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Numerical Analysis & Integral EquationsNumerical Solutions of Nonlinear Equations and Linear Systems

One Newton iteration for f(x)=x²−576, starting at x₀=25, gives x₁ equal to:

f(x)=x²−576 के लिए न्यूटन विधि में x₀=25 से एक चरण के बाद x₁ क्या होगा?
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