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Cilindro

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Cilindro

  1. 1. Recipiente cilíndrico con una atapa (NºL)(120cm^3) 1000 =H Pared AL=2(3.1416)(R)(H) 3.1416(r^2) =H Base AB= (3.1416)(R^2) r AT= AL+ AB 1920 h h r Total ALT= 2(3.1416)(R)(H)+ (3.1416)(R^2) xr altura (H) base pared totalx 4800-2x (x)(4800-2x) 4 38,1971 50,266 959,997 1010,263 8 9,5492 201,062 479,995 681,057 12 4,2441 452,390 319,997 772,387 16 2,3873 804,250 239,997 1044,247 20 1,5278 1256,640 191,989 1448,629 24 1,061 1809,562 159,995 1969,557
  2. 2. 2500,000 PUNTO ADECUADO 2000,000 1500,000 Series1 1000,000 500,000 0,000 0 5 10 15 20 25 302(3.1416)(x)(1920/3.1416r dy 3840x^3+3.1416x^2=0 ---= (3840x^3+3.1416x^2)^2) + (3.1416)(x)^2 dxX= 23.040 Tenemos que dándole un radio de 23.040 obtendríamos el valor esperado Necesitamos la medida de 23.040

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