Sistemas De Numeracion

Métodos computacionales en Ingeniería

Tarea 1

PROBLEMA 1

12,385

|213 |212 |211 |210 |29 |28 |27 |26 |25 |24 |23 |22 |21 |20 | |0 |1 |1 |0 |0 |0 |0 |0 |1 |1 |0 |0 |0 |0 |1 | |

-12,385

1 |0 |0 |1 |1 |1 |1 |1 |0 |0 |1 |1 |1 |1 |0 | |  |  |  |  |  |  |  |  |  |  |  |  |  |  |1 | |1 |0 |0 |1 |1 |1 |1 |1 |0 |0 |1 |1 |1 |1 |1 | |

PROBLEMA 2

7,584,322

Binario

  |222|221 |220 |219 |218 |217 |216 |215 |214 |213 |212 |211 |210 |29 |28 |27 |26 |25 |24 |23 |22 |21 |20 | |0 |1 |1 |1 |0 |0 |1 |1 |1 |0 |1 |1 |1 |0 |1 |0 |0 |1 |0 |0 |0 |0 |1 |0 | |

Octal

|1 |1 |1 |0|0 |1 |1 |1 |0 |1 |1 |1 |0 |1 |0 |0 |1 |0 |0 |0 |0 |1 |0 | | |3 | | |4 | | |7 | | |3 | | |5 | | |1 | | |0 | | |2 | | |
34735102

Hexadecimal

0 |1 |1 |1 |0 |0 |1 |1 |1 |0 |1 |1 |1 |0 |1 |0 |0 |1|0 |0 |0 |0 |1 |0 | | |7 | | | |3 | | | |B | | | |A | | | |4 | | | |2 | | | |
73BA42

PROBLEMA 3

2620 + 3460

2620

0 |1 |0 |1 |0 |0 |0 |1 |1 |1 |1 |0 |0 | |

3460

0 |1 |1 |0 |1 |1 |0|0 |0 |0 |1 |0 |0 | |

0 |1 |0 |1 |0 |0 |0 |1 |1 |1 |1 |0 |0 | |0 |1 |1 |0 |1 |1 |0 |0 |0 |0 |1 |0 |0 | |1 |0 |1 |1 |1 |1 |1 |0 |0 |0 |0 |0 |0 | |+

3460 - 2620

2620 C2

0 |1 |0 |1 |0 |0 |0|1 |1 |1 |1 |0 |0 | |

1 |0 |1 |0 |1 |1 |1 |0 |0 |0 |0 |1 |1 | | | | | | | | | | | | | |1 | |1 |0 |1 |0 |1 |1 |1 |0 |0 |0 |1 |0 |0 | |

Sumando

3460 + C2 2620

0 |1 |1 |0 |1 |1 |0 |0 |0 |0 |1|0 |0 | |1 |0 |1 |0 |1 |1 |1 |0 |0 |0 |1 |0 |0 | |0 |0 |1 |1 |1 |1 |0 |0 |0 |1 |0 |0 |0 | |
PROBLEMA 4

X=-15.0
Y=9.0
IF (.NOT.((SIN(3.0-ALOG(-X))**0.3*4.0+ALOG10(ABS(X/2.0)-5.0)*40.0).GT. 21.0)) & Y=PI*5.0

ALOG(-X) = 2.708050
SIN(3.0-ALOG(-X) = SIN(3.0-2.708050) = 5.095466E-03

ABS(X/2.0) = 7.5
ALOG10(ABS(X/2.0)-5.0)=ALOG10(7.5-5)=0.3979

5.095466E-03**0.3*4.0+0.3979*40 =16.736758

Y = 15.707963

PROBLEMA 5

A) Y = -5.5 , x = 4

Y**X = (-5.5) (-5.5) (-5.5) (-5.5) = 915.06

Al ser un exponente entero, se calcula multiplicando la base por si misma las...