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solmaris [256]
3 years ago
9

5/14 in in simple form

Mathematics
1 answer:
slavikrds [6]3 years ago
7 0

Answer:

Step-by-step explanation:

1. Find the GCD ( or HCF) of numerator and denominator

<u>GCD of 5 and 14 is 1</u>

2. Divide both the numerator and denominator by the GCD

5 divide 1 and 14 divide 1

3 reduced fraction

      5/14

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Evaluate the expression when a= 4, b= - 2, and c= 7 - -&gt; 5a - 9 (b - c)
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A manufacturer of prefabricated homes has decided to subcontract four components of the homes. Several companies are interested
Xelga [282]

Answer:

Following are the solution to the given question:

Step-by-step explanation:

Let x_{ij}=1 when i is the element itself is allocated to 'j' i_0

Min  

185X_{1A}+225X_{1B} +193X_{1C} +207X_{1D} +200X_{2A} +190X_{2B} +175X_{2C}+225X_{2D}+330X_{3A} +320X_{3B} +315X_{3C}+300X_{3D}+375X_{4A}+389X_{4B}+425X_{4C}+ 445X_{4D}

Subject to:

X_{1A} +X_{1B}+X_{1C}+X_{1D}=1\\\\X_{2A} +X_{2B}+X_{2C}+X_{2D}=1\\\\X_{3A} +X_{3B}+X_{3C}+X_{3D}=1\\\\X_{4A} +X_{4B}+X_{4C}+X_{4D}=1\\\\X_{1A} +X_{2A}+X_{3A}+X_{4A}=1\\\\X_{1B} +X_{2B}+X_{3B}+X_{3B}=1\\\\X_{1C} +X_{2C}+X_{3C}+X_{4C}=1\\\\X_{1D} +X_{2D}+X_{3D}+X_{4D}=1

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