Using the exponent properties, which of the following expressions is equivalent to (2–√)3×2−12

?

(1 point)
Responses

22
2 squared

2–√
square root of 2

212
2 raised to the 1 half power

2
2
Question 2
Which of the following expressions is equivalent to 216√32713
when applying the exponent properties?

(1 point)
Responses

2
2

12
1 half

18
18

8

User Icon for Bot GPT 4-o mini Bot GPT 4-o mini answered
1 month ago

Let's analyze the given expressions step by step.

**Question 1: Simplifying \((2^{-\sqrt{}})^3 \times 2^{-12}\)**

1. **Apply exponent multiplication** to the first term:
\[
(2^{-\sqrt{}})^3 = 2^{-3\sqrt{}}
\]

2. **Combine the two terms using exponent addition**:
\[
2^{-3\sqrt{}} \times 2^{-12} = 2^{-3\sqrt{} - 12}
\]

3. Since we have not been given a specific value for \(\sqrt{}\), we can't simplify further. However, we need to analyze the answer choices:
- \(22\) (or \(2^2\)) = \(4\)
- \(2^{-\sqrt{}}\) (not directly obvious what this evaluates to)
- \(212\) (or \(2^{1/2} = \sqrt{2}\))
- \(2\) (or \(2^1 = 2\))

Thus, based on the work we can simplify to an expression in terms of \(2\) but cannot identify one of the options precisely without more information.

**Question 2: Simplifying \(2^{16} \sqrt{3^{2713}}\)**

First, rewrite the square root term using exponent properties:
\[
\sqrt{3^{2713}} = (3^{2713})^{1/2} = 3^{2713/2}
\]

Now combine this with the \(2^{16}\):
\[
2^{16} \times 3^{2713/2}
\]

We want to express this in a concise way:
- Choose which parts we want to keep.

The answer options provided are:
- \(2\)
- \(12\)
- \(18\)
- \(8\)

However, we need to more closely examine based on the values provided. None of the options appear to simplify directly to an exponent without approximation or further manipulation.

So, without precise numerical simplification or values, it appears as though the answers do not fully align. They require either numerical context or address in a broader algebraic method to evaluate.

Feel free to provide additional details or context, so I can further assist!