Answer :
To complete the table representing the area of Melissa's vegetable garden for different lengths of the tomato patch, we can use the given and interpolated values from the problem:
1. We need to find the length of the tomato patch corresponding to an area of 338 square feet.
2. We also need to find the length of the tomato patch corresponding to an area of 151.25 square feet.
Here is the detailed step-by-step solution:
### Finding the Length for the Area of 338 Square Feet
To determine the length corresponding to an area of 338 square feet, we can use linear interpolation between the known lengths and areas:
- Length (L2) for an area of 342.0 square feet is 6.5 feet.
- Length (L1) for an area of 338 square feet is what we're solving for.
Using the linear relationship:
[tex]\( y = y1 + \left(\frac{x-x1}{x2-x1}\right)(y2-y1) \)[/tex]
Plugging in the values, we get:
For the first interpolation calculation:
- [tex]\( x1 = 6.5 \)[/tex]
- [tex]\( y1 = 342.0 \)[/tex]
- [tex]\( y2 = 338 \)[/tex]
The interpolated length (L1) that makes the area 338 square feet is 7.5.
### Finding the Length for the Area of 151.25 Square Feet
To determine the length corresponding to an area of 151.25 square feet, we again use linear interpolation:
- Length (L1) for an area of 147.25 square feet is 7 feet.
- Length (L2) for an area of 151.25 square feet is what we're solving for.
Using the linear relationship again:
For the second interpolation calculation:
- [tex]\( x1 = 7 \)[/tex]
- [tex]\( y1 = 147.25 \)[/tex]
- [tex]\( y2 = 151.25 \)[/tex]
The interpolated length (L2) that makes the area 151.25 square feet is 8.0.
### Completed Table
Now, let's fill in the lengths into the table:
\begin{tabular}{|c|c|}
\hline
\begin{tabular}{c}
Length of Tomato Patch \\
(in feet)
\end{tabular} & \begin{tabular}{c}
Area of Vegetable Garden \\
(in square feet)
\end{tabular} \\
\hline
7.5 & 338 \\
\hline
6.5 & 342.0 \\
\hline
7 & 147.25 \\
\hline
8.0 & 151.25 \\
\hline
\end{tabular}
Thus, the lengths corresponding to the given areas are 7.5 feet and 8.0 feet, and they are filled in the respective boxes in the table as shown.
1. We need to find the length of the tomato patch corresponding to an area of 338 square feet.
2. We also need to find the length of the tomato patch corresponding to an area of 151.25 square feet.
Here is the detailed step-by-step solution:
### Finding the Length for the Area of 338 Square Feet
To determine the length corresponding to an area of 338 square feet, we can use linear interpolation between the known lengths and areas:
- Length (L2) for an area of 342.0 square feet is 6.5 feet.
- Length (L1) for an area of 338 square feet is what we're solving for.
Using the linear relationship:
[tex]\( y = y1 + \left(\frac{x-x1}{x2-x1}\right)(y2-y1) \)[/tex]
Plugging in the values, we get:
For the first interpolation calculation:
- [tex]\( x1 = 6.5 \)[/tex]
- [tex]\( y1 = 342.0 \)[/tex]
- [tex]\( y2 = 338 \)[/tex]
The interpolated length (L1) that makes the area 338 square feet is 7.5.
### Finding the Length for the Area of 151.25 Square Feet
To determine the length corresponding to an area of 151.25 square feet, we again use linear interpolation:
- Length (L1) for an area of 147.25 square feet is 7 feet.
- Length (L2) for an area of 151.25 square feet is what we're solving for.
Using the linear relationship again:
For the second interpolation calculation:
- [tex]\( x1 = 7 \)[/tex]
- [tex]\( y1 = 147.25 \)[/tex]
- [tex]\( y2 = 151.25 \)[/tex]
The interpolated length (L2) that makes the area 151.25 square feet is 8.0.
### Completed Table
Now, let's fill in the lengths into the table:
\begin{tabular}{|c|c|}
\hline
\begin{tabular}{c}
Length of Tomato Patch \\
(in feet)
\end{tabular} & \begin{tabular}{c}
Area of Vegetable Garden \\
(in square feet)
\end{tabular} \\
\hline
7.5 & 338 \\
\hline
6.5 & 342.0 \\
\hline
7 & 147.25 \\
\hline
8.0 & 151.25 \\
\hline
\end{tabular}
Thus, the lengths corresponding to the given areas are 7.5 feet and 8.0 feet, and they are filled in the respective boxes in the table as shown.
