3/12 + 1/6 + 5/36 + 3/4 =

2 n + 1 ; 2 n + 3 ; 2 n + 5 ; 2 n + 7 − f o u r co n sec u t i v e o dd in t e g ers ( 2 n + 1 ) + ( 2 n + 3 ) + ( 2 n + 5 ) + ( 2 n + 7 ) = 56 8 n + 16 = 56 8 n = 56 − 16 8 n = 40 = 40 : 8 = 5 2 n + 1 = 2 ( 5 ) + 1 = 10 + 1 = 11 A n s w er : 11 ; 13 ; 15 ; 17 ​

Answered by Anonymous | 2024-06-10

The four consecutive odd integers whose sum is 56 are 11, 13, 15, and 17. We found these by defining the integers and solving an equation based on their sum. The integers add up to the required total of 56.
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Answered by Anonymous | 2024-10-16

[tex] \frac{3}{12} + \frac{1}{6} + \frac{5}{36 } + \frac{3}{4} \\ \\ = \frac{9}{36} + \frac{6}{36} + \frac{5}{36} + \frac{27}{36} \\ \\ = \frac{9 + 6 + 5 + 27}{36} \\ \\ = \frac{47}{36} = 1 \frac{11}{36} [/tex]

Answered by aquila0152 | 2025-07-11