Question

The owners of a circus are planning a new act. If the path of the trapeze can be modeled by the parabola y = *x + 16 and the path of the zip-line can be modeled by y = 2x + 12, at what point can the trapeze artist grab the second acrobat?

A) (2, 20)
B) (4, 24)
C) (6, 28)
D) (8, 32)

Answer 1

To find where the trapeze artist can grab the second acrobat, we solve the system of equations y = x + 16 and y = 2x + 12 by setting them equal to each other. Solving for x, we get x = 4, which when plugged back into either equation gives us y = 20. Hence, the intersection point is (4, 20).

Step-by-step explanation:

The question involves finding the intersection point of two linear equations which represent the paths of a trapeze artist and a zip-line. To find where the trapeze artist can grab the second acrobat, we need to set the two equations equal to each other (since the intersection point will satisfy both equations) and solve for x:

Setting equation (1) equal to equation (2):

x + 16 = 2x + 12

Now we solve for x:

16 - 12 = 2x - x

4 = x

Plugging x = 4 back into either equation, we get:

y = 4 + 16 = 20 or y = 2(4) + 12 = 20

Thus, the point at which the trapeze artist can grab the second acrobat is (4, 20).