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Home » \( 40\% \times 1.10^8 \) - jntua results

\( 40\% \times 1.10^8 \) - jntua results

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 \tan\phi = \frac{3}{4} \Rightarrow \phi = \arctan\left(\frac{3}{4}\right)

📌 The maximum of $ \sin(\cdot) $ occurs when its argument is $ \frac{\pi}{2} $, so:

📌 \frac{\pi}{6}t + \phi = \frac{\pi}{2} \Rightarrow t = \frac{\frac{\pi}{2} - \phi}{\frac{\pi}{6}} = \frac{6}{\pi} \left( \frac{\pi}{2} - \arctan\left(\frac{3}{4}\right) \right)

📌 Using $ \arctan\left(\frac{3}{4}\right) \approx 0.6435 $ radians:

📌 t \approx \frac{6}{\pi} \left( \frac{\pi}{2} - 0.6435 \right) \approx \frac{6}{\pi} (1.5708 - 0.6435) = \frac{6}{\pi} (0.9273) \approx 1.77 \text{ hours}

📌 So the maximum tide height of $ \boxed{5} $ occurs approximately at $ \boxed{1.77} $ hours after midnight, or around 1:46 AM.

📌 Question: A physiological researcher is analyzing the periodic response of heart rate variability over a 24-hour period, modeled by the function $ f(t) = 5\cos\left(\frac{\pi}{12}t\right) + 12\sin\left(\frac{\pi}{12}t\right) $. What is the amplitude of this function, and at what time does the maximum occur?

📌 This function can be expressed in the form $ R\cos\left(\frac{\pi}{12}t - \phi\right) $, where $ R = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13 $

🔥 Popular Posts

المجموع \( S_5 \) لأول 5 حدود هو:
S_5 = 3 + 6 + 12 + 24 + 48 = 93
S_n = a \frac{r^n - 1}{r - 1}
المخروط له نصف قطر قاعدة 3 وحدات وارتفاع 4 وحدات. ما هو الارتفاع المائ...
يتم إيجاد الارتفاع المائل \( l \) باستخدام نظرية فيثاغورس:
l = \sqrt{r^2 + h^2} = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} =...
مجموع أول \( n \) عدد طبيعي هو 210. ما هي \( n \)؟
يُعطى مجموع أول \( n \) عدد طبيعي بالصيغة:
\frac{n(n + 1)}{2} = 210
n^2 + n - 420 = 0

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