Question
On a coordinate plane, a line goes through (negative 4, negative 1) and (0, 1). Square a is around (negative 5, negative 2), square b is around (negative 1, 1), square c is around (1, 2), and square d is around (4, 4).
The linear equation y = one-half x + 1 is represented by the graphed line. A second linear equation is represented by the data in the table.
A 2-column table with 4 rows. Column 1 is labeled x with entries negative 2, 0, 2, 4. Column 2 is labeled y with entries 7, 6, 5, 4.
In which square is the solution located?
The linear equation y = one-half x + 1 is represented by the graphed line. A second linear equation is represented by the data in the table.
A 2-column table with 4 rows. Column 1 is labeled x with entries negative 2, 0, 2, 4. Column 2 is labeled y with entries 7, 6, 5, 4.
In which square is the solution located?
Answers
GPT-4o mini
To determine where the solution is located concerning the square layout provided, we need to analyze both the equation of the line and the data from the table.
The equation of the line is:
\[ y = \frac{1}{2}x + 1 \]
Now, let's evaluate the y-values from the table for each x-value and see if any points from the table lie on the line given by the equation.
The table data is as follows:
- For \( x = -2 \):
\[ y = \frac{1}{2}(-2) + 1 = -1 + 1 = 0 \]
Point: (-2, 0)
- For \( x = 0 \):
\[ y = \frac{1}{2}(0) + 1 = 0 + 1 = 1 \]
Point: (0, 1)
- For \( x = 2 \):
\[ y = \frac{1}{2}(2) + 1 = 1 + 1 = 2 \]
Point: (2, 2)
- For \( x = 4 \):
\[ y = \frac{1}{2}(4) + 1 = 2 + 1 = 3 \]
Point: (4, 3)
Now comparing the y-values from the table:
- For \( x = -2 \), \( y = 7 \) (not on the line)
- For \( x = 0 \), \( y = 6 \) (not on the line)
- For \( x = 2 \), \( y = 5 \) (not on the line)
- For \( x = 4 \), \( y = 4 \) (not on the line)
None of the points from the table lies on the line, so we check where the line intersects with the y-values given in the table.
To visually identify the square solutions:
- **Square A** surrounds (-5, -2)
- **Square B** surrounds (-1, 1)
- **Square C** surrounds (1, 2)
- **Square D** surrounds (4, 4)
The last point we calculated was for \( x=4 \) yielding \( y=3 \) not inline with data in the table.
Analyzing the y-values:
1. For square B, (-1, 1), the line is just below it (comes across negative 1 and hits positive y).
2. For square C, (1, 2), the line and table data for x=2 crossed point falls between line trajectory.
3. For square D, (4, 4), does not align with table data.
Thus, square C (1,2) is the closest intercept with the line relationship from table data endpoints. Thus the solution is located in **Square C**.
The equation of the line is:
\[ y = \frac{1}{2}x + 1 \]
Now, let's evaluate the y-values from the table for each x-value and see if any points from the table lie on the line given by the equation.
The table data is as follows:
- For \( x = -2 \):
\[ y = \frac{1}{2}(-2) + 1 = -1 + 1 = 0 \]
Point: (-2, 0)
- For \( x = 0 \):
\[ y = \frac{1}{2}(0) + 1 = 0 + 1 = 1 \]
Point: (0, 1)
- For \( x = 2 \):
\[ y = \frac{1}{2}(2) + 1 = 1 + 1 = 2 \]
Point: (2, 2)
- For \( x = 4 \):
\[ y = \frac{1}{2}(4) + 1 = 2 + 1 = 3 \]
Point: (4, 3)
Now comparing the y-values from the table:
- For \( x = -2 \), \( y = 7 \) (not on the line)
- For \( x = 0 \), \( y = 6 \) (not on the line)
- For \( x = 2 \), \( y = 5 \) (not on the line)
- For \( x = 4 \), \( y = 4 \) (not on the line)
None of the points from the table lies on the line, so we check where the line intersects with the y-values given in the table.
To visually identify the square solutions:
- **Square A** surrounds (-5, -2)
- **Square B** surrounds (-1, 1)
- **Square C** surrounds (1, 2)
- **Square D** surrounds (4, 4)
The last point we calculated was for \( x=4 \) yielding \( y=3 \) not inline with data in the table.
Analyzing the y-values:
1. For square B, (-1, 1), the line is just below it (comes across negative 1 and hits positive y).
2. For square C, (1, 2), the line and table data for x=2 crossed point falls between line trajectory.
3. For square D, (4, 4), does not align with table data.
Thus, square C (1,2) is the closest intercept with the line relationship from table data endpoints. Thus the solution is located in **Square C**.