Your answer is A, you can be a carrier and not have the disorder
It would be a straight line
Given :
<span>Triangle fgh is inscribed in circle o
</span>
oh = 6 = radius of the circle
<span>∵ fh is congruent to og
</span>
<span>∴ fh = og = radius og the circle = 6
</span>
<span>∵ of = radius of the circle = 6
</span>
<span>∴ oh = fh = of = radius of the circle = 6
</span>
∴ Δ foh is Equilateral Triangle
<span> </span>∴∠ foh = 60° ⇒⇒⇒ property of <span>the equilateral Triangle
</span>
∵ total area of the circle = π r² and total central angle of the circle = 360°
∴ Area of sector foh = (60°/360°) * π r²
∴ Area of sector foh = (60°/360°) * π * 6² ≈ 18.85
The answer is:
<span>the area of the sector formed by angle foh
= 18.85</span>
Y = mx + b is slope intercept form
x = -10
y = 8
m = 6
1. Solve for b
8 = (6)(-10) + b
b = 68
2. Plug m and b back into your slope intercept equation
y = 6x + 68
