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Home » \(v = \frac40 \pm \sqrt1600 + 19202\). - AMAZONAWS

\(v = \frac40 \pm \sqrt1600 + 19202\). - AMAZONAWS

📅 March 5, 2026 👤 scraface

Mar 05, 2026

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### Question 9
A triangle has sides of lengths 7, 24, and 25. Verify if it is a right...
Check Pythagorean theorem: \(7^2 + 24^2 = 49 + 576 = 625 = 25^2\)
It is a right triangle.
Area = \(rac{1}{2} imes ext{base} imes ext{height} = rac{1}{2}...
### Question 10
The function \(f(x) = 3x^3 - 2x^2 + x - 5\) has a critical point at \(...
First derivative: \(f'(x) = 9x^2 - 4x + 1\)
\(f'(1) = 9(1)^2 - 4(1) + 1 = 6\) (not zero, so not a critical point,...
Second derivative: \(f''(x) = 18x - 4\)

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