Get the answers you need at Westonci.ca, where our expert community is dedicated to providing you with accurate information. Get quick and reliable solutions to your questions from knowledgeable professionals on our comprehensive Q&A platform. Experience the ease of finding precise answers to your questions from a knowledgeable community of experts.
Sagot :
Certainly! Let me guide you through the detailed, step-by-step solution.
### Step 1: Understand the problem and the given data
We have two right circular cylinders:
- A small can of soup with:
- Diameter = 7 cm
- Height = 9 cm
- A large container with:
- Diameter = 9 cm
- Height = 13 cm
### Step 2: Calculate the radius of each cylinder
The radius is half the diameter.
- Small can radius:
[tex]\( r_{\text{small}} = \frac{7}{2} = 3.5 \ \text{cm} \)[/tex]
- Large container radius:
[tex]\( r_{\text{large}} = \frac{9}{2} = 4.5 \ \text{cm} \)[/tex]
### Step 3: Calculate the volume of each cylinder
The volume [tex]\( V \)[/tex] of a right circular cylinder is given by the formula:
[tex]\[ V = \pi r^2 h \][/tex]
#### For the small can:
- Radius [tex]\( r_{\text{small}} = 3.5 \ \text{cm} \)[/tex]
- Height [tex]\( h_{\text{small}} = 9 \ \text{cm} \)[/tex]
[tex]\[ V_{\text{small}} = \pi (3.5)^2 \cdot 9 \][/tex]
After calculating, the volume of the small can is approximately:
[tex]\[ V_{\text{small}} \approx 346 \ \text{cubic cm} \][/tex]
#### For the large container:
- Radius [tex]\( r_{\text{large}} = 4.5 \ \text{cm} \)[/tex]
- Height [tex]\( h_{\text{large}} = 13 \ \text{cm} \)[/tex]
[tex]\[ V_{\text{large}} = \pi (4.5)^2 \cdot 13 \][/tex]
After calculating, the volume of the large container is approximately:
[tex]\[ V_{\text{large}} \approx 827 \ \text{cubic cm} \][/tex]
### Step 4: Determine the minimum number of small cans required
We need to find the number of small cans needed to equal or exceed the volume of the large container.
[tex]\[ \text{Number of small cans} = \left\lceil \frac{V_{\text{large}}}{V_{\text{small}}} \right\rceil \][/tex]
Using the previously calculated volumes:
[tex]\[ \text{Number of small cans} = \left\lceil \frac{827}{346} \right\rceil \][/tex]
[tex]\[ \approx \lceil 2.39 \rceil \][/tex]
Rounding up, we get:
[tex]\[ \text{Number of small cans} = 3 \][/tex]
### Step 5: Present the results
- The volume of the small can is approximately 346 cubic cm.
- The volume of the large container is approximately 827 cubic cm.
- The minimum number of small cans required to fill the large container is 3.
This ensures that the large container is completely filled, taking into account the rounding up to ensure no shortage.
### Step 1: Understand the problem and the given data
We have two right circular cylinders:
- A small can of soup with:
- Diameter = 7 cm
- Height = 9 cm
- A large container with:
- Diameter = 9 cm
- Height = 13 cm
### Step 2: Calculate the radius of each cylinder
The radius is half the diameter.
- Small can radius:
[tex]\( r_{\text{small}} = \frac{7}{2} = 3.5 \ \text{cm} \)[/tex]
- Large container radius:
[tex]\( r_{\text{large}} = \frac{9}{2} = 4.5 \ \text{cm} \)[/tex]
### Step 3: Calculate the volume of each cylinder
The volume [tex]\( V \)[/tex] of a right circular cylinder is given by the formula:
[tex]\[ V = \pi r^2 h \][/tex]
#### For the small can:
- Radius [tex]\( r_{\text{small}} = 3.5 \ \text{cm} \)[/tex]
- Height [tex]\( h_{\text{small}} = 9 \ \text{cm} \)[/tex]
[tex]\[ V_{\text{small}} = \pi (3.5)^2 \cdot 9 \][/tex]
After calculating, the volume of the small can is approximately:
[tex]\[ V_{\text{small}} \approx 346 \ \text{cubic cm} \][/tex]
#### For the large container:
- Radius [tex]\( r_{\text{large}} = 4.5 \ \text{cm} \)[/tex]
- Height [tex]\( h_{\text{large}} = 13 \ \text{cm} \)[/tex]
[tex]\[ V_{\text{large}} = \pi (4.5)^2 \cdot 13 \][/tex]
After calculating, the volume of the large container is approximately:
[tex]\[ V_{\text{large}} \approx 827 \ \text{cubic cm} \][/tex]
### Step 4: Determine the minimum number of small cans required
We need to find the number of small cans needed to equal or exceed the volume of the large container.
[tex]\[ \text{Number of small cans} = \left\lceil \frac{V_{\text{large}}}{V_{\text{small}}} \right\rceil \][/tex]
Using the previously calculated volumes:
[tex]\[ \text{Number of small cans} = \left\lceil \frac{827}{346} \right\rceil \][/tex]
[tex]\[ \approx \lceil 2.39 \rceil \][/tex]
Rounding up, we get:
[tex]\[ \text{Number of small cans} = 3 \][/tex]
### Step 5: Present the results
- The volume of the small can is approximately 346 cubic cm.
- The volume of the large container is approximately 827 cubic cm.
- The minimum number of small cans required to fill the large container is 3.
This ensures that the large container is completely filled, taking into account the rounding up to ensure no shortage.
Thank you for your visit. We are dedicated to helping you find the information you need, whenever you need it. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.