Solucionario completo de Aritmetica de Baldor (Por Leonardo F. Apala T.)

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SOLUCIONARIO DE ARITMETICA DE BALDOR-9. 0.0056 en menos de 0.00001-10. 0.000000349 en menos de 0.00001EJERCICIO 246Hallar la raíz cubica de:-1. 25 con error < 1 425 se multiplica por el cubo de 4:Se halla la raíz cubica de 1 600:25 × 64 = 1 6003√1 600 = 1111 se divide entre 4: 11 4 = 2 3 4-2. 60 con error < 1 5LEONARDO F. APALA TITO 314
SOLUCIONARIO DE ARITMETICA DE BALDOR60 se multiplica por el cubo de 5:Se halla la raíz cubica de 7 500:19 se divide entre 5: 195 = 3 4 5-3. 96 con error < 1 696 se multiplica por el cubo de 6:Se halla la raíz cubica de 20 736:27 se divide entre 6: 27 6 = 9 2 = 4 1 2-4. 120 con error < 1 760 × 125 = 7 5003√7 500 = 1996 × 216 = 20 7363√20 736 = 27120 se multiplica por el cubo de 7:Se halla la raíz cubica de 41 160:34 se divide entre 7: 34 7 = 4 6 7-5. 185 con error < 1 8120 × 343 = 41 1603√41 160 = 34185 se multiplica por el cubo de 8:Se halla la raíz cubica de 94 720:45 se divide entre 8: 45 8 = 5 5 8-6. 300 con error < 1 9185 × 512 = 94 7203√94 720 = 45300 se multiplica por el cubo de 9:Se halla la raíz cubica de 218 700:300 × 729 = 218 7003√218 700 = 6060 se divide entre 9: 609 = 20 3 = 6 2 3-7. 800 con error < 110800 se multiplica por el cubo de 10:800 × 1000 = 800 000Se halla la raíz cubica de 800 000:3√800 000 = 9292 se divide entre 10: 9210 = 46 5 = 9 1 5-8. 1 050 con error < 1 81 050 se multiplica por el cubo de 8:Se halla la raíz cubica de 537 600:81 se divide entre 8: 81 8 = 10 1 8-9. 2 000 con error < 1 42 000 se multiplica por cubo de 4:Se halla la raíz cubica de 128 000:1 050 × 512 = 537 6003√537 600 = 812 000 × 64 = 128 0003√128 000 = 5050 se divide entre 4: 50 4 = 25 2 = 12 1 2-10. 19 con error < 1 919 se multiplica por el cubo de 9:Se halla la raíz cubica de 13 851:24 se divide entre 9: 249 = 8 3 = 2 2 3-11. 0.6 con error < 1 30.6 se multiplica por el cubo de 3:Se halla la raíz cubica de 162/ 10:5/ 2 se divide entre 3:19 × 729 = 13 8513√13 851 = 240.6 × 27 = 6 162× 27 =10 103√162 ∙ 10 2√10 ∙ 10 23√ 162103=3= √1623√103√16 200= 253√10 3 10 = 5 252 ÷ 3 = 5 6LEONARDO F. APALA TITO 315
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Solucionario completo de Aritmetica de Baldor (Por Leonardo F. Apala T.)

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