Otvety Na K-4 Vilenkin P 15 Variant 4 Dlja 6 Klassa May 2026

The final answers for Variant 4 include the common denominator , the comparison result 4/9 > 5/12 , and the equation solution x = 13/36 .

The answers for the Grade 6 mathematics control work based on the Vilenkin textbook (Paragraph 15) focus on fractions, common denominators, and comparing fractional values . 1. Find the Least Common Denominator To find the least common denominator (LCD) for 5125 over 12 end-fraction 7187 over 18 end-fraction , find the Least Common Multiple (LCM) of 12 and 18. Result: 153615 over 36 end-fraction 143614 over 36 end-fraction 2. Compare the Fractions To compare 49four-nineths 5125 over 12 end-fraction , bring them to a common denominator ( 3. Perform Addition and Subtraction A) : The LCD is 24. B) : The LCD is 18. C) : The LCD is 30. 4. Solve the Equation Simplify the right side: (LCD is 36) 5. Word Problem (Example) If a runner covers 25two-fifths of a distance in the first hour and 13one-third in the second: Total distance: Remaining: otvety na k-4 vilenkin p 15 variant 4 dlja 6 klassa

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The final answers for Variant 4 include the common denominator , the comparison result 4/9 > 5/12 , and the equation solution x = 13/36 .

The answers for the Grade 6 mathematics control work based on the Vilenkin textbook (Paragraph 15) focus on fractions, common denominators, and comparing fractional values . 1. Find the Least Common Denominator To find the least common denominator (LCD) for 5125 over 12 end-fraction 7187 over 18 end-fraction , find the Least Common Multiple (LCM) of 12 and 18. Result: 153615 over 36 end-fraction 143614 over 36 end-fraction 2. Compare the Fractions To compare 49four-nineths 5125 over 12 end-fraction , bring them to a common denominator ( 3. Perform Addition and Subtraction A) : The LCD is 24. B) : The LCD is 18. C) : The LCD is 30. 4. Solve the Equation Simplify the right side: (LCD is 36) 5. Word Problem (Example) If a runner covers 25two-fifths of a distance in the first hour and 13one-third in the second: Total distance: Remaining: