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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