The statement that is true is that the median pitches are 5 apart.
10,15,15,15,15,20,20,25,25
Median:15 is in the middle
10,15,15,15,20,25,25,25,25
Median: 20 is in the middle
19 yards divided by 2.5 minutes.
=19/2.5
= 7.6 yards/minute
Answer:
<h2>
<em>3</em><em>9</em><em> </em><em>units</em></h2>
<em>Solution,</em>
<em>Total </em><em>area=</em><em>Area </em><em>of </em><em>ABCD+</em><em> </em><em>Area </em><em>of </em><em>triangle </em><em>ADE</em>
<em>
</em>
<em>hope </em><em>this </em><em>helps.</em><em>.</em><em>.</em>
<em>Good </em><em>luck</em><em> on</em><em> your</em><em> assignment</em><em>.</em><em>.</em><em>.</em>
The path that Gloria follows when she jumped is a path of parabola.
The equation of the parabola that describes the path of her jump is ![\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20-%5Cfrac%7B5%7D%7B49%7D%28x%20-%2014%29%5E2%20%2B%2020%7D)
The given parameters are:
![\mathbf{Height = 20}](https://tex.z-dn.net/?f=%5Cmathbf%7BHeight%20%3D%2020%7D)
![\mathbf{Length = 28}](https://tex.z-dn.net/?f=%5Cmathbf%7BLength%20%3D%2028%7D)
<em>Assume she starts from the origin (0,0)</em>
The midpoint would be:
![\mathbf{Mid = \frac 12 \times Length}](https://tex.z-dn.net/?f=%5Cmathbf%7BMid%20%3D%20%5Cfrac%2012%20%5Ctimes%20Length%7D)
![\mathbf{Mid = \frac 12 \times 28}](https://tex.z-dn.net/?f=%5Cmathbf%7BMid%20%3D%20%5Cfrac%2012%20%5Ctimes%2028%7D)
![\mathbf{Mid = 14}](https://tex.z-dn.net/?f=%5Cmathbf%7BMid%20%3D%2014%7D)
So, the vertex of the parabola is:
![\mathbf{Vertex = (Mid,Height)}](https://tex.z-dn.net/?f=%5Cmathbf%7BVertex%20%3D%20%28Mid%2CHeight%29%7D)
Express properly as:
![\mathbf{(h,k) = (14,20)}](https://tex.z-dn.net/?f=%5Cmathbf%7B%28h%2Ck%29%20%3D%20%2814%2C20%29%7D)
A point on the graph would be:
![\mathbf{(x,y) = (28,0)}](https://tex.z-dn.net/?f=%5Cmathbf%7B%28x%2Cy%29%20%3D%20%2828%2C0%29%7D)
The equation of a parabola is calculated using:
![\mathbf{y = a(x - h)^2 + k}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20a%28x%20-%20h%29%5E2%20%2B%20k%7D)
Substitute
in ![\mathbf{y = a(x - h)^2 + k}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20a%28x%20-%20h%29%5E2%20%2B%20k%7D)
![\mathbf{y = a(x - 14)^2 + 20}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20a%28x%20-%2014%29%5E2%20%2B%2020%7D)
Substitute
in ![\mathbf{y = a(x - 14)^2 + 20}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20a%28x%20-%2014%29%5E2%20%2B%2020%7D)
![\mathbf{0 = a(28 - 14)^2 + 20}](https://tex.z-dn.net/?f=%5Cmathbf%7B0%20%3D%20a%2828%20-%2014%29%5E2%20%2B%2020%7D)
![\mathbf{0 = a(14)^2 + 20}](https://tex.z-dn.net/?f=%5Cmathbf%7B0%20%3D%20a%2814%29%5E2%20%2B%2020%7D)
Collect like terms
![\mathbf{a(14)^2 =- 20}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba%2814%29%5E2%20%3D-%2020%7D)
Solve for a
![\mathbf{a =- \frac{20}{14^2}}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba%20%3D-%20%5Cfrac%7B20%7D%7B14%5E2%7D%7D)
![\mathbf{a =- \frac{20}{196}}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba%20%3D-%20%5Cfrac%7B20%7D%7B196%7D%7D)
Simplify
![\mathbf{a =- \frac{5}{49}}](https://tex.z-dn.net/?f=%5Cmathbf%7Ba%20%3D-%20%5Cfrac%7B5%7D%7B49%7D%7D)
Substitute
in ![\mathbf{y = a(x - 14)^2 + 20}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20a%28x%20-%2014%29%5E2%20%2B%2020%7D)
![\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20-%5Cfrac%7B5%7D%7B49%7D%28x%20-%2014%29%5E2%20%2B%2020%7D)
Hence, the equation of the parabola that describes the path of her jump is ![\mathbf{y = -\frac{5}{49}(x - 14)^2 + 20}](https://tex.z-dn.net/?f=%5Cmathbf%7By%20%3D%20-%5Cfrac%7B5%7D%7B49%7D%28x%20-%2014%29%5E2%20%2B%2020%7D)
See attachment for the graph
Read more about equations of parabola at:
brainly.com/question/4074088