If $m+\frac{1}{m}=8$, then what is the value of $m^{2}+\frac{1}{m^{2}}+4$?
Squaring the equation provided, we get $m^2+2(m)\left(\frac{1}{m}\right) +\frac{1}{m^2}=64,$ so $m^2+\frac{1}{m^2}+4=\boxed{66}$.
66
Level 3
Algebra
Problem: If $m+\frac{1}{m}=8$, then what is the value of $m^{2}+\frac{1}{m^{2}}+4$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of $m^{2}+\frac{1}{m^{2}}+4$, we first need to derive $m^{2}+\frac{1}{m^{2}}$ from the given equation $m+\frac{1}{m}=8$. 

We start by squaring both sides of the equation:

1. $(m + \frac{1}{m})^2 = 8^2$.
2. This expands to $m^2 + 2 + \frac{1}{m^2} = 64$.
3. Rearranging gives us $m^2 + \frac{1}{m^2} = 64 - 2$.
4. Thus, $m^2 + \frac{1}{m^2} = 62$.

Next, we add 4 to this result:

5. $m^2 + \frac{1}{m^2} + 4 = 62 + 4$.
6. Therefore, $m^2 + \frac{1}{m^2} + 4 = 66$.

The final answer is $66$.<|END_OF_LATENT|> The final answer is $66$.