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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?
A:No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 64
B: No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32
C:Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 64
D:Yes, the rectangle can have x = 20 and y = 11 because x + y is less than 32

@taskmasters or anybody

Respuesta :

taskmasters
B. No, the rectangle cannot have x = 20 and y = 11 because x + y ≠ 32

Perimeter = 2(l + w)
Perimeter = 64

Assuming: P = 64; w = 11 ; l = ?
64 = 2(l + 11)
64/2 = l + 11
32 - 11 = l
length = 21.

The answer is B. No, the rectangle cannot have x=20 and y=11 because x + y does not equal 32.

The reasoning is, 20 + 11 = 31. In order to find the perimeter around the whole object, you will need to multiply 31 by 2 and get 62.

But, x is ONLY 20 and y is ONLY 11. It is not saying x(2) and y(2), so your answer has to be B.

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