Is the equation (7^(-4)*7^(3))/(\sqrt(49)) true or false? select the response that correctly answers the question and provides the appropriate justification

Let's analyze the equation step-by-step to determine if it is true or false.

We start with the expression:

\[
\frac{7^{-4} \cdot 7^{3}}{\sqrt{49}}
\]

### Step 1: Simplify the denominator
We know that:

\[
\sqrt{49} = 7
\]

### Step 2: Simplify the numerator
Using the product property of exponents (which states that \(a^m \cdot a^n = a^{m+n}\)), we simplify the numerator:

\[
7^{-4} \cdot 7^{3} = 7^{-4 + 3} = 7^{-1}
\]

### Step 3: Combine the numerator and the denominator
Now we have:

\[
\frac{7^{-1}}{7}
\]

Using the quotient property of exponents (which states that \(\frac{a^m}{a^n} = a^{m-n}\)), we can simplify:

\[
\frac{7^{-1}}{7^1} = 7^{-1 - 1} = 7^{-2}
\]

### Step 4: Result
The final result \(7^{-2}\) means \(\frac{1}{7^2} = \frac{1}{49}\).

### Conclusion
The original equation simplifies to \(7^{-2}\), which is a valid expression.

From the provided options, the correct response is:

**True because the quotient property gives a final exponent of -2.**

However, since none of the options mention this result, the closest accurate justification is that it simplifies through the properties of exponents correctly and yields a defined value. The assertion that it leads to a final exponent of 1 is incorrect; hence, the best answer considering the options would be:

**False when applying the quotient property the final exponent is not 1.**