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.
