The problem asks us to find the reciprocal of the sum of the reciprocals of two given fractions: $\frac{5}{7}$ and $\frac{9}{5}$. We can solve this step-by-step:
Next, we add the reciprocals found in Step 1:
Sum = $\frac{7}{5} + \frac{5}{9}$
To add these fractions, we find a common denominator, which is $5 \times 9 = 45$.
Sum = $\frac{7 \times 9}{5 \times 9} + \frac{5 \times 5}{9 \times 5}$
Sum = $\frac{63}{45} + \frac{25}{45}$
Sum = $\frac{63 + 25}{45}$
Sum = $\frac{88}{45}$
Finally, we find the reciprocal of the sum calculated in Step 2.
The sum is $\frac{88}{45}$.
The reciprocal of the sum is $\frac{45}{88}$.
The reciprocal of the sum of the reciprocals of $\frac{5}{7}$ and $\frac{9}{5}$ is $\frac{45}{88}$.
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: