Identify the type of function.

This is a constant function, which is a type of polynomial. Polynomial functions are defined for all real numbers.

The domain of $f(x)=-5$ is $\boxed{\mathbb{R}}$.

2. $f(x)=3x-4$

This is a linear function, which is a type of polynomial. Polynomial functions are defined for all real numbers.

The domain of $f(x)=3x-4$ is $\boxed{\mathbb{R}}$.

3. $f(x)=-3x^4-7x^2+2x+5$

This is a polynomial function. Polynomial functions are defined for all real numbers.

The domain of $f(x)=-3x^4-7x^2+2x+5$ is $\boxed{\mathbb{R}}$.

4. $f(x)=\frac{2x+1}{(x-1)^2-4}$

For a rational function, the denominator cannot be equal to zero.
Set the denominator to zero and solve for $x$:

Take the square root of both sides:

This gives two possible values for $x$:

So, $x \ne 3$ and $x \ne -1$.

The domain of $f(x)=\frac{2x+1}{(x-1)^2-4}$ is $\boxed{\mathbb{R} \setminus \{-1, 3\}}$.

5. $f(x)=\sqrt{-4x(x+2)}$

For a square root function, the expression under the square root must be greater than or equal to zero.

Set $-4x(x+2) = 0$. This gives $x=0$ or $x+2=0 \implies x=-2$.
The critical points are $x=-2$ and $x=0$.

We test the intervals $(-\infty, -2]$, $[-2, 0]$, and $[0, \infty)$.

• For $x < -2$ (e.g., $x=-3$): $-4(-3)(-3+2) = 12(-1) = -12 < 0$.
• For $-2 \le x \le 0$ (e.g., $x=-1$): $-4(-1)(-1+2) = 4(1) = 4 \ge 0$.
• For $x > 0$ (e.g., $x=1$): $-4(1)(1+2) = -4(3) = -12 < 0$. The inequality $-4x(x+2) \ge 0$ is satisfied when $-2 \le x \le 0$.

The domain of $f(x)=\sqrt{-4x(x+2)}$ is $\boxed{[-2, 0]}$.

6. $f(x)=\frac{-6x}{(x+3)(4-x)}$

For a rational function, the denominator cannot be equal to zero.
Set the denominator to zero and solve for $x$:

This gives two possible values for $x$:

So, $x \ne -3$ and $x \ne 4$.

The domain of $f(x)=\frac{-6x}{(x+3)(4-x)}$ is $\boxed{\mathbb{R} \setminus \{-3, 4\}}$.

7. $f(x)=\frac{x+3}{x^2-7x}$

For a rational function, the denominator cannot be equal to zero.
Set the denominator to zero and solve for $x$:

This gives two possible values for $x$:

So, $x \ne 0$ and $x \ne 7$.

The domain of $f(x)=\frac{x+3}{x^2-7x}$ is $\boxed{\mathbb{R} \setminus \{0, 7\}}$.

8. $f(x)=\sqrt{\frac{2x-5}{3-x}}$

For a square root of a rational function, two conditions must be met:

1. The expression under the square root must be non-negative: $\frac{2x-5}{3-x} \ge 0$.
2. The denominator cannot be zero: $3-x \ne 0$.

Find the critical points by setting the numerator and denominator to zero:
$2x-5 = 0 \implies 2x = 5 \implies x = \frac{5}{2} = 2.5$
$3-x = 0 \implies x = 3$

We test the intervals $(-\infty, 2.5]$, $[2.5, 3)$, and $(3, \infty)$. Note that $x=3$ is excluded due to the denominator.

• For $x < 2.5$ (e.g., $x=0$): $\frac{2(0)-5}{3-0} = \frac{-5}{3} < 0$.
• For $2.5 \le x < 3$ (e.g., $x=2.8$): $\frac{2(2.8)-5}{3-2.8} = \frac{5.6-5}{0.2} = \frac{0.6}{0.2} = 3 \ge 0$.
• For $x > 3$ (e.g., $x=4$): $\frac{2(4)-5}{3-4} = \frac{8-5}{-1} = \frac{3}{-1} = -3 < 0$. The inequality $\frac{2x-5}{3-x} \ge 0$ is satisfied when $2.5 \le x < 3$.

The domain of $f(x)=\sqrt{\frac{2x-5}{3-x}}$ is $\boxed{\left[\frac{5}{2}, 3\right)}$.

9. $f(x)=\frac{x+3}{x^2+1}$

For a rational function, the denominator cannot be equal to zero.
Set the denominator to zero and solve for $x$:

There are no real solutions for $x^2 = -1$, as the square of any real number is non-negative. Therefore, the denominator $x^2+1$ is never zero for any real $x$.

The domain of $f(x)=\frac{x+3}{x^2+1}$ is $\boxed{\mathbb{R}}$.

10. $f(x)=\frac{x+2}{\sqrt{(6-2x)(x+9)}}$

For a square root in the denominator, the expression under the square root must be strictly greater than zero (cannot be zero because it's in the denominator).

Set $(6-2x)(x+9) = 0$. This gives:
$6-2x = 0 \implies 2x = 6 \implies x = 3$
$x+9 = 0 \implies x = -9$
The critical points are $x=-9$ and $x=3$.

We test the intervals $(-\infty, -9)$, $(-9, 3)$, and $(3, \infty)$.

• For $x < -9$ (e.g., $x=-10$): $(6-2(-10))(-10+9) = (6+20)(-1) = 26(-1) = -26 < 0$.
• For $-9 < x < 3$ (e.g., $x=0$): $(6-2(0))(0+9) = (6)(9) = 54 > 0$.
• For $x > 3$ (e.g., $x=4$): $(6-2(4))(4+9) = (6-8)(13) = (-2)(13) = -26 < 0$. The inequality $(6-2x)(x+9) > 0$ is satisfied when $-9 < x < 3$.

The domain of $f(x)=\frac{x+2}{\sqrt{(6-2x)(x+9)}}$ is $\boxed{(-9, 3)}$.