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Create an equation from the model that describes the relationship between the weights on each side of the scale.

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[tex]$\frac{\square}{\square}$[/tex] & [tex]$|a|$[/tex] & [tex]$\square^2$[/tex] & [tex]$\pm$[/tex] & [tex]$=$[/tex] & [tex]$\ \textgreater \ $[/tex] & [tex]$\leq$[/tex] & [tex]$\geq$[/tex] & [tex]$\pi$[/tex] & [tex]$a$[/tex] & [tex]$\sin$[/tex] & [tex]$\cos$[/tex] & [tex]$\tan$[/tex] & [tex]$\sec$[/tex] & [tex]$\cot$[/tex] & [tex]$\sin^{-1}$[/tex] & [tex]$\cos^{-1}$[/tex] & [tex]$\tan^{-1}$[/tex] & [tex]$\log_{\square}$[/tex] & [tex]$\ln$[/tex] & [tex]$\angle$[/tex] & [tex]$\triangle$[/tex] & [tex]$\sim$[/tex] & [tex]$\sum$[/tex] \\
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Sagot :

GinnyAnswer
To solve this problem, we need to create an equation that describes the balance of weights on each side of the scale. Here's the step-by-step solution:

1. Analyze the Left Side of the Scale:
- On the left side, we have three unit weights and a weight with variable value \( y \).
- So, we can represent the total weight on the left side as:
[tex]\[ 1 + 1 + 1 + y = 3 + y \][/tex]

2. Analyze the Right Side of the Scale:
- On the right side, we have three weights, each with variable value \( y \).
- Therefore, the total weight on the right side is:
[tex]\[ y + y + y = 3y \][/tex]

3. Equate the Weights on Both Sides:
- Since the scale is balanced, the total weight on the left side must equal the total weight on the right side.
- Hence, we can set up the following equation:
[tex]\[ 3 + y = 3y \][/tex]

So, the equation that describes the relationship between the weights on each side of the scale is:
[tex]\[ 3 + y = 3y \][/tex]
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