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adll23

  • one year ago

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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  1. narutoboy14
    • one year ago
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    d

  2. mathstudent55
    • one year ago
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    What is the perimeter of a rectangle?

  3. adll23
    • one year ago
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    2 (w + h)

  4. mathstudent55
    • one year ago
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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?

  5. mathstudent55
    • one year ago
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    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.

  6. mathstudent55
    • one year ago
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    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.

  7. adll23
    • one year ago
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    So it would be B?

  8. adll23
    • one year ago
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    @mathstudent55

  9. adll23
    • one year ago
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    Or would it be A? @mathstudent55

  10. narutoboy14
    • one year ago
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    b

  11. mathstudent55
    • one year ago
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    The answer is B.

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