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Home » Stop Losing Races—Mario Kart World Switch 2 Beat Every Track Like a Pro! - Abbey Badges

Stop Losing Races—Mario Kart World Switch 2 Beat Every Track Like a Pro! - Abbey Badges

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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📌 Thus, the only logical resolution is that the minimum condition is **redundant** or **not strictly enforced**, but the problem includes it to suggest concavity. But $ d(t) = t^3 $ has no minimum.

📌 Wait — perhaps the oceanographer uses a cubic that fits the trend but dips? But interpolation at $ t=1,2,3,4 $ with $ d(n)=n^3 $ forces $ d(t)=t^3 $.

📌 Unless the problem has a typo, or we are to use the values to define $ d(t) $, and ignore the minimum as a distractor? But that seems risky.

📌 Another possibility: minimum depth means the depth at which the layer is shallowest over time, but $ d(t) $ could be the rate or amplitude — but the problem says depth ... modeled by a cubic polynomial.

📌 Let’s re-express: perhaps $ d(t) $ is cubic, $ d(1)=1, d(2)=8, d(3)=27, d(4)=64 $, so $ d(t) = t^3 $ is the only cubic fitting. The statement about minimum may be a modeling choice — or perhaps it's a trick: although $ t^3 $ has no minimum, the physical system has a local minimum due to other forces — but mathematically, we must go with the data.

📌 But the problem says achieves its minimum depth exactly once — so it must have one.

📌 Therefore, the only way is if $ d(t) $ is not $ t^3 $, but a different cubic that fits the four points and has a unique minimum. But four points force $ d(t) = t^3 $.

📌 Unless — the values are not $ d(1)=1 $, but rather $ d(1) = a(1)^3 + b(1)^2 + c(1) + d = 1 $, etc. — but that’s the same.

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