Questions: sqrt(9(x+5)^2)=3x+5 Which of the following are TRUE? Choose ALL the correct answers. (2x-1)/(x^2+4)=(2x-1)/x^2+(2x-1)/4 (2x-1)/(x^2+4)=(2x)/(x^2+4)-(1)/(x^2+4) (2x-1)/(x^2+4)=(2-1)/(x+4)=1/(x+4) Which angles have tangent equal to -sqrt(3) in [0,2pi] ? Choose all that apply.

\[ \boxed{\text{The equation is true for all } x.} \]

The first statement is:

\[ \frac{2x-1}{x^{2}+4}=\frac{2x-1}{x^{2}}+\frac{2x-1}{4} \]

To verify, we need to find a common denominator for the right side:

\[ \frac{2x-1}{x^{2}}+\frac{2x-1}{4} = \frac{(2x-1) \cdot 4 + (2x-1) \cdot x^2}{x^2 \cdot 4} \]

Simplifying the numerator:

\[ 4(2x-1) + x^2(2x-1) = 8x - 4 + 2x^3 - x^2 \]

The expression does not simplify to \((2x-1)\), so the statement is false.

The second statement is:

\[ \frac{2x-1}{x^{2}+4}=\frac{2x}{x^{2}+4}-\frac{1}{x^{2}+4} \]

Combine the right side:

\[ \frac{2x}{x^{2}+4} - \frac{1}{x^{2}+4} = \frac{2x - 1}{x^{2}+4} \]

This is equivalent to the left side, so the statement is true.

• First statement: False
• Second statement: True

\[ \boxed{\text{Second statement is true.}} \]

The tangent function is negative in the second and fourth quadrants. The reference angle for \(\tan(\theta) = \sqrt{3}\) is \(\frac{\pi}{3}\).

Thus, the angles in \([0, 2\pi]\) where \(\tan(\theta) = -\sqrt{3}\) are:

• In the second quadrant: \(\pi - \frac{\pi}{3} = \frac{2\pi}{3}\)
• In the fourth quadrant: \(2\pi - \frac{\pi}{3} = \frac{5\pi}{3}\)

\[ \boxed{\theta = \frac{2\pi}{3}, \frac{5\pi}{3}} \]