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)
<u>Answer:</u>
c= -1/2
<u>Step-by-step explanation:</u>
Let's solve your equation step-by-step.
4
/3 = −6c −
5/
3
Step 1: Simplify both sides of the equation.
4
/3 =−6c + −5
/3
Step 2: Flip the equation.
−6c + −5
/3 = 4/3
Step 3: Add 5/3 to both sides.
−6c + −5
/3 + 5
/3 = 4/3 + 5
/3 −6c
=3
Step 4: Divide both sides by -6.
−6c −6 = 3
− 6
c = −1
/2
Answer:
The area would be 48.
Step-by-step explanation:
what is the question it might be an Sony-75"