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
Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga.
Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus.
Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Not the answer you are looking for? Search for more explanations.
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)
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.