Question
Apply the Zero Power Rule to create an equivalent expression to 9⋅(−33) superscript 0+3/2(8−27 superscript 0)
.(2 points)
9⋅(−33) superscript 0+3/2(8−27 superscript 0)
= __ ⋅ __ + 3/2 ( __– __ )
.(2 points)
9⋅(−33) superscript 0+3/2(8−27 superscript 0)
= __ ⋅ __ + 3/2 ( __– __ )
Answers
GPT-4o mini
To apply the Zero Power Rule, we need to remember that any non-zero number raised to the power of zero equals one.
In the expression provided:
1. **Identify the parts affected by the zero exponent:**
- The term \((-33)^0\) will become \(1\) (since any non-zero number to the zero power is 1).
- The term \(27^0\) will also become \(1\) for the same reason.
Now let's rewrite the expression:
Starting with:
\[
9 \cdot (-33)^0 + \frac{3}{2}(8 - 27^0)
\]
Now apply the Zero Power Rule:
\[
9 \cdot 1 + \frac{3}{2}(8 - 1)
\]
Therefore, we can simplify it:
\[
9 + \frac{3}{2}(7)
\]
Now let's rewrite it in the requested format:
\[
9 \cdot 1 + \frac{3}{2}(8 - 1)
\]
Now, to fill in the blanks from your expression:
- For the first part (9 ⋅ __): **9 and 1**
- For the second part (+ 3/2): **3/2**
- For the third and fourth parts (8 – __): **8 and 1**
Thus, the final expression can be filled as follows:
9 ⋅ **1** + **3/2** ( **8** – **1**)
In the expression provided:
1. **Identify the parts affected by the zero exponent:**
- The term \((-33)^0\) will become \(1\) (since any non-zero number to the zero power is 1).
- The term \(27^0\) will also become \(1\) for the same reason.
Now let's rewrite the expression:
Starting with:
\[
9 \cdot (-33)^0 + \frac{3}{2}(8 - 27^0)
\]
Now apply the Zero Power Rule:
\[
9 \cdot 1 + \frac{3}{2}(8 - 1)
\]
Therefore, we can simplify it:
\[
9 + \frac{3}{2}(7)
\]
Now let's rewrite it in the requested format:
\[
9 \cdot 1 + \frac{3}{2}(8 - 1)
\]
Now, to fill in the blanks from your expression:
- For the first part (9 ⋅ __): **9 and 1**
- For the second part (+ 3/2): **3/2**
- For the third and fourth parts (8 – __): **8 and 1**
Thus, the final expression can be filled as follows:
9 ⋅ **1** + **3/2** ( **8** – **1**)