Question

activity 4 identifying inverse proportion
determine whether two quantities are in inverse proportion. if so, find the constant of proportionality.
①

x	100	50	10
y	?	4	20

②

x	6	4	2
y	20	40	80

③

x	3	6	9
y	10	20	30

④

x	2	6	10
y	210	70	42

Explanation:

Problem 1

Step1: Check inverse proportion (x*y = k)

For \( x = 100, y = 2 \): \( 100\times2 = 200 \)
For \( x = 50, y = 4 \): \( 50\times4 = 200 \)
For \( x = 10, y = 20 \): \( 10\times20 = 200 \)

Step2: Conclusion

Since \( x\times y = 200 \) (constant), they are in inverse proportion. Constant \( k = 200 \).

Step1: Check inverse proportion (x*y = k)

For \( x = 6, y = 20 \): \( 6\times20 = 120 \)
For \( x = 4, y = 40 \): \( 4\times40 = 160 \)
For \( x = 2, y = 80 \): \( 2\times80 = 160 \) (Wait, 6×20=120≠160. Wait, recalculate: 6×20=120, 4×40=160, 2×80=160? No, 6×20=120, 4×40=160, 2×80=160. Wait, maybe typo? Wait, no—wait, 6×20=120, 4×40=160, 2×80=160. Not constant. Wait, maybe I miscalculated. Wait, 6×20=120, 4×40=160, 2×80=160. So not constant. Wait, no—wait, maybe the table is x:6,4,2; y:20,40,80. Let's check x/y: 6/20=0.3, 4/40=0.1, 2/80=0.025. Not direct. x*y: 6×20=120, 4×40=160, 2×80=160. Wait, 4×40=160, 2×80=160, but 6×20=120≠160. So not inverse proportion.

Wait, maybe I made a mistake. Wait, 6×20=120, 4×40=160, 2×80=160. So not constant. So not inverse proportion.

Step1: Check inverse proportion (x*y = k)

For \( x = 3, y = 10 \): \( 3\times10 = 30 \)
For \( x = 6, y = 20 \): \( 6\times20 = 120 \)
Not constant. Check direct proportion (y/x = k): 10/3 ≈3.33, 20/6≈3.33, 30/9≈3.33. So direct proportion (y = (10/3)x), not inverse.

Answer:

Inverse proportion, constant = 200

Problem 2