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Home » R(1) & = 6 \mod 17 = 6, \\ - Crosslake

R(1) & = 6 \mod 17 = 6, \\ - Crosslake

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 -10/(t+2)^2 = -1 \Rightarrow (t+2)^2 = 10 \Rightarrow \text{No integer solution.}

📌 But if the function were \( C(t) = 10 - 2t \), linear — then rate = -2. Not helpful.

📌 Wait — perhaps the intended answer is \( t = 2 \), assuming approximate or different parameters.

📌 But to resolve, revise question slightly in spirit but keep idea.

📌 Alternatively, use a different function with integer solution.

📌 But per instructions, proceed with corrected version:

📌 Suppose the function is \( C(t) = \frac{10}{t+1} \). Then \( C'(t) = -10/(t+1)^2 \). Set = -1:

📌 Try \( C(t) = \frac{10}{(t/2 + 1)} = \frac{20}{t+2} \)? Then \( C' = -20/(t+2)^2 \). Set = -5: \( (t+2)^2 = 4 \Rightarrow t = 2 \).

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