anonymous
  • anonymous
help please
Mathematics
chestercat
  • chestercat
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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anonymous
  • anonymous
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anonymous
  • anonymous
thank you so much
anonymous
  • anonymous
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anonymous
  • anonymous
i really need to finish my online class for the summer by 2morrow and i only have 17 questions left
UnkleRhaukus
  • UnkleRhaukus
for the first one, the sector is half a circle
UnkleRhaukus
  • UnkleRhaukus
do you remember the area of a circle?
UnkleRhaukus
  • UnkleRhaukus
anonymous
  • anonymous
oh srry no
UnkleRhaukus
  • UnkleRhaukus
The area of a full circle, (or radius \(r\) and diameter \(d\)) is \[A_\text{circle}=\pi r^2 = \pi\left(\frac d2\right)^2 \]
UnkleRhaukus
  • UnkleRhaukus
Our sector is not a full circle (360°), but only 180° \[A_\text{sector}=\frac{180°}{360°}A_\text{circle }\\ \qquad=\frac12A_\text{circle}\\ \qquad=\tfrac12\pi r^2\\ \qquad=\tfrac12\pi\left(\frac d2\right)^2 \]
anonymous
  • anonymous
what do u think inkyvoyd

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