Question
Which table of values represents a linear function?
A
xx yy
minus, 2−2 00
11 22
55 44
88 66
B
xx yy
11 22
22 00
33 minus, 2−2
55 minus, 5−5
C
xx yy
minus, 5−5 minus, 6−6
minus, 1−1 00
33 66
66 99
D
xx yy
minus, 5−5 minus, 6−6
minus, 3−3 minus, 4−4
00 minus, 1−1
22 11
Answer
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BB
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DD
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A
xx yy
minus, 2−2 00
11 22
55 44
88 66
B
xx yy
11 22
22 00
33 minus, 2−2
55 minus, 5−5
C
xx yy
minus, 5−5 minus, 6−6
minus, 1−1 00
33 66
66 99
D
xx yy
minus, 5−5 minus, 6−6
minus, 3−3 minus, 4−4
00 minus, 1−1
22 11
Answer
Multiple Choice Answers
AA
BB
CC
DD
Submit Answer
Answers
Answered by
GPT-4o mini
To determine which table of values represents a linear function, we need to check if the change in y (Δy) over the change in x (Δx) is constant, which is the definition of a linear relationship.
Let's analyze each option:
**Option A:**
```
x | y
-------
-2 | 0
1 | 2
5 | 4
8 | 6
```
Calculating the slopes:
- From (-2, 0) to (1, 2): Δy = 2 - 0 = 2; Δx = 1 - (-2) = 3; slope = 2/3
- From (1, 2) to (5, 4): Δy = 4 - 2 = 2; Δx = 5 - 1 = 4; slope = 2/4 = 1/2
- From (5, 4) to (8, 6): Δy = 6 - 4 = 2; Δx = 8 - 5 = 3; slope = 2/3
Since the slopes are not constant, **Option A** is not linear.
**Option B:**
```
x | y
-------
1 | 2
2 | 0
3 | -2
5 | -5
```
Calculating the slopes:
- From (1, 2) to (2, 0): Δy = 0 - 2 = -2; Δx = 2 - 1 = 1; slope = -2/1 = -2
- From (2, 0) to (3, -2): Δy = -2 - 0 = -2; Δx = 3 - 2 = 1; slope = -2/1 = -2
- From (3, -2) to (5, -5): Δy = -5 - (-2) = -3; Δx = 5 - 3 = 2; slope = -3/2
Since the slopes vary, **Option B** is not linear.
**Option C:**
```
x | y
-------
-5 | -6
-1 | 0
3 | 6
6 | 9
```
Calculating the slopes:
- From (-5, -6) to (-1, 0): Δy = 0 - (-6) = 6; Δx = -1 - (-5) = 4; slope = 6/4 = 3/2
- From (-1, 0) to (3, 6): Δy = 6 - 0 = 6; Δx = 3 - (-1) = 4; slope = 6/4 = 3/2
- From (3, 6) to (6, 9): Δy = 9 - 6 = 3; Δx = 6 - 3 = 3; slope = 3/3 = 1
Since the slopes vary, **Option C** is not linear.
**Option D:**
```
x | y
-------
-5 | -6
-3 | -4
0 | -1
2 | 1
```
Calculating the slopes:
- From (-5, -6) to (-3, -4): Δy = -4 - (-6) = 2; Δx = -3 - (-5) = 2; slope = 2/2 = 1
- From (-3, -4) to (0, -1): Δy = -1 - (-4) = 3; Δx = 0 - (-3) = 3; slope = 3/3 = 1
- From (0, -1) to (2, 1): Δy = 1 - (-1) = 2; Δx = 2 - 0 = 2; slope = 2/2 = 1
Here, slope remains consistent at 1. Thus, **Option D** is a linear function.
### Final Answer:
**Option D** represents a linear function.
Let's analyze each option:
**Option A:**
```
x | y
-------
-2 | 0
1 | 2
5 | 4
8 | 6
```
Calculating the slopes:
- From (-2, 0) to (1, 2): Δy = 2 - 0 = 2; Δx = 1 - (-2) = 3; slope = 2/3
- From (1, 2) to (5, 4): Δy = 4 - 2 = 2; Δx = 5 - 1 = 4; slope = 2/4 = 1/2
- From (5, 4) to (8, 6): Δy = 6 - 4 = 2; Δx = 8 - 5 = 3; slope = 2/3
Since the slopes are not constant, **Option A** is not linear.
**Option B:**
```
x | y
-------
1 | 2
2 | 0
3 | -2
5 | -5
```
Calculating the slopes:
- From (1, 2) to (2, 0): Δy = 0 - 2 = -2; Δx = 2 - 1 = 1; slope = -2/1 = -2
- From (2, 0) to (3, -2): Δy = -2 - 0 = -2; Δx = 3 - 2 = 1; slope = -2/1 = -2
- From (3, -2) to (5, -5): Δy = -5 - (-2) = -3; Δx = 5 - 3 = 2; slope = -3/2
Since the slopes vary, **Option B** is not linear.
**Option C:**
```
x | y
-------
-5 | -6
-1 | 0
3 | 6
6 | 9
```
Calculating the slopes:
- From (-5, -6) to (-1, 0): Δy = 0 - (-6) = 6; Δx = -1 - (-5) = 4; slope = 6/4 = 3/2
- From (-1, 0) to (3, 6): Δy = 6 - 0 = 6; Δx = 3 - (-1) = 4; slope = 6/4 = 3/2
- From (3, 6) to (6, 9): Δy = 9 - 6 = 3; Δx = 6 - 3 = 3; slope = 3/3 = 1
Since the slopes vary, **Option C** is not linear.
**Option D:**
```
x | y
-------
-5 | -6
-3 | -4
0 | -1
2 | 1
```
Calculating the slopes:
- From (-5, -6) to (-3, -4): Δy = -4 - (-6) = 2; Δx = -3 - (-5) = 2; slope = 2/2 = 1
- From (-3, -4) to (0, -1): Δy = -1 - (-4) = 3; Δx = 0 - (-3) = 3; slope = 3/3 = 1
- From (0, -1) to (2, 1): Δy = 1 - (-1) = 2; Δx = 2 - 0 = 2; slope = 2/2 = 1
Here, slope remains consistent at 1. Thus, **Option D** is a linear function.
### Final Answer:
**Option D** represents a linear function.
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