(35sqrt(x ^ 2 + 10x + 25))/(x + 5) ,but x > - 5

To simplify the expression \(\frac{35\sqrt{x^2 + 10x + 25}}{x + 5}\), we first need to simplify the expression under the square root.

Notice that \(x^2 + 10x + 25\) can be rewritten as a perfect square: \[ x^2 + 10x + 25 = (x + 5)^2 \]

Now we can substitute this back into the original expression: \[ \frac{35\sqrt{(x + 5)^2}}{x + 5} \]

Since \(x > -5\), \(\sqrt{(x + 5)^2} = x + 5\), because the square root of a square is the absolute value, and since \(x + 5\) is positive in this case, we can drop the absolute value.

Thus, we have: \[ \frac{35(x + 5)}{x + 5} \]

For \(x + 5 \neq 0\) (which is true since \(x > -5\)), we can cancel \(x + 5\): \[ 35 \]

So, the simplified expression is: \[ \boxed{35} \]