BoXoN n°1

32

description

Lieu de débauche et de tolérance, Novembre 97 Cabut d'Abrigeon Bobillot Brigaud Dumoulin Fabre Michel Bachelard Robespierre Laforgue...

Transcript of BoXoN n°1

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BoxOn Logistics Software

BoxOn Logistics Software

n 1 n ∫∫ 1 x

n 1 n ∫∫ 1 x

RENAUD DANS TON SAC ALBUM: Marchand de cailloux. En cherchant les clés de l'auto, j'ai fouillé comme un salaud dans ton sac. J'ai mis un sacré boxon,

RENAUD DANS TON SAC ALBUM: Marchand de cailloux. En cherchant les clés de l'auto, j'ai fouillé comme un salaud dans ton sac. J'ai mis un sacré boxon,

3 -1 Chapter 3 Recursion. 3 -2 Iterative algorithm for n factorial n factorial n! = 1 if n = 0 n! = n*(n-1)*(n-2)*...*1 if n>0 Iterative algorithm prod.

3 -1 Chapter 3 Recursion. 3 -2 Iterative algorithm for n factorial n factorial n! = 1 if n = 0 n! = n*(n-1)*(n-2)*...*1 if n>0 Iterative algorithm prod.

Re-Design Boxon : Revue de Presse

Re-Design Boxon : Revue de Presse

Recursion Self-Referencing Functions. Problem # 1 Write a function that, given n, computes n! n! == 1 2... (n-1) n n! == 1 2 ... (n-1) nExample:

Recursion Self-Referencing Functions. Problem # 1 Write a function that, given n, computes n! n! == 1 2... (n-1) n n! == 1 2 ... (n-1) nExample:

BoxOn Logistics...BoxOn, in simple words, is a software which is made to simplify the tasks and help organizations gain full control over everything related to their logistics operations.

BoxOn Logistics...BoxOn, in simple words, is a software which is made to simplify the tasks and help organizations gain full control over everything related to their logistics operations.

117 Cahiers scientifiques · Nombre de combinaisons solo 1 (n = 13) 2 (n = 47) 1 (n = 2) 1 (n = 1) Nombre de combinaisons duo 2 (n = 21) 2 (n = 5) 1 (n = 2) 1 (n = 1) Nombre de combinaisons

117 Cahiers scientifiques · Nombre de combinaisons solo 1 (n = 13) 2 (n = 47) 1 (n = 2) 1 (n = 1) Nombre de combinaisons duo 2 (n = 21) 2 (n = 5) 1 (n = 2) 1 (n = 1) Nombre de combinaisons

Understanding Recursion /1 Powerful computing/problem-solving techniques Examples Factorial: f(n) = 1, if n = 1 f(n) = f(n-1) * n, if n ≥ 1 Quick sort:

Understanding Recursion /1 Powerful computing/problem-solving techniques Examples Factorial: f(n) = 1, if n = 1 f(n) = f(n-1) * n, if n ≥ 1 Quick sort:

Prova Modelo n.º 14 - recursos-para-matematica.webnode.pt · Resposta: D 2. 1 11 331 33 lim lim 1 lim 1 lim 1 1 1 1 1 3 n n n n n ee n n n n ... Resposta: A Página 147 5. Tem-se

Prova Modelo n.º 14 - recursos-para-matematica.webnode.pt · Resposta: D 2. 1 11 331 33 lim lim 1 lim 1 lim 1 1 1 1 1 3 n n n n n ee n n n n ... Resposta: A Página 147 5. Tem-se

1 Beskonačni redovi - UNS5 Za harmonijski red, + + +L+ +L k 1 3 1 2 1 1 Dalamberov kriterium: Košijev kriterijum: 1 1 lim 1 1 1 lim = + = ρ= + →∞ →∞ n n n n n n, pa nema

1 Beskonačni redovi - UNS5 Za harmonijski red, + + +L+ +L k 1 3 1 2 1 1 Dalamberov kriterium: Košijev kriterijum: 1 1 lim 1 1 1 lim = + = ρ= + →∞ →∞ n n n n n n, pa nema

N=n => width = 1/n US = =1/n 3 [1+4+9+….+n 2 ] US =

N=n => width = 1/n US = =1/n 3 [1+4+9+….+n 2 ] US =

Boxon Scanpack

Boxon Scanpack

Boxon Presentaion

Boxon Presentaion

Communiqué de presse : vente aux enchères du Re-Design Boxon

Communiqué de presse : vente aux enchères du Re-Design Boxon

BoxOn PO Box Software Presentation

BoxOn PO Box Software Presentation

PIAnI rOndO | QuAdrO - byok.lighting · 5 n 23 2 1 1 1 1 3 2 6 3 b4 n n n n b7 5 n 23 2 1 1 1 1 3 2 6 3 b4 n n n n b7 b7 5 n 23 2 1 1 1 1 3 2 6 3 b4 n n n n b7 5 n 23 2 1 1 1 1 3

PIAnI rOndO | QuAdrO - byok.lighting · 5 n 23 2 1 1 1 1 3 2 6 3 b4 n n n n b7 5 n 23 2 1 1 1 1 3 2 6 3 b4 n n n n b7 b7 5 n 23 2 1 1 1 1 3 2 6 3 b4 n n n n b7 5 n 23 2 1 1 1 1 3

Den lille tapeskolen - Boxon · 2018. 10. 20. · Den lille tapeskolen Den lille tapeskolen består av tre deler og formålet er at vi vil hjelpe deg med å forstå hvordan tapen

Den lille tapeskolen - Boxon · 2018. 10. 20. · Den lille tapeskolen Den lille tapeskolen består av tre deler og formålet er at vi vil hjelpe deg med å forstå hvordan tapen

n 3 n 2 n 0, 1, 2, 3, n ::: n 1, n 2,n 3,

n 3 n 2 n 0, 1, 2, 3, n ::: n 1, n 2,n 3,

Formulario - UniPa · 8. ∑1 n=1 ( 1)n+1 1 n = log(2) 9. ∑1 n=1 ( 1)n+1 1 2n 1 = ˇ 4 10. ∑1 n=1 ( 1)n+1 1 n2 = ˇ2 12 11. ∑1 n=1 1 (2n 1)2 = ˇ2 8 12. ∑1 n=1 ( 1)n+1 1 (2n

Formulario - UniPa · 8. ∑1 n=1 ( 1)n+1 1 n = log(2) 9. ∑1 n=1 ( 1)n+1 1 2n 1 = ˇ 4 10. ∑1 n=1 ( 1)n+1 1 n2 = ˇ2 12 11. ∑1 n=1 1 (2n 1)2 = ˇ2 8 12. ∑1 n=1 ( 1)n+1 1 (2n