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:
Step-by-step explanation:
n+1/2n=7/2
2n^2+1/2=7/2
4n^2+2=14
4n^2=12
n^2=3
n=radical3
Answer:
1) x < 1, x > 9
2) 2 < x ≤ 6, -4 ≤ x < 0
Step-by-step explanation:
1) lx - 5l > 4
x - 5 = 4
x = 9
-(x - 5) = 4
x = -4 + 5 = 1
x < 1, x > 9
3) 1 < lx-1l ≤ 5
1 < x-1 ≤ 5
2 < x ≤ 6
1 < -(x-1) ≤ 5
-5 ≤ x-1 < -1
-4 ≤ x < 0
Answer:
first one: yes
second one: yes
third one: no
fourth one: yes
Step-by-step explanation:
8=12-8x
-4=8x
-4÷8=8x÷8
-0.5=x
