(let ((x 2) (y 3))
  (let ((x 7)
        (z (+ x y)))
    (* z x)))
  = 35
The initial values in the inner  let are evaluated ``simultaneously''
before the new variables are bound, so  (+ x y) is evaluated with
 x = 2.
(let ((x 2) (y 3))
  (let* ((x 7)
         (z (+ x y)))
    (* z x)))
  = 70
The initial values in the inner let* are evaluated sequentially, so (+ x y) is evaluated with x = 7.
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