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 $\frac{29}{12}$.
What is the value of
$\frac{7}{9} - \frac{11}{12} + \frac{13}{16} - \frac{1}{8}$?
5 \(\frac{3}{4}\) + x + 2 \(\frac{1}{2}\) = 10 \(\frac{1}{8}\) Find the value of x.
Simplify the expression 441 ÷ \(\left[270 \div \frac{3}{7}+\left(17\div \frac{1}{3}\right)-\left(8\frac{1}{2}-\frac{5}{2}\right)\right]\)
Number 0.232323 can be written in rational form as:
Solve: \(\frac{1}{2}\) [{-2(2 + 3)*20}/2]
Match the following.
Column I | Column II | ||
a. | Equivalent fraction of \(\frac{7}{12}\) is | i. | Proper fraction |
b. | Equivalent fraction of \(\frac{9}{15}\) is | ii. | Improper fraction |
c. | \(\frac{7}{11}\) is | iii. | \(\frac{21}{36}\) |
d. | \(\frac{19}{5}\) is | iv. | \(\frac{3}{5}\) |