Answered

Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Experience the convenience of getting accurate answers to your questions from a dedicated community of professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.


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.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope you found this helpful. Feel free to come back anytime for more accurate answers and updated information. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.