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Home » Shocking Facts About Duritos You Won’t Believe Are True! - DNSFLEX

Shocking Facts About Duritos You Won’t Believe Are True! - DNSFLEX

📅 March 12, 2026 👤 scraface

Mar 12, 2026

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📌 $ k=1 $: $ e^{i3\pi/4} = -\frac{\sqrt{2}}{2} + i\frac{\sqrt{2}}{2} $

📌 $ k=2 $: $ e^{i5\pi/4} = -\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} $

📌 $ k=3 $: $ e^{i7\pi/4} = \frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2} $

📌 The imaginary parts are $ \pm \frac{\sqrt{2}}{2} $. The maximum imaginary part is $ \frac{\sqrt{2}}{2} $.

📌 We are told this maximum is $ \sin \theta $ for $ 0 < \theta < \pi $.

📌 \sin \theta = \frac{\sqrt{2}}{2} \quad \Rightarrow \quad \theta = \frac{\pi}{4} \quad \text{or} \quad \theta = \frac{3\pi}{4}

📌 Both are in $ (0, \pi) $, but since the maximum imaginary part is positive and corresponds to $ \frac{\sqrt{2}}{2} $, and the angle is often taken as the acute one in such contexts, but the problem asks for $ \theta $ such that $ \sin \theta = \text{max imag part} $, and $ \theta $ is to be found. However, $ \frac{\sqrt{2}}{2} = \sin \frac{\pi}{4} = \sin \frac{3\pi}{4} $, but $ \frac{3\pi}{4} $ is larger and still in range. But since the root has argument $ \frac{3\pi}{4} $, and the imaginary part is $ \sin \frac{3\pi}{4} = \frac{\sqrt{2}}{2} $, the value $ \theta $ such that $ \sin \theta = \frac{\sqrt{2}}{2} $ and corresponds to the phase is naturally $ \frac{\pi}{4} $, but the problem says "expressed as $ \sin \theta $", and to find $ \theta $. However, since $ \sin \theta = \frac{\sqrt{2}}{2} $,

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Difference:
- 1142.64 = 14.99
#### #### 14.99
A circle with a radius of 7 cm is inscribed in a square. Calculate the...
Side of the square = Diameter of the circle = \(2 \times 7 = 14\) cm.
Area of the square = \(14^2 = 196 \, \text{cm}^2\).
Area of the circle = \(\pi \times r^2 = \pi \times 7^2 = 49\pi \, \tex...
Area of the square not covered by the circle:
- 49\pi \approx 196 - 153.94 = 42.06 \, \text{cm}^2
#### #### 42.06

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