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Ganezh [65]
3 years ago
6

The length of a rectangle is 5 yd more than double the width, and the area of the of the rectangle is 42 yd to the second power

. What is the dimensions of the rectangle
Mathematics
1 answer:
yanalaym [24]3 years ago
7 0

Answer:

The answer to your question is l = 12 and w = 7/2 or 3.5

Step-by-step explanation:

Data

length = l

width = w

Area = 42 yd²

Process

1) Write equations that help to solve this problem

Area = length x width = l x w

Length = 2width + 5  or   l = 2w + 5

2) Substitution

                       42 = lw                               Equation l

                          l = 2w + 5                        Equation ll

                       42 = (2w + 5)w                   Substitute equation ll in l

3.- Expand

                       42 = 2w² + 5w

                       2w² + 5w - 42 = 0

Solve the equation by factoring

-Multiply 2 by 42

                             2 x 42 = 84

-Find the prime factors of 84

                         84 = 2² 3 7

-Find two numbers that added gives +5 using the prime factors

 These numbers are +12 and -7    

-Substitute the values in the equation

                         2w² -7w + 12w - 42

-Factor by grouping

                        w(2w - 7) + 6(2w - 7)                  

                           (2w - 7)(w + 6)

-Find w

                    2w₁ - 7 = 0             w₂ + 6 = 0

                      w₁ = 7/2               w₂ = -6   This value is discarted because

                                                                  there are no negative values

-Find l

                     l = 2(7/2) + 5

                     l = 7 + 5

                     l = 12

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All of these upper-bound/lower-bound problems are worked the same way. The bounds on the measurement are presumed to be half of one unit of the least-significant digit.

Here, the least significant digit of both numbers is in the "units" place, so the maximum error is presumed to be 1/2 unit, or ±0.5 cm.

If the measurement were 7.36 inches, the least significant digit is in the hundredths place, so the maximum error is presumed to be half a hundredth, or ±0.5 × 0.01 inches = ±0.005 inches.

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<em>Comment on measurement bounds</em>

You have to remember that these concepts are mathematical in nature, so only an approximation of the real world. If you have ever used a tape measure to measure a distance of any length, you will have discovered a couple of things:

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