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.
what are the points
How far apart are Dylan and Mya (to two decimal places)?
To find the points of the ball game arena, we can use the equation x/2 + y/2 = 81.
To find the points, we can substitute different values for x and solve for y or vice versa.
Let's substitute some values for x and solve for y:
When x = 0, 0/2 + y/2 = 81
y/2 = 81
y = 162
So, one point of the arena is (0, 162).
When y = 0, x/2 + 0/2 = 81
x/2 = 81
x = 162
So, another point of the arena is (162, 0).
Let's now find the distance between Dylan and Mya.
Using the distance formula:
Distance = sqrt((x2 - x1)^2 + (y2 - y1)^2)
Let Dylan's position be (x1, y1) = (-5, -2) and Mya's position be (x2, y2) = (6, 4).
Distance = sqrt((6 - (-5))^2 + (4 - (-2))^2)
Distance = sqrt((6 + 5)^2 + (4 + 2)^2)
Distance = sqrt((11)^2 + (6)^2)
Distance = sqrt(121 + 36)
Distance = sqrt(157)
Distance ≈ 12.53 meters
Therefore, Dylan and Mya are approximately 12.53 meters apart.
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.
Unfortunately, as a text-based AI, I cannot draw diagrams. However, I can provide you with the information you need to find the distances from Jordyn to Dylan and Jordyn to Mya.
To find the distances from Jordyn to Dylan and Jordyn to Mya, we can use the law of sines. The law of sines states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all three sides of the triangle.
Let's denote the distance from Jordyn to Dylan as d and the distance from Jordyn to Mya as m.
From Task 1, we know that the distance between Dylan and Mya is approximately 12.53 meters.
Using the law of sines, we have:
d/sin(76°) = 12.53/sin(50°)
Therefore, we can solve for d:
d = sin(76°) * (12.53 / sin(50°))
Similarly, we can find m:
m/sin(76°) = 12.53/sin(180° - 76° - 50°)
m = sin(76°) * (12.53 / sin(54°))
Using a calculator to evaluate the sines and perform the calculations, we can find the distances from Jordyn to Dylan (d) and Jordyn to Mya (m) to two decimal places.
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 h = 2t/2 + 8t + 4.9 by 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?
To find how high the ball was off the ground when Dylan's head hit it, we need to substitute the value of t when the ball was hit into the equation h = 2t/2 + 8t + 4.9.
Dylan's head hit the ball, so we need to find the value of t when h = 0 (since the ball hit Dylan's head when it was at the ground level).
0 = 2t/2 + 8t + 4.9
Simplifying the equation:
0 = t + 8t + 4.9
0 = 9t + 4.9
Subtracting 4.9 from both sides:
-4.9 = 9t
Dividing by 9:
t = -4.9/9
Therefore, the ball was at a height of h = 0 meters off the ground when Dylan's head hit it.
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 h = -2t/2 + 8t + 4.9 by 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?
To find how high the ball was off the ground when Dylan's head hit it, we need to substitute the value of t when the ball was hit into the equation h = -2t^2 + 8t + 4.9.
Since Dylan's head hit the ball, we need to find the value of t when h = 0 (since the ball hit Dylan's head when it was at the ground level).
0 = -2t^2 + 8t + 4.9
To solve this quadratic equation, we can either factor it or use the quadratic formula. One way to solve it is by using the quadratic formula:
t = (-b ± sqrt(b^2 - 4ac)) / (2a)
In this case, a = -2, b = 8, and c = 4.9.
Substituting these values into the quadratic formula:
t = (-8 ± sqrt(8^2 - 4(-2)(4.9))) / (2(-2))
t = (-8 ± sqrt(64 + 39.2)) / (-4)
t = (-8 ± sqrt(103.2)) / (-4)
Approximately, t ≈ (-8 ± 10.16) / (-4)
To find the exact value of t when the ball was hit, we need to evaluate the positive root:
t ≈ (-8 + 10.16) / (-4)
t ≈ 2.16 / (-4)
t ≈ -0.54
Therefore, the ball was at a height of h ≈ 0 meters off the ground when Dylan's head hit it.
What was the maximum height the ball reached?
To find the maximum height the ball reached, we need to determine the vertex of the quadratic equation h = -2t^2 + 8t + 4.9.
The vertex of a quadratic equation in the form h = at^2 + bt + c can be found using the formula t = -b / (2a).
In this case, a = -2 and b = 8:
t = -8 / (2(-2))
t = -8 / (-4)
t = 2
Now, we can find the maximum height by substituting this value of t back into the equation:
h = -2(2)^2 + 8(2) + 4.9
h = -2(4) + 16 + 4.9
h = -8 + 16 + 4.9
h = 12.9
Therefore, the maximum height the ball reached was 12.9 meters.