Question 287100
Problem #287100
1.	the length of a rectangle is 5cm. greater than twice its width. if the perimeter is 52 cm., what is the dimensions of the rectangle
2.	Let w=width
            2w+5= length
I am a visual person so I draw the rectangle out and label it.
Next write the formula of the Area of a rectangle.
	A=lw
With the information you have,replace it into the formula:
	52=(2w+5)w (I rewrite the formula for my understanding)
	w(2w+5)=52
Next we need to factor the equation and set it to equal 0:
	2w^2 +5w=52
subtract 52 from the right and rewrite the equation:
	2w^2+5w-52=0

Solve by grouping
2w^2+13w ] [ -8w -52 =0
w(2w+13)       -4(2w+13) =0
(w-4)            (2w+13)

Width = 4

Check:
52= ((2)4+5)4
52= (13)4
52=52

Length = 13

The dimensions of the rectangle are width equals 4cm and the length equals 13cm.