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Home » Ungefähr \( 90 \cdot 3.14 = 282.6 \) m³. - Network4

Ungefähr \( 90 \cdot 3.14 = 282.6 \) m³. - Network4

📅 May 18, 2026 👤 scraface

May 18, 2026

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📌 b_k = M\left( \frac{1}{k} \right) + M\left( \frac{2}{k} \right) + \cdots + M\left( \frac{k}{k} \right).

📌 Find $ b_k $ in simplest form.

📌 M(v) = v - \frac{v^4}{4},

📌 b_k = \sum_{j=1}^{k} M\left( \frac{j}{k} \right) = \sum_{j=1}^{k} \left( \frac{j}{k} - \frac{1}{4} \left( \frac{j}{k} \right)^4 \right).

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- 48x - 30x + 4x^2 = 252
4x^2 - 78x + 360 = 252
4x^2 - 78x + 108 = 0
Dividing the entire equation by 2:
2x^2 - 39x + 54 = 0
Solving this quadratic equation using the quadratic formula \( x = \fr...
x = \frac{39 \pm \sqrt{39^2 - 4 \cdot 2 \cdot 54}}{4}
x = \frac{39 \pm \sqrt{1521 - 432}}{4}
x = \frac{39 \pm \sqrt{1089}}{4}
x = \frac{39 \pm 33}{4}

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