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Marizza181 [45]
3 years ago
12

How long can a go-kart travel on 6 gallons of gas

Mathematics
1 answer:
PilotLPTM [1.2K]3 years ago
6 0

Answer:

It depends in the go kart. But my guess is about 8 miles?

Step-by-step explanation:

I dont know.

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Can you please help me l will give brainliest​
yKpoI14uk [10]

Answer:

question 1 and 3

Step-by-step explanation:

3 0
3 years ago
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What is the length of the segment with endpoints (-3, 4) and (5, 4)? A) 2 B) 8 C) 16 D) 64 E) 2 17
Cerrena [4.2K]

Answer:

The correct answer is B) 8

Step-by-step explanation:

\sqrt{(4-4)^{2}+(5+3)^{2}   } \\\\\\\sqrt{8^{2} } \\\\8

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3 years ago
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What is the volume of the sphere?
poizon [28]

Answer: 113.1

Step-by-step explanation:

The formula is \frac{4}{3}\pir^{3}

The radius is 3, and 3 cubed is 27.

Then, 27 x \pi is about 84.82

Finally, 84.82 x \frac{4}{3} is about 113.1

Hope this helps. Please mark as brainliest, thanks!

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3 years ago
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The questions are in the pic the other answer is 356
Andru [333]
The answer to this question is 356
4 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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