Consider the following algorithm for computing x n for an integer n. If n < 0, x

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Consider the following algorithm for computing xn for an integer n. If n < 0, xn

is 1/x–n. If n is positive and even, then xn = (xn/2 )2. If n is positive and odd, then xn = xn–1 × x. Implement a static method double intPower(double x, int n) that uses this algorithm. Add it to a class called Numeric.

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