Question
A sharpened pencil is actually a composite solid
It is made from 2 figures:
A cylinder with height of 6 inches and radius of 0.5 inches.
Using 3.14 for pi, and not rounding, the volume of the cylinder is: *blank 1* inches
A cone with height of 1.5 inches and radius of 0.5 inches. Using 3.14 for pi, and not rounding, the volume of the cone is: *blank 2* inches
Without rounding, the volume of the entire pencil is *blank 3* inches.
Make sure no one can find this
It is made from 2 figures:
A cylinder with height of 6 inches and radius of 0.5 inches.
Using 3.14 for pi, and not rounding, the volume of the cylinder is: *blank 1* inches
A cone with height of 1.5 inches and radius of 0.5 inches. Using 3.14 for pi, and not rounding, the volume of the cone is: *blank 2* inches
Without rounding, the volume of the entire pencil is *blank 3* inches.
Make sure no one can find this
Answers
GPT-4o mini
To find the volumes of both the cylinder and the cone that make up the sharpened pencil, we can use the formulas for the volumes of these shapes.
**1. Volume of the Cylinder:**
The formula for the volume of a cylinder is:
\[
V = \pi r^2 h
\]
where:
- \( r \) is the radius,
- \( h \) is the height.
Given:
- \( r = 0.5 \) inches,
- \( h = 6 \) inches,
- \( \pi = 3.14 \).
Substituting in the values:
\[
V = 3.14 \times (0.5)^2 \times 6
\]
\[
V = 3.14 \times 0.25 \times 6
\]
\[
V = 3.14 \times 1.5
\]
\[
V = 4.71 \text{ cubic inches.}
\]
So, for *blank 1*, the volume of the cylinder is **4.71** inches³.
---
**2. Volume of the Cone:**
The formula for the volume of a cone is:
\[
V = \frac{1}{3} \pi r^2 h
\]
where:
- \( r \) is the radius,
- \( h \) is the height.
Given:
- \( r = 0.5 \) inches,
- \( h = 1.5 \) inches.
Substituting in the values:
\[
V = \frac{1}{3} \times 3.14 \times (0.5)^2 \times 1.5
\]
\[
V = \frac{1}{3} \times 3.14 \times 0.25 \times 1.5
\]
\[
V = \frac{1}{3} \times 3.14 \times 0.375
\]
\[
V = \frac{1.1775}{3}
\]
\[
V = 0.3925 \text{ cubic inches.}
\]
So, for *blank 2*, the volume of the cone is **0.3925** inches³.
---
**3. Total Volume of the Pencil:**
To find the total volume of the pencil, we simply add the volumes of the cylinder and the cone together:
\[
\text{Total Volume} = \text{Volume of Cylinder} + \text{Volume of Cone}
\]
\[
\text{Total Volume} = 4.71 + 0.3925
\]
\[
\text{Total Volume} = 5.1025 \text{ cubic inches.}
\]
So, for *blank 3*, the volume of the entire pencil is **5.1025** inches³.
---
In summary:
- Blank 1: **4.71**
- Blank 2: **0.3925**
- Blank 3: **5.1025**
**1. Volume of the Cylinder:**
The formula for the volume of a cylinder is:
\[
V = \pi r^2 h
\]
where:
- \( r \) is the radius,
- \( h \) is the height.
Given:
- \( r = 0.5 \) inches,
- \( h = 6 \) inches,
- \( \pi = 3.14 \).
Substituting in the values:
\[
V = 3.14 \times (0.5)^2 \times 6
\]
\[
V = 3.14 \times 0.25 \times 6
\]
\[
V = 3.14 \times 1.5
\]
\[
V = 4.71 \text{ cubic inches.}
\]
So, for *blank 1*, the volume of the cylinder is **4.71** inches³.
---
**2. Volume of the Cone:**
The formula for the volume of a cone is:
\[
V = \frac{1}{3} \pi r^2 h
\]
where:
- \( r \) is the radius,
- \( h \) is the height.
Given:
- \( r = 0.5 \) inches,
- \( h = 1.5 \) inches.
Substituting in the values:
\[
V = \frac{1}{3} \times 3.14 \times (0.5)^2 \times 1.5
\]
\[
V = \frac{1}{3} \times 3.14 \times 0.25 \times 1.5
\]
\[
V = \frac{1}{3} \times 3.14 \times 0.375
\]
\[
V = \frac{1.1775}{3}
\]
\[
V = 0.3925 \text{ cubic inches.}
\]
So, for *blank 2*, the volume of the cone is **0.3925** inches³.
---
**3. Total Volume of the Pencil:**
To find the total volume of the pencil, we simply add the volumes of the cylinder and the cone together:
\[
\text{Total Volume} = \text{Volume of Cylinder} + \text{Volume of Cone}
\]
\[
\text{Total Volume} = 4.71 + 0.3925
\]
\[
\text{Total Volume} = 5.1025 \text{ cubic inches.}
\]
So, for *blank 3*, the volume of the entire pencil is **5.1025** inches³.
---
In summary:
- Blank 1: **4.71**
- Blank 2: **0.3925**
- Blank 3: **5.1025**