I don't get what you want me to explain.
Just add 303.28+295.73 and you will get $599.01 as your answer!!! I hope this helps!!!
Using quadratic function concepts, it is found that:
1) The equation is ![y = -0.08x^2 + 4x](https://tex.z-dn.net/?f=y%20%3D%20-0.08x%5E2%20%2B%204x)
2) The minimum distance is of 2.64 yards, and the maximum is of 47.36 yards.
A quadratic equation has the following format:
![y = ax^2 + bx + c](https://tex.z-dn.net/?f=y%20%3D%20ax%5E2%20%2B%20bx%20%2B%20c)
In this problem:
- The ball was kicked from the ground, thus
, which means that c = 0.
Item 1:
The vertex is:
, in which:
![x_v = -\frac{b}{2a}](https://tex.z-dn.net/?f=x_v%20%3D%20-%5Cfrac%7Bb%7D%7B2a%7D)
![y_v = -\frac{\Delta}{4a} = -\frac{b^2 - 4ac}{4a}](https://tex.z-dn.net/?f=y_v%20%3D%20-%5Cfrac%7B%5CDelta%7D%7B4a%7D%20%3D%20-%5Cfrac%7Bb%5E2%20-%204ac%7D%7B4a%7D)
In this problem, it is (25,50). Thus:
![-\frac{b}{2a} = 25](https://tex.z-dn.net/?f=-%5Cfrac%7Bb%7D%7B2a%7D%20%3D%2025)
![-b = 50a](https://tex.z-dn.net/?f=-b%20%3D%2050a)
![b = -50a](https://tex.z-dn.net/?f=b%20%3D%20-50a)
![-\frac{b^2 - 4ac}{4a} = 50](https://tex.z-dn.net/?f=-%5Cfrac%7Bb%5E2%20-%204ac%7D%7B4a%7D%20%3D%2050)
![-b^2 = 200a](https://tex.z-dn.net/?f=-b%5E2%20%3D%20200a)
![-(50a)^2 = 200a](https://tex.z-dn.net/?f=-%2850a%29%5E2%20%3D%20200a)
![-2500a^2 - 200a = 0](https://tex.z-dn.net/?f=-2500a%5E2%20-%20200a%20%3D%200)
![200a(-12.5a - 1) = 0](https://tex.z-dn.net/?f=200a%28-12.5a%20-%201%29%20%3D%200)
Then, as ![a \neq 0](https://tex.z-dn.net/?f=a%20%5Cneq%200)
![a = -\frac{1}{12.5} = -0.08](https://tex.z-dn.net/?f=a%20%3D%20-%5Cfrac%7B1%7D%7B12.5%7D%20%3D%20-0.08)
![b = -50a = -50(-0.08) = 4](https://tex.z-dn.net/?f=b%20%3D%20-50a%20%3D%20-50%28-0.08%29%20%3D%204)
Thus, the equation, in standard form, is:
![y = -0.08x^2 + 4x](https://tex.z-dn.net/?f=y%20%3D%20-0.08x%5E2%20%2B%204x)
Item 2:
The distances are the values of x for which:
![y = 10](https://tex.z-dn.net/?f=y%20%3D%2010)
Then
![-0.08x^2 + 4x = 10](https://tex.z-dn.net/?f=-0.08x%5E2%20%2B%204x%20%3D%2010)
![-0.08x^2 + 4x - 10 = 0](https://tex.z-dn.net/?f=-0.08x%5E2%20%2B%204x%20-%2010%20%3D%200)
Which is a quadratic equation with
. Then:
![\Delta = 4^2 - 4(-0.08)(-10) = 12.8](https://tex.z-dn.net/?f=%5CDelta%20%3D%204%5E2%20-%204%28-0.08%29%28-10%29%20%3D%2012.8)
The minimum distance is of 2.64 yards, and the maximum is of 47.36 yards.
A similar problem is given at brainly.com/question/24713268