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Newton's method for cube roots is based on the fact that if y is an approximation to the cube root of x, then a better approximation is given by the value:
(x/y2+2y)/3
(a) Use this formula to implement a cube-root procedure analogous to the square-root procedure from the lecture notes.
(b) Modify your original implementation to allow the good-enough? procedure to be included as a argument to the cube-root procedure; i.e., (cube-root 3 good-enough?). In this way, roots of arbitrary precision should be possible by defining new forms of good-enough?
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I Want a answer for solving the big M method in the topic of simplex method...
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A: (a) int foo = 123; (b) int bar(123);
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