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Home » 8a + 4b + 2c + d = 10 \quad \text(3) - Aurero

8a + 4b + 2c + d = 10 \quad \text(3) - Aurero

📅 March 13, 2026 👤 scraface

Mar 13, 2026

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📌 r = \sqrt{34}

📌 The radius of the lake is $ oxed{\sqrt{34}} $ km.

📌 Question: An astronomer observes a protoplanetary disk rotating uniformly, where a planetesimal orbits along a circular path of radius $ R $. If the arc length between two opposite points on the orbit is $ 120^\circ $ of the circle, what is the ratio of the chord length (distance between these two points) to the radius $ R $?

📌 Solution: The central angle corresponding to the arc is $ 120^\circ $, or $ rac{2\pi}{3} $ radians. The chord length $ c $ subtended by a central angle $ heta $ in a circle of radius $ R $ is given by:

📌 c = 2R \sin\left(rac{ heta}{2}

📌 Substituting $ heta = rac{2\pi}{3} $:

📌 c = 2R \sin\left(rac{\pi}{3}

📌 ight) = 2R \cdot rac{\sqrt{3}}{2} = R\sqrt{3}

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Final answer: $ \boxed{(2x - 3y)(2x + 3y)(4x^2 + 9y^2)} $
Question: Compute the square of $ 3\sqrt{2} - 2\sqrt{3} $.
Use the identity $ (a - b)^2 = a^2 - 2ab + b^2 $:
Compute each term:
2(3\sqrt{2})(2\sqrt{3}) = 12\sqrt{6}
Final answer: $ \boxed{30 - 12\sqrt{6}} $
Question: Rationalize the denominator of $ \frac{5}{\sqrt{7} + \sqrt{2...
Multiply numerator and denominator by the conjugate $ \sqrt{7} - \sqrt...
\frac{5}{\sqrt{7} + \sqrt{2}} \cdot \frac{\sqrt{7} - \sqrt{2}}{\sqrt{7...
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