The goal is to identify the rational number from the given options that falls strictly between $\frac{1}{4}$ and $\frac{1}{2}$.
To compare the fractions effectively, we first find a common denominator for the boundary fractions $\frac{1}{4}$ and $\frac{1}{2}$. The least common multiple of 4 and 2 is 4. We can express both fractions with a denominator of 8 for easier comparison with the options.
Now, we need to find the option that is greater than $\frac{2}{8}$ and less than $\frac{4}{8}$.
Let's examine each option:
Based on the comparison, the only rational number that lies between $\frac{1}{4}$ and $\frac{1}{2}$ is $\frac{3}{8}$.
If \(\sqrt{1+\frac{\sqrt{3}}{2}}- \sqrt{1-\frac{\sqrt{3}}{2}}= c\) , then the value of c is:
If \(\frac{\sqrt{38-5\sqrt{3} } }{\sqrt{26+7\sqrt{3} } }= \frac{a+b\sqrt{3} }{23} \) , b > 0, then the value of (b – a) is:
If \( \frac{5}{4{\sqrt 2 }} + \frac{{3 + 2\sqrt 2 }}{{3 - 2\sqrt 2 }} - \frac{{3 - 2\sqrt 2 }}{{3 + 2\sqrt 2 }} = a + b\sqrt 2 \) , then what is the value of (3a + 4b)?
If \(\frac {8 + 2\sqrt 3}{3\sqrt 3 + 5} = a\sqrt 3 - b,\) then the value of a + b is equal to:
If \(\frac{\sqrt{26-7\sqrt{3} } }{\sqrt{14+5\sqrt{3} } } = \frac{b+a\sqrt{3} }{11}\) , b > 0, then what is the value of \(\sqrt{(b-a)} \) ?