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Home » $ \Rightarrow 7a + 3b + c = -4 \quad \text(5) $ - Dyverse

$ \Rightarrow 7a + 3b + c = -4 \quad \text(5) $ - Dyverse

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 x^2 + y^2 + z^2 = c \sqrt{x^2 + y^2}

📌 Now let \(r = \sqrt{x^2 + y^2}\), so:

📌 r^2 + z^2 = c r \quad \Rightarrow \quad r^2 - c r + z^2 = 0

📌 r^2 - c r + rac{c^2}{4} + z^2 = rac{c^2}{4} \quad \Rightarrow \quad \left(r - rac{c}{2}

📌 ight)^2 + z^2 = \left(rac{c}{2}

📌 This is the equation of a **circle** in the \( (r,z) \)-plane, rotated about the \(z\)-axis, forming a **torus**? No â actually, since \(r\) is radial, this describes a **sphere** of radius \(c/2\) centered at \((x, y, z) = (0, 0, 0, c/2)\) in cylindrical coordinates.

📌 Wait: \(\left(r - rac{c}{2}

📌 ight)^2\) is a circle in the \(r\)-axis plane, centered at \(r = c/2, z = 0\), radius \(c/2\). Rotated about the \(z\)-axis, this forms a **torus**? No â it forms a **sphere**.

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\text{LCM} = 2^3 \times 3^1 = 8 \times 3 = 24
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What is the sum of all values of \(b\) for which \(\sqrt{(b-3)^2} = 5\...
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\(b - 3 = -5\)
Solving the first equation:
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