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xÂ³ + 4xÂ² - 192 = 0 - 500apps

📅 March 14, 2026 👤 scraface

Mar 14, 2026

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📌 The solutions are \( t = 2 \) and \( t = rac{2}{3} \).

📌 oxed{2, rac{2}{3}}

📌 An industrial hygienist is evaluating exposure levels to a new chemical in a factory, modeled by a polynomial \( h(x) = 2x^3 - 9x^2 + 12x - 4 \). Calculate the product of the non-zero roots of this polynomial.

📌 We use Vieta's formulas to find the product of the roots. For \( h(x) = 2x^3 - 9x^2 + 12x - 4 \), the product of all roots is \( -d/a = -(-4)/2 = 2 \). Assuming the roots \( r_1, r_2, r_3 \) include a non-zero root \( r_1 \) and potentially zero roots, we check if zero is a root by substituting \( x = 0 \). Since \( h(0) = -4

📌 eq 0 \), zero is not a root. Therefore, the product of all roots is the product of the non-zero roots, which is \( 2 \).

📌 These questions and solutions demonstrate various applications of algebraic concepts in real-world contexts, providing a rich learning experience.

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\cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A...
Let $A = \theta + 60^\circ$, $B = \theta - 60^\circ$. Then:
\cos(\theta + 60^\circ) + \cos(\theta - 60^\circ) = 2 \cos\theta \cos...
Since $\cos 60^\circ = \frac{1}{2}$, we have:
\cos\theta \cdot \frac{1}{2} = \cos\theta
\cos\theta = \sqrt{3}
But $\cos\theta = \sqrt{3}$ has no real solution since $|\cos\theta| \...
\cos(\theta + 60^\circ) + \cos(\theta - 60^\circ) = 2 \cos\theta \cos...
So the equation is $\cos\theta = \sqrt{3}$, which is impossible. Thus,...
\cos(\theta + 60^\circ) + \cos(\theta - 60^\circ) = 1

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