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:
b=9.
Step-by-step explanation:
Answer:
From the given, we establish this relationship:
RS + ST = RT
RS = ST since S is the midpoint of RT
Thus, let's use RS = ST to find x,
By substituting the given expressions:
RS = ST
8x + 11 = 14x - 1
12 = 6x
x = 12/6
thus,
x = 2
Answer:
no I have no idea I justwant points
Answer: 1/9
Explanation: these types of repeating decimals are equivalent to the fraction that has the number doing the repeating in its numerator and the number 9 in its denominator.
Hope this helps!
