were added for this assignment.

a could be a partial set of values for a linear function?

| x | y |
|----|----|
| 0 | 1 |
| 1 | 2 |
| 2 | 5 |
| 3 | 10 |

| x | y |
|----|----|
| 0 | 9 |
| 1 | 8 |
| 2 | 5 |
| 3 | 0 |

| x | y |
|----|----|
| 0 | 3 |
| 1 | 5 |

| x | y |
|----|----|
| 0 | 0 |
| 1 | 2 |

Question

were added for this assignment.

a could be a partial set of values for a linear function?

| x | y |
|----|----|
| 0 | 1 |
| 1 | 2 |
| 2 | 5 |
| 3 | 10 |

| x | y |
|----|----|
| 0 | 9 |
| 1 | 8 |
| 2 | 5 |
| 3 | 0 |

| x | y |
|----|----|
| 0 | 3 |
| 1 | 5 |

| x | y |
|----|----|
| 0 | 0 |
| 1 | 2 |

Accepted Answer

To determine which table represents a linear function, we check if the rate of change (slope) between consecutive points is constant. The slope $ m $ between two points $(x_1, y_1)$ and $(x_2, y_2)$ is calculated as $ m = \frac{y_2 - y_1}{x_2 - x_1} $.

### Analyzing the First Table (Top - Left):
Points: $(0, 1)$, $(1, 2)$, $(2, 5)$, $(3, 10)$
- Slope between $(0, 1)$ and $(1, 2)$: $ \frac{2 - 1}{1 - 0} = 1 $
- Slope between $(1, 2)$ and $(2, 5)$: $ \frac{5 - 2}{2 - 1} = 3 $
- Slopes are not constant ($1
eq 3$), so this is not linear.

### Analyzing the Second Table (Top - Right):
Points: $(0, 9)$, $(1, 8)$, $(2, 5)$, $(3, 0)$
- Slope between $(0, 9)$ and $(1, 8)$: $ \frac{8 - 9}{1 - 0} = -1 $
- Slope between $(1, 8)$ and $(2, 5)$: $ \frac{5 - 8}{2 - 1} = -3 $
- Slopes are not constant ($-1
eq -3$), so this is not linear.

### Analyzing the Third Table (Bottom - Left):
Points: $(0, 3)$, $(1, 5)$ (only two points shown, but we can check the slope).
- Slope between $(0, 3)$ and $(1, 5)$: $ \frac{5 - 3}{1 - 0} = 2 $.
- If we assume more points follow this slope, the rate of change is constant. However, we need to check the fourth table too.

### Analyzing the Fourth Table (Bottom - Right):
Points: $(0, 0)$, $(1, 2)$ (only two points shown).
- Slope between $(0, 0)$ and $(1, 2)$: $ \frac{2 - 0}{1 - 0} = 2 $.
- Wait, this also has a slope of $2$. But the third table (bottom - left) has two points, and the fourth table (bottom - right) also has two points. Wait, maybe the third table is incomplete, but let’s re - check.

Wait, the fourth table (bottom - right) as shown has only two points: $(0, 0)$ and $(1, 2)$. The third table (bottom - left) has $(0, 3)$ and $(1, 5)$ (slope $2$). But the problem might have a typo, or we missed points. Wait, no—let’s re - evaluate.

Wait, the **bottom - left table** (with $x = 0, y = 3$ and $x = 1, y = 5$) has a constant slope of $2$ for the two points. If we assume it’s a linear function (since only two points are given, and two points define a line), this could be linear. However, the **bottom - right table** (with $x = 0, y = 0$ and $x = 1, y = 2$) also has a slope of $2$. But maybe the original problem had more points, but based on the visible data:

Wait, no—let’s re - check the first two tables. The first table has changing slopes, the second too. The third table (bottom - left) with $(0, 3)$ and $(1, 5)$ has a slope of $2$. The fourth table (bottom - right) with $(0, 0)$ and $(1, 2)$ has a slope of $2$. But maybe the intended answer is the **bottom - left table** (or bottom - right, but let’s confirm).

Wait, no—maybe the third table (bottom - left) is the one with a constant slope for the visible points. Wait, no, the bottom - right table: $(0, 0)$ and $(1, 2)$ have a slope of $2$. But the third table: $(0, 3)$ and $(1, 5)$ also slope $2$. However, the problem likely expects the table with a constant slope. Let’s re - check the first two tables:

First table: slopes $1, 3, 5$ (not constant).
Second table: slopes $-1, -3, -5$ (not constant).
Third table: slope $2$ (constant for two points).
Fourth table: slope $2$ (constant for two points).

But maybe the original problem had a different setup. Wait, perhaps the **bottom - left table** (with $x = 0, y = 3$ and $x = 1, y = 5$) is the answer, or the bottom - right. But let’s assume the bottom - left (or bottom - right) has a constant slope. Wait, no—let’s re - check the problem statement. The question is “Which could be a partial set of values for a linear function?”

The key is that for a linear function, the slope between any two points must be constant.

- Top - left: slopes $1, 3, 5$ (not constant).
- Top - right: slopes $-1, -3, -5$ (not constant).
- Bottom - left: slope $2$ (constant for two points).
- Bottom - right: slope $2$ (constant for two points).

But maybe the intended answer is the **bottom - left table** (or bottom - right). However, based on typical problems, the table with $x = 0, y = 3$ and $x = 1, y = 5$ (bottom - left) or $x = 0, y = 0$ and $x = 1, y = 2$ (bottom - right) has a constant slope. But let’s confirm with the first two:

Top - left: $x$ increases by $1$, $y$ increases by $1$, then $3$, then $5$ (not linear).
Top - right: $x$ increases by $1$, $y$ decreases by $1$, then $3$, then $5$ (not linear).
Bottom - left: $x$ increases by $1$, $y$ increases by $2$ (linear for two points).
Bottom - right: $x$ increases by $1$, $y$ increases by $2$ (linear for two points).

Assuming the problem expects the table with a constant slope (and the bottom - left or bottom - right), but likely the **bottom - left table** (with $x = 0, y = 3$; $x = 1, y = 5$) or the **bottom - right table** (with $x = 0, y = 0$; $x = 1, y = 2$) is correct. However, based on the initial analysis, the table with $x = 0, 3$ and $x = 1, 5$ (bottom - left) has a constant slope of $2$, so it could represent a linear function.

# Answer:
The table with $ x $ and $ y $ values:

| $ x $ | $ y $ |
|--------|--------|
| $ 0 $ | $ 3 $ |
| $ 1 $ | $ 5 $ |

(or the table with $ x = 0, y = 0 $; $ x = 1, y = 2 $, depending on the full problem, but the bottom - left table is a valid candidate for a linear function’s partial values.)

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QUESTION IMAGE

Question

Explanation:

Analyzing the First Table (Top - Left):

Analyzing the Second Table (Top - Right):

Analyzing the Third Table (Bottom - Left):

Analyzing the Fourth Table (Bottom - Right):

Answer: