Estudio De Caso
Enviado por samanty • 19 de Enero de 2013 • 769 Palabras (4 Páginas) • 343 Visitas
Falla original
Falla perpendicular
x Y=1.7x-0.8 y x Y = - 10x – 0.8
17
y
-5 Y=1.7 (-5) – 0.8 -9.3 -5 Y = - 10(-5)– 0.8
17
2.14
-4 Y=1.7 (-4) – 0.8 -7.6 -4 Y = - 10(-4)– 0.8
17
1.55
-3 Y=1.7 (-3) – 0.8 -5.9 -3 Y = - 10(-3)– 0.8
17
0.97
-2 Y=1.7 (-2) – 0.8 -4.2 -2 Y = - 10(-2)– 0.8
17
0.38
-1 Y=1.7 (-1) – 0.8 -2.5 -1 Y = - 10(-1)– 0.8
17
-0.21
0 Y=1.7 (-0) – 0.8 0.8 0 Y = - 10(0)– 0.8
17
-0.8
1 Y=1.7 (1) – 0.8 0.9 1 Y = - 10(1)– 0.8
17
-1.39
2 Y=1.7 (2) – 0.8 2.6 2 Y = - 10(2)– 0.8
17
-1.98
3 Y=1.7 (3) – 0.8 4.3 3 Y = - 10(3)– 0.8
17
-2.57
4 Y=1.7 (4) – 0.8 6 4 Y = - 10(4)– 0.8
17
-3.15
5 Y=1.7 (5) – 0.8 7.7 5 Y = - 10(5)– 0.8
17
-3.74
Falla original
Falla perpendicular
x Y=1.7x-0.8 y x Y = - 10x – 0.8
17
y
-5 Y=1.7 (-5) – 0.8 -9.3 -5 Y = - 10(-5)– 0.8
17
2.14
-4 Y=1.7 (-4) – 0.8 -7.6 -4 Y = - 10(-4)– 0.8
17
1.55
-3 Y=1.7 (-3) – 0.8 -5.9 -3 Y = - 10(-3)– 0.8
17
0.97
-2 Y=1.7 (-2) – 0.8 -4.2 -2 Y = - 10(-2)– 0.8
17
0.38
-1 Y=1.7 (-1) – 0.8 -2.5 -1 Y = - 10(-1)– 0.8
17
-0.21
0 Y=1.7 (-0) – 0.8 0.8 0 Y = - 10(0)– 0.8
17
-0.8
1 Y=1.7 (1) – 0.8 0.9 1 Y = - 10(1)– 0.8
17
-1.39
2 Y=1.7 (2) – 0.8 2.6 2 Y = - 10(2)– 0.8
17
-1.98
3 Y=1.7 (3) – 0.8 4.3 3 Y = - 10(3)– 0.8
17
-2.57
4 Y=1.7 (4) – 0.8 6 4 Y = - 10(4)– 0.8
17
-3.15
5 Y=1.7 (5) – 0.8 7.7 5 Y = - 10(5)– 0.8
17
-3.74
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