since it has a diameter of 28, then its radius must be half that or 14.
![\textit{area of a circle}\\\\ A=\pi r^2~~ \begin{cases} r=radius\\[-0.5em] \hrulefill\\ r=14 \end{cases}\implies A=\pi (14)^2\implies A=196\pi ~\hfill \stackrel{\stackrel{semi-circle}{half~that}}{98\pi }](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20circle%7D%5C%5C%5C%5C%20A%3D%5Cpi%20r%5E2~~%20%5Cbegin%7Bcases%7D%20r%3Dradius%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20r%3D14%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Cpi%20%2814%29%5E2%5Cimplies%20A%3D196%5Cpi%20~%5Chfill%20%5Cstackrel%7B%5Cstackrel%7Bsemi-circle%7D%7Bhalf~that%7D%7D%7B98%5Cpi%20%7D)
Answer:
The correct answer is x = 6.
Step-by-step explanation:
To solve this problem, we first must recognize that DE + EF = DF.
From this information, we can set up the following equation when we substitute in the given values:
DE + EF = DF
(4x-1) + 9 = 9x - 22
Our first step is going to be to combine like terms on the left side of the equation. This is modeled below:
4x - 1 + 9 = 9x - 22
4x + 8 = 9x - 22
Then, we should subtract 4x from both sides of the equation.
8 = 5x - 22
Now, we can add 22 to both sides of the equation.
30 = 5x
Finally, we can divide both sides by 5.
x = 6
Therefore, the correct answer is x = 6.
Hope this helps!
Either way you go 180 degrees clockwise or counter clockwise it will land in the same spot
Answer:
Step-by-step explanation:
