Question

is 1/2 (9 - 7^0) + (-29)^0 equivalent to 1/2 (9-1) - 1

Answer 1

To find out if \( \frac{1}{2}(9 - 7^0) + (-29)^0 \) is equivalent to \( \frac{1}{2}(9 - 1) - 1 \), let's simplify both expressions step by step.

### Step 1: Simplifying the left side

1. **Calculate \( 7^0 \)**:
\[
7^0 = 1
\]

2. **Substitute \( 7^0 \) into the expression**:
\[
\frac{1}{2}(9 - 1) + (-29)^0
\]

3. **Next, calculate \( (-29)^0 \)**:
\[
(-29)^0 = 1
\]

4. **Now plug in \( (-29)^0 \) into the expression**:
\[
\frac{1}{2}(9 - 1) + 1
\]

5. **Continue simplifying**:
\[
9 - 1 = 8 \quad \text{so,}
\]
\[
\frac{1}{2}(8) + 1 = 4 + 1 = 5
\]

### Step 2: Simplifying the right side

1. Start with the expression:
\[
\frac{1}{2}(9 - 1) - 1
\]

2. **Calculate \( 9 - 1 \)**:
\[
9 - 1 = 8
\]

3. **Now substitute this back into the expression**:
\[
\frac{1}{2}(8) - 1
\]

4. **Continue simplifying**:
\[
\frac{1}{2}(8) = 4 \quad \text{so,}
\]
\[
4 - 1 = 3
\]

### Conclusion

Now we have:

- Left side: \( 5 \)
- Right side: \( 3 \)

Since \( 5 \neq 3 \), the two expressions are not equivalent.

Thus, the answer is **no**, \( \frac{1}{2}(9 - 7^0) + (-29)^0 \) is **not** equivalent to \( \frac{1}{2}(9 - 1) - 1 \).