sicp

Yes I am reading sicp.

(define ex1.2 (/ (+ 5 4 (- 2 (- 3 (+ 6 (/ 4 5)))))
                 (* 3 (- 6 2)(- 2 7)))
  )

;sumofsquaresoflargest: # # # -> #
;takes three #'s & returns sum of the squares of the two largest
(define (sumofsquaresoflargest a b c)
  (+ (* (largest1 a b c)(largest1 a b c))
     (* (largest2 a b c)(largest2 a b c))))

;largest1: # # # -> #
;takes three numbers and returns the largest
(define (largest1 a b c)
  (cond [(and (> a b)(> a c)) a]
        [(and (> b a)(> b c)) b]
        [(and (> c a)(> c b)) c]
        [else a]))

;largest2: # # # -> #
;takes three numbers and returns the 2nd largest
(define (largest2 a b c)
  (cond [(and (> a b)(< a c)) a]
        [(and (> b a)(< b c)) b]
        [(and (> c a)(< c b)) c]
        [else a]))

;In normal order it would just loop infinitely trying to substitute (p) for (p) looking for a primitive operation to perform. Applicative would evaluate it as 0 by just evaluating (= x 0) as 0 and return 0.

>>1
I think you got the normal and applicative orders backwards.