Elementary Technical Mathematics

Published by Brooks Cole
ISBN 10: 1285199197
ISBN 13: 978-1-28519-919-1

Chapter 6 - Section 6.6 - Applications Involving Equations - Exercises - Page 251: 1


$x=10$ $inches$

Work Step by Step

Step 1: Since we know the thickness of the shelves, we can add them to find the space taken by the shelves themselves. We assign the variables: $a=$ thickness of each shelf $b =$ total thickness occupied by the shelves The total thickness occupied by the shelves is then $b=8a$ $\rightarrow$ because it is a set of eight. $b=8(1 in)=8$ $inches$ Step 2: Find the space remaining in the clearance. We assign the variables: $c=$ floor to ceiling clearance $= 8ft$ $2 in$ $d=$ floor to ceiling clearance minus total thickness occupied by the shelves. So, $d=$ $ 8ft$ $2 in$ $-8in$ $\rightarrow$ remember 1 ft = 12 inches. $d=98in - 8in=90in$ Step 3: For this step, refer to the diagram in the answer. Since no shelf is against the floor or the ceiling, then we have $9$ empty spaces between the shelves. Let x = space between each shelf and the next. So, $x=\frac{d}{9}=\frac{90in}{9}=10in$
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