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Step Into Style with the Pink New Balance 574—Limited Drops Are Arriving NOW! - Crankk.io

📅 February 28, 2026 👤 scraface

Feb 28, 2026

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📌 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.

📌 Perhaps d(1) = 1 means the depth is 1 at time 1, which is exactly $ t^3 $.

📌 We are stuck.

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To find the ratio of the areas, we calculate the area of each shape.
The area \(A\) of a circle with radius \(r\) is given by:
A_{\text{pond}} = \pi r^2
The area \(A\) of a semicircle with radius \(R\) is half the area of a...
A_{\text{semicircle}} = \frac{1}{2} \pi R^2
For the walkway with radius \(5r\):
A_{\text{walkway}} = \frac{1}{2} \pi (5r)^2 = \frac{1}{2} \pi \times 2...
The area of the walkway is the area of the full circle with radius \(5...
A_{\text{walkway only}} = \pi (5r)^2 - \pi r^2 = 25\pi r^2 - \pi r^2 =...
The ratio of the area of the pond to the area of the walkway is:

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