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aalyn [17]
3 years ago
9

Solve by factoring: X^2-13x-14=0

Mathematics
1 answer:
AURORKA [14]3 years ago
4 0

Answer:

x = -1 and x= 14

Step-by-step explanation:

x^2 -13x -14=0

x^2 + x- 14x -14 = 0

x(x+1) -14(x+1) = 0

(x+1)(x-14) =0

x+1 = 0

x= -1

x-14=0

x= 14

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Una lancha que viaja a 10 m/s pasa por debajo de un puente 3 segundos después que ha pasado un bote que viaja a 7 m/s, ¿después
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Answer:

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

Step-by-step explanation:

Sea el punto debajo del puente el punto de referencia y que ambas lanchas se desplazan a velocidad a continuación, las ecuaciones cinemáticas para cada embarcación son presentadas a continuación:

Bote a 7 metros por segundo

x_{A} = x_{o}+v_{A}\cdot t (Ec. 1)

Lancha a 10 metros por segundo

x_{B} = x_{o}+v_{B}\cdot (t-3\,s) (Ec. 2)

Donde:

x_{o} - Posición debajo del puente, medido en metros.

x_{A}, x_{B} - Posición final de cada embarcación, medido en metros.

v_{A}, v_{B} - Velocidad de cada embarcación, medida en metros por segundo.

t - Tiempo, medido en segundos.

Para determinar la posición en la que ambas embarcaciones se encuentran, se debe determinar el instante en que ocurre a partir de la siguiente condición: x_{A} = x_{B}

Igualando (Ec. 1) y (Ec. 2) se tiene que:

v_{A}\cdot t = v_{B}\cdot (t-3\,s)

Ahora despejamos el tiempo:

3\cdot v_{B} = (v_{B}-v_{A})\cdot t

t = \frac{3\cdot v_{B}}{v_{B}-v_{A}}

Si sabemos que v_{B} = 10\,\frac{m}{s} y v_{A} = 7\,\frac{m}{s}, entonces:

t = \frac{3\cdot \left(10\,\frac{m}{s} \right)}{10\,\frac{m}{s}-7\,\frac{m}{s}}

t = 10\,s

Ahora, la posición de encuentro es: (x_{o} = 0\,m, v_{A} = 7\,\frac{m}{s} y t = 10\,s)

x_{A} = 0\,m + \left(7\,\frac{m}{s} \right)\cdot (10\,s)

x_{A} = 70\,m

La lancha y el bote se encontrarán a 70 metros de distancia del puente.

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