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)
On your graph, draw a horizontal line at y=5. Basically a line over (-3,5) to (3,5) or whatever the greatest x's are. Since it is greater than or equal to, it is a solid line. Greater than, you shade above the line
Answer:
x = 6
Step-by-step explanation:
Pardon the bad handwriting
Answer:
hey help me too lol
Step-by-step explanation:
its so importnt
Given:
In triangle ABC, AB = AC, AD is angle bisector and measure of angle C is 49 degrees.
To find:
The value of x and y.
Solution:
In triangle ABC,
(Given)
So, triangle ABC is an isosceles triangle and by the definition of base angles the base angles of isosceles triangle are congruent.
In isosceles triangle ABC,



The angle bisector of an isosceles triangle is the median and altitude of the triangle. So, the angle bisector is perpendicular to the base.


In triangle ABD,
[Angle sum property]
Therefore, the correct option is B.