The zero power rule states that any non-zero number raised to the power of zero equals one. That is, \( a^0 = 1 \) for any \( a \neq 0 \).
Now, let's apply this rule to the expression you provided:
- The term \( (-40)^0 \) becomes 1 because any non-zero number raised to the power of zero is 1.
- The term \( 32^0 \) also becomes 1 for the same reason.
Now substitute these values into the expression:
\[ (-40)^0 + \frac{1}{2}(11 - 32^0) = 1 + \frac{1}{2}(11 - 1) \]
This simplifies to:
\[ 1 + \frac{1}{2}(11 - 1) = 1 + \frac{1}{2}(10) \]
Now calculate \( \frac{1}{2}(10) \):
\[ \frac{1}{2}(10) = 5 \]
So the entire expression simplifies to:
\[ 1 + 5 = 6 \]
Filling in the blanks, you have:
\[ (-40)^0 + \frac{1}{2}(11 - 32^0) = 1 + \frac{1}{2}(11 - 1) \]
Thus, the completed expression is:
\[ 1 + 1/2(10) \]
So the filled blanks would be:
- First blank: 1
- Second blank: 11
- Third blank: 1
The final expression looks like:
\[ 1 + \frac{1}{2}(11 - 1) \]
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