These lines are perpendicular!
y = x
y = -x–2
The slopes are reciprocals of each other, so they are perpendicular.
If you do not mind me asking, what did Seth write? Us helpers cannot answer it if we do not have the full question. I apologize if this seems rude.
We know that
<span>the regular hexagon can be divided into 6 equilateral triangles
</span>
area of one equilateral triangle=s²*√3/4
for s=3 in
area of one equilateral triangle=9*√3/4 in²
area of a circle=pi*r²
in this problem the radius is equal to the side of a regular hexagon
r=3 in
area of the circle=pi*3²-----> 9*pi in²
we divide that area into 6 equal parts------> 9*pi/6----> 3*pi/2 in²
the area of a segment formed by a side of the hexagon and the circle is equal to <span>1/6 of the area of the circle minus the area of 1 equilateral triangle
</span>so
[ (3/2)*pi in²-(9/4)*√3 in²]
the answer is
[ (3/2)*pi in²-(9/4)*√3 in²]
The midpoint between -4 and 8 is answer B) 2
4.36 I think I hope this helps