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

February 22, 2026

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

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A right triangle has legs of lengths \( \sqrt{5} \) and \( \sqrt{11} \). Find the length of the altitude to the hypotenuse.
Hypotenuse: \( c = \sqrt{(\sqrt{5})^2 + (\sqrt{11})^2} = \sqrt{5 + 11} = \sqrt{16} = 4 \).
Area: \( rac{1}{2} \cdot \sqrt{5} \cdot \sqrt{11} = rac{1}{2} \sqrt{55} \).
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} $.

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