<p>The length of a rectangle is 2 ft. more than twice the width. The area is 40 ft^2. Find the length of the rectangle. </p>
<p>A.6
B.8
C.10
D.12
E.14</p>
<p>Thanks.</p>
<p>The length of a rectangle is 2 ft. more than twice the width. The area is 40 ft^2. Find the length of the rectangle. </p>
<p>A.6
B.8
C.10
D.12
E.14</p>
<p>Thanks.</p>
<p>Bump anyone? Or are all fast a sleep,</p>
<p>It’s C.</p>
<p>Am I right?</p>
<p>Yea though could you tell me how you did it?^</p>
<p>I’m not good with explaining but ill give it a try.
Area= L * W</p>
<p>Just use trial and error.</p>
<p>Length=2x+2</p>
<p>Start with C:</p>
<p>10=2x+2</p>
<p>10-2–> 8/2—> 4=Width</p>
<p>10*4=40</p>
<p>You can also solve this problem algebraically. </p>
<p>length = 2x + 2 (“the length of a rectangle is 2 ft. more than twice the width”
width = x </p>
<p>length * width = area = 40</p>
<p>x(2x + 2) = 40</p>
<p>2x^2 + 2x = 40</p>
<p>2x^2 + 2x - 40 = 0</p>
<p>x^2 + x - 20 = 0 (I divided everything by 2)</p>
<p>Solve the quadratic:</p>
<p>(x+5)(x-4)</p>
<p>x = -5, x = 4</p>
<p>x cannot be negative; the side of a rectangle cannot be negative. So x can only = 4.</p>
<p>Back to the length:</p>
<p>x = 4</p>
<p>Length = 2x + 2 = 8 + 2 = 10.</p>
<p>Yea thats right thanks ICe</p>