Respuesta :
Answer:
[tex]15.2+2w\leq 28[/tex]
Step-by-step explanation:
Let w represent width of the rope-off section.
We have been given that a manager needs to rope off a rectangular section for a private party the length of the section must be 7.6 m the manager can use no more than 28 m of the rope.
We will use perimeter of rectangle formula to solve our given problem. We know that perimeter of a rectangle is equal to 2 times the sum of length and width.
[tex]\text{Perimeter}=2l+2w[/tex]
Upon substituting our given values, we will get:
[tex]\text{Perimeter}=2(7.6)+2w\\\\\text{Perimeter}=15.2+2w[/tex]
Since the manager can use no more than 28 m of the rope, so perimeter of rope-off section should be less than or equal to 28 meters.
We can represent this information in an inequality as:
[tex]15.2+2w\leq 28[/tex]
Therefore, our required inequality would be [tex]15.2+2w\leq 28[/tex].
Let us find width as:
[tex]15.2-15.2+2w\leq 28-15.2[/tex]
[tex]2w\leq 12.8[/tex]
[tex]\frac{2w}{2}\leq \frac{12.8}{2}\\\\w\leq6.4[/tex]
Therefore, the width of the rope-off section should be less than or equal to 6.4 meters.
