Set $ C(t) = \frac10(t+3) $. Then \( C' = -10/(t+3)^2 = -1 \Rightarrow (t+3)^2 = 10 \Rightarrow t \approx 1.16 — no. - AMAZONAWS

📅 March 6, 2026 👤 scraface

Mar 06, 2026

$Set $ C(t) = \frac{10}{(t+3)} $. Then \( C' = -10/(t+3)^2 = -1 \Rightarrow (t+3)^2 = 10 \Rightarrow t \approx 1.16 — no.$