(-1 $\frac{2}{7}$) + (-3 $\frac{5}{7}$) + (6 $\frac{4}{7}$) is ______.
$\frac{11}{7}$
To solve the problem \((-1 \frac{2}{7}) + (-3 \frac{5}{7}) + (6 \frac{4}{7})\), we need to follow these steps:
- Convert the mixed fractions to improper fractions:
- \(-1 \frac{2}{7} = -\frac{7 \times 1 + 2}{7} = -\frac{9}{7}\)
- \(-3 \frac{5}{7} = -\frac{7 \times 3 + 5}{7} = -\frac{26}{7}\)
- \(6 \frac{4}{7} = \frac{7 \times 6 + 4}{7} = \frac{46}{7}\)
- Add the improper fractions:
- Combine the fractions: \(-\frac{9}{7} - \frac{26}{7} + \frac{46}{7}\)
- Add the numerators: \((-9) + (-26) + 46 = 11\)
- The sum is: \(\frac{11}{7}\).
Thus, the correct answer is \(\frac{11}{7}\), which matches option A.
Important Questions from Fractions
Which fraction among the following is the least ?
\(\frac{5}{11}, \frac{7}{12}, \frac{8}{13}, \frac{9}{17}\)
Find the value of the following expression:
\(\frac{{3 \div 1 \times 2 + 5 - 2}}{{3 \times 3 - 2}}\)
Simplify the expression 441 ÷ \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)
If the sum of two positive numbers is 65 and the square root of their product is 26, then the sum of their reciprocals is:
The value of \(9 \div [\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}\div(\frac{3}{4}-\frac{1}{3})\;of\;\frac{2}{9}]\) is:
