Simplify this expression to a common fraction:  $\frac{1}{\frac{1}{(\frac{1}{2})^{1}}+\frac{1}{(\frac{1}{2})^{2}}+\frac{1}{(\frac{1}{2})^{3}}}$
$\frac{1}{\frac{1}{(\frac{1}{2})^1}+\frac{1}{(\frac{1}{2})^2}+\frac{1}{(\frac{1}{2})^3}}=\frac{1}{\frac{2}{1}+\frac{4}{1}+\frac{8}{1}}=\boxed{\frac{1}{14}}$.
\frac{1}{14}
Level 4
Prealgebra
Problem: Simplify this expression to a common fraction:  $\frac{1}{\frac{1}{(\frac{1}{2})^{1}}+\frac{1}{(\frac{1}{2})^{2}}+\frac{1}{(\frac{1}{2})^{3}}}$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To simplify the expression, first evaluate the denominator: $\frac{1}{(\frac{1}{2})^{1}}+\frac{1}{(\frac{1}{2})^{2}}+\frac{1}{(\frac{1}{2})^{3}}$. Each term can be rewritten as follows:

1. The first term is $\frac{1}{(\frac{1}{2})^{1}} = 2$.
2. The second term is $\frac{1}{(\frac{1}{2})^{2}} = 4$.
3. The third term is $\frac{1}{(\frac{1}{2})^{3}} = 8$.

Now, sum these values: $2 + 4 + 8 = 14$. 

Next, substitute this sum back into the original expression. The expression now becomes $\frac{1}{14}$. 

Thus, the simplification leads directly to the final answer, which is $\frac{1}{14}$.<|END_OF_LATENT|> The final answer is $\frac{1}{14}$.