$(0.\overline{5} + 0.\overline{6} + 0.\overline{7} + 0.\overline{8}) = ?$
The question asks to find the sum of four repeating decimals: $0.\overline{5}$, $0.\overline{6}$, $0.\overline{7}$, and $0.\overline{8}$.
First, convert each repeating decimal into its fractional form. A repeating decimal of the form $0.\overline{d}$ is equivalent to the fraction $\frac{d}{9}$.
Next, add these fractions together. Since they have a common denominator (9), we simply add the numerators:
$ \frac{5}{9} + \frac{6}{9} + \frac{7}{9} + \frac{8}{9} = \frac{5 + 6 + 7 + 8}{9} $
Calculate the sum of the numerators:
$ 5 + 6 + 7 + 8 = 26 $
The sum is $\frac{26}{9}$.
Finally, convert the improper fraction $\frac{26}{9}$ into a mixed number. Divide 26 by 9:
$ 26 \div 9 = 2 \text{ with a remainder of } 8 $
Therefore, the mixed number is $2\frac{8}{9}$.
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: