Step 1: Find $f(x)$. We are given $f(g(x)) = \frac{2x-1}{x}$ and $g(x) = 5x+2$. Let $y = g(x)$. Then $y = 5x+2$. To find $x$ in terms of $y$, we rearrange the equation: $$y - 2 = 5x$$ $$x = \frac{y-2}{5}$$ Now substitute $x$ in the expression for $f(g(x))$: $$f(y) = \frac{2\left(\frac{y-2}{5}\right)-1}{\frac{y-2}{5}}$$ $$f(y) = \frac{\frac{2y-4}{5}-1}{\frac{y-2}{5}}$$ To simplify the numerator, find a common denominator: $$f(y) = \frac{\frac{2y-4-5}{5}}{\frac{y-2}{5}}$$ $$f(y) = \frac{\frac{2y-9}{5}}{\frac{y-2}{5}}$$ $$f(y) = \frac{2y-9}{y-2}$$ Replace $y$ with $x$ to find $f(x)$: $$f(x) = \frac{2x-9}{x-2}$$ *a) $\boxed{f(x) = \frac{2x-9}{x-2}}$ Step 2: Find $g^{-1}(x)$. Let $y = g(x) = 5x+2$. To find the inverse, swap $x$ and $y$: $$x = 5y+2$$ $$x-2 = 5y$$ $$y = \frac{x-2}{5}$$ So, $g^{-1}(x) = \frac{x-2}{5}$. Step 3: Find $g \circ f(x)$. We have $f(x) = \frac{2x-9}{x-2}$ and $g(x) = 5x+2$. $$g(f(x)) = g\left(\frac{2x-9}{x-2}\right)$$ $$g(f(x)) = 5\left(\frac{2x-9}{x-2}\right) + 2$$ $$g(f(x)) = \frac{5(2x-9)}{x-2} + \frac{2(x-2)}{x-2}$$ $$g(f(x)) = \frac{10x-45+2x-4}{x-2}$$ $$g(f(x)) = \frac{12x-49}{x-2}$$ Step 4: Solve the equation $g^{-1}(x) + 3 = g \circ f(x)$. Substitute the expressions for $g^{-1}(x)$ and $g \circ f(x)$: $$\frac{x-2}{5} + 3 = \frac{12x-49}{x-2}$$ Simplify the left side: $$\frac{x-2}{5} + \frac{15}{5} = \frac{x-2+15}{5} = \frac{x+13}{5}$$ Now the equation is: $$\frac{x+13}{5} = \frac{12x-49}{x-2}$$ Cross-multiply: $$(x+13)(x-2) = 5(12x-49)$$ $$x^2 - 2x + 13x - 26 = 60x - 245$$ $$x^2 + 11x - 26 = 60x - 245$$ Rearrange into a quadratic equation $ax^2+bx+c=0$: $$x^2 + 11x - 60x - 26 + 245 = 0$$ $$x^2 - 49x + 219 = 0$$ Use the quadratic formula $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$: $$x = \frac{-(-49) \pm \sqrt{(-49)^2 - 4(1)(219)}}{2(1)}$$ $$x = \frac{49 \pm \sqrt{2401 - 876}}{2}$$ $$x = \frac{49 \pm \sqrt{1525}}{2}$$ Simplify the square root: $\sqrt{1525} = \sqrt{25 \times 61} = 5\sqrt{61}$. $$x = \frac{49 \pm 5\sqrt{61}}{2}$$ The values of $x$ must not make any denominator zero. For $f(x)$ and $g(f(x))$, $x-2 \neq 0$, so $x \neq 2$. The solutions found are not equal to 2. *b) The truth set is $\boxed{\left\{\frac{49 - 5\sqrt{61}}{2}, \frac{49 + 5\sqrt{61}}{2}\right\}}$ Step 5: Find the slopes of the lines. The first line is $3x - 5y + 2 = 0$. Rearrange to slope-intercept form $y = mx+c$: $$5y = 3x + 2$$ $$y = \frac{3}{5}x + \frac{2}{5}$$ The slope of the first line is $m_1 = \frac{3}{5}$. The second line is $x + 4y + 5 = 0$. Rearrange to slope-intercept form $y = mx+c$: $$4y = -x - 5$$ $$y = -\frac{1}{4}x - \frac{5}{4}$$ The slope of the second line is $m_2 = -\frac{1}{4}$. Step 6: Calculate the angle between the lines. The formula for the angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is: $$\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$$ Substitute the slopes: $$\tan \theta = \left| \frac{\frac{3}{5} - \left(-\frac{1}{4}\right)}{1 + \left(\frac{3}{5}\right)\left(-\frac{1}{4}\right)} \right|$$ $$\tan \theta = \left| \frac{\frac{3}{5} + \frac{1}{4}}{1 - \frac{3}{20}} \right|$$ Calculate the numerator: $$\frac{3}{5} + \frac{1}{4} = \frac{12}{20} + \frac{5}{20} = \frac{17}{20}$$ Calculate the denominator: $$1 - \frac{3}{20} = \frac{20}{20} - \frac{3}{20} = \frac{17}{20}$$ Substitute these values back into the formula: $$\tan \theta = \left| \frac{\frac{17