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$\\text{(2) 1 on die-1 and 3 on die-2.} \\;\\;\\;P(\\text{1 die-1}) \\,=\\,\\frac{1}{6} \\;\\;\\;P(\\text{3 die-2}) \\,=\\,\\frac{1}{6} P(\\text{1 die-1 }\\wedge\\text{ 3 die-2}) \\:=\\:\\frac{1}{6}\\,\\cdot\\frac{1}{6}\\:=\\:\\frac{1}{36 }$ $V=\\frac{\\pi h}{r-R}\\int_R\\,^{r}u^2\\,du=\\frac{\\pi h}{3(r-R)}\\[u^3\\]_R^r=\\frac{\\pi h}{3(r-R)}\$r^3-R^3\$=\\frac{\\pi h}{3}\$r^2+rR+R^2\$$ $\\begin{align*} (A-B) \\cup (B-C) \\ &=\\ (A \\cap B^c) \\cup (B \\cap C^c) \\quad &\\text{(definition of difference)}\\\\ &=\\ (A\\cup B)\\cap (B^c\\cup B) \\cap (A\\cup C^c)\\cap (B^c\\cup C^c) \\quad &\\text{(distributive law)}\\\\ &=\\ (A\\cup B)\\cap\\left((A\\cup C^c)\\cap C^c\\right) \\quad &\\text{($A \\cup A^c = X$, our observation, and associativity)}\\\\ &=\\ (A\\cup B)\\cap C^c \\quad &\\text{($(A\\cup B)\\cap B = B$)}. \\end{align*}$ $X_L=2\\pi fL=2\\pi\\frac{500}{\\pi}\\:\\text{ \\frac{1}{s}}\\cdot0.9\\text{ H}=9000\\text{ \\Omega}$ $y(x)=\\sum\\frac{{(-1)}^n}{(2n+1)!}x^n$ $V=\\int_0\\,^2\\frac{1}{2}\\pi \$\\frac{2-x}{2}\$^2\\,dx=\\frac{\\pi}{8}\\int_0\\,^2(2-x)^2\\,dx=\\frac{\\pi}{8}\\int_0\\,^2 u^2\\,du$

$\\int \\frac{6 - 5t + t^2} {6}dt = \\int (1 - \\frac{5}{6}t + \\frac{1}{6}t^2) dt = t - \\frac{5}{6}\\cdot \\frac{1}{2}t^2 + \\frac{1}{6}\\cdot \\frac{1}{3}t^3 + C = t - \\frac{5}{12}t^2 + \\frac{1}{18}t^3 + C$

$\\lim_{u\\to0}\\,\\frac{\\cos(u)\\,-\\,1}{u} \\\\ =\\,\\lim_{u\\to0}\\,\$\\frac{\\cos(u)\\,-\\,1}{u}\\,\\cdot\\,\\frac{\\cos(u)\\,+\\,1}{\\cos(u)\\,+\\,1 }\$ \\\\ \\,=\\,\\lim_{u\\to0}\\,\\frac{\\cos^2(u)\\,-\\,1}{u\$\\cos(u)\\,+\\,1\$} \\\\ =\\,\\lim_{u\\to0}\\,\\frac{-\\sin^2(u)}{u\$\\cos(u)\\,+\\,1\$} \\\\ =\\,\\lim_{u\\to0}\\,-\$\\frac{\\sin(u)}{u}\\,\\cdot\\,\\frac{\\sin(u)}{\\cos(u) \\,+\\,1}\$ \\\\ =\\,-1\\,\\cdot\\,\\frac{0}{1\\,+\\,1}\\,=\\,0$

$\\lim_{x\\to a}\\,\\frac{\\cos(x)\\,-\\,\\cos(a)}{x\\,-\\,a} \\\\ u\\,=\\,x\\,-\\,a \\\\ \\lim_{u\\to0}\\,\\frac{\\cos(u\\,+\\,a)\\,-\\,\\cos(a)}{u} \\\\ =\\,\\lim_{u\\to0}\\,\\frac{\\cos(u)\\cos(a)\\,-\\,\\sin(u)\\sin(a)\\,-\\,\\cos(a)}{u} \\\\ =\\,\\lim_{u\\to0}\\,\\cos(a)\\frac{\$\\cos(u)\\,-\\,1\$}{u}\\,-\\,\\lim_{u\\to0}\\,\\sin(a)\\frac{\\sin(u)}{u} \\\\ =\\,\\cos(a)\\,\\cdot\\,0\\,-\\,\\sin(a)\\,\\cdot\\,1\\,=\\,-\\sin(a)$

$\\begin{Bmatrix} \\frac{2}{600}\\text{ of the time, he wins \\$250} \\\\ \\\\ \\\\ \\frac{1}{600}\\text{ of the time, he wins \\$125} \\\\ \\\\ \\\\ \\frac{25}{600}\\text{ of the time, he wins \\$75} \\\\ \\\\ \\\\ \\frac{50}{600}\\text{ of the time, he wins \\$10} \\\\ \\\\ \\\\ \\frac{100}{600}\\text{ of the time, he wins \\$5} \\\\ \\\\ \\\\ \\frac{600}{600}\\text{ of the time, he pays \\$5} \\end{Bmatrix}$ $\\tan\$\\frac{\\theta}{2}\$\\,=\\,\\frac{\\sin(\\theta)}{1 \\,+\\,\\cos(\\theta)} \\\\ \\pm\\sqrt{\\frac{1}{\\cos^2\$\\frac{\\theta}{2}\$}\\,-\\,1}\\,=\\,\\frac{\\sin(\\theta)}{1\\,+\\,\\cos(\\theta)}\\, \\leftarrow\\,\\sin^2(x)\\,+\\,\\cos^2(x)\\,=\\,1\\,\\Leftri ghtarrow\\,\\tan^2(x)\\,+\\,1\\,=\\,\\frac{1}{\\cos^2(x)}\\ \\ \\frac{1}{\\cos^2\$\\frac{\\theta}{2}\$}\\,-\\,1\\,=\\,\\frac{\\sin^2(\\theta)}{1\\,+\\,2\\cos(\\theta)\\ ,+\\,\\cos^2(\\theta)} \\\\ \\frac{1}{\\cos^2\$\\frac{\\theta}{2}\$}\\,=\\,\\frac{\\si n^2(\\theta)}{1\\,+\\,2\\cos(\\theta)\\,+\\,\\cos^2(\\theta )}\\,+\\,1 \\\\ \\frac{1}{\\cos^2\$\\frac{\\theta}{2}\$}\\,=\\,\\frac{\\si n^2(\\theta)\\,+\\,1\\,+\\,2cos(\\theta)\\,+\\,\\cos^2(\\the ta)}{(1\\,+\\,\\cos(\\theta))^2} \\\\ \\frac{1}{\\cos^2\$\\frac{\\theta}{2}\$}\\,=\\,\\frac{2(1 \\,+\\,\\cos(\\theta))}{(1\\,+\\,\\cos(\\theta))^2}\\,=\\,\\f rac{2}{1\\,+\\,\\cos(\\theta)} \\\\ \\cos^2\$\\frac{\\theta}{2}\$\\,=\\,\\frac{1\\,+\\,\\cos(\\t heta)}{2} \\\\ \\cos\$\\frac{\\theta}{2}\$\\,=\\,\\pm\\sqrt{\\frac{1\\,+\\, \\cos(\\theta)}{2}}$

$\\text{ 1 1} \\\\ \\text{ 1 \\frac32 1} \\\\ \\text{ 1 \\frac64 \\frac64 1} \\\\ \\text{ 1 \\frac{10}{7} \\frac{10}{6} \\frac{10}{7} 1 \\\\ \\text{1 \\frac{15}{11} \\frac{15}{9} \\frac{15}{9} \\frac{15}{11} 1$

$8a^3+27=0 (2a)^3+3^3=0 (2a+3)(4a^2-6a+9)=0 2a+3=0 2a=-3 a_1=\\frac{-3}{2} 4a^2-6a+9=0 a_{2,3}=\\frac{6\\pm i \\sqrt{108}}{8} = \\frac{6\\pm i \\sqrt{36*3}}{8} = \\frac{6\\pm 6i\\sqrt{3}}{8} = \\frac{3(1\\pm i \\sqrt{3)}}{4}$

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