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Vlad1618 [11]
3 years ago
13

Help me answer question 2 ​

Mathematics
1 answer:
andrew-mc [135]3 years ago
7 0

Answer:

x + x + x

Step-by-step explanation:

We have 3y and are asked to express it equivalently with the answers provided.

I will go over the flaws of each answer and single it down to the correct one.

x * x * x would equal x^3, since this is being multiplied.

x^3 is equal to x * x * x, therefore it is not correct.

x + 3 is not correct because you cannot add numbers or variables that are not like terms, so it would not equal to 3y.

x + x + x would equal to 3y because they are all like terms and do not form exponents.

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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
ExtremeBDS [4]

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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