If fog(x) = (2x-1)/x and g(x) = 5x + 2 find; a.) f(x) b.) The truth set of g^-1(x) + 3 = gof(x). 2. Find the measure of the obtuse angle between the lines 3x - 5y + 2 = 0 and x + 4y + 5 = 0

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: