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The length of a rectangle is 24 units. Can the perimeter x of the rectangle be 60 units when its width y is 11 units? . a. No, the rectangle cannot have x = 60 and y = 11 because x = 48 + 2y. . . b. No, the rectangle cannot have x = 60 and y = 11 because x = 24 + 2y. . . c. Yes, the rectangle can have x = 60 and y = 11 because x + y is greater than 48. . . d. Yes, the rectangle can have x = 60 and y = 11 because x + y is greater than 24.

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taskmasters
No the rectangle cannot have x = 60 and y = 11 because x = 24 + 2y. The correct option among all the options that are given in the question is the first option or option "A". The perimeter of the rectangle can be found by adding all four sides of the rectangle. I a rectangle the opposite sides are equal and so on adding we get
Perimeter (x) = 24 + 24 + 11 + 11 = 70. 
There is no chance that the perimeter of the given rectangle will be 60.
calculista

Let

L--------> the length of the rectangle

y------> the width of the rectangle

x------> the perimeter of the rectangle

we know that

The perimeter of the rectangle is equal to

[tex]x=2L+2y[/tex]

In this problem we have

[tex]L=24\ units[/tex]

For a width [tex]y=11\ units[/tex]

Find the value of the perimeter

[tex]x=2*24+2*11=48+22=70\ units[/tex]

so

[tex]70\ units \neq 60\ units[/tex]

therefore

the answer is the option A

No, the rectangle cannot have x = 60 and y = 11 because x = 48 + 2y. .

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