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Home » Standard deviation: \( \sqrt30 \approx 5.477 \) → rounds to 5.5 - Tacotoon

Standard deviation: \( \sqrt30 \approx 5.477 \) → rounds to 5.5 - Tacotoon

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 A_{\text{circle}} = \pi r^2 = \pi \cdot 25 = 25\pi

📌 \frac{A_{\text{circle}}}{A_{\text{triangle}}} = \frac{25\pi}{150} = \frac{\pi}{6}

📌 Thus, the ratio is $ \boxed{\frac{\pi}{6}} $.Question: A plant biologist is modeling the growth pattern of a genetically modified crop using the function $ f(t) = A \cos(\omega t + \phi) $, where $ t $ represents time in days. If the maximum rate of change of growth occurs at $ t = \frac{\pi}{4\omega} $ and equals $ 5A\omega $, find the value of $ A $ given that $ \cos(\phi) = \frac{3}{5} $ and $ \sin(\phi) = \frac{4}{5} $.

📌 The rate of change of $ f(t) $ is given by its derivative:

📌 f'(t) = -A\omega \sin(\omega t + \phi)

📌 The maximum rate of change occurs when $ \sin(\omega t + \phi) = \pm 1 $, so:

📌 |f'(t)|_{\text{max}} = A\omega

📌 However, the problem states that the maximum rate of change is $ 5A\omega $, which suggests the given value corresponds to the amplitude of the derivative, i.e.,

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