College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter P, Prerequisites - Section P.5 - Algebraic Expressions - P.5 Exercises - Page 36: 90

Answer

See the explanation

Work Step by Step

$A=(x-20)(y-20),$ a. -The width of the lot is $y,$ to get the width of the building envelope, we substract it from the width of the lot. Therefore, by substracting $10ft$ from two sides we get the width of the building envelope $y-20$. -The length of the lot is $x,$ to get the length of the building envelope, we substract it from the length of the lot. Therefore, by substracting $10ft$ from two sides we get the length of the building envelope $x-20$. b.$A=xy-20x-20y+400$ c. $A_1=(100-20)(400-20)=30,400ft^2,$ $A_2=(200-20)(200-20)=32,400ft^2$ Therefore, the second lot has larger building envelope.
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.