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VladimirAG [237]
3 years ago
6

two trains leave the station at the same time, one heading east and the other west. The eastbound train travels at 95 miles per

hour. The westbound train travels at 105 miles per hour. How long will it take for the two trains to be 320 miles apart? Do not do any rounding.
Mathematics
1 answer:
laila [671]3 years ago
4 0

Answer:

  1.6 hours

Step-by-step explanation:

Their separation speed is 95 +105 = 200 miles per hour. The time required for the given distance is ...

  time = distance/speed

  time = (320 mi)/(200 mi/h) = 1.6 h

It will take 1.6 hours for the trains to be 320 miles apart.

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Devide 7/24 by 35/48and reduce the quotient to the lowest fraction
liubo4ka [24]
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7/24 ÷ 35/48 = 7/24 * 48/35

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Hope this helps!


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3 years ago
What is the slope of the line that contains the points in the table ?​
zubka84 [21]

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Step-by-step explanation:

6 0
3 years ago
¿Para cuál(es) valor(es) de p las rectas de ecuación x − 1/p = 2 − y/p y x − 1/1 − p = y − 2/2 son perpendiculares? A) Solo para
Akimi4 [234]

Respuesta:

C) Solo para el -1

Explicación paso a paso:

Para resolver este problema, debemos de determinar la pendiente en cada una de las ecuaciones provistas:

\frac{x-2}{p}=\frac{2-y}{p}

y

\frac{x-1}{1-p}=\frac{y-2}{2}

ahora bien, necesitamos conocer el valor de la pendiente de una de las dos ecuaciones. Tomemos la primera ecuación y resolvámosla para y:

\frac{x-2}{p}=\frac{2-y}{p}

Multiplicamos ambos lados para p y obtenemos:

x-1=2-y

volteamos la ecuación y nos da:

2-y=x-1

pasamos el 2 a restar al otro lado y nos da:

-y=x-1-2

-y=x-3

y dividimos ambos lados de la ecuación dentro de -1

y=-x+3

esta ecuación ya tiene la forma pendiente intercepto:

y=mx+b

donde m es nuestra pendiente:

m_{1}=-1

Esta es la pendiente de una de las dos ecuaciones, para que la segunda ecuación sea perpendicular a la primera, su pendiente debe de ser el recíproco negativo de la pendiente de la primera ecuación, entonces la pendiente de la segunda ecuación debe ser:

m_{2}=-\frac{1}{m_{1}}

m_{2}=-\frac{1}{-1}

m_{2}=1

ahora tomamos la segunda ecuación y encontramos su pendiente. Tomemos la ecuación:

\frac{x-1}{1-p}=\frac{y-2}{2}

y despejemos y, comenzamos multiplicando ambos lados de la ecuación por 2, así que obtenemos:

2\frac{x-1}{1-p}=y-2

Multiplicamos el 2 por cada término de la fracción, entonces obtenemos:

\frac{2x-2}{1-p}=y-2

ahora pasamos el 2 a sumar al lado izquierdo y obtenemos:

\frac{2x-2}{1-p}+2=y

Ahora podemos separar la fracción del lado izquierdo en dos fracciones para obtener:

\frac{2x}{1-p}-\frac{2}{1-p}+2=y

volteamos la ecuación y nos da:

y=\frac{2x}{1-p}-\frac{2}{1-p}+2

Ahora nuestra ecuación ya tiene la forma y=mx+b

de aquí podemos determinar nuestra pendiente:

m=\frac{2}{1-p}

Con la primera ecuación determinamos que esta pendiente debería de ser igual a 1, entonces igualamos esa segunda pendiente a 1 para obtener:

\frac{2}{1-p}=1

y despejamos p

Pasamos a multiplicat el 1-p al lado derecho de la ecuación para obtener:

2=1-p

volteamos la ecuación:

1-p=2

pasamos el 1 a restar al lado derecho:

-p=2-1

-p=1

y multiplicamos ambos lados de la ecuación por -1 para obtener:

p=-1

Entonces la respuesta es C) solo para el -1

4 0
3 years ago
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