Answer:
[tex]2\times (x+y) > 70\\x \leq 30\\y\leq 20[/tex]
c) (35,20) and (25,15)
Step-by-step explanation:
We are given the following information in the question:
Let x be the length of the rectangle and y be the width.
Perimeter of rectangle = [tex]2\times \text{(Length + Width)}[/tex]
a) Then, we can have the following inequalities:
[tex]2\times (x+y) > 70\\x \leq 30\\y\leq 20[/tex]
b) The attached image shows the graph for the three inequalities.
c) The two possible combination of length and width of rectangle could be:
(35,20) and (25,15)
The points are shown in the graph and satisfies all the three inequalities.
