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Home » Time to empty the tank is \( \frac90\pi4.5\pi = 20 \) hours. - Get link 4share

Time to empty the tank is \( \frac90\pi4.5\pi = 20 \) hours. - Get link 4share

📅 March 10, 2026 👤 scraface

Mar 10, 2026

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📌 Hypotenuse: \( c = \sqrt{(\sqrt{5})^2 + (\sqrt{11})^2} = \sqrt{5 + 11} = \sqrt{16} = 4 \).

📌 Area: \( rac{1}{2} \cdot \sqrt{5} \cdot \sqrt{11} = rac{1}{2} \sqrt{55} \).

📌 Also: \( rac{1}{2} \cdot 4 \cdot h = 2h \).

📌 Set equal: \( 2h = rac{1}{2} \sqrt{55} \) â \( h = rac{\sqrt{55}}{4} \).

📌 #### \(rac{\sqrt{55}}{4}\)Question: Let $ f(u) = u - rac{u^4}{4} $ for every real number $ u $. If $ b_1 = 2 $ and $ b_{n+1} = f(b_n) $ for $ n \geq 1 $, find $ b_3 $.

📌 Solution: Compute $ b_2 = f(2) = 2 - rac{2^4}{4} = 2 - rac{16}{4} = 2 - 4 = -2 $. Then $ b_3 = f(-2) = -2 - rac{(-2)^4}{4} = -2 - rac{16}{4} = -2 - 4 = -6 $. Thus, $ b_3 = oxed{-6} $.

📌 Question: A science communicator designs an exhibit where the number of interactive stations $ S $ at time $ t $ (in hours) satisfies $ S(t) = t^2 - 5t + 6 $. Find all times $ t $ when $ S(t) = 0 $.

📌 Solution: Solve $ t^2 - 5t + 6 = 0 $. Factor: $ (t - 2)(t - 3) = 0 $. The solutions are $ t = 2 $ and $ t = 3 $. Thus, the times are $ oxed{2} $ and $ oxed{3} $.

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