The perimeter of a rectangle is 64 units. Can the length x of the rectangle can be 20 units when its width y is 11 units? (4 points) No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 64 No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32 Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 64 Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 32

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The perimeter of a rectangle is 64 units. Can the length x of the rectangle can be 20 units when its width y is 11 units? (4 points) No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 64 No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32 Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 64 Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 32

Mathematics
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d
What is the perimeter of a rectangle?
2 (w + h)

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Great. The problem is calling the length x and the width y. According to the variables of the problem, then, the perimeter is P = 2(x + y) Ok?
We are told the length is 20 and the width is 11. What would the perimeter be? P = 2(x + y) P = 2(20 + 11) P = 2(31) P = 62 With the length 20 and width 11, the perimeter is 62, not 64. This means the perimeter cannot be 64.
Notice that when you calculate the perimeter, you add the length and width, and then you multiply that by 2. That means the length + the width must equal half the perimeter. In our case, the length + the width = 20 + 11 = 31 Since we are told the perimeter is 64, half the perimeter is 32. 31 is not half the perimeter, so 64 cannot be the perimeter.
So it would be B?
Or would it be A? @mathstudent55
b
The answer is B.

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