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Sergeu [11.5K]
2 years ago
6

Least common denominator of 3/5 and 11/20

Mathematics
1 answer:
podryga [215]2 years ago
6 0
Solution:

Rewriting input as fractions if necessary:
3/5, 11/20

For the denominators (5, 20) the least common multiple (LCM) is 20.

LCM(5, 20)

Therefore, the least common denominator (LCD) is 20.
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The length of a rectangle is 5 meters less than twice the width. If the area of the rectangle is 273 meters, find the dimensions
AleksandrR [38]

The length of the rectangle = 16 meters

The width of the rectangle = 10.5 meters

<u>Step-by-step explanation</u>:

Given that,

The Area of the rectangle is 273.

<u><em>Step 1</em></u><em> :</em>

Let, the width of the rectangle be 'x'

The length of the rectangle is 2x - 5

<u><em>Step 2</em></u><em> :</em>

Area of the rectangle = length \times width

                             273 = (2x - 5) x

                             273 = 2x^2 - 5x

<u><em>Step 3</em></u><em> :</em>

The equation formed is 2x^2 - 5x -273 = 0

a= coefficient of x^2 = 2

b= coefficient of x = -5

c = constant = -273

Find the roots of the equation, to find the width.

<u><em>Step 4</em></u><em> :</em>

Find the roots using factorizing method,

ac = 2\times-273 = -546 and b= -5

Factorizing -546 as -26\times21

The product of -26\times21 = -546

The sum of -26 + 21 = -5

<u><em>Step 5</em></u><em> :</em>

The roots of the equation 2x^2 - 5x -273 = 0 are x= -26/2 and x= 21/2

The values of x are x= -13 and x= 10.5

Since the width cannot be a negative value, neglect x= -13

<u><em>Step 6</em></u><em> :</em>

Therefore,

The width of the rectangle = 10.5 meters

The length of the rectangle = 2x-5

                                                = 2(10.5) - 5

                                                = 21 - 5

                                    length = 16 meters

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Step-by-step explanation:

a)

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