At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Ask your questions and receive accurate answers from professionals with extensive experience in various fields on our platform. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.


The length of a rectangle is 3 yd more than twice the width, and the area of the rectangle is 35 yd2
. Find the dimensions of the rectangle.


Sagot :

naǫ
The width of the rectangle = x.
The length of the rectangle is 3 yd more than twice the width = 2x+3.
The area is 35 yd².

The area is the width times the length.
[tex]35=x(2x+3) \\ 35=2x^2+3x \\ 0=2x^2+3x-35 \\ 0=2x^2+10x-7x-35 \\ 0=2x(x+5)-7(x+5) \\ 0=(2x-7)(x+5) \\ 2x-7=0 \ \lor \ x+5=0 \\ 2x=7 \ \lor \ x=-5 \\ x=\frac{7}{2} \ \lor \ x=-5[/tex]

The length and width must be positive so [tex]x=\frac{7}{2}=3.5[/tex].
[tex]2x+3=2 \times \frac{7}{2} + 3=7+3=10[/tex]

The dimension of the rectangle are 3.5 yd and 10 yd.
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.