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Home » We require \( n_k \leq 50 \): - Ready Digital AB

We require \( n_k \leq 50 \): - Ready Digital AB

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Setze die Werte ein: \( L = -\left[ 1 \cdot \log(0.8) + 0 \cdot \log(0.2) \right] = -\log(0.8) \).

📌 Berechne \( \log(0.8) \approx -0.2231 \), also \( L \approx 0.2231 \).

📌 #### 0.2231

📌 ### Frage 8

📌 Ein Biomedizinischer Ingenieur modelliert die Spannungskonzentration \( S \) an einem Implantat als \( S = \frac{2 \gamma}{\pi a} \), wobei \( \gamma = 50 \) MPa und \( a \) der Rissradius in mm ist. Wenn die maximal zulässige Spannung 20 MPa beträgt, wie groß ist der maximale zulässige Rissradius?

📌 Setze \( S = 20 \): \( 20 = \frac{2 \times 50}{\pi a} \).

📌 Vereinfache: \( 20 = \frac{100}{\pi a} \).

📌 Löse nach \( a \): \( a = \frac{100}{20 \pi} = \frac{5}{\pi} \approx 1.5915 \) mm.

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Therefore, the answer is $oxed{1}$.
Question: What is the greatest common divisor of $3^{15} - 1$ and $3^9...
Solution: Use the property:
\gcd(a^m - 1, a^n - 1) = a^{\gcd(m,n)} - 1, \quad ext{for } a > 1.
\gcd(3^{15} - 1, 3^9 - 1) = 3^{\gcd(15,9)} - 1.
Now compute $\gcd(15,9) = 3$.
\gcd(3^{15} - 1, 3^9 - 1) = 3^3 - 1 = 27 - 1 = 26.
Therefore, the greatest common divisor is $oxed{26}$.**Question:** An...
Solution:** Start with the given equation:
rac{x+y}{x-y} + rac{x-y}{x+y} = 3.

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