SOLUTION: The length of a picture frame is 3 in. greater than the width. The perimiter is less than 52in. Describe the dimentions of the frame.
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Question 218757: The length of a picture frame is 3 in. greater than the width. The perimiter is less than 52in. Describe the dimentions of the frame.
Found 2 solutions by drj, MathTherapy:
Answer by drj(1380) (Show Source): You can put this solution on YOUR website!
The length of a picture frame is 3 in. greater than the width. The perimeter is less than 52in. Describe the dimensions of the frame.
Step 1. Let w be the width.
Step 2. Let w+3 be the length.
Step 3. Perimeter P means adding up all four sides of the triangle. That is,
Step 4. Solving yields the following steps
Subtract 6 to both sides of the equation
Divide 4 to both sides of the equation
{{4w/4=46/4}}}
and which is a true statement
Step 5. ANSWER: The width is in. and length is in.
I hope the above steps and explanation were helpful.
For Step-By-Step videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry please visit http://www.FreedomUniversity.TV/courses/Trigonometry.
Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs.
Respectfully,
Dr J
http://www.FreedomUniversity.TV
Answer by MathTherapy(10549) (Show Source): You can put this solution on YOUR website!
The length of a picture frame is 3 in. greater than the width. The perimiter is less than 52in. Describe the dimentions of the frame.
Let the width of the frame be W
Since its length is 3 inches greater than its width, then its length = W + 3
Since the perimeter is less than 52 inches, then we'll have:
2(W) + 2(W + 3) < 52
2W + 4W + 6 < 52
6W + 6 < 52
6W < 46
W < or <, or <
Since width < 11.5 inches, and length is 3 more then width, then length < (11.5 + 3), or <
--------
Check
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Since width, or W < 11.5, and since length, or L < 14.5, then let width, or W = 11, and length, or L = 14
2W + 2L < Perimeter
2(11) + 2(14) < 52
22 + 28 < 52
50 < 52 (TRUE)
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