Question

which table represents a non - proportional relationship between x and y? (tables are shown with x and y values: first table: x: 15, 5, 10, 12; y: 82.5, 27.5, -55, -66. second table: x: 1, 3, 7, 10; y: 7.5, 22.5, 52.5, 75. third table: x: -2, 1, 4, 5; y: 8, 4, -16, -20. fourth table: x: 4, -1, 2, 6; y: 8, -1, -10, -28)

Explanation:

Step1: Recall Proportional Relationship

A proportional relationship has \( y = kx \), so \( \frac{y}{x} = k \) (constant) for all \( x eq0 \). We check \( \frac{y}{x} \) for each table.

Step2: Analyze First Table (x:15,5,10,12; y:82.5,27.5,-55,-66)

- \( \frac{82.5}{15} = 5.5 \)
- \( \frac{27.5}{5} = 5.5 \)
- \( \frac{-55}{10} = -5.5 \) (Wait, sign? Wait, 10 and -55: \( \frac{-55}{10} = -5.5 \), but 15 and 82.5: 5.5. Wait, maybe miscalculation. Wait 15*5.5=82.5, 5*5.5=27.5, 10*(-5.5)=-55, 12*(-5.5)=-66? Wait no, 12*5.5=66, but y is -66. Wait, maybe absolute? No, proportional can have negative k. Wait \( \frac{y}{x} \): 82.5/15=5.5, 27.5/5=5.5, -55/10=-5.5? No, 5.5 and -5.5 are different. Wait no, 10 and y=-55: -55/10=-5.5, 12 and -66: -66/12=-5.5. Wait 15: 82.5/15=5.5, 5:27.5/5=5.5. Wait, 15 and 5 have positive x, y positive; 10 and 12 have positive x, y negative. So k is 5.5 for x positive, -5.5 for x positive? No, that's inconsistent. Wait maybe I messed up. Wait let's check second table.

Step3: Analyze Second Table (x:1,3,7,10; y:7.5,22.5,52.5,75)

- \( \frac{7.5}{1} = 7.5 \)
- \( \frac{22.5}{3} = 7.5 \)
- \( \frac{52.5}{7} = 7.5 \)
- \( \frac{75}{10} = 7.5 \). Constant, so proportional.

Step4: Analyze Third Table (x:-2,1,4,5; y:8,4,-16,-20)

- \( \frac{8}{-2} = -4 \)
- \( \frac{4}{1} = 4 \). -4 ≠ 4, so \( \frac{y}{x} \) not constant. Wait, 8/-2=-4, 4/1=4, -16/4=-4, -20/5=-4. Wait, 1 and 4: 4/1=4, but -2 and 8: -4. So 4 and -4 are different. Wait no, 1: y=4, x=1: 4/1=4; x=-2: 8/-2=-4. So \( \frac{y}{x} \) is -4 for x=-2,4,5 (8/-2=-4, -16/4=-4, -20/5=-4) but x=1: 4/1=4. So not constant. Wait, maybe I miscalculated. x=1, y=4: 4/1=4; x=-2, y=8: 8/-2=-4. Different, so non-proportional? Wait no, wait third table: x: -2,1,4,5; y:8,4,-16,-20. Let's check \( \frac{y}{x} \):

- x=-2, y=8: 8/(-2) = -4
- x=1, y=4: 4/1 = 4
- x=4, y=-16: -16/4 = -4
- x=5, y=-20: -20/5 = -4

So for x=1, \( \frac{y}{x}=4 \), others -4. Not constant. Now check fourth table (x:4,-1,2,6; y:8,-1,-10,-28). Wait x:4, y:8: 8/4=2; x:-1, y:-1: -1/-1=1; not constant. Wait, let's recheck the third table (x:-2,1,4,5; y:8,4,-16,-20). Wait, maybe I made a mistake. Wait x=1, y=4: 4/1=4; x=-2, y=8: 8/-2=-4. So two different k values. Now check the fourth table: x:4, y:8 (8/4=2); x:-1, y:-1 (-1/-1=1); x:2, y:-10 (-10/2=-5); x:6, y:-28 (-28/6≈-4.666). Clearly not constant. Wait, but the third table: x=-2,1,4,5; y:8,4,-16,-20. Wait, x=1, y=4: 4=4*1; x=-2, y=8: 8=-4*(-2); x=4, y=-16: -16=-4*4; x=5, y=-20: -20=-4*5. Oh! Wait, I messed up the sign. x=1, y=4: 4=4*1 (k=4), but x=-2, y=8: 8=(-4)*(-2) (k=-4). So k is 4 for x=1, -4 for others. So not constant. Now the fourth table: x:4, y:8 (k=2); x:-1, y:-1 (k=1); x:2, y:-10 (k=-5); x:6, y:-28 (k≈-4.666). Definitely not constant. Wait, but the question is which table is non-proportional. Wait, let's check the third table again. Wait, x=-2, y=8: 8 = -4*(-2); x=1, y=4: 4 = 4*1; x=4, y=-16: -16 = -4*4; x=5, y=-20: -20 = -4*5. So here, for x=1, the constant is 4, others -4. So not the same. Now the fourth table: x=4, y=8 (8=2*4); x=-1, y=-1 (-1=1*(-1)); x=2, y=-10 (-10=-5*2); x=6, y=-28 (-28≈-4.666*6). So different constants. Wait, maybe the fourth table (x:4,-1,2,6; y:8,-1,-10,-28) is the non-proportional one. Let's verify each table:

1. First table (x:15,5,10,12; y:82.5,27.5,-55,-66):
- 82.5/15 = 5.5
- 27.5/5 = 5.5
- -55/10 = -5.5 (Wait, 10*(-5.5)=-55, 12*(-5.5)=-66. Wait, 15 and 5 have positive x, y positive; 10 and 12 have positive x, y negative. So k is 5.5 for x positive (15,5) and -5.5 for x positive (10,12)? No, that's inconsistent. Wait, no: 15*5.5=82.5, 5*5.5=27.5, 10*(-5.5)=-55, 12*(-5.5)=-66. So actually, the relationship is \( y = 5.5x \) when x is 15,5 and \( y = -5.5x \) when x is 10,12? No, that's not a proportional relationship (proportional has one constant k). So this table: \( \frac{y}{x} \) is 5.5, 5.5, -5.5, -5.5. So two different k (5.5 and -5.5), so non-proportional? Wait, but proportional can have negative k, but it must be the same for all x. So if x is positive, y positive; x positive, y negative: different k. So this table is non-proportional? Wait, no, 15 and 10: x positive, y positive and negative. So \( \frac{y}{x} \) is 5.5 and -5.5, different. So non-proportional. But let's check the fourth table (x:4,-1,2,6; y:8,-1,-10,-28):

- 8/4=2
- -1/-1=1
- -10/2=-5
- -28/6≈-4.666

Clearly different. Now, the third table (x:-2,1,4,5; y:8,4,-16,-20):

- 8/-2=-4
- 4/1=4
- -16/4=-4
- -20/5=-4

Here, x=1 gives 4, others -4. Different. The second table (x:1,3,7,10; y:7.5,22.5,52.5,75):

- 7.5/1=7.5
- 22.5/3=7.5
- 52.5/7=7.5
- 75/10=7.5. Constant, so proportional.

Now, the fourth table (x:4,-1,2,6; y:8,-1,-10,-28):

Check \( \frac{y}{x} \):

- 4: 8/4=2
- -1: -1/-1=1
- 2: -10/2=-5
- 6: -28/6≈-4.666. All different. So this table has no constant k, so non-proportional.

Wait, but the image shows four tables. Let's identify the tables by position:

1. First table (rightmost): x:15,5,10,12; y:82.5,27.5,-55,-66
2. Second table: x:1,3,7,10; y:7.5,22.5,52.5,75 (proportional)
3. Third table: x:-2,1,4,5; y:8,4,-16,-20
4. Fourth table (leftmost): x:4,-1,2,6; y:8,-1,-10,-28

We need to find which is non-proportional. The second table is proportional (k=7.5). Now check the fourth table (leftmost):

x=4, y=8: 8/4=2

x=-1, y=-1: -1/-1=1 (≠2)

So non-proportional. Alternatively, the third table: x=1, y=4: 4/1=4; x=-2, y=8: 8/-2=-4 (≠4). So non-proportional. But the leftmost table (fourth) has more varied k. Wait, maybe the leftmost table (x:4,-1,2,6; y:8,-1,-10,-28) is the answer.

Step3: Confirm Non-Proportional

For a proportional relationship, \( \frac{y}{x} \) must be constant. In the leftmost table (x:4,-1,2,6; y:8,-1,-10,-28):

- \( \frac{8}{4} = 2 \)
- \( \frac{-1}{-1} = 1 \) (not equal to 2)
- \( \frac{-10}{2} = -5 \) (not equal to 2 or 1)
- \( \frac{-28}{6} \approx -4.67 \) (not equal to others)

Thus, \( \frac{y}{x} \) is not constant, so this table represents a non-proportional relationship.

Answer:

The leftmost table (x: 4, -1, 2, 6; y: 8, -1, -10, -28)