Algebra: A Combined Approach (4th Edition)

Published by Pearson
ISBN 10: 0321726391
ISBN 13: 978-0-32172-639-1

Chapter 5 - Test - Page 409: 28


$A=(4x^{2}-9)$ square inches

Work Step by Step

The top of the table is a rectangle, so its area $(A)$ will be $A= L \times W,$ where $L=(2x+3)$ inches and $W=(2x-3)$ inches. Therefore, $A=(2x+3)(2x-3)$ $A=2x(2x-3)+3(2x-3)$ $A=4x^{2}-6x+6x-9$ $A=4x^{2}-9$ Therefore, the area of the table is $A=(4x^{2}-9)$ square inches.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.