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____ [38]
3 years ago
14

The length of a rectangle is 5 in. more than the four times the width. The perimeter of the rectangle is 80 in. Find the length

and width.
Mathematics
1 answer:
kogti [31]3 years ago
5 0

L=33 and W=7

Step one, What do we know?

We are told that the perimeter is  80  feet; since the perimeter of a rectangle is 2 times the length plus the width. We are also told that the length is 5 feet longer than four times the width.

Step two, Start solving it.

Combining these two we have 4w + 5=40 - w. We put the w on the right side and the 5 on the left side by adding and subtracting. Now we have 5w=35. We divide both sides by 5 and we get W=7. In order to figure out L, we do

40 - W. We get L=33.

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Answer with explanation:

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x_{3}=0.2012-\frac{\log(0.6036)+0.2024072}{\frac{1}{0.6036}+10 \times 0.2012}\\\\x_{3}=0.2012- \frac{-0.2192+0.2025}{1.6567+2.012}\\\\x_{3}=0.2012-\frac{-0.0167}{3.6687}\\\\x_{3}=0.2012+0.0045\\\\x_{3}=0.2057

x_{4}=0.2057-\frac{\log(0.6171)+0.21156}{\frac{1}{0.6171}+10 \times 0.2057}\\\\x_{4}=0.2057- \frac{-0.2096+0.21156}{1.6204+2.057}\\\\x_{4}=0.2057-\frac{0.0019}{3.6774}\\\\x_{4}=0.2057-0.0005\\\\x_{4}=0.2052

So, root of the equation =0.205 (Approx)

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 Approximate relative error in terms of Percentage

   =0.41 × 100

   = 41 %

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