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Home » These Tiny Name Tags Could Be Your Secret Weapon in Minecraft—Click to Discover! - Upplift

These Tiny Name Tags Could Be Your Secret Weapon in Minecraft—Click to Discover! - Upplift

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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📌 We count the number of such assignments where the maximum R position < the minimum L position.

📌 Let \( r_{\max} = \max(\text{R positions}) \), \( l_{\min} = \min(\text{L positions}) \). We require \( r_{\max} < l_{\min} \).

📌 Since positions are from 1 to 5, the possible values are limited.

📌 We can iterate over possible \( r_{\max} = m \) and \( l_{\min} = k \) with \( m < k \).

📌 Possible values: \( m \) ranges from 2 to 4, and for each \( m \), \( k \) ranges from \( m+1 \) to 5.

📌 But \( k \geq m+1 \), and \( m \geq 2 \), \( k \leq 5 \).

📌 For each such pair (m, k), we count the number of ways:

📌 Choose 2 distinct positions for R’s: both < m? No: R’s must be all \( < k \), and max \( R \leq m \), and \( m < k \), so all R positions ∈ [1, m-1]? No: if \( k \geq m+1 \), we need R’s all < k, but \( m < k \), so R’s can be in [1, k−1], but they must all be < k and their max < k, but we require max R < k — always true if R’s are in [1, k−1]. But we need **both R’s < k**, which is automatic if we choose from 1 to k−1, but the condition is only upper bound on L’s.

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\boxed{4}
Question:** A cylinder has a height equal to twice its radius. If the...
Let the original radius be \( r \), so the original height is \( h = 2...
Original volume:
V_1 = \pi r^2 h = \pi r^2 (2r) = 2\pi r^3
New dimensions:
New radius: \( 0.8r \)
New height: \( 1.5 \times 2r = 3r \)
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V_2 = \pi (0.8r)^2 (3r) = \pi (0.64r^2)(3r) = 1.92\pi r^3

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Denominator:
\frac{2x^4 + 2}{x^4 - 1} = 2 \cdot \frac{x^4 + 1}{x^4 - 1}
Let’s simplify $ x^4 + 1 $ and $ x^4 - 1 $. Since $ x + \frac{1}{x} =...
x^2 + \frac{1}{x^2} = 23 \Rightarrow x^4 + \frac{1}{x^4} = (x^2 + \fra...
x^4 + 1 = x^4 + \frac{1}{x^4} \cdot x^4 = x^4 + \frac{x^4}{x^4} = x^4...
Instead, observe:
\frac{x^2 + 1}{x^2 - 1} + \frac{x^2 - 1}{x^2 + 1} = y + \frac{1}{y} =...
Let’s compute this directly using the identity:
\frac{a + b}{a - b} + \frac{a - b}{a + b} = \frac{(a + b)^2 + (a - b)^...
\frac{2(x^2 + 1)}{x^2 - 1}

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