The correct description of a sphere from the given options is:
- A sphere has a center.
A sphere is a three-dimensional shape where every point on its surface is equidistant from its center. It does not have faces, edges, or vertices.
1. A sphere has a face
2. A sphere has an edge
3. A sphere has a vertex
4, A sphere has a center
7 answers
The correct description of a sphere from the given options is:
A sphere is a three-dimensional shape where every point on its surface is equidistant from its center. It does not have faces, edges, or vertices.
The statement that is true about the diameter of a sphere is:
Additionally, I would like to clarify that:
Therefore, the correct answer is statement 3.
To find the surface area of a sphere, we use the formula:
\[ \text{Surface Area} = 4\pi r^2 \]
where \( r \) is the radius of the sphere. Given that the diameter of the sphere is 30 inches, we find the radius \( r \) as follows:
\[ r = \frac{\text{diameter}}{2} = \frac{30}{2} = 15 \text{ inches} \]
Now we can calculate the surface area:
\[ \text{Surface Area} = 4\pi (15)^2 \]
Calculating \( (15)^2 \):
\[ (15)^2 = 225 \]
Now substituting back into the surface area formula:
\[ \text{Surface Area} = 4\pi \times 225 = 900\pi \text{ in}^2 \]
Therefore, the surface area of the sphere is:
\[ \boxed{900\pi \text{ in}^2} \]
To find the dimensions of the cube box that can snugly fit the spherical soap with a surface area of 63.585 inĀ², we first need to determine the radius of the sphere.
We use the surface area formula for a sphere:
\[ \text{Surface Area} = 4\pi r^2 \]
Plugging in the surface area:
\[ 63.585 = 4 \times 3.14 \times r^2 \]
Now, calculating \( 4 \times 3.14 \):
\[ 4 \times 3.14 = 12.56 \]
Substituting this back into the equation:
\[ 63.585 = 12.56 r^2 \]
Next, we divide both sides by 12.56 to solve for \( r^2 \):
\[ r^2 = \frac{63.585}{12.56} \approx 5.06 \]
Now, taking the square root to find \( r \):
\[ r \approx \sqrt{5.06} \approx 2.25 \text{ in.} \]
Since the diameter of the sphere (which determines the size of the cube) is twice the radius, we calculate the diameter:
\[ \text{Diameter} = 2r \approx 2 \times 2.25 = 4.5 \text{ in.} \]
Since the cube must fit the sphere snugly, the dimensions of the cube box should also be equal to the diameter of the sphere.
Thus, the dimensions of the cube box should be:
\[ \boxed{4.5 \text{ in.} \times 4.5 \text{ in.} \times 4.5 \text{ in.}} \]