The local orchestra has been invited to play at a festival. there are 111 members of the orchestra and 6 are licensed to drive large multi-passenger vehicles. busses hold 25 people but are much more expensive to rent. passenger vans hold 12 people. make a system of equations to find the smallest number of busses the orchestra can rent. let x= the number of busses and y= the number of vans make an equation representing the number of vehicles needed. make an equation representing the total number of seats in vehicles for the orchestra members. solve the system of equations. how many busses does the orchestra need to rent? how many 12-passenger vans does the orchestra need to rent?

Using a System of equations, the number of buses and vans required are 3 respectively.

Using the system of equations :

Total number of buses :

Total number of passengers :

From (1)

Substitute (3) into (2)

25(6-v) + 12v = 111

150 - 25v + 12v = 111

-13v = 111 - 150

-13v = - 39

v = 39/13

v = 3

From (3)

b = 6 - 3

b = 3

Hence, the number of vans and buses required are 3 and 3 respectively.

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- Make an equation representing the number of vehicles needed.

We have six drivers so

x + y ≤ 6

That's not really an equation; it's an inequality.  We want to use all our drivers so we can use the small vans, so

x + y = 6

- Make an equation representing the total number of seats in vehicles for the orchestra members.

s = 25x + 12y

That's how many seats total; it has to be at least 111 so again an inequality,

25x + 12y ≥ 111

We solve it like a system of equations.  

x + y = 6

y = 6 - x

111 = 25x + 12y = 25x + 12(6-x)

111 = 25x + 72 - 12x

111 - 72 = 13 x

39 = 13 x

x = 3

Look at that,  it worked out exactly.  It didn't have to.

y = 6 - x = 3

3 buses, 3 vans

la parábola es el lugar

geométrico en el cual los puntos tienen una distancia idéntica, equidistantes,

de un punto denominado foco, f, y una recta denominada directriz

el radio vector de una

parábola es el segmento que existe entre un punto cualquiera y el foco.

por la ecuación se observa

que la parábola abre hacia la derecha, por lo cual podemos usar las siguientes fórmulas

la ecuación ordinaria de la

parábola es y^2 = 4px, donde p es la longitud desde el vértice de la parábola al

foco.

la fórmula de la directriz

es x + p = 0

resolviendo:

y^2-9x = 0

y^2 = 9x

4p = 9, entonces p = 9/4

entonces el foco tendrá la

coordenada f(9/4,0)

la directriz vendrá

expresada por

x + p = 0

x = -p

x = -9/4

el problema indica que la

ordenada es igual a 6, entonces

(6)^2=9x

x = 36/9

x = 4

entonces el punto al cual

hay que obtener la distancia será p(4,6)

calculando se tendrá: pf = 

pf = 6,25 u

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step-by-step explanation:

the answer is b