Recall that the length a spring streches varies directly with the amount of weight attached to it. a certain spring stretches 5cm 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.




\( This\ spring\ stretches\ in\ proportion\ to\ weight,\ that\ is,\ for\ every \\ 10\ grams\ of\ weight\ the\ amount\ of\ stretch\ is\ 5\ cm. \\ \\x\ [g]\ \ \ |\ \ 10\ \ |\ \ 20\ \ |\ \ 30\ \ |\ \ 40\ \ |.|\ \ \ a\ \ \ |\\-\\y\ [cm]\ |\ \ 5\ \ \ |\ \ 10\ \ |\ \ 15\ \ |\ \ 20\ \ |.|\ \ \frac{1}{2} a\ \ |\\\\a\ variation\ equation\ is\ y= \frac{1}{2} x \)


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