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:
69 Pieces of candy
Step-by-step explanation
So you get 3 pieces of candy at every house. And you do that 23 more times. We can right that out as 23*3. Which is 69.
<b and the angle 79 degrees are exterior alternate angles which are equal
b = 79 degrees.
Ok so for number 5, AB is proportional to BC just like PQ is proportional to QR. AB = 9, BC = 15, PQ = 3, and QR = x. 9/15 = 3/x. Cross Multiply: This gives you - 9x = 45. Divide both sides by 9, and you get x = 5. QR = 5
