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Home » 4—Can Ron Jerry Really Rewrite Success? The Bold Truth Revealed! - Appcentric

4—Can Ron Jerry Really Rewrite Success? The Bold Truth Revealed! - Appcentric

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 But if \( y < -1 \), \( \frac{y+1}{y-1} > 0 \), valid. However, \( x^2 = \frac{y+1}{y-1} \geq 0 \) requires this fraction non-negative, which it is.

📌 But we must check if all such \( y \) are attainable. From \( x^2 = \frac{y+1}{y-1} \), for any \( y \ne 1 \), there's a real \( x \) if the right-hand side is non-negative — which holds when \( y > 1 \) or \( y \leq -1 \).

📌 But when \( y = -1 \), \( x^2 = 0 \Rightarrow x = 0 \), valid.

📌 As \( y \to 1^- \), \( \frac{y+1}{y-1} \to -\infty \), invalid.

📌 When \( y < -1 \), \( \frac{y+1}{y-1} > 0 \), e.g., \( y = -2 \Rightarrow x^2 = \frac{-1}{-3} = \frac{1}{3} \), real \( x \).

📌 But can \( y \) be arbitrary? Let’s solve for \( y \):

📌 y = \frac{x^2 + 1}{x^2 - 1} = 1 + \frac{2}{x^2 - 1}.

📌 As \( x^2 \to 1^+ \), \( y \to +\infty \); \( x^2 \to 1^- \), \( y \to -\infty \); \( x^2 \to \infty \

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1.08⁶ = 1.3605 × 1.1664 ≈ 1.5869
Rounded to nearest tenth: 23.8 cm
A = 300 × e^(-0.04×6) = 300 × e^(-0.24)
A ≈ 300 × 0.7866 = 235.98
At t =
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