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Home » One root near \( x \approx 0.542 \) - Ready Digital AB

One root near \( x \approx 0.542 \) - Ready Digital AB

📅 March 11, 2026 👤 scraface

Mar 11, 2026

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📌 Solution: Prime factorizations: $15 = 3 imes 5$, $20 = 2^2 imes 5$. LCM is $2^2 imes 3 imes 5 = 60$. The synchronized cycle length is $oxed{60}$.Question: A civil engineer designs a parabolic arch with the equation $ y = -ax^2 + bx + c $. If the arch's highest point is at $ (3, 10) $ and passes through $ (0, 4) $, find the value of $ a $.

📌 Solution: The vertex form of a parabola is $ y = -a(x - h)^2 + k $, where $ (h, k) $ is the vertex. Given the vertex $ (3, 10) $, the equation becomes $ y = -a(x - 3)^2 + 10 $. Substituting the point $ (0, 4) $:

📌 $$ 4 = -a(0 - 3)^2 + 10 \implies 4 = 9a + 10 \implies 9a = -6 \implies a = -rac{2}{3}. $$

📌 However, since the coefficient of $ x^2 $ is given as $ -a $, and parabolic arches open downward, $ a $ must be positive. Correcting the sign:

📌 $$ 4 = -a(9) + 10 \implies -9a = 6 \implies a = -rac{2}{3} \quad ext{(invalid)}. $$

📌 Re-evaluating, the standard form $ y = -ax^2 + bx + c $ with vertex $ (3, 10) $ gives $ b = 2a \cdot 3 = 6a $ and $ 10 = -a(9) + b(3) + c $. Substituting $ b = 6a $ and $ c = 4 $ (from $ (0, 4) $):

📌 $$ 10 = -9a + 18a + 4 \implies 10 = 9a + 4 \implies 9a = 6 \implies a = rac{2}{3}. $$

📌 Thus, the value of $ a $ is $ oxed{\dfrac{2}{3}} $.

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