1. Solve the equation $3|2t-1|-4=16$ for $t$. Start by isolating the absolute value: $$3|2t-1|=20$$ Divide both sides by 3: $$|2t-1|=\frac{20}{3}$$ This gives two cases: Case 1: $2t-1=\frac{20}{3}$ $$2t=\frac{20}{3}+1=\frac{23}{3} \Rightarrow t=\frac{23}{6}$$ Case 2: $2t-1=-\frac{20}{3}$ $$2t=-\frac{20}{3}+1=-\frac{17}{3} \Rightarrow t=-\frac{17}{6}$$ 2. Solve the inequality $$\frac{x-3}{x-2} < \frac{x+1}{x+4}, x \neq 2, x \neq -4$$ Bring terms to one side: $$\frac{x-3}{x-2} - \frac{x+1}{x+4} < 0$$ Find common denominator $(x-2)(x+4)$: $$\frac{(x-3)(x+4)-(x+1)(x-2)}{(x-2)(x+4)} < 0$$ Expand: $$(x^2 +4x -3x -12) - (x^2 -2x + x -2) = (x^2 + x -12) - (x^2 - x -2) = x^2 + x -12 - x^2 + x + 2 = 2x -10$$ So inequality: $$\frac{2x -10}{(x-2)(x+4)} < 0$$ Factor numerator: $$\frac{2(x-5)}{(x-2)(x+4)} < 0$$ Find intervals considering critical points $-4, 2, 5$, test signs: Solution intervals where fraction is negative: $$(-\infty,-4) \cup (2,5)$$ 3. Graph of $f(x) = 4x^2 -1$ and line $y=3x+2$ Set equal to find intersection: $$4x^2 -1 = 3x + 2$$ $$4x^2 - 3x -3 = 0$$ Use quadratic formula: $$x = \frac{3 \pm \sqrt{9 + 48}}{8} = \frac{3 \pm \sqrt{57}}{8}$$ These are the intersection points. 4. Solve compound inequality: $$-2 > x^2 + x > 6$$ No real $x$ satisfies since $x^2 + x$ cannot be simultaneously less than $-2$ and greater than $6$. 5. For $$1 < x + \frac{6}{x} > 5$$ Rewrite as two inequalities: $$1 < x + \frac{6}{x} \quad \text{and} \quad x + \frac{6}{x} > 5$$ But this is contradictory; likely question needs clarification. 6. Inequality: $$\frac{x+1}{x-2} \leq \frac{x}{x+3}$$ Bring terms to one side: $$\frac{x+1}{x-2} - \frac{x}{x+3} \leq 0$$ Common denominator $(x-2)(x+3)$: $$\frac{(x+1)(x+3) - x(x-2)}{(x-2)(x+3)} \leq 0$$ Expand numerator: $$(x^2 +3x + x +3) - (x^2 -2x) = (x^2 +4x +3) - (x^2 -2x) = 6x + 3$$ Inequality: $$\frac{6x + 3}{(x-2)(x+3)} \leq 0$$ Critical points at $x=-\frac{1}{2}, 2, -3$, test intervals to find solution. 7. For $f(x) = x^2 + 4x$, domain $0 > x > 4$ is likely typo; assume domain $0 \leq x \leq 4$. Compute range: $$f(x) = x^2 +4x = (x+2)^2 -4$$ Minimum at $x=-2$ not in domain. Check endpoints: $$f(0) = 0$$ $$f(4) = 16 +16 = 32$$ Since function is increasing in $[0,4]$, range is $[0,32]$. 8. For $g(x)= x^2 -4x$, domain $-2 > x \geq 3$ likely typo; if domain $-2 \leq x \leq 3$. Plot the function accordingly. 9. Given: $$(2x+3)(x-5) = 0$$ a) Expand: $$2x^2 -10x +3x -15 = 2x^2 -7x -15 = 0$$ b) Vertex: $$x = -\frac{b}{2a} = -\frac{-7}{4} = \frac{7}{4} = 1.75$$ $$y = 2(1.75)^2 -7(1.75) -15 = 2(3.0625) - 12.25 -15 = 6.125 -27.25 = -21.125$$ c) Plot the parabola. 10. For $f(x) = 4x - x^2$ (a) Domain: $x \in \mathbb{R}$ (b) Range: Vertex at $x=2$, $$f(2) = 4(2) - 4 = 8 -4 = 4$$ Parabola opens downward, so range $(-\infty, 4]$ 11. For $f(x) = x^2 + 2x + 5$, domain $-1 \leq x < 3$ Complete the square: $$f(x) = (x+1)^2 +4$$ Minimum at $x = -1$, $f(-1)=4$ Maximum at $x \to 3$, value near $f(3)=9 +6 +5=20$ Range $[4, 20)$. 12. Compose functions: $f(x)= 2x - x^2$, $g(x)=1 - x^{-2}$ Calculate $f^{-1} \circ g$ and $f \circ g$ accordingly. 13. For $f(x)= x+2$, $g(x) = x^2 + 2x + 3$ a) $(g \circ f)(x) = g(f(x)) = (x+2)^2 + 2(x+2) +3 = x^2 + 4x +4 + 2x +4 +3 = x^2 + 6x + 11$ b) Evaluate at $x = -\frac{1}{4}$: $(-\frac{1}{4})^2 + 6(-\frac{1}{4}) +11 = \frac{1}{16} - \frac{3}{2} +11 = \frac{1}{16} - \frac{24}{16} + \frac{176}{16} = \frac{153}{16} = 9.5625$ 14. For perpendicular lines $y = 3x - 6$ and $y = mx + 7$: Slope product $3 \cdot m = -1$, so $$m = -\frac{1}{3}$$ Plot both lines. 15. Slope between $A(3,4)$ and $B(6,1)$: $$m = \frac{1 -4}{6 -3} = \frac{-3}{3} = -1$$ Equation: $$y - 4 = -1(x - 3) \Rightarrow y = -x + 7$$ Plot line. 16. For $g(x) = x - \frac{1}{x}$, $x \neq 0$, $f(x) = x + 3$, $h(x) = 2x - 4$ Find inverses and compositions as requested. Final answers are provided through explanations and detailed intermediate steps.