To find \( s(0) \) from the given function \( f(x) = 5(\text{sgn}(4)) + \left[\frac{2x - 1}{2}\right] \), we will first evaluate \( f(0) \):
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Calculate \( \text{sgn}(4) \):
- The sign function \( \text{sgn}(x) \) returns \( 1 \) if \( x > 0 \), \( 0 \) if \( x = 0 \), and \( -1 \) if \( x < 0 \).
- Since \( 4 > 0 \), we have \( \text{sgn}(4) = 1 \).
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Substitute \( \text{sgn}(4) \) into \( f(x) \): \[ f(x) = 5(1) + \left[\frac{2x - 1}{2}\right] = 5 + \left[\frac{2x - 1}{2}\right] \]
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Evaluate \( f(0) \): \[ f(0) = 5 + \left[\frac{2(0) - 1}{2}\right] = 5 + \left[\frac{-1}{2}\right] \] The floor function \( \left[ y \right] \) gives the greatest integer less than or equal to \( y \): \[ \left[\frac{-1}{2}\right] = -1 \] Therefore, \[ f(0) = 5 - 1 = 4 \]
Thus, \( s(0) = f(0) = 4 \).
The answer is \( \boxed{4} \).