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2025 tax refund delay reasons - Network4

📅 May 18, 2026 👤 scraface

May 18, 2026

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📌 #### 30Question: A satellite dish is shaped like a parabola with a focus at (0, 5). If the dish's opening is 20 units wide, find the equation of the parabola in standard form.

📌 Solution: The standard form of a parabola with vertical axis is $ y = \frac{1}{4p}x^2 $. The focus is at $ (0, p) $, so $ p = 5 $. The width of 20 units implies the parabola passes through $ (10, 5) $. Substituting: $ 5 = \frac{1}{20}(100) \Rightarrow 5 = 5 $, which holds. Thus, the equation is $ y = \frac{1}{20}x^2 $. \boxed{y = \dfrac{1}{20}x^2}

📌 Question: An ichthyologist models fish population growth with a quadratic function $ h(x) $ satisfying $ h(1) = 4 $, $ h(2) = 11 $, and $ h(3) = 22 $. Find $ h(4) $.

📌 Solution: Assume $ h(x) = ax^2 + bx + c $. Using $ h(1) = a + b + c = 4 $, $ h(2) = 4a + 2b + c = 11 $, $ h(3) = 9a + 3b + c = 22 $. Subtract first equation from second: $ 3a + b = 7 $. Subtract second from third: $ 5a + b = 11 $. Subtract these: $ 2a = 4 \Rightarrow a = 2 $. Then $ 3(2) + b = 7 \Rightarrow b = 1 $. From $ 2 + 1 + c = 4 \Rightarrow c = 1 $. Thus, $ h(x) = 2x^2 + x + 1 $, and $ h(4) = 32 + 4 + 1 = 37 $. \boxed{37}

📌 Question: Compute $ \sum_{k=1}^{50} \frac{1}{k(k+2)} $.

📌 Solution: Use partial fractions: $ \frac{1}{k(k+2)} = \frac{1}{2}\left( \frac{1}{k} - \frac{1}{k+2} \right) $. The sum becomes $ \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=1}^{50} \frac{1}{k+2} \right) = \frac{1}{2} \left( \sum_{k=1}^{50} \frac{1}{k} - \sum_{k=3}^{52} \frac{1}{k} \right) $. Telescoping gives $ \frac{1}{2} \left( \frac{1}{1} + \frac{1}{2} - \frac{1}{51} - \frac{1}{52} \right) = \frac{1}{2} \left( \frac{3}{2} - \frac{103}{2652} \right) = \frac{1}{2} \cdot \frac{3975}{2652} = \frac{1325}{1768} $. \boxed{\dfrac{1325}{1768}}

📌 Question: Let $ p $ and $ q $ be complex numbers such that $ p + q = 3 + 4i $ and $ p^2 + q^2 = 10 - 24i $. Find $ pq $.

📌 Solution: Use identity $ p^2 + q^2 = (p + q)^2 - 2pq $. Substituting: $ 10 - 24i = (3 + 4i)^2 - 2pq $. Compute $ (3 + 4i)^2 = 9 + 24i - 16 = -7 + 24i $. Then $ 10 - 24i = -7 + 24i - 2pq $. Rearranging: $ 17 - 48i = -2pq \Rightarrow pq = \frac{-17 + 48i}{2} $. \boxed{-\dfrac{17}{2} + 24i}

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