Question

Apply to zero power roll to create an equivalent expression to nine times in parentheses -33 power of 0+3 over two and parentheses 8-27 power of zero so we need the answer for that equation

Answer 1

To solve the expression \( 9 \times \left(-33^0 + \frac{3}{2}\right) \times (8 - 27^0) \), we first need to evaluate the components step by step.

1. **Calculate \( -33^0 \)**:
Any non-zero number raised to the power of 0 is 1. Therefore,
\[
-33^0 = -1
\]

2. **Calculate \( 27^0 \)**:
Similarly,
\[
27^0 = 1
\]

3. **Now substitute these values back into the expression**:
\[
9 \times \left(-1 + \frac{3}{2}\right) \times (8 - 1)
\]

4. **Calculate \( -1 + \frac{3}{2} \)**:
To combine these, convert -1 to a fraction:
\[
-1 = -\frac{2}{2} \quad \text{so} \quad -1 + \frac{3}{2} = -\frac{2}{2} + \frac{3}{2} = \frac{1}{2}
\]

5. **Now calculate \( 8 - 1 \)**:
\[
8 - 1 = 7
\]

6. **Substitute these results back into the expression**:
\[
9 \times \frac{1}{2} \times 7
\]

7. **Now perform the calculations**:
- First, calculate \( 9 \times \frac{1}{2} = \frac{9}{2} \).
- Then, calculate \( \frac{9}{2} \times 7 = \frac{9 \times 7}{2} = \frac{63}{2} \).

Thus, the final result of the expression is:
\[
\frac{63}{2}
\]

That's the equivalent expression and final answer.