(=~ is interesting. Thanks.
On 22 October 2013 13:53, Matthias Felleisen wrote:
>
> ;; equation3 : number -> boolean
> ;; to determine whether n is a solution for 2n^2 = 102
> (define (equation3 n)
> (=~ (* 2 n n) 102 .001))
>
>
> -- Matthias
>
>
>
> On Oct 22, 2013, at 8:40 AM, Bo Gus
Hi BoGus,
fwiw, this exercise is NOT in HTDP 2nd edition but HTDP 1st edition, 1-4th
printing.
Just in case you really want to work through HtDP 2e.
-- Matthias
On Oct 22, 2013, at 8:53 AM, Matthias Felleisen wrote:
>
> ;; equation3 : number -> boolean
> ;; to determine whether n
;; equation3 : number -> boolean
;; to determine whether n is a solution for 2n^2 = 102
(define (equation3 n)
(=~ (* 2 n n) 102 .001))
-- Matthias
On Oct 22, 2013, at 8:40 AM, Bo Gus wrote:
> equation 2 is 2n^2 = 102 so I implement like this:
>
> ;; equation3 : number -> boolean
>
equation 2 is 2n^2 = 102 so I implement like this:
;; equation3 : number -> boolean
;; to determine whether n is a solution for 2n^2 = 102
(define (equation3 n)
(= (* 2 n n) 102))
And my answer is the same as per the online answer. so great.
But how can I check a valid answer.
Eg if I do
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