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

Given,
Perimeter of rectangle = 64 units.
Width = y = 11 units

Let,
Length = 20 units, as discussed in the question

Perimeter of rectangle = 2( length + width )

Solving Left Hand Side,

= > Perimeter of rectangle

= > 64 units [ Given ]

Solving Right Hand Side ,

= > 2( 20 + 11 )

= > 2( 31 )

= > 2 × 31



As we can see that 64 ≠ 62

64 ≠ 2( x + y )

Divide by 2 on both sides,

32 ≠ x + y

Hence, option B is correct.

Answercounterclockwise rotation about the origin by 180 degrees followed by reflection about the y-axisexplanationthe given figure has verticesrecall that for counterclockwise rotation about the origin, when we rotate figure pqrs counterclockwise through an angle of 180° about the origin, the coordinates will becomenext, we reflect in the y-axis by negating the x-coordinates of the resulting figure to obtain,therefore the correct answer is c