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kolbaska11 [484]
2 years ago
11

Pls show working 3 friends shared the driving on a long trip. Marla drove 7 miles more than Guido. Guido drove five times as far

as Juanita did Juanita drove 112 miles. How Long was the trip thank you
Mathematics
2 answers:
olga2289 [7]2 years ago
5 0

Step-by-step explanation:

Marla drove 567

Gudio drove 560

The trip was 679

GarryVolchara [31]2 years ago
3 0

Answer:

1239

Step-by-step explanation:

Juanita: x

Guido: 5x

Marla: 5x+7

We know Juanita drove 112 miles. This means that x=112.

Juanita: 112

Guido: 5*112=560

Marla: 560+7= 567

Now we can add up the number of miles each friend drove to get the total distance of the trip.

112+560+567=1239

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Mary sells toys the selling price of each toy is $10.16 she sold 15 toys for the day. How much money did she collect from sellin
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A linear function is parallel to the line y= 4x + 8 and also goes through the point (2,15).What is the linear equation for this
horrorfan [7]

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Now, the slope of the known line is 4, so our line's slope will be four as well.


In general, when you know the slope m of a line and one of its points (x_0,y_0), the equation of the line can be derived from the following formula:


y-y_0 = m(x-x_0)


Which in your case becomes


y-15 = 4(x-2)


Expand the right hand side and solve for y:


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3 0
3 years ago
1) UN MOVIL A SE MUEVE DESDE UN PUNTO CON VELOCIDAD CONSTANTE DE 20m/s EN EL MISMO INSTANTE A UNA DISTANCIA DE 1200m, OTRO MOVIL
alisha [4.7K]

Answer:

El móvil B necesita 60 segundos para alcanzar al móvil A y le alcanza una distancia de 2400 metros con respecto al punto de referencia.

Step-by-step explanation:

Supóngase que cada movil viaja en el mismo plano y que el móvil B se localiza inicialmente en la posición x = 0\,m, mientras que el móvil A se encuentra en la posición x = 1200\,m. Ambos móviles viajan a rapidez constante. Si el móvil B alcanza al móvil A después de cierto tiempo, el sistema de ecuaciones cinemáticas es el siguiente:

Móvil A

x_{A} = 1200\,m+\left(20\,\frac{m}{s} \right)\cdot t

Móvil B

x_{B} = \left(40\,\frac{m}{s} \right)\cdot t

Donde:

x_{A}, x_{B} - Posiciones finales de cada móvil, medidas en metros.

t - Tiempo, medido en segundos.

Si x_{A} = x_{B}, el tiempo requerido por el móvil B para alcanzar al móvil A es:

1200\,m+\left(20\,\frac{m}{s} \right)\cdot t = \left(40\,\frac{m}{s} \right)t

1200\,m = \left(20\,\frac{m}{s} \right)\cdot t

t = \frac{1200\,m}{20\,\frac{m}{s} }

t = 60\,s

El móvil B necesita 60 segundos para alcanzar al móvil A.

Ahora, la distancia se obtiene por sustitución directa en cualquiera de las ecuaciones cinemáticas:

x_{B} = \left(40\,\frac{m}{s} \right)\cdot (60\,s)

x_{B} = 2400\,m

El móvil B alcanza al móvil A a una distancia de 2400 metros con respecto al punto de referencia.

3 0
3 years ago
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