Answer :
To determine if the given formula [tex]$\pi(8.5 \, \text{ft})^2 (2 \, \text{ft}) + \frac{1}{2} \pi (2 \, \text{ft})^2 (13 \, \text{ft} - 8.5 \, \text{ft})$[/tex] calculates the total volume of grains that can be stored in a silo with a conical base, we need to break down and analyze the formula.
First, understand the components of the formula:
1. [tex]$\pi(8.5 \, \text{ft})^2 (2 \, \text{ft})$[/tex]:
- This appears to be the volume of a cylindrical part of the silo.
- Here, [tex]$8.5 \, \text{ft}$[/tex] is likely the radius of the cylindrical part's base, and [tex]$2 \, \text{ft}$[/tex] is its height.
- The formula for the volume of a cylinder is [tex]\(V = \pi r^2 h\)[/tex], where [tex]\(r\)[/tex] is the radius and [tex]\(h\)[/tex] is the height.
2. [tex]$\frac{1}{2} \pi (2 \, \text{ft})^2 (13 \, \text{ft} - 8.5 \, \text{ft})$[/tex]:
- This looks like the volume formula for a conical part of the silo.
- Here, [tex]$2 \, \text{ft}$[/tex] is likely the radius of the conical part's base.
- [tex]\(13 \,\text{ft} - 8.5 \,\text{ft}\)[/tex] calculates the height of the cone by subtracting the height of the cylindrical part (8.5 ft) from the total height (13 ft).
- The formula for the volume of a cone is [tex]\(V = \frac{1}{3} \pi r^2 h\)[/tex].
However, notice that while the standard cone volume formula is used (with [tex]\(\frac{1}{3}\)[/tex]), the expression shows [tex]\(\frac{1}{2}\)[/tex]. This discrepancy means the formula does not precisely match the standard geometric ones.
Since we are only verifying the suitability of the formula and not performing a detailed computation of the silo's actual volume, we can conclude:
The given formula has the right form and components (cylinder and cone structures) but uses [tex]\(\frac{1}{2}\)[/tex] instead of [tex]\(\frac{1}{3}\)[/tex] in calculating the cone volume. Thus, it might be a mistake for this specific calculation.
So, while the given expression is close, it's not entirely correct due to the incorrect factor for the cone's volume.
First, understand the components of the formula:
1. [tex]$\pi(8.5 \, \text{ft})^2 (2 \, \text{ft})$[/tex]:
- This appears to be the volume of a cylindrical part of the silo.
- Here, [tex]$8.5 \, \text{ft}$[/tex] is likely the radius of the cylindrical part's base, and [tex]$2 \, \text{ft}$[/tex] is its height.
- The formula for the volume of a cylinder is [tex]\(V = \pi r^2 h\)[/tex], where [tex]\(r\)[/tex] is the radius and [tex]\(h\)[/tex] is the height.
2. [tex]$\frac{1}{2} \pi (2 \, \text{ft})^2 (13 \, \text{ft} - 8.5 \, \text{ft})$[/tex]:
- This looks like the volume formula for a conical part of the silo.
- Here, [tex]$2 \, \text{ft}$[/tex] is likely the radius of the conical part's base.
- [tex]\(13 \,\text{ft} - 8.5 \,\text{ft}\)[/tex] calculates the height of the cone by subtracting the height of the cylindrical part (8.5 ft) from the total height (13 ft).
- The formula for the volume of a cone is [tex]\(V = \frac{1}{3} \pi r^2 h\)[/tex].
However, notice that while the standard cone volume formula is used (with [tex]\(\frac{1}{3}\)[/tex]), the expression shows [tex]\(\frac{1}{2}\)[/tex]. This discrepancy means the formula does not precisely match the standard geometric ones.
Since we are only verifying the suitability of the formula and not performing a detailed computation of the silo's actual volume, we can conclude:
The given formula has the right form and components (cylinder and cone structures) but uses [tex]\(\frac{1}{2}\)[/tex] instead of [tex]\(\frac{1}{3}\)[/tex] in calculating the cone volume. Thus, it might be a mistake for this specific calculation.
So, while the given expression is close, it's not entirely correct due to the incorrect factor for the cone's volume.