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Home » \( s = \frac160019.6 \approx 81.63 \, \textm \) - Appfinity Technologies

\( s = \frac160019.6 \approx 81.63 \, \textm \) - Appfinity Technologies

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Solution: The probability of neither test being positive is $ \left(1 - rac{2}{3}

📌 ight)\left(1 - rac{3}{4}

📌 ight) = rac{1}{3} imes rac{1}{4} = rac{1}{12} $. Thus, the probability of at least one positive is $ 1 - rac{1}{12} = rac{11}{12} $. Final answer: $oxed{\dfrac{11}{12}}$.Question: Define $M(u) = u - rac{u^2}{2}$ for all real numbers $u$. If $a_1 = 1$ and $a_{n+1} = M(a_n)$ for $n \geq 1$, find $a_3$.

📌 Solution: Start with $a_1 = 1$. Compute $a_2 = M(a_1) = 1 - rac{1^2}{2} = rac{1}{2}$. Next, compute $a_3 = M(a_2) = rac{1}{2} - rac{(rac{1}{2})^2}{2} = rac{1}{2} - rac{1}{8} = rac{3}{8}$. Thus, the value of $a_3$ is $oxed{\dfrac{3}{8}}$.

📌 Question: Let $x, y, z$ be positive real numbers such that $2x + 3y + 4z = 12$. Find the minimum value of $x^2 + y^2 + z^2$.

📌 Solution: Use the Cauchy-Schwarz inequality: $(2^2 + 3^2 + 4^2)(x^2 + y^2 + z^2) \geq (2x + 3y + 4z)^2$. This gives $29(x^2 + y^2 + z^2) \geq 144$, so $x^2 + y^2 + z^2 \geq rac{144}{29}$. Equality holds when $rac{x}{2} = rac{y}{3} = rac{z}{4} = k$, leading to $x = 2k$, $y = 3k$, $z = 4k$. Substituting into $2x + 3y + 4z = 12$ gives $4k + 9k + 16k = 29k = 12$, so $k = rac{12}{29}$. Thus, the minimum value is $oxed{\dfrac{144}{29}}$.

📌 Question: Find the center of the hyperbola $9x^2 - 18x - 16y^2 + 64y = 144$.

📌 Solution: Complete the square for $x$ and $y$. For $x$: $9(x^2 - 2x) = 9[(x - 1)^2 - 1] = 9(x - 1)^2 - 9$. For $y$: $-16(y^2 - 4y) = -16[(y - 2)^2 - 4] = -16(y - 2)^2 + 64$. Substitute back: $9(x - 1)^2 - 9 - 16(y - 2)^2 + 64 = 144$. Simplify: $9(x - 1)^2 - 16(y - 2)^2 = 89$. The center is at $(1, 2)$. Thus, the center is $oxed{(1, 2)}$.

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