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Natasha2012 [34]
3 years ago
11

The length of a rectangle is 2cm more twice the width. If each dimension were increased by 2, the area would be at least 180 cm^

2 greater. Find the smallest possible dimensions for the rectangle.
Mathematics
2 answers:
julsineya [31]3 years ago
7 0

Answer:ok I see

Step-by-step explanation:

Now

LenaWriter [7]3 years ago
5 0
You ugly my guy change your pfp son
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The length of a rectangle is 5959 inches greatergreater than twice the width. if the diagonal is 2 inches more than the​ length,
lorasvet [3.4K]

Length: 2w + 59

width: w

diagonal: (2w + 59) + 2  = 2w + 61

Length² + width² = diagonal²

(2w + 59)² + (w)² = (2w + 61)²

(4w² + 118w + 3481) + w²  = 4w² + 122w + 3721

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w = [-(b) +/- \sqrt{(b)^{2}  - 4(a)(c) }]/2(a)

 = [-(-4) +/- \sqrt{(-4)^{2}  - 4(1)(-240) }]/2(1)

   =  [4 +/- \sqrt{(16  + 960) }]/2        

  = [4 +/- \sqrt{(976) }]/2  

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  = 1 +/- 2\sqrt{(61) }

since width cannot be negative, disregard  1 - 2√61

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