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Home » Number of doubling periods = 18 ÷ 3 = 6 - High Altitude Science

Number of doubling periods = 18 ÷ 3 = 6 - High Altitude Science

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Solution: Original radius $ r = \sqrt{\frac{100\pi}{\pi}} = 10 \, \text{km} $. New radius $ r' = 8 \, \text{km} $. Original area $ 100\pi $, new area $ \pi (8)^2 = 64\pi $. The decrease is $ 100\pi - 64\pi = \boxed{36\pi} \, \text{km}^2 $.

📌 Question: A zoologist tracks an animal’s path forming a right triangle with hypotenuse $ 25 \, \text{m} $ and inradius $ 6 \, \text{m} $. What is the ratio of the inscribed circle’s area to the triangle’s area?

📌 Solution: For a right triangle, the inradius $ r = \frac{a + b - c}{2} $, where $ c $ is the hypotenuse. Let $ a $ and $ b $ be the legs. Then $ r = \frac{a + b - 25}{2} = 6 \Rightarrow a + b = 37 $. The area of the triangle is $ \frac{ab}{2} $, and the inradius formula $ r = \frac{a + b - c}{2} $ gives consistency. Using $ a^2 + b^2 = 625 $ and $ a + b = 37 $, solve for $ ab $: $ (a + b)^2 = 1369 = a^2 + b^2 + 2ab = 625 + 2ab \Rightarrow ab = 372 $. Area $ = 186 \, \text{m}^2 $. Circle area $ \pi r^2 = 36\pi $. Ratio $ \frac{36\pi}{186} = \boxed{\frac{6\pi}{31}} $.

📌 Question: A hydrologist calculates the angle of a groundwater flow slope using trigonometry. If the slope rises 1 meter over a horizontal distance of $ \sqrt{3} $ meters, what is $ \tan \theta $?

📌 Solution: The slope forms a right triangle with opposite side 1 and adjacent side $ \sqrt{3} $. Thus, $ \tan \theta = \frac{1}{\sqrt{3}} = \boxed{\frac{\sqrt{3}}{3}} $.Question: What is the remainder when the sum $ 1^3 + 2^3 + 3^3 + \dots + 10^3 $ is divided by 9?

📌 Solution: The formula for the sum of the first $ n $ cubes is $ \left( \frac{n(n+1)}{2} \right)^2 $. For $ n = 10 $, we compute:

📌 \left( \frac{10 \cdot 11}{2} \right)^2 = (55)^2 = 3025

📌 Now, compute $ 3025 \mod 9 $. Since $ 55 \equiv 1 \mod 9 $, $ 55^2 \equiv 1^2 = 1 \mod 9 $.

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Soit \( x \) le nombre de gadgets A et \( y \) le nombre de gadgets B.
À partir du temps d'assemblage :
\[ 2x + y = 10 \]
À partir du temps de test :
\[ x + 2y = 8 \]
Résoudre le système d'équations :
À partir de la première équation, résoudre pour \( y \) :
\[ y = 10 - 2x \]
\[ x + 2(10 - 2x) = 8 \]
\[ x + 20 - 4x = 8 \]

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