Need !
a ride at an amusement park has two options for cars, one that holds 6 passengers and one that holds 3 passengers. if the ride designer wants to make sure that each run holds at least 12 passengers, which graph shows the possible combinations of the number of 6-passenger cars, x, and the number of 3-passenger cars y?

yep i agree it's the 4th one i took the test and i got that one correct

Step-by-step explanation:

Fourth option is correct.

Step-by-step explanation:

Since we have given that

Number of passengers for first car can hold = 6

Number of passengers for second car can hold = 3

According to question, If the ride designer wants to make sure that each run holds at least 12 passengers.

Let the number of 6 passengers cars be 'x'.

Let the number of 3-passengers cars be 'y'.

So, our equation becomes,

Put x = 0, y= 0

So, it becomes 0≥4 which is false.

So, it is shaded away from the origin.

At x= 0, y≥4

at y = 0, x≥6

Hence, Fourth option is correct.

-2/

simplifying

5(x + -6) = 2(x + 3)

reorder the terms:

5(-6 + x) = 2(x + 3)

(-6 * 5 + x * 5) = 2(x + 3)

(-30 + 5x) = 2(x + 3)

reorder the terms:

-30 + 5x = 2(3 + x)

-30 + 5x = (3 * 2 + x * 2)

-30 + 5x = (6 + 2x)

solving

-30 + 5x = 6 + 2x

solving for variable 'x'.

move all terms containing x to the left, all other terms to the right.

add '-2x' to each side of the equation.

-30 + 5x + -2x = 6 + 2x + -2x

combine like terms: 5x + -2x = 3x

-30 + 3x = 6 + 2x + -2x

combine like terms: 2x + -2x = 0

-30 + 3x = 6 + 0

-30 + 3x = 6

add '30' to each side of the equation.

-30 + 30 + 3x = 6 + 30

combine like terms: -30 + 30 = 0

0 + 3x = 6 + 30

3x = 6 + 30

combine like terms: 6 + 30 = 36

3x = 36

divide each side by '3'.

x = 12

simplifying

x = 12

this should .