Answer:
Step-by-step explanation:
y ÷ 3(4 - 2) + 5.5
= y ÷ 3(2) + 5.5
= y ÷ 6 + 5.5
= y/6 + 5.5
= (y + 33)/6
Given data:
The diameter of the cut sphere, D=14 in.
The radius of the cut sphere is,
![\begin{gathered} r=\frac{D}{2} \\ r=\frac{14}{2} \\ r=7\text{ in} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20r%3D%5Cfrac%7BD%7D%7B2%7D%20%5C%5C%20r%3D%5Cfrac%7B14%7D%7B2%7D%20%5C%5C%20r%3D7%5Ctext%7B%20in%7D%20%5Cend%7Bgathered%7D)
The cut sphere is called a hemisphere.
The surface area of a sphere is
![A_1=4\pi r^2](https://tex.z-dn.net/?f=A_1%3D4%5Cpi%20r%5E2)
So, the lateral surface area of a hemisphere is half the surface area of sphere. Therefore, the lateral surface area of a hemisphere is,
![\begin{gathered} A_2=\frac{4\pi r^2}{2} \\ A_2=2\pi r^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A_2%3D%5Cfrac%7B4%5Cpi%20r%5E2%7D%7B2%7D%20%5C%5C%20A_2%3D2%5Cpi%20r%5E2%20%5Cend%7Bgathered%7D)
The hemisphere has a lateral surface and a circular surface. The area of the circular surface is,
![A_3=\pi r^2](https://tex.z-dn.net/?f=A_3%3D%5Cpi%20r%5E2)
Therefore, the total area of the hemisphere is,
![\begin{gathered} A=A_2+A_3 \\ A=2\text{ }\pi r^2+\pi r^2 \\ A=3\text{ }\pi r^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A%3DA_2%2BA_3%20%5C%5C%20A%3D2%5Ctext%7B%20%7D%5Cpi%20r%5E2%2B%5Cpi%20r%5E2%20%5C%5C%20A%3D3%5Ctext%7B%20%7D%5Cpi%20r%5E2%20%5Cend%7Bgathered%7D)
The total surface area of a hemisphere is,
![\begin{gathered} A_{}=3\text{ }\pi r^2 \\ A=3\text{ }\pi\times7^2 \\ A=461.8in^2 \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20A_%7B%7D%3D3%5Ctext%7B%20%7D%5Cpi%20r%5E2%20%5C%5C%20A%3D3%5Ctext%7B%20%7D%5Cpi%5Ctimes7%5E2%20%5C%5C%20A%3D461.8in%5E2%20%5Cend%7Bgathered%7D)
Therefore, the total surface area of the cut sphere is 461.8 square inches.
Pentagon - 5 sides = 5 x 4 = 20
Hexagon - 6 sides = 6 x 5 = 30
Nanogon - 9 sides = 9
20 + 30 + 9 = 59 IN TOTAL
Get rid of parenthesis.
z + z + 6
Add common terms together.
2z + 6 is an expresssion equivalent to z + (z+6)
Hope this helps!