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Home » So the equation holds. The smallest positive $ heta $ is $ - Redraw

So the equation holds. The smallest positive $ heta $ is $ - Redraw

📅 May 5, 2026 👤 scraface

May 05, 2026

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📌 3(y^3 - 6y^2 + 12y - 8) - 4(y^2 - 4y + 4) + 5y - 10 - 2 = 3y^3 - 18y^2 + 36y - 24 - 4y^2 + 16y - 16 + 5y - 12 = 3y^3 - 22y^2 + 57y - 52.

📌 Thus, $ g(x^2 - 2) = 3(x^2 - 2)^3 - 22(x^2 - 2)^2 + 57(x^2 - 2) - 52 $. Expanding:

📌 3(x^6 - 6x^4 + 12x^2 - 8) - 22(x^4 - 4x^2 + 4) + 57x^2 - 114 - 52 = 3x^6 - 18x^4 + 36x^2 - 24 - 22x^4 + 88x^2 - 88 + 57x^2 - 166 = 3x^6 - 40x^4 + 181x^2 - 278.

📌 The final answer is $ \boxed{3x^6 - 40x^4 + 181x^2 - 278} $.

📌 Question: A climatologist models temperature anomalies with a cubic polynomial $ g(x) $ satisfying $ g(1) = -3 $, $ g(2) = 5 $, $ g(3) = 21 $, and $ g(4) = 55 $. Find $ g(x^2 - 1) $.

📌 Solution: Assume $ g(x) = ax^3 + bx^2 + cx + d $. Using the given values, we form:

📌 a + b + c + d = -3 \\

📌 8a + 4b + 2c + d = 5 \\

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S = 2\pi r^2 + 2\pi r(2r) = 2\pi r^2 + 4\pi r^2 = 6\pi r^2
We know that \(S = 54\pi\), so:
6\pi r^2 = 54\pi
Dividing both sides by \(6\pi\), we get:
r^2 = 9
Taking the square root of both sides, we find:
r = 3
Thus, the radius of the cylinder is \(\boxed{3}\).
Question:** A cone has a slant height of \(10\) units and a base radiu...
The lateral surface area \(A\) of a cone is given by:

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