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Home » \fracV_\textcylV_\textcube = \frac36\pi r^3216r^3 = \frac36\pi216 = \frac\pi6 - jntua results

\fracV_\textcylV_\textcube = \frac36\pi r^3216r^3 = \frac36\pi216 = \frac\pi6 - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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rac{p+q}{p-q} + rac{p-q}{p+q} \geq 2 \Rightarrow R = \left( rac{p+q}...
ight) - 2 \geq 2 - 2 = 0,
with equality only if $ rac{p+q}{p-q} = rac{p-q}{p+q} $, which impli...
R = rac{x + 1}{x - 1} + rac{x - 1}{x + 1} - 2.
Let $ f(x) = rac{x+1}{x-1} + rac{x-1}{x+1} - 2 $. Combine:
f(x) = rac{(x+1)^2 + (x-1)^2}{(x-1)(x+1)} - 2 = rac{2x^2 + 2}{x^2 -...
ight) = 2\left( rac{2}{x^2 - 1}
ight) = rac{4}{x^2 - 1}.
Since $ x > 1 $, $ x^2 - 1 > 0 $, so $ f(x) > 0 $, and as $ x o 1^+ $...
rac{a}{b} + rac{b}{a} - 2 \geq 0,

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This means $ x + 3 $ is a common multiple of $ 7, 8, $ and $ 9 $.
First, compute the least common multiple (LCM) of $ 7, 8, $ and $ 9 $:
\text{lcm}(7, 8, 9) = \text{lcm}(7, \text{lcm}(8, 9)) = \text{lcm}(7,...
So $ x + 3 = 504k $ for some integer $ k $.
We want $ x $ to be a two-digit number, so:
x = 504k - 3
Try $ k = 1 $:
x = 504(1) - 3 = 501 \quad \text{(too large)}
Try $ k = 0 $:
x = 504(0) - 3 = -3 \quad \text{(not positive)}

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