Given that \(x * y = \frac{x + y}{2}, x \circ y = \frac{x^{2}}{y}\) and \((3 * b) \circ 48 = \frac{1}{3}\), find b, where b > 0.
\((x * y) = \frac{x+y}{2}\)
\((3 * b) = \frac{3+b}{2}\)
\(x \circ y = \frac{x^{2}}{y}\)
\((\frac{3+b}{2}) \circ 48 = \frac{(\frac{3+b}{2})^{2}}{48} = \frac{1}{3}\)
\(\frac{(3+b)^{2}}{48 \times 4} = \frac{1}{3}\)
\((3 + b)^{2} = \frac{48 \times 4}{3} = 64\)
\(b^{2} + 6b + 9 = 64 \implies b^{2} + 6b + 9 - 64 = 0\)
\(b^{2} + 6b - 55 = 0 \implies b^{2} - 5b + 11b - 55 = 0\)
\(b(b - 5) + 11(b - 5) = 0 \implies (b - 5) = \text{0 or (} b + 11) = 0\)
Since b > 0, b - 5 = 0
b = 5.
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