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Home » Quotient: \( x^2 - 6x + 8 \). - Midis

Quotient: \( x^2 - 6x + 8 \). - Midis

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 a_n = \frac{1}{n} \cdot \frac{n(n+1)}{2} - \frac{1}{3n^3} \cdot \left( \frac{n(n+1)}{2} \right)^2

📌 = \frac{n+1}{2} - \frac{1}{3n^3} \cdot \frac{n^2(n+1)^2}{4}

📌 Simplify the second term:

📌 \frac{1}{3n^3} \cdot \frac{n^2(n+1)^2}{4} = \frac{(n+1)^Question: A historian studying ancient navigation techniques discovers a triangular chart used by early mariners, with side lengths of $13$, $14$, and $15$ units. What is the length of the shortest altitude in this triangle?

📌 Solution: To find the shortest altitude, we first compute the area using Heron’s formula. Let the triangle have sides $a = 13$, $b = 14$, and $c = 15$. The semi-perimeter is:

📌 A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{21(21 - 13)(21 - 14)(21 - 15)} = \sqrt{21 \cdot 8 \cdot 7 \cdot 6} = \sqrt{7056} = 84

📌 The altitude corresponding to side $a$ is $h_a = \frac{2A}{a} = \frac{168}{13} \approx 12.92$, for $b$ it is $h_b = \frac{168}{14} = 12$, and for $c$ it is $h_c = \frac{168}{15} = 11.2$. The shortest altitude corresponds to the longest side, which is $15$, yielding:

📌 Question: A glaciologist models a cross-section of a glacier as a right triangle, where the hypotenuse represents the slope surface of length $z$, and the inscribed circle has radius $c$. If the ratio of the area of the circle to the area of the triangle is $\frac{\pi c^2}{\frac{1}{2}ab}$, express this ratio in terms of $z$ and $c$.

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