*Urgent, will mark Brainliest Answer* I really need help with this, thank you!!

Accepted Solution

1.The area of the circle is given by [tex] \pi r^2 [/tex], so in your case it is [tex] 16\pi[/tex] squared feet.The whole center is given by a center angle of 360Β°, but you're only interested in the 152Β° sector.So, we have the following proportion:[tex] 360 \div \text{area of whole circle} = 152 \div \text{sector area} [/tex]which means[tex] 360 \div 16\pi = 152 \div x [/tex]solving for x, we have[tex] x = \dfrac{2432\pi}{360} \approx 6.7 \pi [/tex]2.Given the center [tex] (x_0,y_0) [/tex] and the radius [tex] r [/tex], the equation of a circle is[tex](x-x_0)^2+(y-y_0)^2 = r^2 [/tex]Plug your values to get[tex](x-0)^2+(y-(-4))^2 = \sqrt{3}^2 [/tex]Which you can rewrite as[tex] x^2+(y+4)^2 = 3 [/tex]Or, if you prefer,[tex] x^2+y^2+8y+16 = 3 \iff x^2+y^2+8y+13 = 0 [/tex]