Let's solve the equation \( \text{RQD} = 100e^{-0.1\lambda}(0.1\lambda + 1) \) for \( \lambda = 20 \).

1. **Substitute** \( \lambda = 20 \) into the equation:

\[
\text{RQD} = 100e^{-0.1(20)}(0.1(20) + 1)
\]

2. **Simplify** the exponents and the terms:

\[
\text{RQD} = 100e^{-2}(2 + 1)
\]

\[
\text{RQD} = 100e^{-2}(3)
\]

3. **Calculate** the value of \( e^{-2} \):

\[
e^{-2} \approx 0.1353
\]

4. **Multiply** the values:

\[
\text{RQD} = 100 \times 0.1353 \times 3
\]

\[
\text{RQD} \approx 100 \times 0.4059
\]

\[
\text{RQD} \approx 40.59
\]

So, the value of RQD when \( \lambda = 20 \) is approximately **40.59**.

Would you like further details or have any questions?

Here are 5 related questions you might find interesting:
1. How does the exponential decay factor \( e^{-0.1\lambda} \) affect the RQD value as \( \lambda \) increases?
2. What would RQD be if \( \lambda = 10 \)?
3. Can you generalize the behavior of RQD as \( \lambda \) approaches very large values?
4. How does the constant term 0.1 in front of \( \lambda \) influence the rate at which RQD decreases?
5. What happens to RQD if \( \lambda \) is very small, such as \( \lambda = 1 \)?

**Tip:** Exponential decay functions, such as \( e^{-x} \), rapidly decrease as \( x \) increases, which can significantly reduce the outcome of equations where such terms are present.

Learn how to solve the equation RQD = 100e^{-0.1\lambda}(0.1\lambda + 1) for \( \lambda = 20 \) with a detailed step-by-step solution. Understand the impact of exponential decay on the RQD value and the calculation process involved. Suitable for advanced students and professionals.

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