Tres aviones salen a la misma hora de un mismo aeropuerto, el primero sale cada 8 dias, el segundo cada 10 dias, y el tercero cada 20 dias. Si salen el 2 de enero, ¿cuales seran las fechas proximas en que volveran a salir juntos? ( el año es de 365 dias)

Responder:

Después de 40 días (11 de febrero)

Explicación paso a paso:

Dado que :

El avión 1 sale cada 8 días

El avión 2 sale cada 10 días

El avión 3 sale cada 20 días

Si todos se van el 2 de enero; La próxima fecha en que se irán juntos se puede obtener obteniendo el mínimo común múltiplo de los días de cada avión:

___81020

2 | _4510

2 | _ 25 5

2 | _ 1 5 5

5 | _ 1 1 1

El mínimo común múltiplo es, por tanto;

2 * 2 * 2 * 5 = 40

Todos se irán juntos nuevamente después de 40 días:

(2 de enero + 40 días) = 11 de febrero

step-by-step explanation:

12: 00 p.m. to 7: 00 p.m. is a total of 7 hours.

4: 00 p.m. to 7: 00 p.m. is a total of 3 hours.

this means that from 4: 00 p.m to 7: 00 p.m. this represents 3/7 of what was lost from 12: 00 p.m. to 7: 00 p.m.

after knowing this, after doing 11,600 - 6,800 = 4,800

4,800 represents 3/7 of what was lost for 7 hours.

we then divide 4,800 by 3 which gets 1,600.

because we know that 1,600 represents 1/7 of how much water was lost, we just have to multiply by 7, which gets 11,200.   after adding 11,200 to how much water was left at 7: 00 p.m. which was 6,800. 11,200 + 6,800 = 18,000

this means that the pool originally had 18,000 gallons in the pool.

you can also check your work by doing 18,000 - 4(1600), which becomes           1800 - 6400 which gets you 11,600.

look at the picture.

y-intercept for the function on a picture is 4 → point (0, 4).

y-intercept for the function

is for x = 0. substitute:

y-intercept is 8 → point (0, 8).