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Home » Solve for \( t \): \( t = 20 / 9.8 \approx 2.04 \) seconds - ECD Germany

Solve for \( t \): \( t = 20 / 9.8 \approx 2.04 \) seconds - ECD Germany

📅 May 2, 2026 👤 scraface

May 02, 2026

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📌 #### 150.8Question: A resilience coordinator is modeling the tidal patterns in New Orleans to plan flood mitigation strategies. The height of the tide at a particular location is modeled by the function $ h(t) = 4\sin\left(\frac{\pi}{6}t\right) + 3\cos\left(\frac{\pi}{6}t\right) $, where $ t $ is the time in hours after midnight. What is the maximum height of the tide, and at what time does it occur?

📌 To find the maximum height of the tide, we rewrite the function in the form $ R\sin\left(\frac{\pi}{6}t + \phi\right) $, where $ R $ is the amplitude and gives the maximum height.

📌 h(t) = 4\sin\left(\frac{\pi}{6}t\right) + 3\cos\left(\frac{\pi}{6}t\right)

📌 This can be expressed as:

📌 h(t) = R\sin\left(\frac{\pi}{6}t + \phi\right)

📌 where $ R = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5 $

📌 So the maximum height is $ \boxed{5} $.

📌 To find the time at which this maximum occurs, we use:

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Therefore, the truth-weight parameter is $ oxed{5} $.
Question: A pharmacologist compares two drug efficacy models: $ f(x) =...
Solution: Compute $ f(6) $ and $ g(6) $:
f(6) = (6)^2 - 5(6) + 3m = 36 - 30 + 3m = 6 + 3m
g(6) = (6)^2 - 5(6) + 7m = 36 - 30 + 7m = 6 + 7m
Given $ f(6) = g(6) - 12 $:
+ 3m = (6 + 7m) - 12 = -6 + 7m
Now solve:
+ 3m = -6 + 7m
Subtract $ 3m $ from both sides:

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