A ball game between Dylan and Mya is about to begin. Using sidewalk chalk, they draw their “arena” on the ground. The arena can be modeled by the equation x/2 + y/2 + 81 . Dylan has taken his position at (-5, -2) and Mya has taken her position at (6, 4). Each graph unit represents 1 meter.

Question

Task 1: Graph the arena and the positions of Dylan and Mya on the graph below.
ANSWER:

Task 2: How far apart are Dylan and Mya (to two decimal places)?
ANSWER:

Task 3: Jordyn joins the game so that she forms a triangle with Dylan and Mya. Her exact coordinates are NOT important, but we do know that from Jordyn’s position, the angle formed to Dylan and Mya is 76°. From Dylan’s position, the angle formed to Mya and Jordyn is 50°.
Draw a diagram below with this information and the information from Tasks 1 and 2.
Use this diagram to find the distances from Jordan to Dylan and Jordyn to Mya (to two decimal places).
Remember: DO NOT assign Jordyn coordinates. Place Jordyn somewhere so that the angles above make sense.
ANSWER:

Task 4: Dylan, Mya and Jordyn proceed to play the ball game which consists of running around the inside of the arena while throwing the ball back and forth to each other. The game starts to get out of control when Dylan head-butts the ball in a path modeled by ​᠎​h = -2t/2 + 8t + 4.9 ​᠎​​᠎​​᠎​​᠎​the relation where t is time in seconds and h is height in meters.
How high was the ball off the ground when Dylan’s head hit it?
ANSWER:

Task 5: What was the maximum height the ball reached?
ANSWER:

Task 6: How long did it take the ball to reach its maximum height?
ANSWER:

Answer 1

I'm sorry, but I can't generate a solution to this problem as it involves drawing and graphing.

Answer 2

ignore the drawing and graphing.

Answer 3

Task 1: The arena equation is x/2 + y/2 + 81 = 0.
Dylan's position is (-5, -2) and Mya's position is (6, 4).
Plot these points on a graph.

Task 2: To find the distance between Dylan and Mya, we can use the distance formula.
Distance = √((x2 - x1)^2 + (y2 - y1)^2)
Distance = √((6 - (-5))^2 + (4 - (-2))^2)
Distance = √(11^2 + 6^2)
Distance ≈ 12.04 meters

Task 3: Since we do not have the exact coordinates of Jordyn, we cannot find the distances from Jordyn to Dylan and Mya.

Task 4: To find the height of the ball when Dylan's head hit it, we need to plug in the time t into the height equation.
h = -2t/2 + 8t + 4.9
Since we do not have a specific value of t, we cannot determine the exact height.

Task 5: To find the maximum height reached by the ball, we can use the vertex form of the quadratic equation.
The general equation is h = at^2 + bt + c, where a = -2, b = 8, and c = 4.9.
The maximum height is reached at the vertex, which can be found using the formula t = -b/2a.
t = -(8) / (2(-2))
t = 2 seconds

To find the maximum height, substitute the value of t = 2 into the equation:
h = -2(2)^2 + 8(2) + 4.9
h ≈ 17.8 meters

Task 6: The ball takes 2 seconds to reach its maximum height.

Answer 4

How long was the ball in the air (to two decimal places)?

Answer 5

To determine how long the ball was in the air, we need to find the time at which the height, h, is equal to zero. We can set the height equation to zero and solve for t:

0 = -2t/2 + 8t + 4.9

Simplifying the equation:

0 = -t + 8t + 4.9
0 = 7t + 4.9
-4.9 = 7t
t = -4.9/7
t ≈ -0.7

Since time cannot be negative in this context, we disregard the negative solution. The ball was in the air for approximately 0.7 seconds.

Answer 6

To add more excitement to the game, a second ball is added. Dylan kicks his ball along a path modeled by the equation 12x + 9y = -54 and Mya’s ball travels on a path modeled by the equation 2x - y - 4 =0 .
If the game were set on a Cartesian Plane, to what coordinates should Jordyn go to trap BOTH balls at the same time? Use an algebraic method to solve this problem.

Answer 7

To find the coordinates where Jordyn should go to trap both balls at the same time, we need to solve the system of equations formed by the paths of Dylan's ball and Mya's ball.

The equations are:
12x + 9y = -54
2x - y - 4 = 0

We can solve this system using the method of substitution or elimination. Let's use the substitution method:

From the second equation, we can isolate y:
y = 2x - 4

Substitute this value of y into the first equation:
12x + 9(2x - 4) = -54
12x + 18x - 36 = -54
30x - 36 = -54
30x = -54 + 36
30x = -18
x = -18/30
x = -3/5

Substitute this value of x back into the second equation to find y:
2(-3/5) - y - 4 = 0
-6/5 - y - 4 = 0
-6/5 - y = 4
-6/5 - 4 = y
y = -6/5 - 20/5
y = -26/5

Therefore, Jordyn should go to the coordinates (-3/5, -26/5) to trap both balls at the same time.

Answer 8

Dylan kicks his ball over Jordyn’s head. The path of his kick is shown. Determine the equation of the path of his ball in standard form.

Answer 9

To determine the equation of the path of Dylan's ball, we need to convert the given equation, 12x + 9y = -54, into standard form (Ax + By + C = 0).

Starting with the given equation:
12x + 9y = -54

First, we want to get rid of the coefficient of 12 in front of x. We can achieve this by dividing the entire equation by 12:
(12x + 9y)/12 = -54/12
x + (9/12)y = -54/12

Next, we can simplify the equation by simplifying the fraction (9/12):
x + (3/4)y = -9/4

To eliminate the fraction, we can multiply the entire equation by 4:
4x + 3y = -9

Therefore, the equation of the path of Dylan's ball in standard form is 4x + 3y = -9.

Answer 10

The kids decide their arena is too small. They use rope to enclose a new, rectangular arena. The area of this new enclosure is given by A = 99w - 3w/2 , where A is the area in square meters and w is width in meters. What dimensions should they make the arena in order to have an arena with an area of 726 m2? Use your knowledge of quadratic relations to solve this problem. Do not use a guess and check method.

Answer 11

To find the dimensions of the new arena that will have an area of 726 m^2, we can set the equation A = 99w - 3w/2 equal to 726, and then solve for the width, w.

99w - 3w/2 = 726

First, let's get rid of the fraction by multiplying the entire equation by 2:
2(99w - 3w/2) = 2(726)
198w - 3w = 1452
195w = 1452
w = 1452/195
w = 7.46

Therefore, the width of the new arena should be approximately 7.46 meters. To find the length, we can substitute this value of w back into the equation A = 99w - 3w/2:

A = 99(7.46) - 3(7.46)/2
A = 734.54 - 11.19
A = 723.35

So, the dimensions of the new arena should be approximately 7.46 meters by 723.35 meters to have an area of 726 m^2.