Recall that the length a spring stretches varies directly with the amount of weight attached to it. A certain spring stretches 5 cm when a 10-gram weight is attached.

Write a direct variation equation relating the weight \( x \) and the amount of stretch \( y \).

Direct variation is y=kx
In this case, y=1/2x. The spring stretches 5 cm for every 10 grams. Therefore if y=5 and x=10, k would have to be 1/2. That's how we come up with our direct variation model.

Answered by runescapememo8 | 2024-06-10

T hi s s p r in g s t re t c h es in p ro p or t i o n t o w e i g h t , t ha t i s , f or e v ery 10 g r am s o f w e i g h t t h e am o u n t o f s t re t c h i s 5 c m . x [ g ] ∣ 10 ∣ 20 ∣ 30 ∣ 40 ∣...∣ a ∣ − − − − − − − − − − − − − − − − − − y [ c m ] ∣ 5 ∣ 10 ∣ 15 ∣ 20 ∣...∣ 2 1 ​ a ∣ a v a r ia t i o n e q u a t i o n i s y = 2 1 ​ x

Answered by kate200468 | 2024-06-24

The direct variation equation relating the weight and the amount of stretch in the spring is y = 0.5 x . This indicates that the stretch of the spring in cm is half the weight in grams attached to it. Thus, for every gram of weight, the spring stretches 0.5 cm.
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Answered by kate200468 | 2024-09-26