Calculate $(-1)^{47} + 2^{(3^3+4^2-6^2)}$.
Recall that $(-a)^n=-a^n$ when $n$ is odd. Because 47 is odd, $(-1)^{47}=-1^{47}=-1$. Evaluating the remaining exponents, we get \begin{align*}
(-1)^{47} + 2^{(3^3+4^2-6^2)}&= -1 + 2^{(3^3+4^2-6^2)} \\
&=-1 + 2^{(27+16-36)} \\
&=-1 + 2^{(43-36)} \\
&=-1 + 2^{7} \\
&=-1+128 \\
&=\boxed{127}
\end{align*}
127
Level 2
Prealgebra
Problem: Calculate $(-1)^{47} + 2^{(3^3+4^2-6^2)}$.
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the problem, first evaluate the term $(-1)^{47}$. Since the exponent 47 is odd, $(-1)^{47} = -1$.

Next, calculate the term $2^{(3^3 + 4^2 - 6^2)}$. Start by evaluating the components inside the exponent:

1. Calculate $3^3$: 
   $3^3 = 27$.

2. Calculate $4^2$: 
   $4^2 = 16$.

3. Calculate $6^2$: 
   $6^2 = 36$.

Now substitute these values into the expression for the exponent:

$3^3 + 4^2 - 6^2 = 27 + 16 - 36$.

Perform the addition and subtraction step-by-step:

1. Add $27 + 16 = 43$.
2. Subtract $36$ from $43$: 
   $43 - 36 = 7$.

Now we have the exponent simplified to 7. Therefore, we can evaluate $2^7$:

$2^7 = 128$.

Now combine the results from both terms:

$(-1)^{47} + 2^{(3^3 + 4^2 - 6^2)} = -1 + 128$.

Perform the final addition:

$-1 + 128 = 127$.

Thus, the final answer is $127$.<|END_OF_LATENT|> The final answer is $127$.