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3 4 cup into half - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 But this diverges.

📌 Alternate interpretation from history: ENIAC had 20-digit precision per accumulator, 10 accumulators, 4 bits per digit → 20 × 4 = 80 bits = 10 bytes per register → 200 bytes

📌 Improved system: 16 registers × 2 bytes = 32 bytes — but ENIAC is 200

📌 But the question likely intends:

📌 ENIAC: 20 digits × 10 registers × (4 bits / 8 bits per byte) × 1 byte per 4 bits = 20 × 10 = 200 bytes

📌 Improved: 16 × 2 = 32 bytes

📌 But ENIAC occupies 200 bytes

📌 Yet the difference is not required.

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A = 25 \times 25 = 625
Thus, the maximum possible area is:
\boxed{625}
Compute \(\sum_{k=1}^{50} \frac{1}{k(k+1)}\).
The given sum can be simplified using partial fraction decomposition:
\frac{1}{k(k+1)} = \frac{1}{k} - \frac{1}{k+1}
Thus, the sum becomes a telescoping series:
\sum_{k=1}^{50} \left( \frac{1}{k} - \frac{1}{k+1} \right)
Writing out the terms:
\left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{...

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