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Home » \(x^3 - 3x^2 + 2x = x((5x - 6) - 3x + 2) = x(2x - 4) = 2x^2 - 4x\). - Abu Waleed Tea

\(x^3 - 3x^2 + 2x = x((5x - 6) - 3x + 2) = x(2x - 4) = 2x^2 - 4x\). - Abu Waleed Tea

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 \]Question: An anthropologist discovers ancient carvings depicting a regular hexagon with an area of \( 54\sqrt{3} \) square units. If each side is reduced by 2 units, by how many square units does the area decrease?

📌 Solution: The area \( A \) of a regular hexagon with side length \( s \) is \( A = \frac{3\sqrt{3}}{2}s^2 \). Given \( 54\sqrt{3} = \frac{3\sqrt{3}}{2}s^2 \), solving for \( s^2 \) yields \( s^2 = 36 \), so \( s = 6 \). The new side length is \( 6 - 2 = 4 \). The new area is \( \frac{3\sqrt{3}}{2}(4)^2 = 24\sqrt{3} \). The decrease in area is \( 54\sqrt{3} - 24\sqrt{3} = 30\sqrt{3} \). \boxed{30\sqrt{3}}

📌 Question: A renewable energy engineer designs a tidal turbine foundation shaped as a circular sector with radius 12 meters and central angle \( 90^\circ \). If the radius is increased by 4 meters, by how many square meters does the area increase?

📌 Solution: The area of a circular sector is \( A = \frac{\theta}{360} \pi r^2 \). Original area: \( \frac{90}{360} \pi (12)^2 = 36\pi \). New radius: \( 12 + 4 = 16 \) meters. New area: \( \frac{90}{360} \pi (16)^2 = 64\pi \). The increase is \( 64\pi - 36\pi = 28\pi \). \boxed{28\pi}

📌 Question: A glaciologist models a glacier's movement using a right triangle with legs measuring 15 km and 20 km. If the shorter leg is increased by 5 km, by how many square kilometers does the area increase?

📌 Solution: Original area: \( \frac{1}{2} \times 15 \times 20 = 150 \) km². New shorter leg: \( 15 + 5 = 20 \) km. New area: \( \frac{1}{2} \times 20 \times 20 = 200 \) km². The increase is \( 200 - 150 = 50 \) km². \boxed{50}

📌 Question: Compute \( \sin 225^\circ $.

📌 Solution: \( 225^\circ \) lies in the third quadrant, where sine is negative. Reference angle: \( 225^\circ - 180^\circ = 45^\circ \). Thus, \( \sin 225^\circ = -\sin 45^\circ = -\frac{\sqrt{2}}{2} \). \boxed{-\dfrac{\sqrt{2}}{2}}

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