The question asks for the sum of three fractions: $\frac{1}{3}$, $\frac{4}{3}$, and $\frac{3}{4}$.
First, add the fractions that share a common denominator:
$ \frac{1}{3} + \frac{4}{3} = \frac{1+4}{3} = \frac{5}{3} $
Now, add the result ($\frac{5}{3}$) to the remaining fraction ($\frac{3}{4}$):
$ \frac{5}{3} + \frac{3}{4} $
To add these fractions, find a common denominator. The least common multiple (LCM) of 3 and 4 is 12.
Convert the fractions to equivalent fractions with the denominator 12:
Add the numerators of the equivalent fractions:
$ \frac{20}{12} + \frac{9}{12} = \frac{20+9}{12} = \frac{29}{12} $
Therefore, the sum of $\frac{1}{3}$, $\frac{4}{3}$, and $\frac{3}{4}$ is $\frac{29}{12}$.
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: