Given this Scheme procedure: (define foo (lambda (x) (if (= x 0) 1 (* x (foo (- x 1))) ) ))

Given this Scheme procedure: (define foo (lambda (x) (if (= x 0) 1 (* x (foo (- x 1))) ) ))

   Given this Scheme procedure:


    (define foo (lambda (x)
      (if (= x 0)
          1
          (* x (foo (- x 1)))
      )
    ))


    What does this procedure do?

 
    Answers:
    A.
    square(x)
   
    B.
    absolute(x)
   
    C.
    x!
   
    D.
    cube(x)


Answer. C.