} = e^{-105k}
\]

Take the natural logarithm of both sides:

\[
\ln \left( \frac{10.875}{348} \right) = \ln \left( e^{-105k} \right)
\]

\[
\ln \left( \frac{10.875}{348} \right) = -105k
\]

Now, calculate the left side:

\[
\ln \left( \frac{10.875}{348} \right) \approx \ln(0.03125) \approx -3.4657
\]

So:

\[
-3.4657 = -105k
\]

\[
k \approx \frac{3.4657}{105} \approx 0.03301
\]

Now, use the relationship between the half-life \(t_{1/2}\) and the decay constant \(k\):

\[
t_{1/2} = \frac{\ln(2)}{k}
\]

\[
t_{1/2} = \frac{0.693}{0.03301} \approx 21 \, \text{minutes}.
\]

---

### 2. **Find a formula for \(G(t)\):**

Using the formula:

\[
G(t) = G_0 \cdot e^{-kt}
\]

We found \(k \approx 0.03301\), so the equation becomes:

\[
G(t) = 348 \cdot e^{-0.03301t}.
\]

---

### 3. **How many grams of goo will remain after 80 minutes?**

Now, substitute \(t = 80\) into the formula:

\[
G(80) = 348 \cdot e^{-0.03301 \cdot 80}
\]

First, calculate the exponent:

\[
-0.03301 \cdot 80 \approx -2.6408
\]

Now:

\[
G(80) = 348 \cdot e^{-2.6408} \approx 348 \cdot 0.0711 \approx 24.75 \, \text{grams}.
\]

---

### Summary:
1. **Half-life**: \(21 \, \text{minutes}\).
2. **Formula for \(G(t)\)**: \(G(t) = 348 \cdot e^{-0.03301t}\).
3. **Remaining grams after 80 minutes**: \(24.75 \, \text{grams}\).

---

Would you like further details on any step or have more questions?

---

Here are 5 related questions to expand on this problem:
1. How does the decay constant change if the half-life decreases?
2. What is the amount of goo after 50 minutes?
3. How would the formula change if the initial amount was 500 grams?
4. What is the general form of exponential growth versus decay?
5. How do you interpret the value of \(k\) in real-life scenarios?

**Tip:** Exponential decay is often used to model radioactive substances, cooling processes, and even population decline in biology!

At the beginning of an experiment, a scientist has 348 grams of radioactive goo. After 105 minutes, the sample has decayed to 10.875 grams. What is the half-life of the goo in minutes? Find a formula for G(t), the amount of goo remaining at time t. How many grams of goo will remain after 80 minutes?

Learn how to calculate the half-life of radioactive goo starting with 348 grams that decays to 10.875 grams in 105 minutes. Step-by-step solutions include the exponential decay formula and calculations for remaining grams after 80 minutes. Suitable for high school students in Grades 10-12.

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