Datasets:

chenhegu
/

MATH

Dataset card Data Studio Files Files and versions Community

Split (2)

train · 7.5k rows

messages listlengths 2 2
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $\\omega$ be a complex number such that $\\omega^7 = 1$ and $\\omega \\ne 1.$ Compute\n\\[\\omega^{16} + \\omega^{18} + \\omega^{20} + \\dots + \\omega^{54}.\\]", "role": "user" }, { "content": "First, we can take out a factor of $\\omega^{16}$:\n\\[\\omega^{16} + \\omega^{18} + \\omega^{20} + \\dots + \\omega^{54} = \\omega^{16} (1 + \\omega^2 + \\omega^4 + \\dots + \\omega^{38}).\\]By the formula for a geometric series,\n\\[\\omega^{16} (1 + \\omega^2 + \\omega^4 + \\dots + \\omega^{38}) = \\omega^{16} \\cdot \\frac{1 - \\omega^{40}}{1 - \\omega^2}.\\](Note that this expression is valid, because $\\omega \\neq 1$ and $\\omega \\neq -1.$)\n\nSince $\\omega^7 = 1,$\n\\[\\omega^{16} \\cdot \\frac{1 - \\omega^{40}}{1 - \\omega^2} = \\omega^2 \\cdot \\frac{1 - \\omega^5}{1 - \\omega^2} = \\frac{\\omega^2 - \\omega^7}{1 - \\omega^2} = \\frac{\\omega^2 - 1}{1 - \\omega^2} = \\boxed{-1}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The sequence $(a_n)$ satisfies\n\\[a_1 + a_2 + a_3 + \\dots + a_n = n^2 a_n\\]for all $n \\ge 2.$ If $a_{63} = 1,$ find $a_1.$", "role": "user" }, { "content": "From $a_1 + a_2 + a_3 + \\dots + a_n = n^2 a_n,$\n\\[(n^2 - 1) a_n = a_1 + a_2 + \\dots + a_{n - 2} + a_{n - 1}.\\]Likewise,\n\\[((n - 1)^2 - 1) a_{n - 1} = a_1 + a_2 + \\dots + a_{n - 2}.\\]Subtracting these equations, we get\n\\[(n^2 - 1) a_n - ((n - 1)^2 - 1) a_{n - 1} = a_{n - 1},\\]so\n\\[(n^2 - 1) a_n = (n - 1)^2 a_{n - 1}.\\]Then $(n - 1)(n + 1) a_n = (n - 1)^2 a_{n - 1},$ so\n\\[a_n = \\frac{n - 1}{n + 1} \\cdot a_{n - 1}\\]for all $n \\ge 2.$\n\nTherefore,\n\\begin{align}\na_n &= \\frac{n - 1}{n + 1} \\cdot a_{n - 1} \\\\\n&= \\frac{n - 1}{n + 1} \\cdot \\frac{n - 2}{n} \\cdot a_{n - 2} \\\\\n&= \\frac{n - 1}{n + 1} \\cdot \\frac{n - 2}{n} \\cdot \\frac{n - 3}{n - 1} \\cdot a_{n - 3} \\\\\n&= \\dotsb \\\\\n&= \\frac{n - 1}{n + 1} \\cdot \\frac{n - 2}{n} \\cdot \\frac{n - 3}{n - 1} \\dotsb \\frac{2}{4} \\cdot \\frac{1}{3} \\cdot a_1 \\\\\n&= \\frac{2a_1}{n(n + 1)}.\n\\end{align}We are told that $a_{63} = 1,$ so\n\\[\\frac{2a_1}{63 \\cdot 64} = 1.\\]Thus, $a_1 = \\boxed{2016}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Suppose that $f(x)$ and $g(x)$ are functions which satisfy $f(g(x)) = x^2$ and $g(f(x)) = x^3$ for all $x \\ge 1.$ If $g(16) = 16,$ then compute $[g(4)]^3.$", "role": "user" }, { "content": "Setting $x = 4$ in $f(g(x)) = x^2,$ we get\n\\[f(g(4)) = 16.\\]Then\n\\[g(f(g(4)) = g(16) = 16.\\]But $g(f(g(4)) = [g(4)]^3,$ so $[g(4)]^3 = \\boxed{16}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $z$ be a nonreal complex number such that $\|z\| = 1.$ Find the real part of $\\frac{1}{1 - z}.$", "role": "user" }, { "content": "Let $z = x + yi,$ where $x$ and $y$ are real numbers. Then from the condition $\|z\| = 1,$ $\\sqrt{x^2 + y^2} = 1,$ so $x^2 + y^2 = 1.$\n\nNow,\n\\begin{align}\n\\frac{1}{1 - z} &= \\frac{1}{1 - x - yi} \\\\\n&= \\frac{1 - x + yi}{(1 - x - yi)(1 - x + yi)} \\\\\n&= \\frac{1 - x + yi}{(1 - x)^2 + y^2} \\\\\n&= \\frac{1 - x + yi}{1 - 2x + x^2 + y^2} \\\\\n&= \\frac{1 - x + yi}{2 - 2x}.\n\\end{align}The real part of this complex number is $\\frac{1 - x}{2 - 2x} = \\frac{1 - x}{2(1 - x)} = \\boxed{\\frac{1}{2}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Compute $(1 + i)^4.$", "role": "user" }, { "content": "We have that\n\\[(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i,\\]so $(1 + i)^4 = (2i)^2 = 4i^2 = \\boxed{-4}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The graph of the rational function $\\frac{q(x)}{2x^5+x^4-7x^2+1}$ has a horizontal asymptote. What is the largest possible degree of $q(x)$?", "role": "user" }, { "content": "For the given function to have a horizontal asymptote, it can't go to infinity as $x$ goes to infinity. This is only possible if the numerator has the same or smaller degree than the denominator. Since the denominator has degree 5, the largest possible degree of $q(x)$ that will allow the function to have a horizontal asymptote is $\\boxed{5}.$\n\nWe note that 5 is in fact possible, because if we take $q(x) = x^5,$ then the rational function has horizontal asymptote $y = \\frac 12.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find all the integer roots of\n\\[x^3 - 3x^2 - 13x + 15 = 0.\\]Enter all the integer roots, separated by commas.", "role": "user" }, { "content": "By the Integer Root Theorem, the possible integer roots are all the divisors of 15 (including negative divisors), which are $-15,$ $-5,$ $-3,$ $-1,$ $1,$ $3,$ $5,$ and $15.$ Checking, we find that the only integer roots are $\\boxed{-3,1,5}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The graph of $y = f(x)$ is shown below.\n\n[asy]\nunitsize(0.5 cm);\n\nreal func(real x) {\n real y;\n if (x >= -3 && x <= 0) {y = -2 - x;}\n if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}\n if (x >= 2 && x <= 3) {y = 2(x - 2);}\n return(y);\n}\n\nint i, n;\n\nfor (i = -5; i <= 5; ++i) {\n draw((i,-5)--(i,5),gray(0.7));\n draw((-5,i)--(5,i),gray(0.7));\n}\n\ndraw((-5,0)--(5,0),Arrows(6));\ndraw((0,-5)--(0,5),Arrows(6));\n\nlabel(\"$x$\", (5,0), E);\nlabel(\"$y$\", (0,5), N);\n\ndraw(graph(func,-3,3),red);\n\nlabel(\"$y = f(x)$\", (3,-2), UnFill);\n[/asy]\n\nWhich is the graph of $y = f(x - 1)$?\n\n[asy]\nunitsize(0.5 cm);\n\npicture[] graf;\nint i, n;\n\nreal func(real x) {\n real y;\n if (x >= -3 && x <= 0) {y = -2 - x;}\n if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}\n if (x >= 2 && x <= 3) {y = 2(x - 2);}\n return(y);\n}\n\nfor (n = 1; n <= 5; ++n) {\n graf[n] = new picture;\n for (i = -5; i <= 5; ++i) {\n draw(graf[n],(i,-5)--(i,5),gray(0.7));\n draw(graf[n],(-5,i)--(5,i),gray(0.7));\n }\n draw(graf[n],(-5,0)--(5,0),Arrows(6));\n draw(graf[n],(0,-5)--(0,5),Arrows(6));\n\n label(graf[n],\"$x$\", (5,0), E);\n label(graf[n],\"$y$\", (0,5), N);\n}\n\ndraw(graf[1],shift(0,1)graph(func,-3,3),red);\ndraw(graf[2],shift(-1,0)graph(func,-3,3),red);\ndraw(graf[3],reflect((1,0),(1,1))graph(func,-3,3),red);\ndraw(graf[4],shift(1,0)graph(func,-3,3),red);\ndraw(graf[5],shift(0,-1)graph(func,-3,3),red);\n\nlabel(graf[1], \"A\", (0,-6));\nlabel(graf[2], \"B\", (0,-6));\nlabel(graf[3], \"C\", (0,-6));\nlabel(graf[4], \"D\", (0,-6));\nlabel(graf[5], \"E\", (0,-6));\n\nadd(graf[1]);\nadd(shift((12,0))(graf[2]));\nadd(shift((24,0))(graf[3]));\nadd(shift((6,-12))(graf[4]));\nadd(shift((18,-12))*(graf[5]));\n[/asy]\n\nEnter the letter of the graph of $y = f(x - 1).$", "role": "user" }, { "content": "The graph of $y = f(x - 1)$ is produced by taking the graph of $y = f(x)$ and shifting one unit to the right. The correct graph is $\\boxed{\\text{D}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a,$ $b,$ $c$ be real numbers such that $a + b + c = 0.$ Find the set of all possible values of $ab + ac + bc.$", "role": "user" }, { "content": "Squaring the equation $a + b + c = 0,$ we get\n\\[a^2 + b^2 + c^2 + 2(ab + ac + bc) = 0.\\]Hence, $2(ab + ac + bc) = -(a^2 + b^2 + c^2) \\le 0,$ so\n\\[ab + ac + bc \\le 0.\\]Equality occurs when $a = b = c = 0.$\n\nNow, set $c = 0,$ so $a + b = 0,$ or $b = -a.$ Then\n\\[ab + ac + bc = ab = -a^2\\]can take on all nonpositive values. Therefore, the set of all possible values of $ab + ac + bc$ is $\\boxed{(-\\infty,0]}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $x$ and $y$ be real numbers greater than 1 such that\n\\[(\\log_2 x)^4 + (\\log_3 y)^4 + 8 = 8 (\\log_2 x)(\\log_3 y).\\]Compute $x^{\\sqrt{2}} + y^{\\sqrt{2}}.$", "role": "user" }, { "content": "Let $a = \\log_2 x$ and $b = \\log_3 y.$ Since $x > 1$ and $y > 1,$ $a > 0$ and $b > 0.$\n\nBy AM-GM,\n\\begin{align}\na^4 + b^4 + 8 &= a^4 + b^4 + 4 + 4 \\\\\n&\\ge 4 \\sqrt[4]{(a^4)(b^4)(4)(4)} \\\\\n&= 8ab.\n\\end{align}Since $a^4 + b^4 + 8 = 8ab,$ we have equality. Therefore, $a^4 = 4$ and $b^4 = 4.$ Then $a = \\sqrt[4]{4} = \\sqrt{2},$ so\n\\[x = 2^a = 2^{\\sqrt{2}}.\\]Similarly, $b = \\sqrt[4]{4} = \\sqrt{2},$ so\n\\[y = 3^b = 3^{\\sqrt{2}}.\\]Hence, $x^{\\sqrt{2}} + y^{\\sqrt{2}} = 2^2 + 3^2 = \\boxed{13}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Compute\n\\[\\frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}.\\]", "role": "user" }, { "content": "Each of the terms is of the form $x^4 + 324$. To factor, we write: \\[\\begin{aligned} x^4 + 324 &= (x^4 + 36x^2 + 324) - 36x^2\\\\& = (x^2+18)^2 - 36x^2 \\\\& = (x^2-6x+18)(x^2+6x+18) \\\\ &= (x(x-6)+18)(x(x+6)+18). \\end{aligned}\\]Therefore, the given expression equals \\[\\frac{(10\\cdot4+18)(10\\cdot16+18)(22\\cdot16+18)(22\\cdot28+18) \\dotsm (58\\cdot52+18)(58\\cdot64+18)}{(4\\cdot(-2)+18)(4\\cdot10+18)(16\\cdot10+18)(16\\cdot22+18) \\dotsm (52\\cdot46+18)(52\\cdot58+18)}.\\]Nearly all the terms cancel, leaving just \\[\\frac{58 \\cdot 64 + 18}{4 \\cdot (-2) + 18} = \\boxed{373}.\\]Remark. The factorization $x^4+324 = (x^2-6x+18)(x^2+6x+18)$ is a special case of the Sophie Germain identity, which is derived in the same way; it states that \\[a^4 + 4b^4 = (a^2-2ab+2b^2)(a^2+2ab+2b^2).\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a$ and $b$ be real numbers. One of the roots of $x^3 + ax + b = 0$ is $1 + i \\sqrt{3}.$ Find $a + b.$", "role": "user" }, { "content": "Since the coefficients are real, another root is $1 - i \\sqrt{3}.$ By Vieta's formulas, the sum of the roots is 0, so the third root is $-2.$ Hence, the cubic polynomial is\n\\begin{align}\n(x - 1 - i \\sqrt{3})(x - 1 + i \\sqrt{3})(x + 2) &= ((x - 1)^2 - (i \\sqrt{3})^2)(x + 2) \\\\\n&= ((x - 1)^2 + 3)(x + 2) \\\\\n&= x^3 + 8.\n\\end{align}Thus, $a = 0$ and $b = 8,$ so $a + b = \\boxed{8}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: There exists a constant $c,$ so that among all chords $\\overline{AB}$ of the parabola $y = x^2$ passing through $C = (0,c),$\n\\[t = \\frac{1}{AC} + \\frac{1}{BC}\\]is a fixed constant. Find the constant $t.$\n\n[asy]\nunitsize(1 cm);\n\nreal parab (real x) {\n return(x^2);\n}\n\npair A, B, C;\n\nA = (1.7,parab(1.7));\nB = (-1,parab(-1));\nC = extension(A,B,(0,0),(0,1));\n\ndraw(graph(parab,-2,2));\ndraw(A--B);\ndraw((0,0)--(0,4));\n\ndot(\"$A$\", A, E);\ndot(\"$B$\", B, SW);\ndot(\"$(0,c)$\", C, NW);\n[/asy]", "role": "user" }, { "content": "To get a handle on the constant $t,$ we can look at some particular cases.\n\nSuppose we let $AB$ approach a vertical line. Then $\\frac{1}{AC}$ approaches 0, and $B$ approaches $(0,0),$ so $\\frac{1}{AC} + \\frac{1}{BC}$ approaches $c.$ Hence,\n\\[t = \\frac{1}{c}.\\]Now, suppose we take $A = (\\sqrt{c},c)$ and $B = (-\\sqrt{c},c).$ Then\n\\[t = \\frac{1}{AC} + \\frac{1}{BC} = \\frac{1}{\\sqrt{c}} + \\frac{1}{\\sqrt{c}} = \\frac{2}{\\sqrt{c}}.\\]Hence, $\\frac{1}{c} = \\frac{2}{\\sqrt{c}},$ so $\\sqrt{c} = \\frac{1}{2},$ and $c = \\frac{1}{4}.$ Therefore, $t = \\boxed{4}.$ (Note that this makes $C$ the focus of the parabola.)\n\nFor a full solution, let's check that this value works. Let $y = mx + \\frac{1}{4}$ be the equation of line $AB.$ Setting $y = x^2,$ we get\n\\[x^2 = mx + \\frac{1}{4},\\]or $x^2 - mx - c = 0.$ Let $x_1$ and $x_2$ be the roots of this equation. By Vieta's formulas, $x_1 + x_2 = m$ and $x_1 x_2 = -\\frac{1}{4}.$\n\nAlso, $A$ and $B$ are $(x_1,x_1^2)$ and $(x_2,x_2^2)$ in some order, so\n\\begin{align}\n\\frac{1}{AC} + \\frac{1}{BC} &= \\frac{1}{\\sqrt{x_1^2 + (x_1^2 - \\frac{1}{4})^2}} + \\frac{1}{\\sqrt{x_2^2 + (x_2^2 - \\frac{1}{4})^2}} \\\\\n&= \\frac{1}{\\sqrt{x_1^2 + x_1^4 - \\frac{1}{2} x_1^2 + \\frac{1}{16}}} + \\frac{1}{\\sqrt{x_2^2 + x_2^4 - \\frac{1}{2} x_2^2 + \\frac{1}{16}}} \\\\\n&= \\frac{1}{\\sqrt{x_1^4 + \\frac{1}{2} x_1^2 + \\frac{1}{16}}} + \\frac{1}{\\sqrt{x_2^4 + \\frac{1}{2} x_2^2 + \\frac{1}{16}}} \\\\\n&= \\frac{1}{\\sqrt{(x_1^2 + \\frac{1}{4})^2}} + \\frac{1}{\\sqrt{(x_2^2 + \\frac{1}{4})^2}} \\\\\n&= \\frac{1}{x_1^2 + \\frac{1}{4}} + \\frac{1}{x_2^2 + \\frac{1}{4}}.\n\\end{align}We have that $x_1^2 x_2^2 = (x_1 x_2)^2 = \\left( -\\frac{1}{4} \\right)^2 = \\frac{1}{16}$ and\n\\[x_1^2 + x_2^2 = (x_1 + x_2)^2 - 2x_1 x_2 = m^2 + \\frac{1}{2}.\\]Hence,\n\\begin{align}\n\\frac{1}{x_1^2 + \\frac{1}{4}} + \\frac{1}{x_2^2 + \\frac{1}{4}} &= \\frac{x_1^2 + \\frac{1}{4} + x_2^2 + \\frac{1}{4}}{(x_1^2 + \\frac{1}{4})(x_2^2 + \\frac{1}{4})} \\\\\n&= \\frac{x_1^2 + x_2^2 + \\frac{1}{2}}{x_1^2 x_2^2 + \\frac{1}{4} (x_1^2 + x_2^2) + \\frac{1}{16}} \\\\\n&= \\frac{m^2 + 1}{\\frac{1}{16} + \\frac{1}{4} (m^2 + \\frac{1}{2}) + \\frac{1}{16}} \\\\\n&= \\frac{m^2 + 1}{\\frac{1}{4} m^2 + \\frac{1}{4}} \\\\\n&= 4.\n\\end{align}", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $x,$ $y,$ $z$ be positive real number such that $xyz = \\frac{2}{3}.$ Compute the minimum value of\n\\[x^2 + 6xy + 18y^2 + 12yz + 4z^2.\\]", "role": "user" }, { "content": "We may think of trying to apply AM-GM directly to all five terms. Ignoring the constants, this give us a term of\n\\[\\sqrt[5]{x^2 \\cdot xy \\cdot y^2 \\cdot yz \\cdot z^2} = \\sqrt[5]{x^3 y^4 z^3}.\\]This doesn't work, because the condition is $xyz = \\frac{2}{3},$ so we want a power of $xyz.$ So, to get more one power of $y,$ relative to $x$ and $z,$ we split every term except $y^2$ in half:\n\\[\\frac{x^2}{2} + \\frac{x^2}{2} + 3xy + 3xy + 18y^2 + 6yz + 6yz + 2z^2 + 2z^2.\\]Then by AM-GM,\n\\begin{align}\n&\\frac{x^2}{2} + \\frac{x^2}{2} + 3xy + 3xy + 18y^2 + 6yz + 6yz + 2z^2 + 2z^2 \\\\\n&\\ge 9 \\sqrt[9]{\\frac{x^2}{2} \\cdot \\frac{x^2}{2} \\cdot 3xy \\cdot 3xy \\cdot 18y^2 \\cdot 6yz \\cdot 6yz \\cdot 2z^2 \\cdot 2z^2} \\\\\n&= 9 \\sqrt[9]{5832x^6 y^6 z^6} \\\\\n&= 18.\n\\end{align}Equality occurs when $\\frac{x^2}{2} = 3xy = 18y^2 = 6yz = 2z^2.$ Along with the condition $xyz = \\frac{2}{3},$ we can solve to get $x = 2,$ $y = \\frac{1}{3},$ $z = 1,$ so the minimum value is $\\boxed{18}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $z$ be a complex number such that\n\\[z^2 + \|z\|^2 = 3 - 5i.\\]Find $\|z\|^2.$", "role": "user" }, { "content": "Let $z = a + bi,$ where $a$ and $b$ are real numbers. Then $z^2 = (a + bi)^2 = a^2 + 2abi - b^2$ and $\|z\|^2 = a^2 + b^2,$ so\n\\[a^2 + 2abi - b^2 + a^2 + b^2 = 3 - 5i.\\]Equating real and imaginary parts, we get\n\\begin{align}\n2a^2 &= 3, \\\\\n2ab &= -5.\n\\end{align}From the first equation, $a^2 = \\frac{3}{2}.$ From the second equation,\n\\[b = -\\frac{5}{2a},\\]so\n\\[b^2 = \\frac{25}{4a^2} = \\frac{25}{4 \\cdot 3/2} = \\frac{25}{6}.\\]Therefore,\n\\[\|z\|^2 = a^2 + b^2 = \\frac{3}{2} + \\frac{25}{6} = \\boxed{\\frac{17}{3}}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: How many integers $-11 \\leq n \\leq 11$ satisfy $(n-2)(n+4)(n + 8)<0$?", "role": "user" }, { "content": "Since $(n-2)(n+4)(n+8)=0$ when $n=2, -4,$ or $-8$, we will consider the four cases $-11 \\leq n < -8$, $-8<n<-4$, $-4<n<2$, and $2<n\\leq 11$ separately. If $n=2$, $n=-4$, or $n=-8$, then all three factors are 0. If $n>2$, then all three factors are positive. If $-4<n<2$, then $n-2$ is negative, while the other two factors are positive, so the product is negative. If $-8<n<-4$, then $n+8$ is positive, while the other two factors are negative, so the product is positive. If $n<-8$, then all three factors are negative, so the product is negative. In total, there are $\\boxed{8}$ solutions: $-11,-10,-9,-3,-2,-1,0,1$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: A circle passes through the point $(0,1),$ and is tangent to the parabola $y = x^2$ at $(2,4).$ Find the center of the circle.\n\n[asy]\nunitsize(0.4 cm);\n\nreal parab (real x) {\n return(x^2);\n}\n\ndraw(graph(parab,-3.5,3.5));\ndraw(Circle((-16/5,53/10),13sqrt(17)/10));\n\ndot((0,1));\ndot(\"$(2,4)$\", (2,4), E);\n[/asy]", "role": "user" }, { "content": "First, consider the tangent line to the parabola at $(2,4).$ The equation of this tangent is of the form\n\\[y - 4 = m(x - 2).\\]Setting $y = x^2,$ we get $x^2 - 4 = m(x - 2),$ or $x^2 - mx + 2m - 4 = 0.$ Since we have a tangent, $x = 2$ is a double root of this quadratic. In other words, this quadratic is identical to $(x - 2)^2 = x^2 - 4x + 4.$ Hence, $m = 4.$\n\nLet the center of the circle be $(a,b).$ The line joining the center $(a,b)$ and $(2,4)$ is the perpendicular to the tangent line, which means its slope is $-\\frac{1}{4}.$ This gives us the equation\n\\[\\frac{b - 4}{a - 2} = -\\frac{1}{4}.\\]Since the points $(2,4)$ and $(0,1)$ are on the circle, they must be equidistant from its center. The set of all points equidistant from $(2,4)$ and $(0,1)$ is the perpendicular bisector of the line segment joining $(2,4)$ and $(0,1)$. Therefore, the center of the circle must lie on the perpendicular bisector of the line segment joining $(2,4)$ and $(0,1)$. The midpoint of this line segment is $\\left( 1, \\frac{5}{2} \\right),$ and its slope is\n\\[\\frac{4 - 1}{2 - 0} = \\frac{3}{2}.\\]Hence, $(a,b)$ must satisfy\n\\[\\frac{b - 5/2}{a - 1} = -\\frac{2}{3}.\\]So,\n\\begin{align}\nb - 4 &= -\\frac{1}{4} (a - 2), \\\\\nb - \\frac{5}{2} &= -\\frac{2}{3} (a - 1).\n\\end{align*}Solving this system, we find $(a,b) = \\boxed{\\left( -\\frac{16}{5}, \\frac{53}{10} \\right)}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The graph of $y = ax^2 + bx + c$ has a maximum value of 54, and passes through the points $(-2,0)$ and $(4,0).$ Find $a + b + c.$", "role": "user" }, { "content": "Since the graph passes through the points $(-2,0)$ and $(4,0),$ the equation is of the form $a(x + 2)(x - 4).$\n\nThe graph has a maximum, and this maximum value occurs at the average of $-2$ and 4, namely $x = \\frac{-2 + 4}{2} = 1.$ But $a + b + c$ is exactly the value of $y = ax^2 + bx + c$ at $x = 1,$ so $a + b + c = \\boxed{54}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The circles $C_1$ and $C_2$ are defined by the equations $x^2 + y^2 = 1$ and $(x - 2)^2 + y^2 = 16,$ respectively. Find the locus of the centers $(a,b)$ of all circles externally tangent to $C_1$ and internally tangent to $C_2.$ Enter your answer in the form\n\\[Pa^2 + Qb^2 + Ra + Sb + T = 0,\\]where all the coefficients are integers, $P$ is positive, and $\\gcd(\|P\|,\|Q\|,\|R\|,\|S\|,\|T\|) = 1.$\n\nNote: The word \"locus\" is a fancy word for \"set\" in geometry, so \"the locus of the centers\" means \"the set of the centers\".", "role": "user" }, { "content": "Let $(a,b)$ be the center of a circle that is tangent to $C_1$ and $C_2,$ and let $r$ be the radius.\n\n[asy]\nunitsize(1 cm);\n\npair A, B, O, P, Q;\n\nO = (0,0);\nP = (2,0);\nQ = (1,sqrt(21)/2);\nA = intersectionpoint(O--Q,Circle(Q,1.5));\nB = intersectionpoint(Q--interp(P,Q,2),Circle(Q,1.5));\n\ndraw(Circle(O,1));\ndraw(Circle(P,4));\ndraw(Circle(Q,1.5));\ndraw(O--Q);\ndraw(P--B);\n\nlabel(\"$r$\", (Q + A)/2, NW);\nlabel(\"$r$\", (Q + B)/2, SW);\nlabel(\"$1$\", (O + A)/2, NW);\nlabel(\"$4 - r$\", (P + Q)/2, NE, UnFill);\nlabel(\"$C_1$\", dir(225), dir(225));\nlabel(\"$C_2$\", P + 4*dir(70), dir(70));\n\ndot(\"$(0,0)$\", O, S);\ndot(\"$(2,0)$\", P, S);\ndot(A);\ndot(B);\ndot(\"$(a,b)$\", Q, NE);\n[/asy]\n\nThen the square of the distance of the center of this circle from the center of $C_1$ is $a^2 + b^2 = (r + 1)^2$ and the square of the distance of the center of this circle from the center of $C_2$ is $(a - 2)^2 + b^2 = (4 - r)^2.$ Subtracting these equations, we get\n\\[a^2 - (a - 2)^2 = (r + 1)^2 - (4 - r)^2.\\]This simplifies to $4a - 4 = 10r - 15,$ so $r = \\frac{4a + 11}{10}.$\n\nSubstituting into the equation $a^2 + b^2 = (r + 1)^2,$ we get\n\\[a^2 + b^2 = \\left( \\frac{4a + 21}{10} \\right)^2.\\]This simplifies to $\\boxed{84a^2 + 100b^2 - 168a - 441 = 0}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The polynomial $f(x) = x^3 + x^2 + 2x + 3$ has three distinct roots. Let $g(x) = x^3+bx^2+cx+d$ be a cubic polynomial with leading coefficient $1$ such that the roots of $g(x)$ are the squares of the roots of $f(x)$. Find the ordered triple $(b,c,d)$.", "role": "user" }, { "content": "If $r$ is a root of $f(x) = 0$, then $r^3+r^2+2r+3=0$. Rearranging, we have \\[r^3+2r=-r^2-3,\\]and squaring this equation gives \\[r^6+4r^4+4r^2=r^4+6r^2+9,\\]or \\[r^6+3r^4-2r^2-9=0.\\]Rewriting this equation in the form $(r^2)^3 + 3(r^2)^2 - 2r^2 - 9 =0$, we see that the polynomial $x^3+3x^2-2x-9$ has $r^2$ as a root, so three of its roots are the squares of the roots of $f(x)$. But this polynomial is cubic, so these are its only roots. Thus, $g(x)=x^3+3x^2-2x-9$, and so $(b,c,d) = \\boxed{(3,-2,-9)}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the minimum value of\n\\[\\frac{x^2}{y - 1} + \\frac{y^2}{x - 1}\\]for real numbers $x > 1$ and $y > 1.$", "role": "user" }, { "content": "Let $a = x - 1$ and $b = y - 1.$ Then $x = a + 1$ and $y = b + 1,$ so\n\\begin{align}\n\\frac{x^2}{y - 1} + \\frac{y^2}{x - 1} &= \\frac{(a + 1)^2}{b} + \\frac{(b + 1)^2}{a} \\\\\n&= \\frac{a^2 + 2a + 1}{b} + \\frac{b^2 + 2b + 1}{a} \\\\\n&= 2 \\left( \\frac{a}{b} + \\frac{b}{a} \\right) + \\frac{a^2}{b} + \\frac{1}{b} + \\frac{b^2}{a} + \\frac{1}{a}.\n\\end{align}By AM-GM,\n\\[\\frac{a}{b} + \\frac{b}{a} \\ge 2 \\sqrt{\\frac{a}{b} \\cdot \\frac{b}{a}} = 2\\]and\n\\[\\frac{a^2}{b} + \\frac{1}{b} + \\frac{b^2}{a} + \\frac{1}{a} \\ge 4 \\sqrt[4]{\\frac{a^2}{b} \\cdot \\frac{1}{b} \\cdot \\frac{b^2}{a} \\cdot \\frac{1}{a}} = 4,\\]so\n\\[2 \\left( \\frac{a}{b} + \\frac{b}{a} \\right) + \\frac{a^2}{b} + \\frac{1}{b} + \\frac{b^2}{a} + \\frac{1}{a} \\ge 2 \\cdot 2 + 4 = 8.\\]Equality occurs when $a = b = 1,$ or $x = y = 2,$ so the minimum value is $\\boxed{8}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Suppose that $(u_n)$ is a sequence of real numbers satisfying\n\\[u_{n+2}=2u_{n+1}+u_n\\]and that $u_3=9$ and $u_6=128$. What is $u_5$?", "role": "user" }, { "content": "Let $u_4 = a.$ Then $u_5 = 2u_4 + u_3 = 2a + 9$ and $u_6 = 2u_5 + u_4 = 2(2a + 9) + a = 5a + 18 = 128.$ Solving for $a,$ we find $a = 22,$ so $u_5 = 2 \\cdot 22 + 9 = \\boxed{53}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a,$ $b,$ $c$ be the roots of the cubic polynomial $x^3 - x - 1 = 0.$ Find\n\\[a(b - c)^2 + b(c - a)^2 + c(a - b)^2.\\]", "role": "user" }, { "content": "By Vieta's formulas,\n\\begin{align}\na + b + c &= 0, \\\\\nab + ac + bc &= -1, \\\\\nabc &= 1.\n\\end{align}Then\n\\begin{align}\na(b - c)^2 + b(c - a)^2 + c(a - b)^2 &= a(b^2 - 2bc + c^2) + b(c^2 - 2ac + a^2) + c(a^2 - 2ab + b^2) \\\\\n&= (ab^2 - 2abc + ac^2) + (bc^2 - 2abc + ba^2) + (ca^2 - 2abc + cb^2) \\\\\n&= (ab^2 - 2 + ac^2) + (bc^2 - 2 + ba^2) + (ca^2 - 2 + cb^2) \\\\\n&= ab^2 + ac^2 + bc^2 + ba^2 + ca^2 + cb^2 - 6 \\\\\n&= a^2 (b + c) + b^2 (a + c) + c^2 (a + b) - 6.\n\\end{align}From $a + b + c = 0,$ $b + c = -a.$ Simillarly, $a + c = -b$ and $a + b = -c,$ so\n\\[a^2 (b + c) + b^2 (a + c) + c^2 (a + b) - 6 = -a^3 - b^3 - c^3 - 6.\\]Since $a$ is a root of $x^3 - x - 1 = 0,$ $a^3 - a - 1 = 0,$ so $-a^3 = -a - 1.$ Similarly, $-b^3 = -b - 1$ and $-c^3 = -c - 1,$ so\n\\begin{align}\n-a^3 - b^3 - c^3 - 6 &= (-a - 1) + (-b - 1) + (-c - 1) - 6 \\\\\n&= -(a + b + c) - 9 \\\\\n&= \\boxed{-9}.\n\\end{align}", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Is the function $f(x) = \\lfloor x \\rfloor + \\frac{1}{2}$ even, odd, or neither?\n\nEnter \"odd\", \"even\", or \"neither\".", "role": "user" }, { "content": "Since $f \\left( \\frac{1}{2} \\right) = \\left\\lfloor \\frac{1}{2} \\right\\rfloor + \\frac{1}{2} = \\frac{1}{2}$ and $f \\left( -\\frac{1}{2} \\right) = \\left\\lfloor -\\frac{1}{2} \\right\\rfloor + \\frac{1}{2} = -\\frac{1}{2},$ so if $f$ is either even or odd, it must be odd.\n\nBut $f(0) = \\lfloor 0 \\rfloor + \\frac{1}{2}.$ Every odd function $f(x)$ satisfies $f(0) = 0,$ so $f(x)$ is $\\boxed{\\text{neither}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Determine the complex number $z$ satisfying the equation $2z-3\\bar{z}=-2-30i$. Note that $\\bar{z}$ denotes the conjugate of $z$.", "role": "user" }, { "content": "Let $z=a+bi$, where $a$ and $b$ are real numbers representing the real and imaginary parts of $z$, respectively. Then $\\bar{z}=a-bi$, so that $-3\\bar{z}=-3a+3ib$. We now find that \\[2z-3\\bar{z} = (2a-3a) + (2b +3b)i. \\]So if $2z-3\\bar{z}=-2-30i$ then we must have $2a-3a=-2$ and $2b+3b=-30$. This immediately gives us $a=2$ and $b=-6$. Therefore the complex number we are seeking is $z=\\boxed{2-6i}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $f(n)$ be the integer closest to $\\sqrt[4]{n}.$ Find $\\sum_{k=1}^{1995}\\frac 1{f(k)}.$", "role": "user" }, { "content": "We have $f(n) = m$ if and only if \\[m - \\frac{1}{2} < \\sqrt[4]{n} < m + \\frac{1}{2},\\]or \\[\\left(m - \\frac{1}{2}\\right)^4 < n < \\left(m + \\frac{1}{2}\\right)^4.\\]Expanding the fourth powers, we get \\[m^4 - 2m^3 + \\frac{3}{2}m^2 - \\frac{1}{2}m + \\frac{1}{16} < n < m^4+ 2m^3 + \\frac{3}{2}m^2 + \\frac{1}{2}m + \\frac{1}{16}.\\]The leftmost and rightmost expressions are both non-integers, and their difference is $4m^3 + m$. Therefore, there are exactly $4m^3 + m$ values of $n$ that satisfy this inequality.\n\nFor each $m$, there are $4m^3 + m$ terms of the form $\\frac{1}{m}$ in the sum, so those terms contribute $(4m^3+m) \\cdot \\frac{1}{m} = 4m^2 + 1$ to the sum. Thus, from $m=1$ to $m=6$, we get $4(1+4+9+16+25+36) + 6 = 370$.\n\nThe remaining terms have $m=7$. Since $6.5^4 = 1785 \\frac{1}{16}$, these are the terms from $n=1786$ to $n=1995$, inclusive. There are $1995 - 1786 + 1 = 210$ such terms, so they contribute $210 \\cdot \\frac{1}{7} = 30$ to the sum. Therefore, the final answer is $370 + 30 = \\boxed{400}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $f(x) = x^2-3x$. For what values of $x$ is $f(f(x)) = f(x)$? Enter all the solutions, separated by commas.", "role": "user" }, { "content": "Expanding $f(f(x)) = f(x)$ gives us $$(x^2-3x)^2-3(x^2-3x)=x^2-3x.$$Rather than expanding, we can subtract $x^2-3x$ from both sides to get $$(x^2-3x)^2-4(x^2-3x)=0.$$Factoring out $x^2-3x$ gives $(x^2-3x)(x^2-3x-4)=0$. Factoring each quadratic separately, we get $$x(x-3)(x+1)(x-4)=0.$$Thus the values of $x$ are $\\boxed{0, 3, -1, 4}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $A = (1,0),$ $B = (4,3),$ and $C = (p,q)$ be three points on the parabola $y = -x^2 + 6x - 5,$ where $1 \\le p \\le 4.$ Find the largest possible area of triangle $ABC.$", "role": "user" }, { "content": "We have that $q = -p^2 + 6p - 5,$ so by the Shoelace Theorem, the area of triangle $ABC$ is\n\\begin{align}\n&\\frac{1}{2} \|(1)(3) + (4)(-p^2 + 6p - 5) + (p)(0) - (0)(4) - (3)(p) - (-p^2 + 6p - 5)(1)\| \\\\\n&= \\frac{1}{2} \|-3p^2 + 15p - 12\| \\\\\n&= \\frac{3}{2} \|p^2 - 5p + 4\| \\\\\n&= \\frac{3}{2} \|(p - 1)(p - 4)\|.\n\\end{align}Since $1 \\le p \\le 4,$ $\|(p - 1)(p - 4)\| = (p - 1)(4 - p),$ so we want to maximize\n\\[\\frac{3}{2} (p - 1)(4 - p).\\]The maximum value occurs at $p = \\frac{5}{2},$ so the maximum area is\n\\[\\frac{3}{2} \\left( \\frac{5}{2} - 1 \\right) \\left( 4 - \\frac{5}{2} \\right) = \\boxed{\\frac{27}{8}}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: A polynomial with integer coefficients is of the form\n\\[x^3 + a_2 x^2 + a_1 x - 11 = 0.\\]Enter all the possible integer roots of this polynomial, separated by commas.", "role": "user" }, { "content": "By the Integer Root Theorem, the possible integer roots are all the divisors of 11 (including negative divisors), so they are $\\boxed{-11, -1, 1, 11}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find all real numbers $a$ such that the roots of the polynomial\n$$x^3 - 6x^2 + 21x + a$$form an arithmetic progression and are not all real.", "role": "user" }, { "content": "Let the three roots be $r-d$, $r$, and $r+d$, for some complex numbers $r$ and $d$. Then Vieta's formulas give\n$$(r-d)+r+(r+d)=6 \\qquad\\text{and}\\qquad (r-d)r+(r-d)(r+d)+r(r+d)=21.$$Simplifying these equations, we have\n$$3r=6 \\qquad\\text{and}\\qquad 3r^2-d^2=21.$$From $3r=6$, we deduce $r=2$. Substituting this into our second equation gives $12-d^2=21$, so $d^2=-9$ and $d=\\pm 3i$. Therefore, the roots of the cubic are $2-3i$, $2$, and $2+3i$, so\n$$a = -2(2-3i)(2+3i) = -2\\left(2^2-(3i)^2\\right) = -2(4+9) = \\boxed{-26}.$$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a,$ $b,$ $c,$ $d,$ $e,$ $f$ be positive real numbers such that $a + b + c + d + e + f = 7.$ Find the minimum value of\n\\[\\frac{1}{a} + \\frac{4}{b} + \\frac{9}{c} + \\frac{16}{d} + \\frac{25}{e} + \\frac{36}{f}.\\]", "role": "user" }, { "content": "By Cauchy-Schwarz,\n\\[(a + b + c + d + e + f) \\left( \\frac{1}{a} + \\frac{4}{b} + \\frac{9}{c} + \\frac{16}{d} + \\frac{25}{e} + \\frac{36}{f} \\right) \\ge (1 + 2 + 3 + 4 + 5 + 6)^2 = 441,\\]so\n\\[\\frac{1}{a} + \\frac{4}{b} + \\frac{9}{c} + \\frac{16}{d} + \\frac{25}{e} + \\frac{36}{f} \\ge \\frac{441}{7} = 63.\\]Equality occurs when $a^2 = \\frac{b^2}{4} = \\frac{c^2}{9} = \\frac{d^2}{16} = \\frac{e^2}{25} = \\frac{f^2}{36}$ and $a + b + c + d + e + f = 7.$ Solving, we find $a = \\frac{1}{3},$ $b = \\frac{2}{3},$ $c = 1,$ $d = \\frac{4}{3},$ $e = \\frac{5}{3},$ and $f = 2,$ so the minimum value is $\\boxed{63}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $\\omega$ be a nonreal root of $z^3 = 1.$ Let $a_1,$ $a_2,$ $\\dots,$ $a_n$ be real numbers such that\n\\[\\frac{1}{a_1 + \\omega} + \\frac{1}{a_2 + \\omega} + \\dots + \\frac{1}{a_n + \\omega} = 2 + 5i.\\]Compute\n\\[\\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \\frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \\dots + \\frac{2a_n - 1}{a_n^2 - a_n + 1}.\\]", "role": "user" }, { "content": "We have that $\\omega^3 = 1.$ Then $\\omega^3 - 1 = 0,$ which factors as $(\\omega - 1)(\\omega^2 + \\omega + 1) = 0.$ Since $\\omega$ is nonreal, $\\omega^2 + \\omega + 1 = 0.$ By the quadratic formula,\n\\[\\omega = -\\frac{1}{2} \\pm \\frac{\\sqrt{3}}{2} i.\\]Taking the conjugate of the given equation, we get\n\\[\\frac{1}{a_1 + \\overline{\\omega}} + \\frac{1}{a_2 + \\overline{\\omega}} + \\dots + \\frac{1}{a_n + \\overline{\\omega}} = 2 - 5i.\\]Note that if $a$ is a real number, then\n\\begin{align}\n\\frac{1}{a + \\omega} + \\frac{1}{a + \\overline{\\omega}} &= \\frac{a + \\omega + a + \\overline{\\omega}}{(a + \\omega)(a + \\overline{\\omega})} \\\\\n&= \\frac{2a + \\omega + \\overline{\\omega}}{a^2 + (\\omega + \\overline{\\omega}) a + \\omega \\overline{\\omega}} \\\\\n&= \\frac{2a - 1}{a^2 - a + 1}.\n\\end{align}Therefore,\n\\begin{align}\n\\sum_{k = 1}^n \\frac{2a_k - 1}{a_k^2 - a_k + 1} &= \\sum_{k = 1}^n \\left( \\frac{1}{a_k + \\omega} + \\frac{1}{a_k + \\overline{\\omega}} \\right) \\\\\n&= 2 + 5i + 2 - 5i \\\\\n&= \\boxed{4}.\n\\end{align}", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: If $\\left( r + \\frac{1}{r} \\right)^2 = 3,$ then find $r^3 + \\frac{1}{r^3}.$", "role": "user" }, { "content": "Expanding, we get $r^2 + 2 + \\frac{1}{r^2} = 3,$ so\n\\[r^2 - 1 + \\frac{1}{r^2} = 0.\\]Then\n\\[r^3 + \\frac{1}{r^3} = \\left( r + \\frac{1}{r} \\right) \\left( r^2 - 1 + \\frac{1}{r^2} \\right) = \\boxed{0}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a$ and $b$ be real numbers such that\n\\[a^3 - 15a^2 + 20a - 50 = 0 \\quad \\text{and} \\quad 8b^3 - 60b^2 - 290b + 2575 = 0.\\]Compute $a + b.$", "role": "user" }, { "content": "Let $x = a - 5.$ Then $a = x + 5,$ so\n\\[(x + 5)^3 - 15(x + 5)^2 + 20(x + 5) - 50 = 0,\\]which simplifies to $x^3 - 55x - 200 = 0.$\n\nLet $y = b - \\frac{5}{2}.$ Then $b = y + \\frac{5}{2},$ so\n\\[8 \\left( y + \\frac{5}{2} \\right)^3 - 60 \\left( y + \\frac{5}{2} \\right)^2 - 290 \\left( y + \\frac{5}{2} \\right) + 2575 = 0,\\]which simplifies to $y^3 - 55y + 200 = 0.$ (Note that through these substitutions, we made the quadratic term vanish in each of these cubic equations.)\n\nConsider the function $f(t) = t^3 - 55t.$ Observe that the polynomial $f(t)$ has three roots 0, $\\sqrt{55},$ and $-\\sqrt{55}.$ Its graph is shown below.\n\n[asy]\nunitsize (0.2 cm);\n\nreal cubic (real x) {\n return ((x^3 - 55*x)/12);\n}\n\ndraw(graph(cubic,-8.5,8.5));\ndraw((-18,0)--(18,0));\ndraw((0,-18)--(0,18));\n\ndot(\"$\\sqrt{55}$\", (sqrt(55),0), SE);\ndot(\"$-\\sqrt{55}$\", (-sqrt(55),0), SW);\n[/asy]\n\nLet $0 \\le t \\le \\sqrt{55}.$ Then\n\\[[f(t)]^2 = (t^3 - 55t)^2 = t^2 (t^2 - 55)^2 = t^2 (55 - t^2)^2 = t^2 (55 - t^2)(55 - t^2).\\]By AM-GM,\n\\[2t^2 (55 - t^2)(55 - t^2) \\le \\left( \\frac{(2t^2) + (55 - t^2) + (55 - t^2)}{3} \\right)^3 = \\left( \\frac{110}{3} \\right)^3 < 40^3,\\]so\n\\[[f(t)]^2 < 32000 < 32400,\\]which means $\|f(t)\| < 180.$\n\nSince $f(t)$ is an odd function, $\|f(t)\| < 180$ for $-\\sqrt{55} \\le t \\le 0$ as well. This means that the equation $f(t) = 200$ has exactly one real root. Similarly, $f(t) = -200$ has exactly one real root. Furthermore, since $f(t)$ is an odd function, these roots add up to 0.\n\nThen\n\\[a - 5 + b - \\frac{5}{2} = 0,\\]so $a + b = 5 + \\frac{5}{2} = \\boxed{\\frac{15}{2}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The function $f$ is not defined for $x = 0,$ but for all non-zero real numbers $x,$\n\\[f(x) + 2f \\left( \\frac{1}{x} \\right) = 3x.\\]Find the real solutions to $f(x) = f(-x).$ Enter the real solutions, separated by commas.", "role": "user" }, { "content": "We are given\n\\[f(x) + 2f \\left( \\frac{1}{x} \\right) = 3x.\\]Replacing $x$ with $\\frac{1}{x},$ we get\n\\[f \\left( \\frac{1}{x} \\right) + 2f(x) = \\frac{3}{x}.\\]We can view these equations as a system in $f(x)$ and $f \\left( \\frac{1}{x} \\right).$ Solving for $f(x),$ we find\n\\[f(x) = \\frac{2 - x^2}{x}.\\]Then the equation $f(x) = f(-x)$ becomes\n\\[\\frac{2 - x^2}{x} = \\frac{2 - x^2}{-x}.\\]Then $2 - x^2 = x^2 - 2,$ so $x^2 = 2.$ The solutions are $\\boxed{\\sqrt{2},-\\sqrt{2}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The three roots of the cubic $ 30 x^3 - 50x^2 + 22x - 1$ are distinct real numbers strictly between $ 0$ and $ 1$. If the roots are $p$, $q$, and $r$, what is the sum\n\\[ \\frac{1}{1-p} + \\frac{1}{1-q} +\\frac{1}{1-r} ?\\]", "role": "user" }, { "content": "Since $p,q, r$ are roots of $ 30 x^3 - 50x^2 + 22x - 1$, $ {1-p},{1-q}, {1-r} $ are roots of $ 30 (1-x)^3 - 50(1-x)^2 + 22(1-x) - 1$.\n\nIf we consider only the constant terms in the expansion of the above polynomial, we find that the constant coefficient is $30 - 50 +22 -1 = 1$. Similarly, the linear coefficient of the above polynomial is $30(-3)+50(2)-22=-12$\n\nHence, $\\frac{1}{1-p} , \\frac{1}{1-q} ,\\frac{1}{1-r} $ are the roots of a cubic in the reversed form $1x^3-12x^2+\\dotsb$. Using Vieta's formula,\n\\[\\frac{1}{1-p} + \\frac{1}{1-q} +\\frac{1}{1-r} = - \\frac{-12}{1} = \\boxed{12}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the product of the roots of the equation \\[(2x^3 + x^2 - 8x + 20)(5x^3 - 25x^2 + 19) = 0.\\]", "role": "user" }, { "content": "The left-hand side, when multiplied out, is a polynomial of degree $6.$ By Vieta's formulas, the product of the roots is determined by its $x^6$ coefficient and its constant term. The $x^6$ coefficient is $2 \\cdot 5 = 10$ and the constant term is $20 \\cdot 19 = 380,$ so the product of the roots is $\\tfrac{380}{10} = \\boxed{38}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the roots of $6x^4+19x^3-51x^2+20x$. Enter your answer as a list of numbers separated by commas.", "role": "user" }, { "content": "Since the polynomial has no constant term, we can immediately factor out an $x$ from every term\n$$x(6x^3+19x^2-51x+20),$$and our first root $x=0$. Let $g(x) = 6x^3+19x^2-51x+20$. Then the remaining roots of our original polynomial are the roots of $g(x)$. By trying out simple values, we can see that $g(0) = 20 > 0$ and $g(1) = 6+19-51+20 = -6<0$. Thus, there must be a root of $g(x)$ between $0$ and $1$. From the Rational Root Theorem, we know that if $g(p/q) = 0$ then $p$ must divide $20$ and $q$ must divide $6$.\n\nChecking rational numbers of the form $p/q$, where $p$ divides $20$ and $q$ divides $6$, and $p/q$ is between $0$ and $1$, we find that\n$$\\begin{aligned} g\\left(\\frac{1}{2}\\right) &= 6\\cdot\\frac{1}{8}+19\\cdot\\frac{1}{4}-51\\cdot\\frac{1}{2}+20 = 0.\n\\end{aligned}$$This means that $2x - 1$ is a factor of $g(x).$ Dividing by $2x-1$ gives us $g(x) = (2x-1)(3x^2+11x-20)$.\n\nThe quadratic $3x^2+11x-20$ factors as $(3x-4)(x+5),$ so our last two roots are $4/3$ and $-5$.\n\nThus, the roots of $6x^4+19x^3-51x^2+20x$ are $\\boxed{0, \\frac{1}{2}, \\frac{4}{3}, -5}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $x_1,$ $x_2,$ $x_3,$ $\\dots,$ $x_{100}$ be positive real numbers such that $x_1^2 + x_2^2 + x_3^2 + \\dots + x_{100}^2 = 1.$ Find the minimum value of\n\\[\\frac{x_1}{1 - x_1^2} + \\frac{x_2}{1 - x_2^2} + \\frac{x_3}{1 - x_3^2} + \\dots + \\frac{x_{100}}{1 - x_{100}^2}.\\]", "role": "user" }, { "content": "Note that $x_i < 1$ for all $i.$\n\nWe claim that\n\\[\\frac{x}{1 - x^2} \\ge \\frac{3 \\sqrt{3}}{2} x^2\\]for all $0 < x < 1.$ This is equivalent to $2x \\ge 3 \\sqrt{3} x^2 (1 - x^2) = 3x^2 \\sqrt{3} - 3x^4 \\sqrt{3},$ or\n\\[3 \\sqrt{3} x^4 - 3x^2 \\sqrt{3} + 2x \\ge 0.\\]We can factor this as\n\\[x (x \\sqrt{3} - 1)^2 (x \\sqrt{3} + 2) \\ge 0,\\]which clearly holds. Thus,\n\\[\\frac{x}{1 - x^2} \\ge \\frac{3 \\sqrt{3}}{2} x^2.\\]It follows that\n\\[\\frac{x_1}{1 - x_1^2} + \\frac{x_2}{1 - x_2^2} + \\frac{x_3}{1 - x_3^2} + \\dots + \\frac{x_{100}}{1 - x_{100}^2} \\ge \\frac{3 \\sqrt{3}}{2} (x_1^2 + x_2^2 + x_3^2 + \\dots + x_{100}^2) = \\frac{3 \\sqrt{3}}{2}.\\]Equality occurs when $x_1 = x_2 = x_3 = \\frac{1}{\\sqrt{3}}$ and $x_4 = x_5 = \\dots = x_{100} = 0,$ so the minimum value is $\\boxed{\\frac{3 \\sqrt{3}}{2}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: What is the sum of the 2009 fractions of the form $\\frac{2}{n(n+2)}$ if the values of $n$ are the positive integers from 1 through 2009? Express your answer as a decimal to the nearest thousandth.", "role": "user" }, { "content": "We are asked to find \\[\n\\frac{2}{1\\cdot3}+\\frac{2}{2\\cdot4}\n+\\frac{2}{3\\cdot5}\n+\\frac{2}{4\\cdot6}+\\cdots+\\frac{2}{2009\\cdot2011}.\n\\] Observe that $\\frac{2}{n(n+2)}$ may be written as $\\frac{1}{n}-\\frac{1}{n+2}$. Applying this identity, our sum becomes \\[\n\\frac{1}{1}-\\frac{1}{3}+\\frac{1}{2}-\\frac{1}{4}\n+\\frac{1}{3}-\\frac{1}{5}\n+\\frac{1}{4}-\\frac{1}{6}+\\cdots+\\frac{1}{2009}-\\frac{1}{2011}.\n\\] Every negative term cancels with the term three places to the right. The only terms which remain are \\[\n1+\\frac{1}{2}-\\frac{1}{2010}-\\frac{1}{2011}.\n\\] To the nearest thousandth, the sum is $\\boxed{1.499}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $x$ and $y$ be real numbers such that\n\\[4x^2 + 8xy + 5y^2 = 1.\\]Let $m$ and $M$ be the minimum and maximum values of $2x^2 + 3xy + 2y^2,$ respectively. Find the product $mM.$", "role": "user" }, { "content": "Let $k = 2x^2 + 3xy + 2y^2.$ Then\n\\[2x^2 + 3xy + 2y^2 = k = k(4x^2 + 8xy + 5y^2) = 4kx^2 + 8kxy + 5ky^2 = 0,\\]so $(4k - 2) x^2 + (8k - 3) xy + (5k - 2) y^2 = 0.$\n\nIf $y = 0,$ then $4x^2 = 1,$ so\n\\[2x^2 + 3xy + 2y^2 = \\frac{1}{2}.\\]Otherwise, we can divide both sides of $(4k - 2) x^2 + (8k - 3) xy + (5k - 2) y^2 = 0$ by $y^2,$ to get\n\\[(4k - 2) \\left( \\frac{x}{y} \\right)^2 + (8k - 3) \\frac{x}{y} + (5k - 2) = 0.\\]This is a quadratic in $\\frac{x}{y},$ so and its discriminant must be nonnegative:\n\\[(8k - 3)^2 - 4 (4k - 2)(5k - 2) \\ge 0.\\]This simplifies to $-16k^2 + 24k - 7 \\ge 0,$ or $16k^2 - 24k + 7 \\le 0.$ The roots of the quadratic $16k^2 - 24k + 7 = 0$ are $\\frac{3 \\pm \\sqrt{2}}{4},$ so the solution to $16k^2 - 24k + 7 \\le 0$ is\n\\[\\frac{3 - \\sqrt{2}}{4} \\le k \\le \\frac{3 + \\sqrt{2}}{4}.\\]For any value of $k$ in this interval, we can take $x = ky,$ then substitute into $4x^2 + 8xy + 5y^2 = 1,$ and obtain solutions in $x$ and $y.$ Thus, $m = \\frac{3 - \\sqrt{2}}{4}$ and $M = \\frac{3 + \\sqrt{2}}{4},$ so $mM = \\boxed{\\frac{7}{16}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The graph of $y = f(x)$ is shown below.\n\n[asy]\nunitsize(0.5 cm);\n\nreal func(real x) {\n real y;\n if (x >= -3 && x <= 0) {y = -2 - x;}\n if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}\n if (x >= 2 && x <= 3) {y = 2(x - 2);}\n return(y);\n}\n\nint i, n;\n\nfor (i = -5; i <= 5; ++i) {\n draw((i,-5)--(i,5),gray(0.7));\n draw((-5,i)--(5,i),gray(0.7));\n}\n\ndraw((-5,0)--(5,0),Arrows(6));\ndraw((0,-5)--(0,5),Arrows(6));\n\nlabel(\"$x$\", (5,0), E);\nlabel(\"$y$\", (0,5), N);\n\ndraw(graph(func,-3,3),red);\n\nlabel(\"$y = f(x)$\", (3,-2), UnFill);\n[/asy]\n\nWhich is the graph of $y = \\frac{1}{2} f(x) + 3$?\n\n[asy]\nunitsize(0.5 cm);\n\npicture[] graf;\nint i, n;\n\nreal func(real x) {\n real y;\n if (x >= -3 && x <= 0) {y = -2 - x;}\n if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}\n if (x >= 2 && x <= 3) {y = 2(x - 2);}\n return(y);\n}\n\nreal funcc(real x) {\n return(1/2func(x) + 3);\n}\n\nfor (n = 1; n <= 5; ++n) {\n graf[n] = new picture;\n for (i = -5; i <= 5; ++i) {\n draw(graf[n],(i,-5)--(i,5),gray(0.7));\n draw(graf[n],(-5,i)--(5,i),gray(0.7));\n }\n draw(graf[n],(-5,0)--(5,0),Arrows(6));\n draw(graf[n],(0,-5)--(0,5),Arrows(6));\n\n label(graf[n],\"$x$\", (5,0), E);\n label(graf[n],\"$y$\", (0,5), N);\n}\n\ndraw(graf[1],shift((0,3))xscale(1/2)graph(func,-3,3),red);\ndraw(graf[2],shift((0,-2))yscale(1/2)graph(func,-3,3),red);\ndraw(graf[3],graph(funcc,-3,3),red);\ndraw(graf[4],yscale(1/2)graph(func,-3,3),red);\ndraw(graf[5],xscale(1/2)graph(func,-3,3),red);\n\nlabel(graf[1], \"A\", (0,-6));\nlabel(graf[2], \"B\", (0,-6));\nlabel(graf[3], \"C\", (0,-6));\nlabel(graf[4], \"D\", (0,-6));\nlabel(graf[5], \"E\", (0,-6));\n\nadd(graf[1]);\nadd(shift((12,0))(graf[2]));\nadd(shift((24,0))(graf[3]));\nadd(shift((6,-12))(graf[4]));\nadd(shift((18,-12))*(graf[5]));\n[/asy]\n\nEnter the letter of the graph of $y = \\frac{1}{2} f(x) + 3.$", "role": "user" }, { "content": "The graph of $y = \\frac{1}{2} f(x)$ is produced by taking the graph of $y = f(x)$ and compressing it vertically by a factor of $\\frac{1}{2}.$ We then get the graph of $y = \\frac{1}{2} f(x) + 3$ by shifting upward by three units. The correct graph is $\\boxed{\\text{C}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: If $x+7$ is a factor of $cx^3 + 19x^2 - 3cx + 35$, find the constant $c$.", "role": "user" }, { "content": "Though it is possible to solve this problem using polynomial long division, it is quicker to use the Factor Theorem.\n\nLet $f(x) = cx^3 + 19x^2 - 3cx + 35$. If $x+7$ is a factor of $f(x)$, the factor theorem tells us that $f(-7) = 0.$ Then\n\\[c(-7)^3 + 19(-7)^2 - 3c(-7) + 35 = 0,\\]which simplifies to $-322c + 966 = 0.$ We can solve for $c$ to get $c = \\boxed{3}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the sum of all the solutions to $2^{\|x\|} + 3\|x\| = 18.$", "role": "user" }, { "content": "If $x$ is a solution, then $-x$ is a also a solution. Thus, we can pair all the solutions, and their sum is $\\boxed{0}.$\n\nLet $f(x) = 2^{\|x\|} + 3\|x\|.$ Since $f(0) = 0$ and $f(4) = 28,$ the equation $f(x) = 18$ has at least one solution in the interval $0 \\le x \\le 4.$ This ensures that the sum that the problem asks for is not an \"empty\" sum.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: A region $S$ in the complex plane is defined by \\begin{align}\nS = \\{x + iy: - 1\\le x\\le1, - 1\\le y\\le1\\}.\n\\end{align}A complex number $z = x + iy$ is chosen uniformly at random from $S$. What is the probability that $\\left(\\frac34 + \\frac34i\\right)z$ is also in $S$?", "role": "user" }, { "content": "We can directly compute\n\\[\\left(\\frac34 + \\frac34i\\right)z = \\left(\\frac34 + \\frac34i\\right)(x + iy) = \\frac{3(x-y)}4 + \\frac{3(x+y)}4 \\cdot i.\\]This number is in $S$ if and only if $-1 \\leq \\frac{3(x-y)}4 \\leq 1$ and at the same time $-1 \\leq \\frac{3(x+y)}4 \\leq 1$. This simplifies to $\|x-y\|\\leq\\frac 43$ and $\|x+y\|\\leq\\frac 43$.\n\nLet $T = \\{ x + iy : \|x-y\|\\leq\\frac 43 \\ \\text{and} \\ \|x+y\|\\leq\\frac 43 \\}$, and let $[X]$ denote the area of the region $X$. Then, the probability we seek is $\\frac {[S\\cap T]}{[S]} = \\frac{[S\\cap T]}4$. All we need to do is to compute the area of the intersection of $S$ and $T$. It is easiest to do this graphically:\n\n[asy]\nunitsize(2cm);\ndefaultpen(0.8);\npath s = (-1,-1) -- (-1,1) -- (1,1) -- (1,-1) -- cycle;\npath t = (4/3,0) -- (0,4/3) -- (-4/3,0) -- (0,-4/3) -- cycle;\npath s_cap_t = (1/3,1) -- (1,1/3) -- (1,-1/3) -- (1/3,-1) -- (-1/3,-1) -- (-1,-1/3) -- (-1,1/3) -- (-1/3,1) -- cycle;\nfilldraw(s, lightred, black);\nfilldraw(t, lightgreen, black);\nfilldraw(s_cap_t, lightyellow, black);\ndraw( (-5/3,0) -- (5/3,0), dashed );\ndraw( (0,-5/3) -- (0,5/3), dashed );\n[/asy]\n\nCoordinate axes are dashed, $S$ is shown in red, $T$ in green and their intersection is yellow. The intersections of the boundary of $S$ and $T$ are obviously at $(\\pm 1,\\pm 1/3)$ and at $(\\pm 1/3,\\pm 1)$.\n\nHence, each of the four red triangles is an isosceles right triangle with legs of length $\\frac 23$, and the area of a single red triangle is $\\frac 12 \\cdot \\left( \\frac 23 \\right)^2 = \\frac 29$. Then, the area of all four is $\\frac 89$, and therefore the area of $S\\cap T$ is $4 - \\frac 89$. Thus, the probability we seek is $\\frac{ [S\\cap T]}4 = \\frac{ 4 - \\frac 89 }4 = 1 - \\frac 29 = \\boxed{\\frac 79}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the equation of the directrix of the parabola $y = \\frac{x^2 - 6x + 5}{12}.$", "role": "user" }, { "content": "Recall that a parabola is defined as the set of all points that are equidistant to the focus $F$ and the directrix. Completing the square on $x,$ we get\n\\[y = \\frac{1}{12} (x - 3)^2 - \\frac{1}{3}.\\]To make the algebra a bit easier, we can find the directrix of the parabola $y = \\frac{1}{12} x^2,$ shift the parabola right by 3 units to get $y = \\frac{1}{12} (x - 3)^2$ (which does not change the directrix), and then shift it downward $\\frac{1}{3}$ units to find the directrix of the parabola $y = \\frac{1}{12} (x - 3)^2 - \\frac{1}{3}.$\n\nSince the parabola $y = \\frac{1}{12} x^2$ is symmetric about the $y$-axis, the focus is at a point of the form $(0,f).$ Let $y = d$ be the equation of the directrix.\n\n[asy]\nunitsize(1.5 cm);\n\npair F, P, Q;\n\nF = (0,1/4);\nP = (1,1);\nQ = (1,-1/4);\n\nreal parab (real x) {\n return(x^2);\n}\n\ndraw(graph(parab,-1.5,1.5),red);\ndraw((-1.5,-1/4)--(1.5,-1/4),dashed);\ndraw(P--F);\ndraw(P--Q);\n\ndot(\"$F$\", F, NW);\ndot(\"$P$\", P, E);\ndot(\"$Q$\", Q, S);\n[/asy]\n\nLet $\\left( x, \\frac{1}{12} x^2 \\right)$ be a point on the parabola $y = \\frac{1}{12} x^2.$ Then\n\\[PF^2 = x^2 + \\left( \\frac{1}{12} x^2 - f \\right)^2\\]and $PQ^2 = \\left( \\frac{1}{12} x^2 - d \\right)^2.$ Thus,\n\\[x^2 + \\left( \\frac{1}{12} x^2 - f \\right)^2 = \\left( \\frac{1}{12} x^2 - d \\right)^2.\\]Expanding, we get\n\\[x^2 + \\frac{1}{144} x^4 - \\frac{f}{6} x^2 + f^2 = \\frac{1}{144} x^4 - \\frac{d}{6} x^2 + d^2.\\]Matching coefficients, we get\n\\begin{align}\n1 - \\frac{f}{6} &= -\\frac{d}{6}, \\\\\nf^2 &= d^2.\n\\end{align}From the first equation, $f - d = 6.$ Since $f^2 = d^2,$ $f = d$ or $f = -d.$ We cannot have $f = d,$ so $f = -d.$ Then $-2d = 6,$ so $d = -3.$\n\nThus, the equation of the directrix of $y = \\frac{1}{12} x^2$ is $y = -3,$ so the equation of the directrix of $y = \\frac{1}{12} (x - 3)^2 - \\frac{1}{3}$ is $\\boxed{y = -\\frac{10}{3}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the domain of the function $z(x) = \\sqrt[3]{x - 1} + \\sqrt[3]{8 - x}.$", "role": "user" }, { "content": "Since we can take the cube root of any real number (positive or negative), $z(x) = \\sqrt[3]{x - 1} + \\sqrt[3]{8 - x}$ is defined for all real numbers $x.$ Thus, the domain of $z(x)$ is $\\boxed{(-\\infty,\\infty)}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a,$ $b,$ and $c$ be constants, and suppose that the inequality \\[\\frac{(x-a)(x-b)}{x-c} \\le 0\\]is true if and only if either $x < -4$ or $\|x-25\| \\le 1.$ Given that $a < b,$ find the value of $a + 2b + 3c.$", "role": "user" }, { "content": "We first unpack the statement $x < -4$ or $\|x- 25 \| \\le 1.$ The inequality $\|x-25\| \\le 1$ is equivalent to $-1 \\le x-25 \\le 1,$ which is in turn equivalent to $24 \\le x \\le 26.$ Therefore, we have either $x < -4$ or $24 \\le x \\le 26,$ so the solution set for $x$ is \\[(-\\infty, -4) \\cup [24, 26].\\]The sign of the expression $\\frac{(x-a)(x-b)}{x-c}$ changes at $x = a,$ $x = b,$ and $x = c,$ which means that $a,$ $b,$ and $c$ must be the numbers $-4,$ $24,$ and $26,$ in some order. Furthermore, since $24$ and $26$ are endpoints of a closed interval (that is, they are included in the solution set), it must be the case that $a$ and $b$ are $24$ and $26$ in some order, because the inequality is true when $x=a$ or $x=b,$ but is not true when $x=c$ (since that would make the denominator zero). Since $a < b,$ we have $a = 24$ and $b = 26,$ and then $c = -4.$\n\nIn conclusion, the given inequality must be \\[\\frac{(x-24)(x-26)}{x+4} \\le 0.\\]To check that the solution to this inequality is $(-\\infty, -4) \\cup [24, 26],$ we can build a sign table, where $f(x)$ is the expression on the left-hand side: \\begin{tabular}{c\|ccc\|c} &$x-24$ &$x-26$ &$x+4$ &$f(x)$ \\\\ \\hline$x<-4$ &$-$&$-$&$-$&$-$\\\\ [.1cm]$-4<x<24$ &$-$&$-$&$+$&$+$\\\\ [.1cm]$24<x<26$ &$+$&$-$&$+$&$-$\\\\ [.1cm]$x>26$ &$+$&$+$&$+$&$+$\\\\ [.1cm]\\end{tabular}This shows that $f(x) < 0$ when $x \\in (-\\infty, -4) \\cup (24, 26),$ and since $f(x) = 0$ for $x \\in \\{24, 26\\},$ we indeed have the solution set \\[x \\in (-\\infty, -4) \\cup [24, 26].\\]Thus, $a+2b+3c=24+2(26) + 3(-4) = \\boxed{64}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $f(x) = 4x^4+12x^3-9x^2+x+3$ and $d(x) = x^2+3x-2$. If $f(x) = q(x)d(x) + r(x)$ for some polynomials $q(x)$ and $r(x)$ where $\\deg r < \\deg d$, calculate $q(1)+r(-1)$.", "role": "user" }, { "content": "\\[\n\\begin{array}{c\|cc ccc}\n\\multicolumn{2}{r}{4x^2} & -1 \\\\\n\\cline{2-6}\nx^2+3x-2 & 4x^4 & +12x^3&-9x^2&+x&+3 \\\\\n\\multicolumn{2}{r}{-4x^4} & -12x^3&+8x^2 \\\\ \n\\cline{2-4}\n\\multicolumn{2}{r}{0} & 0 & -x^2 &+x & +3 \\\\\n\\multicolumn{2}{r}{} & & +x^2 &+3x&-2 \\\\ \n\\cline{4-6}\n\\multicolumn{2}{r}{} & & 0 & 4x &+1 \\\\\n\\end{array}\n\\]Since $\\deg d > \\deg (4x+1)$ we cannot divide any further. So, $q(x) = 4x^2-1$ and $r(x)=4x+1$. Then\n$$q(1)+r(-1) = 4(1)^2+1+4(-1)-1=\\boxed{0}.$$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The solution to the inequality\n\\[y = -x^2 + ax + b \\le 0\\]is $(-\\infty,-3] \\cup [5,\\infty).$ Find the vertex of the parabola $y = -x^2 + ax + b.$", "role": "user" }, { "content": "The roots of the quadratic are $-3$ and 5, so\n\\[y = -x^2 + ax + b = -(x + 3)(x - 5) = -x^2 + 2x + 15 = -(x - 1)^2 + 16.\\]Thus, the vertex is $\\boxed{(1,16)}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Two positive real numbers have geometric mean $\\sqrt{3}$ and harmonic mean $\\frac{3}{2}.$ Enter the two numbers, separated by commas.", "role": "user" }, { "content": "Let the two numbers be $a$ and $b.$ Then $\\sqrt{ab} = \\sqrt{3},$ so $ab = 3.$ Also,\n\\[\\frac{2}{\\frac{1}{a} + \\frac{1}{b}} = \\frac{2ab}{a + b} = \\frac{3}{2},\\]so $a + b = \\frac{4}{3} ab = 4.$\n\nThen by Vieta's formulas, $a$ and $b$ are the roots of the quadratic\n\\[x^2 - 4x + 3 = (x - 1)(x - 3),\\]so the two numbers are $\\boxed{1,3}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The graph of $y = f(x)$ is shown below.\n\n[asy]\nunitsize(0.5 cm);\n\nreal func(real x) {\n real y;\n if (x >= -3 && x <= 0) {y = -2 - x;}\n if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}\n if (x >= 2 && x <= 3) {y = 2(x - 2);}\n return(y);\n}\n\nint i, n;\n\nfor (i = -5; i <= 5; ++i) {\n draw((i,-5)--(i,5),gray(0.7));\n draw((-5,i)--(5,i),gray(0.7));\n}\n\ndraw((-5,0)--(5,0),Arrows(6));\ndraw((0,-5)--(0,5),Arrows(6));\n\nlabel(\"$x$\", (5,0), E);\nlabel(\"$y$\", (0,5), N);\n\ndraw(graph(func,-3,3),red);\n\nlabel(\"$y = f(x)$\", (3,-2), UnFill);\n[/asy]\n\nWhich is the graph of $y = f(\|x\|)$?\n\n[asy]\nunitsize(0.5 cm);\n\npicture[] graf;\nint i, n;\n\nreal func(real x) {\n real y;\n if (x >= -3 && x <= 0) {y = -2 - x;}\n if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}\n if (x >= 2 && x <= 3) {y = 2(x - 2);}\n return(y);\n}\n\nreal funca(real x) {\n return(func(abs(x)));\n}\n\nreal funcb(real x) {\n real y = max(0,func(x));\n return(y);\n}\n\nreal funcd(real x) {\n return(abs(func(x)));\n}\n\nreal funce(real x) {\n return(func(-abs(x)));\n}\n\nfor (n = 1; n <= 5; ++n) {\n graf[n] = new picture;\n for (i = -5; i <= 5; ++i) {\n draw(graf[n],(i,-5)--(i,5),gray(0.7));\n draw(graf[n],(-5,i)--(5,i),gray(0.7));\n }\n draw(graf[n],(-5,0)--(5,0),Arrows(6));\n draw(graf[n],(0,-5)--(0,5),Arrows(6));\n\n label(graf[n],\"$x$\", (5,0), E);\n label(graf[n],\"$y$\", (0,5), N);\n}\n\ndraw(graf[1],graph(funca,-3,3),red);\ndraw(graf[2],graph(funcb,-3,3),red);\ndraw(graf[3],reflect((0,0),(0,1))graph(func,-3,3),red);\ndraw(graf[4],graph(funcd,-3,3),red);\ndraw(graf[5],graph(funce,-3,3),red);\n\nlabel(graf[1], \"A\", (0,-6));\nlabel(graf[2], \"B\", (0,-6));\nlabel(graf[3], \"C\", (0,-6));\nlabel(graf[4], \"D\", (0,-6));\nlabel(graf[5], \"E\", (0,-6));\n\nadd(graf[1]);\nadd(shift((12,0))(graf[2]));\nadd(shift((24,0))(graf[3]));\nadd(shift((6,-12))(graf[4]));\nadd(shift((18,-12))*(graf[5]));\n[/asy]\n\nEnter the letter of the graph of $y = f(\|x\|).$", "role": "user" }, { "content": "If $x \\ge 0,$ then $f(\|x\|) = f(x).$ And if $x < 0,$ then $f(\|x\|) = f(-x).$ Thus, the graph of $y = \|f(x)\|$ is obtained by taking the part of the graph of $y = f(x)$ that is to the right of the $y$-axis, and making a copy by reflecting it across the $y$-axis. The correct graph is $\\boxed{\\text{A}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $f(x)=x^4+14x^3+52x^2+56x+16$. Let $z_1,z_2,z_3,z_4$ be the four roots of $f$. Find the smallest possible value of $\|z_{a}z_{b}+z_{c}z_{d}\|$ where $\\{a,b,c,d\\}=\\{1,2,3,4\\}$.", "role": "user" }, { "content": "Note that\n\\[\\frac{1}{16}f(2x)=x^4+7x^3+13x^2+7x+1.\\]Because the coefficients of this polynomial are symmetric, if $r$ is a root of $f(x)$ then $\\frac{4}{r}$ is as well. Further, $f(-1)=-1$ and $f(-2)=16$ so $f(x)$ has two distinct roots on $(-2,0)$ and two more roots on $(-\\infty,-2)$. Now, if $\\sigma$ is a permutation of $\\{1,2,3,4\\}$:\n\\[\|z_{\\sigma(1)} z_{\\sigma(2)} + z_{\\sigma(3)} z_{\\sigma(4)}\| \\le \\frac{1}{2} (z_{\\sigma(1)} z_{\\sigma(2)} + z_{\\sigma(3)} z_{\\sigma(4)} + z_{\\sigma(4)} z_{\\sigma(3)} + z_{\\sigma(2)}z_{\\sigma(1)}).\\]Let the roots be ordered $z_1 \\le z_2 \\le z_3 \\le z_4$, then by rearrangement the last expression is at least:\n\\[\\frac{1}{2}(z_1z_4+z_2z_3+z_3z_2+z_4z_1).\\]Since the roots come in pairs $z_1z_4=z_2z_3=4$, our expression is minimized when $\\sigma(1)=1,\\sigma(2)=4,\\sigma(3)=3,\\sigma(4)=2$ and its minimum value is $\\boxed{8}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the smallest positive integer $a$ such that $x^4 + a^2$ is not prime for any integer $x.$", "role": "user" }, { "content": "For $1 \\le a \\le 7,$ we give a value of $x$ for which $x^4 + a^2$ is prime:\n\\[\n\\begin{array}{c\|c\|c}\na & x & a^4 + x^2 \\\\ \\hline\n1 & 1 & 2 \\\\\n2 & 1 & 5 \\\\\n3 & 10 & 10009 \\\\\n4 & 1 & 17 \\\\\n5 & 2 & 41 \\\\\n6 & 1 & 37 \\\\\n7 & 20 & 160049\n\\end{array}\n\\]For $a = 8,$\n\\begin{align}\nx^4 + a^2 &= x^4 + 64 \\\\\n&= x^4 + 16x^2 + 64 - 16x^2 \\\\\n&= (x^2 + 8)^2 - (4x)^2 \\\\\n&= (x^2 + 4x + 8)(x^2 - 4x + 8).\n\\end{align}For any positive integer, both factors $x^2 + 4x + 8$ and $x^2 - 4x + 8$ are greater than 1, so $x^4 + 64$ is always composite. Thus, the smallest such $a$ is $\\boxed{8}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Given that $\\alpha$ and $\\beta$ are the roots of $x^2 - 2x - 1 = 0,$ find $5 \\alpha^4 + 12 \\beta^3.$", "role": "user" }, { "content": "If $x$ satisfies $x^2 - 2x - 1 = 0,$ then\n\\begin{align}\nx^2 &= 2x + 1, \\\\\nx^3 &= x(2x + 1) = 2x^2 + x = 2(2x + 1) + x = 5x + 2, \\\\\nx^4 &= x(5x + 2) = 5x^2 + 2x = 5(2x + 1) + 2x = 12x + 5.\n\\end{align}Hence,\n\\begin{align}\n5 \\alpha^4 + 12 \\beta^3 &= 5(12 \\alpha + 5) + 12 (5 \\beta + 2) \\\\\n&= 60 \\alpha + 25 + 60 \\beta + 24 \\\\\n&= 60 (\\alpha + \\beta) + 49 \\\\\n&= 60 \\cdot 2 + 49 \\\\\n&= \\boxed{169}.\n\\end{align}", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The parabola $y=ax^2+bx+c$ has vertex $(p,p)$ and $y$-intercept $(0,-p)$, where $p\\ne 0$. What is $b$?", "role": "user" }, { "content": "Since the vertex is $(p,p),$ the parabola is of the form\n\\[y = a(x - p)^2 + p.\\]Setting $x = 0,$ we get $y = ap^2 + p = -p,$ so $a = -\\frac{2}{p}.$ Then\n\\[y = -\\frac{2}{p} (x^2 - 2px + p^2) + p = -\\frac{2}{p} x^2 + 4x - p,\\]so $b = \\boxed{4}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $f(x) = \\frac{3}{9^x + 3}.$ Find\n\\[f \\left( \\frac{1}{1001} \\right) + f \\left( \\frac{2}{1001} \\right) + f \\left( \\frac{3}{1001} \\right) + \\dots + f \\left( \\frac{1000}{1001} \\right).\\]", "role": "user" }, { "content": "Note that\n\\begin{align}\nf(x) + f(1 - x) &= \\frac{3}{9^x + 3} + \\frac{3}{9^{1 - x} + 3} \\\\\n&= \\frac{3}{9^x + 3} + \\frac{3 \\cdot 9^x}{9 + 3 \\cdot 9^x} \\\\\n&= \\frac{3}{9^x + 3} + \\frac{9^x}{3 + 9^x} \\\\\n&= \\frac{3 + 9^x}{9^x + 3} \\\\\n&= 1.\n\\end{align}Thus, we can pair the 1000 terms in the sum into 500 pairs, such that the sum of the terms in each pair is 1. Therefore, the sum is equal to $\\boxed{500}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find all values of $k$ so that\n\\[x^2 - (k - 3) x - k + 6 > 0\\]for all $x.$", "role": "user" }, { "content": "If we graph $y = x^2 - (k - 3) x - k + 6,$ then we obtain an upward-facing parabola. Thus, the inequality\n\\[x^2 - (k - 3) x - k + 6 > 0\\]holds as long as the discriminant of the quadratic is negative.\n\nThis gives us\n\\[(k - 3)^2 - 4(-k + 6) < 0.\\]This simplifies to $k^2 - 2k - 15 < 0,$ which factors as $(k + 3)(k - 5) < 0.$ Thus, $k \\in \\boxed{(-3,5)}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $A := \\mathbb{Q} \\setminus \\{0,1\\}$ denote the set of all rationals other than 0 and 1. A function $f : A \\rightarrow \\mathbb{R}$ has the property that for all $x \\in A$,\n\\[\nf\\left( x\\right) + f\\left( 1 - \\frac{1}{x}\\right) = \\log\\lvert x\\rvert.\n\\]Compute the value of $f(2007)$. Enter your answer in the form \"$\\log(a)$\", where $a$ is some number.", "role": "user" }, { "content": "Let $g : A \\to A$ be defined by $g(x) := 1-1/x$; the key property is that \\[\ng(g(g(x))) = 1-\\frac{1}{1-\\frac{1}{1-\\frac{1}{x}}} = x.\n\\]The given equation rewrites as $f(x) + f(g(x)) = \\log\|x\|$. Substituting $x=g(y)$ and $x=g(g(z))$ gives the further equations $f(g(y)) + f(g) g(y)) = \\log\|g(x)\|$ and $f(g) g(z)) + f(z) = \\log\|g(g(x))\|.$ Setting $y$ and $z$ to $x$ and solving the system of three equations for $f(x)$ gives \\[\nf(x) = \\frac{1}{2} \\cdot \\left (\\log\|x\| - \\log\|g(x)\| + \\log\|g(g(x))\| \\right).\n\\]For $x=2007$, we have $g(x) = \\frac{2006}{2007}$ and $g(g(x)) = \\frac{-1}{2006}$, so that \\[\nf(2007) = \\frac{\\log\|2007\| - \\log\\left\|\\frac{2006}{2007}\\right\| + \\log\\left\|\\frac{-1}{2006}\\right\|}{2} = \\boxed{\\log\\left(\\frac{2007}{2006}\\right)}.\n\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: A real number $a$ is chosen randomly and uniformly from the interval $[-20, 18]$. Find the probability that the roots of the polynomial\n\\[x^4 + 2ax^3 + (2a - 2)x^2 + (-4a + 3)x - 2\\]are all real.", "role": "user" }, { "content": "Let $p(x)$ be the given polynomial. Notice that \\[p(1) = 1 + (2a) + (2a-2) - (4a+3) - 2 = 0,\\]so $1$ is a root of $p(x).$ Performing polynomial division, we then have \\[p(x) = (x-1)(x^3+(2a+1)x^2+(4a-1)x+2).\\]Notice that \\[p(-2) = 1 \\cdot (-8 + 4(2a+1) - 2(4a-1) + 2) = 0,\\]so $-2$ is a root of $p(x)$ as well. Dividing the cubic term by $x+2,$ we then have \\[p(x) = (x-1)(x+2)(x^2+(2a-1)x+1).\\]Therefore, we want to find the probability that the roots of $x^2 + (2a-1)x + 1$ are all real. This occurs if and only if the discriminant is nonnegative: \\[(2a-1)^2 - 4 \\ge 0,\\]or $(2a-1)^2 \\ge 4.$ Thus, either $2a-1 \\ge 2$ or $2a-1 \\le -2.$ The first inequality is equivalent to $a \\ge \\tfrac{3}{2},$ and the second is equivalent to $a \\le -\\tfrac{1}{2}.$ This shows that all values of $a$ except those in the interval $\\left(-\\tfrac12, \\tfrac32\\right)$ satisfy the condition. This interval has length $2,$ and the given interval $[-20, 18],$ which contains it completely, has length $18 - (-20) = 38,$ so the probability is \\[1 - \\frac{2}{38} = \\boxed{\\frac{18}{19}}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Compute the length of the segment tangent from the origin to the circle that passes through the points $(3,4),$ $(6,8),$ and $(5,13).$", "role": "user" }, { "content": "Let $O = (0,0),$ $A = (3,4),$ $B = (6,8),$ and $C = (5,13).$ Let $T$ be a point on the circumcircle of triangle $ABC,$ so that $\\overline{OT}$ is tangent to the circumcircle. Note that $O,$ $A,$ and $B$ are collinear.\n\n[asy]\nunitsize(0.4 cm);\n\npair A, B, C, O, T;\n\nA = (3,4);\nB = (6,8);\nC = (5,13);\nO = circumcenter(A,B,C);\nT = intersectionpoints(Circle(O/2,abs(O)/2),circumcircle(A,B,C))[1];\n\ndraw(circumcircle(A,B,C));\ndraw((0,0)--(6,8));\ndraw((0,0)--T);\ndraw((-10,0)--(10,0));\ndraw((0,-2)--(0,18));\n\nlabel(\"$O = (0,0)$\", (0,0), SW);\n\ndot(\"$A = (3,4)$\", A, SE);\ndot(\"$B = (6,8)$\", B, E);\ndot(\"$C = (5,13)$\", C, NE);\ndot(\"$T$\", T, SW);\n[/asy]\n\nThen by power of a point, $OT^2 = OA \\cdot OB = 5 \\cdot 10 = 50,$ so $OT = \\sqrt{50} = \\boxed{5 \\sqrt{2}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Multiply $(2x^3-5y^2)(4x^6+10x^3y^2+25y^4)$.", "role": "user" }, { "content": "The given product can be rewritten in the form $(a-b)(a^2+ab+b^2)$, which is the factorization of $a^3-b^3$ for $a=2x^3$ and $b=5y^2$. Therefore, the expression can be rewritten as $a^3-b^3=(2x^3)^3-(5y^2)^3=\\boxed{8x^9-125y^6}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find all values of $x$ that satisfy \\[\\frac{x^2}{x+1} \\ge \\frac{2}{x-1} + \\frac{5}{4}.\\]", "role": "user" }, { "content": "Combining the terms on the right-hand side, we have \\[\\frac{x^2}{x+1} \\ge \\frac{5x+3}{4(x-1)}.\\]Then, moving all the terms to the left-hand side and combining denominators again, we get \\[\\begin{aligned} \\frac{x^2}{x+1} - \\frac{5x+3}{4(x-1)} &\\ge 0 \\\\ \\frac{4x^2(x-1)-(x+1)(5x+3)}{(x+1)(x-1)} &\\ge 0 \\\\ \\frac{4x^3-9x^2-8x-3}{(x+1)(x-1)} &\\ge 0. \\end{aligned}\\]We try to factor the numerator. Using the rational root theorem to test for rational roots, we see that $x=3$ is a root of $4x^3-9x^2-8x-3.$ Then, doing the polynomial division gives \\[4x^3-9x^2-8x-3 = (x-3)(4x^2+3x+1),\\]so we have \\[\\frac{(x-3)(4x^2+3x+1)}{(x+1)(x-1)} \\ge 0.\\]Since $4x^2+3x+1$ has a positive $x^2$ coefficient, and its discriminant is $3^2 - 4 \\cdot 4= -7,$ which is negative, it follows that $4x^2 + 3x + 1 > 0$ for all $x.$ Thus, the above inequality is equivalent to \\[f(x) = \\frac{x-3}{(x+1)(x-1)} \\ge 0.\\]We make a sign table for $f(x)$: \\begin{tabular}{c\|ccc\|c} &$x-3$ &$x+1$ &$x-1$ &$f(x)$ \\\\ \\hline$x<-1$ &$-$&$-$&$-$&$-$\\\\ [.1cm]$-1<x<1$ &$-$&$+$&$-$&$+$\\\\ [.1cm]$1<x<3$ &$-$&$+$&$+$&$-$\\\\ [.1cm]$x>3$ &$+$&$+$&$+$&$+$\\\\ [.1cm]\\end{tabular}We see that $f(x) > 0$ when $-1 < x < 1$ or $x > 3.$ Since the inequality is nonstrict, we also include the values of $x$ such that $f(x) = 0,$ that is, only $x = 3.$ Therefore, the solution to the inequality is \\[x \\in \\boxed{(-1, 1) \\cup [3, \\infty)}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find all real values of $x$ that satisfy $\\frac{1}{x+1} + \\frac{3}{x+7} \\ge \\frac23.$ (Give your answer in interval notation.)", "role": "user" }, { "content": "Moving all the terms to the left-hand side, we have \\[\\frac{1}{x+1} + \\frac{3}{x+7} -\\frac23 \\ge 0.\\]To solve this inequality, we find a common denominator: \\[\\frac{3(x+7) + 3 \\cdot 3(x+1) - 2(x+1)(x+7)}{3(x+1)(x+7)} \\ge 0,\\]which simplifies to \\[-\\frac{2(x+4)(x-2)}{3(x+1)(x+7)} \\ge 0.\\]Therefore, we want the values of $x$ such that \\[f(x) = \\frac{(x+4)(x-2)}{(x+1)(x+7)} \\le 0.\\]To do this, we make the following sign table: \\begin{tabular}{c\|cccc\|c} &$x+4$ &$x-2$ &$x+1$ &$x+7$ &$f(x)$ \\\\ \\hline$x<-7$ &$-$&$-$&$-$&$-$&$+$\\\\ [.1cm]$-7<x<-4$ &$-$&$-$&$-$&$+$&$-$\\\\ [.1cm]$-4<x<-1$ &$+$&$-$&$-$&$+$&$+$\\\\ [.1cm]$-1<x<2$ &$+$&$-$&$+$&$+$&$-$\\\\ [.1cm]$x>2$ &$+$&$+$&$+$&$+$&$+$\\\\ [.1cm]\\end{tabular}Because the inequality $f(x) \\le 0$ is nonstrict, we must also include the values of $x$ such that $f(x) = 0,$ which are $x=-4$ and $x=2.$ Putting it all together, the solutions to the inequality are \\[x \\in \\boxed{(-7, -4] \\cup (-1, 2]}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Is\n\\[f(x) = \\frac{1}{2^x - 1} + \\frac{1}{2}\\]an even function, odd function, or neither?\n\nEnter \"odd\", \"even\", or \"neither\".", "role": "user" }, { "content": "We have that\n\\begin{align}\nf(-x) &= \\frac{1}{2^{-x} - 1} + \\frac{1}{2} \\\\\n&= \\frac{2^x}{1 - 2^x} + \\frac{1}{2} \\\\\n&= \\frac{1 - (1 - 2^x)}{1 - 2^x} + \\frac{1}{2} \\\\\n&= \\frac{1}{1 - 2^x} - 1 + \\frac{1}{2} \\\\\n&= \\frac{1}{1 - 2^x} - \\frac{1}{2} \\\\\n&= -\\frac{1}{2^x - 1} - \\frac{1}{2} \\\\\n&= -f(x),\n\\end{align}so $f(x)$ is an $\\boxed{\\text{odd}}$ function.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $F_1 = (0,1)$ and $F_ 2= (4,1).$ Then the set of points $P$ such that\n\\[PF_1 + PF_2 = 6\\]form an ellipse. The equation of this ellipse can be written as\n\\[\\frac{(x - h)^2}{a^2} + \\frac{(y - k)^2}{b^2} = 1.\\]Find $h + k + a + b.$", "role": "user" }, { "content": "We have that $2a = 6,$ so $a = 3.$ The distance between the foci is $2c = 4,$ so $c = 2.$ Hence, $b = \\sqrt{a^2 - c^2} = \\sqrt{5}.$\n\nThe center of the ellipse is the midpoint of $\\overline{F_1 F_2},$ which is $(2,1).$ Thus, the equation of the ellipse is\n\\[\\frac{(x - 2)^2}{3^2} + \\frac{(y - 1)^2}{(\\sqrt{5})^2} = 1.\\]Hence, $h + k + a + b = 2 + 1 + 3 + \\sqrt{5} = \\boxed{6 + \\sqrt{5}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Complex numbers $a,$ $b,$ and $c$ are zeros of a polynomial $P(z) = z^3 + qz + r,$ and $\|a\|^2 + \|b\|^2 + \|c\|^2 = 250.$ The points corresponding to $a,$ $b,$ and $c$ in the complex plane are the vertices of a right triangle with hypotenuse $h.$ Find $h^2.$", "role": "user" }, { "content": "By Vieta's formula, the sum of the roots is equal to 0, or $a+b+c=0$. Therefore, $\\frac{a+b+c}{3}=0$. Since the centroid of any triangle is the average of its vertices, the centroid of this triangle is the origin.\n\nWithout loss of generality, let the right angle be at $b.$ Let $x = \|b - c\|$ and $y = \|a - b\|.$ The magnitudes of $a$, $b$, and $c$ are just $\\frac{2}{3}$ of the medians because the origin, or the centroid in this case, cuts the median in a ratio of $2:1$.\n\nHence,\n\\[\|a\|^2=\\frac{4}{9}\\cdot \\left( \\left(\\frac{x}{2} \\right)^2+y^2 \\right)=\\frac{x^2}{9}+\\frac{4y^2}{9}\\]because $\|a\|$ is two thirds of the median from $a$. Similarly,\n\\[\|c\|^2=\\frac{4}{9}\\cdot \\left(x^2 + \\left( \\frac{y}{2} \\right)^2 \\right)=\\frac{4x^2}{9}+\\frac{y^2}{9}.\\]Furthermore,\n\\[\|b\|^2=\\frac{4}{9}\\cdot\\frac{x^2+y^2}{4}=\\frac{x^2}{9}+\\frac{y^2}{9}.\\]Hence,\n\\[\|a\|^2+\|b\|^2+\|c\|^2=\\frac{6x^2+6y^2}{9}=\\frac{2x^2+2y^2}{3}=250.\\]Thus, $h^2=x^2+y^2=\\frac{3}{2}\\cdot 250=\\boxed{375}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The graph of $y = f(x)$ is shown below.\n\n[asy]\nunitsize(0.5 cm);\n\nreal func(real x) {\n real y;\n if (x >= -3 && x <= 0) {y = -2 - x;}\n if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}\n if (x >= 2 && x <= 3) {y = 2(x - 2);}\n return(y);\n}\n\nint i, n;\n\nfor (i = -5; i <= 5; ++i) {\n draw((i,-5)--(i,5),gray(0.7));\n draw((-5,i)--(5,i),gray(0.7));\n}\n\ndraw((-5,0)--(5,0),Arrows(6));\ndraw((0,-5)--(0,5),Arrows(6));\n\nlabel(\"$x$\", (5,0), E);\nlabel(\"$y$\", (0,5), N);\n\ndraw(graph(func,-3,3),red);\n\nlabel(\"$y = f(x)$\", (3,-2), UnFill);\n[/asy]\n\nWhich is the graph of $y = f(x + 2)$?\n\n[asy]\nunitsize(0.5 cm);\n\npicture[] graf;\nint i, n;\n\nreal func(real x) {\n real y;\n if (x >= -3 && x <= 0) {y = -2 - x;}\n if (x >= 0 && x <= 2) {y = sqrt(4 - (x - 2)^2) - 2;}\n if (x >= 2 && x <= 3) {y = 2(x - 2);}\n return(y);\n}\n\nreal funcc(real x) {\n return(func(x - 2));\n}\n\nreal funce(real x) {\n return(func(x + 2));\n}\n\nfor (n = 1; n <= 5; ++n) {\n graf[n] = new picture;\n for (i = -5; i <= 5; ++i) {\n draw(graf[n],(i,-5)--(i,5),gray(0.7));\n draw(graf[n],(-5,i)--(5,i),gray(0.7));\n }\n draw(graf[n],(-5,0)--(5,0),Arrows(6));\n draw(graf[n],(0,-5)--(0,5),Arrows(6));\n\n label(graf[n],\"$x$\", (5,0), E);\n label(graf[n],\"$y$\", (0,5), N);\n}\n\ndraw(graf[1],shift((2,-1))graph(func,-3,3),red);\ndraw(graf[2],shift((2,1))graph(func,-3,3),red);\ndraw(graf[3],graph(funcc,-1,5),red);\ndraw(graf[4],shift((-2,-1))graph(func,-3,3),red);\ndraw(graf[5],graph(funce,-5,1),red);\n\nlabel(graf[1], \"A\", (0,-6));\nlabel(graf[2], \"B\", (0,-6));\nlabel(graf[3], \"C\", (0,-6));\nlabel(graf[4], \"D\", (0,-6));\nlabel(graf[5], \"E\", (0,-6));\n\nadd(graf[1]);\nadd(shift((12,0))(graf[2]));\nadd(shift((24,0))(graf[3]));\nadd(shift((6,-12))(graf[4]));\nadd(shift((18,-12))*(graf[5]));\n[/asy]\n\nEnter the letter of the graph of $y = f(x + 2).$", "role": "user" }, { "content": "The graph of $y = f(x + 2)$ is produced by taking the graph of $y = f(x)$ and shifting two units to the left. The correct graph is $\\boxed{\\text{E}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the only real number that can be expressed in the form \\[(a + bi)^3 - 107i,\\]where $i^2 = -1,$ and $a$ and $b$ are positive integers.", "role": "user" }, { "content": "We have \\[\\begin{aligned} (a+bi)^3 - 107i &= (a^3 + 3a^2bi - 3ab^2 - b^3i) - 107i \\\\ &=(a^3 - 3ab^2) + (3a^2b-b^3-107)i. \\end{aligned}\\]If this is a real number, then we must have \\[0 = 3a^2b-b^3-107\\]or \\[107 = b(3a^2-b^2).\\]Since $107$ is prime, either $b=1$ or $b=107.$ If $b=1,$ then we have $107 = 3a^2-1,$ so $a^2 = 36$ and $a=6.$ If $b = 107,$ then we have $1 = 3a^2 - 107^2,$ so $a^2 = \\frac{1 + 107^2}{3}.$ But $107^2 \\equiv 2^2 \\equiv 1 \\pmod{3},$ so the right-hand side is not an integer. Thus, $(a, b) = (6, 1)$ is the only possibility. Then the answer is \\[a^3 - 3ab^2 = 6^3-3 \\cdot 6 \\cdot 1^2 = \\boxed{198}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Solve\n\\[-1 < \\frac{x^2 - 14x + 11}{x^2 - 2x + 3} < 1.\\]", "role": "user" }, { "content": "We consider both inequalities separately.\n\nThe left inequality is equivalent to\n\\[\\frac{x^2 - 14x + 11}{x^2 - 2x + 3} + 1 > 0,\\]or\n\\[\\frac{2x^2 - 16x + 14}{x^2 - 2x + 3} > 0.\\]Then\n\\[\\frac{x^2 - 8x + 7}{x^2 - 2x + 3} > 0.\\]The numerator factors as\n\\[\\frac{(x - 1)(x - 7)}{x^2 - 2x + 3} > 0.\\]The denominator $x^2 - 2x + 3 = (x - 1)^2 + 2$ is always positive.\n\nThe quadratic $(x - 1)(x - 7)$ is positive precisely when $x < 1$ or $x > 7.$\n\nThe right inequality is equivalent to\n\\[1 - \\frac{x^2 - 14x + 11}{x^2 - 2x + 3} > 0,\\]or\n\\[\\frac{12x - 8}{x^2 - 2x + 3} > 0.\\]Then\n\\[\\frac{3x - 2}{x^2 - 2x + 3} > 0.\\]Since the denominator is always positive, this inequality holds if and only if $x > \\frac{2}{3}.$\n\nThe solution is then\n\\[x \\in \\boxed{\\left( \\frac{2}{3}, 1 \\right) \\cup (7,\\infty)}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find\n\\[\\binom{100}{0} - \\binom{100}{1} + \\binom{100}{2} - \\dots + \\binom{100}{100}.\\]", "role": "user" }, { "content": "By the Binomial Theorem,\n\\[(x + y)^{100} = \\binom{100}{0} x^{100} + \\binom{100}{1} x^{99} y + \\binom{100}{2} x^{98} y^2 + \\dots + \\binom{100}{100} y^{100}.\\]Setting $x = 1$ and $y = -1,$ we get\n\\[\\binom{100}{0} - \\binom{100}{1} + \\binom{100}{2} - \\dots + \\binom{100}{100} = \\boxed{0}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: For what values of the constant $c$ does the graph of $f(x) = \\frac{x^2-x+c}{x^2+x-20}$ have exactly one vertical asymptote?\n\nEnter all possible values, separated by commas.", "role": "user" }, { "content": "We can factor the denominator to get $$f(x) = \\frac{x^2-x+c}{(x-4)(x+5)}.$$Hence, the graph of $f(x)$ has vertical asymptotes at $x=-5$ and $x=4$, unless there is a factor of $x-4$ or $x+5$ in the numerator that cancels out the corresponding factor in the denominator (in this case there will be a hole at that point rather than an asymptote). So, we need to find $c$ such that $x^2 - x + c$ has a factor of $x-4$ or $x + 5,$ but not both.\n\nThat is to say, we need $c$ such that either $4$ or $-5$ is a root. If $x = 4$ is a root, we must have $(4)^2-4+c=0$ which gives us $c=-12.$ If $-5$ is a root, then we must have $(-5)^2 - (-5) + c = 0,$ or $c = - 30.$\n\nThus, the values that work are $c = \\boxed{-12 \\text{ or } -30}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the value of $k$ so that\n\\[3 + \\frac{3 + k}{4} + \\frac{3 + 2k}{4^2} + \\frac{3 + 3k}{4^3} + \\dotsb = 8.\\]", "role": "user" }, { "content": "We have that\n\\[3 + \\frac{3 + k}{4} + \\frac{3 + 2k}{4^2} + \\frac{3 + 3k}{4^3} + \\dotsb = 8.\\]Multiplying this equation by 4, we get\n\\[12 + (3 + k) + \\frac{3 + 2k}{4} + \\frac{3 + 3k}{4^2} + \\dotsb = 32.\\]Subtracting these equations, we get\n\\[12 + k + \\frac{k}{4} + \\frac{k}{4^2} + \\frac{k}{4^3} + \\dotsb = 24.\\]Then\n\\[12 + \\frac{k}{1 - 1/4} = 24.\\]Solving for $k,$ we find $k = \\boxed{9}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The sequence $\\{a_n\\}$ satisfies $a_1 = 1$ and $5^{a_{n + 1} - a_n} - 1 = \\frac {1}{n + \\frac {2}{3}}$ for $n \\geq 1$. Find the least integer $k$ greater than $1$ for which $a_k$ is an integer.", "role": "user" }, { "content": "We rewrite the given equation as \\[5^{a_{n+1} - a_n} = 1 + \\frac{1}{n +\\frac{2}{3}} = \\frac{3n+5}{3n+2}.\\]Then, we observe a telescoping product: \\[\\begin{aligned} 5^{a_n - a_1} &= 5^{a_2 - a_1} \\cdot 5^{a_3-a_2} \\cdots 5^{a_n - a_{n-1}} \\\\ &= \\frac{8}{5} \\cdot \\frac{11}{8} \\cdots \\frac{3n+2}{3n-1} \\\\ &= \\frac{3n+2}{5}. \\end{aligned}\\]Since $a_1 = 1$, we have \\[5^{a_n} = 3n+2\\]for all $n \\ge 1$. Thus, $a_k$ is an integer if and only if $3k+2$ is a power of $5$. The next power of $5$ which is of the form $3k+2$ is $5^3 = 125$, which is $3(41) + 2$. Thus $k = \\boxed{41}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Write $x^{10} + x^5 + 1$ as the product of two polynomials with integer coefficients.", "role": "user" }, { "content": "Let $\\omega$ satisfy $x^2 + x + 1 = 0,$ so $\\omega^2 + \\omega + 1 = 0.$ Then $(\\omega - 1)(\\omega^2 + \\omega + 1) = \\omega^3 - 1 = 0,$ so $\\omega^3 = 1.$ Also,\n\\begin{align}\n\\omega^{10} + \\omega^5 + 1 &= \\omega^9 \\cdot \\omega + \\omega^3 \\cdot \\omega^2 + 1 \\\\\n&= \\omega + \\omega^2 + 1 \\\\\n&= 0.\n\\end{align}Therefore, $x^2 + x + 1$ is a factor of $x^{10} + x^5 + 1.$\n\nTo bring out this factorization, we can write\n\\begin{align}\nx^{10} + x^5 + 1 &= x^{10} - x + x^5 - x^2 + x^2 + x + 1 \\\\\n&= x(x^9 - 1) + x^2 (x^3 - 1) + x^2 + x + 1 \\\\\n&= x(x^3 - 1)(x^6 + x^3 + 1) + x^2 (x - 1)(x^2 + x + 1) + x^2 + x + 1 \\\\\n&= x(x - 1)(x^2 + x + 1)(x^6 + x^3 + 1) + x^2 (x - 1)(x^2 + x + 1) + x^2 + x + 1 \\\\\n&= \\boxed{(x^2 + x + 1)(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)}.\n\\end{align}", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The line with equation $y = x$ is an axis of symmetry of the curve with equation\n\\[y = \\frac{px + q}{rx + s},\\]where $p,$ $q,$ $r,$ $s$ are all nonzero. Which of the following statements must hold?\n\n(A) $p + q = 0$\n\n(B) $p + r = 0$\n\n(C) $p + s = 0$\n\n(D) $q + r = 0$\n\n(E) $q + s = 0$\n\n(F) $r + s = 0$", "role": "user" }, { "content": "Since $y = x$ is an axis of symmetry, if point $(a,b)$ lies on the graph, then so does $(b,a).$ Thus, the equation of the graph can also be written as\n\\[x = \\frac{py + q}{ry + s}.\\]Substituting $y = \\frac{px + q}{rx + s},$ we get\n\\[x = \\frac{p \\cdot \\frac{px + q}{rx + s} + q}{r \\cdot \\frac{px + q}{rx + s} + s} = \\frac{p(px + q) + q(rx + s)}{r(px + q) + s(rx + s)}.\\]Cross-multiplying, we get\n\\[x[r(px + q) + s(rx + s)] = p(px + q) + q(rx + s).\\]Expanding, we get\n\\[(pr + rs) x^2 + (s^2 - p^2) x - (pq + qs) = 0.\\]We can take out a factor of $p + s$:\n\\[(p + s)(rx^2 + (s - p) x - q) = 0.\\]This equation must holds for all $x.$ Since $r \\neq 0,$ the quadratic $rx^2 + (s - p) x - q$ cannot be 0 for all $x,$ so we must have $p + s = 0.$ The correct statement is $\\boxed{\\text{(C)}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Compute\n\\[\\frac{2 + 6}{4^{100}} + \\frac{2 + 2 \\cdot 6}{4^{99}} + \\frac{2 + 3 \\cdot 6}{4^{98}} + \\dots + \\frac{2 + 98 \\cdot 6}{4^3} + \\frac{2 + 99 \\cdot 6}{4^2} + \\frac{2 + 100 \\cdot 6}{4}.\\]", "role": "user" }, { "content": "Let\n\\[S = \\frac{2 + 6}{4^{100}} + \\frac{2 + 2 \\cdot 6}{4^{99}} + \\frac{2 + 3 \\cdot 6}{4^{98}} + \\dots + \\frac{2 + 98 \\cdot 6}{4^3} + \\frac{2 + 99 \\cdot 6}{4^2} + \\frac{2 + 100 \\cdot 6}{4}.\\]Then\n\\[4S = \\frac{2 + 6}{4^{99}} + \\frac{2 + 2 \\cdot 6}{4^{98}} + \\frac{2 + 3 \\cdot 6}{4^{97}} + \\dots + \\frac{2 + 98 \\cdot 6}{4^2} + \\frac{2 + 99 \\cdot 6}{4} + \\frac{2 + 100 \\cdot 6}{1}.\\]Subtracting these equations, we get\n\\[3S = 602 - \\frac{6}{4} - \\frac{6}{4^2} - \\dots - \\frac{6}{4^{98}} - \\frac{6}{4^{99}} - \\frac{8}{4^{100}}.\\]From the formula for a geometric series,\n\\begin{align}\n\\frac{6}{4} + \\frac{6}{4^2} + \\dots + \\frac{6}{4^{98}} + \\frac{6}{4^{99}} &= \\frac{6}{4^{99}} (1 + 4 + \\dots + 4^{97} + 4^{98}) \\\\\n&= \\frac{6}{4^{99}} \\cdot \\frac{4^{99} - 1}{4 - 1} \\\\\n&= 2 \\cdot \\frac{4^{99} - 1}{4^{99}} \\\\\n&= 2 - \\frac{2}{4^{99}}.\n\\end{align}Therefore,\n\\[3S = 602 - 2 + \\frac{2}{4^{99}} - \\frac{8}{4^{100}} = 602 - 2 + \\frac{2}{4^{99}} - \\frac{2}{4^{99}} = 600,\\]so $S = \\boxed{200}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a,$ $b,$ $c$ be complex numbers such that\n\\[a + b + c = ab + ac + bc = abc = 1.\\]Enter the values $a,$ $b,$ $c,$ separated by commas, in any order.", "role": "user" }, { "content": "By Vieta's formulas, $a,$ $b,$ and $c$ are the roots of\n\\[x^3 - x^2 + x - 1 = 0.\\]We can write this as $x^2 (x - 1) + (x - 1) = 0,$ or $(x - 1)(x^2 + 1) = 0.$ The roots are $\\boxed{1,i,-i}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Given any two positive real numbers $x$ and $y$, then $x \\, \\Diamond \\, y$ is a positive real number defined in terms of $x$ and $y$ by some fixed rule. Suppose the operation $x \\, \\Diamond \\, y$ satisfies the equations $(xy) \\, \\Diamond \\, y=x(y \\, \\Diamond \\, y)$ and $(x \\, \\Diamond \\, 1) \\, \\Diamond \\, x = x \\, \\Diamond \\, 1$ for all $x,y>0$. Given that $1 \\, \\Diamond \\, 1=1$, find $19 \\, \\Diamond \\, 98$.", "role": "user" }, { "content": "Setting $y = 1$ in the first equation, we get\n\\[x \\, \\Diamond \\, 1 = x (1 \\, \\Diamond \\, 1) = x.\\]Then from the second equation,\n\\[x \\, \\Diamond \\, x = x \\, \\Diamond \\, 1 = x.\\]Then from the first equation,\n\\[(xy) \\, \\Diamond \\, y=x(y \\, \\Diamond \\, y) = xy.\\]Therefore,\n\\[19 \\, \\Diamond \\, 98 = \\left( \\frac{19}{98} \\cdot 98 \\right) \\, \\Diamond \\, 98 = \\frac{19}{98} \\cdot 98 = \\boxed{19}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a_0=-2,b_0=1$, and for $n\\geq 0$, let\n\\begin{align}a_{n+1}&=a_n+b_n+\\sqrt{a_n^2+b_n^2},\\\\b_{n+1}&=a_n+b_n-\\sqrt{a_n^2+b_n^2}.\\end{align}Find $\\frac{1}{a_{2012}} + \\frac{1}{b_{2012}}.$", "role": "user" }, { "content": "We have that\n\\begin{align}\n\\frac{1}{a_{n + 1}} + \\frac{1}{b_{n + 1}} &= \\frac{1}{a_n + b_n + \\sqrt{a_n^2 + b_n^2}} + \\frac{1}{a_n + b_n - \\sqrt{a_n^2 + b_n^2}} \\\\\n&= \\frac{a_n + b_n - \\sqrt{a_n^2 + b_n^2} + a_n + b_n + \\sqrt{a_n^2 + b_n^2}}{(a_n + b_n)^2 - (a_n^2 + b_n^2)} \\\\\n&= \\frac{2a_n + 2b_n}{2a_n b_n} \\\\\n&= \\frac{1}{a_n} + \\frac{1}{b_n}.\n\\end{align}Thus, $\\frac{1}{a_n} + \\frac{1}{b_n}$ is a constant, which means\n\\[\\frac{1}{a_{2012}} + \\frac{1}{b_{2012}} = \\frac{1}{a_0} + \\frac{1}{b_0} = \\boxed{\\frac{1}{2}}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the positive root of\n\\[x^3 - 3x^2 - x - \\sqrt{2} = 0.\\]", "role": "user" }, { "content": "Since we have a coefficient of $\\sqrt{2},$ we can guess that the positive root is of the form $a + b \\sqrt{2},$ where $a$ and $b$ are integers. So, let $x = a + b \\sqrt{2}.$ Substituting, we get\n\\[(a + b \\sqrt{2})^3 - 3(a + b \\sqrt{2})^2 - (a + b \\sqrt{2}) - \\sqrt{2} = 0.\\]This expands as\n\\[(a^3 + 3a^2 b \\sqrt{2} + 6ab^2 + 2b^3 \\sqrt{2}) - 3(a^2 + 2ab \\sqrt{2} + 2b^2) - (a + b \\sqrt{2}) - \\sqrt{2} = 0,\\]so\n\\[(a^3 + 6ab^2 - 3a^2 - 6b^2 - a) + (3a^2 b + 2b^3 - 6ab - b - 1) \\sqrt{2} = 0.\\]Hence,\n\\begin{align}\na^3 + 6ab^2 - 3a^2 - 6b^2 - a &= 0, \\\\\n3a^2 b + 2b^3 - 6ab - b - 1 &= 0.\n\\end{align}From the first equation,\n\\[6ab^2 - 6b^2 = -a^3 + 3a^2 + a,\\]so\n\\[6b^2 (a - 1) = -(a^3 - 3a^2 - a).\\]Thus, $a - 1$ divides $a^3 - 3a^2 - a.$ Since $a - 1$ divides $(a - 1)(a - 3)(a + 1) = a^3 - 3a^2 - a + 3,$ $a - 1$ divides 3. This means $a - 1$ can be $-3,$ $-1,$ 1, or 3, so $a$ is $-2$, 0, 2, or 4.\n\nIf $a = -2,$ then $b^2 = -1,$ which has no solutions.\n\nIf $a = 0,$ then $b^2 = 0,$ so $b = 0,$ which does not work.\n\nIf $a = 2,$ then $b^2 = 1,$ so $b = -1$ or $b = 1.$ Only $a = 2$ and $b = 1$ satisfy the second equation.\n\nIf $a = 4,$ then $b^2 = -\\frac{2}{3},$ which has no solutions.\n\nTherefore, $a = 2$ and $b = 1$ works, so $x = \\boxed{2 + \\sqrt{2}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Consider the function\n\\[f(x) = \\max \\{-11x - 37, x - 1, 9x + 3\\}\\]defined for all real $x.$ Let $p(x)$ be a quadratic polynomial tangent to the graph of $f$ at three distinct points with $x$-coordinates $x_1,$ $x_2,$ $x_3.$ Find $x_1 + x_2 + x_3.$", "role": "user" }, { "content": "Since a parabola can be tangent to a given line in at most one point, the parabola must be tangent to all three lines $y = -11x - 37,$ $y = x - 1,$ and $y = 9x + 3.$ Thus, if $a$ is the leading coefficient of $p(x),$ then\n\\begin{align}\np(x) - (-11x - 37) &= a(x - x_1)^2, \\\\\np(x) - (x - 1) &= a(x - x_2)^2, \\\\\np(x) - (9x + 3) &= a(x - x_3)^2.\n\\end{align}Subtracting the first two equations, we get\n\\begin{align}\n12x + 36 &= a(x - x_1)^2 - a(x - x_2)^2 \\\\\n&= a(x - x_1 + x - x_2)(x_2 - x_1) \\\\\n&= 2a(x_2 - x_1) x + a(x_1^2 - x_2^2).\n\\end{align}Matching coefficients, we get\n\\begin{align}\n2a(x_2 - x_1) &= 12, \\\\\na(x_1^2 - x_2^2) &= 36.\n\\end{align}Dividing these equations, we get $-\\frac{1}{2} (x_1 + x_2) = 3,$ so $x_1 + x_2 = -6.$\n\nSubtracting other pairs of equations gives us $x_1 + x_3 = -4$ and $x_2 + x_3 = -1.$ Then $2x_1 + 2x_2 + 2x_3 = -11,$ so\n\\[x_1 + x_2 + x_3 = \\boxed{-\\frac{11}{2}}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the number of functions $f : \\mathbb{R} \\to \\mathbb{R}$ such that\n\\[f(x + f(y)) = x + y\\]for all real numbers $x$ and $y.$", "role": "user" }, { "content": "Setting $x = -f(y),$ we get\n\\[f(0) = -f(y) + y,\\]so $f(y) = y - f(0)$ for all real numbers $x.$ Then the given functional equation becomes\n\\[f(x + y - f(0)) = x + y,\\]or $x + y - f(0) - f(0) = x + y.$ Then $f(0) = 0,$ so $f(x) = x$ for all real numbers $x.$ This function does satisfy the given functional equation, giving us $\\boxed{1}$ solution.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a$ and $b$ be real numbers. Find the maximum value of $a \\cos \\theta + b \\sin \\theta$ in terms of $a$ and $b.$", "role": "user" }, { "content": "By the Cauchy-Schwarz Inequality,\n\\[(a \\cos \\theta + b \\sin \\theta)^2 \\le (a^2 + b^2)(\\cos^2 \\theta + \\sin^2 \\theta) = a^2 + b^2,\\]so $a \\cos \\theta + b \\sin \\theta \\le \\sqrt{a^2 + b^2}.$\n\nIf $a = b = 0,$ then $a \\cos \\theta + b \\sin \\theta = 0$ for all $\\theta.$ Otherwise, $a^2 + b^2 > 0,$ and we can find an angle $\\theta$ such that\n\\[\\cos \\theta = \\frac{a}{\\sqrt{a^2 + b^2}} \\quad \\text{and} \\quad \\sin \\theta = \\frac{b}{\\sqrt{a^2 + b^2}},\\]which makes $a \\cos \\theta + b \\sin \\theta = \\sqrt{a^2 + b^2}.$ Thus, the maximum value is $\\boxed{\\sqrt{a^2 + b^2}}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: When the polynomial $x^4 - 6x^3 + 16x^ 2 - 25x + 10$ is divided by $x^2 - 2x + k,$ the remainder is $x + a.$ Enter the ordered pair $(k,a).$", "role": "user" }, { "content": "The long division is shown below.\n\n\\[\n\\begin{array}{c\|ccccc}\n\\multicolumn{2}{r}{x^2} & -4x & +(8 - k) & \\\\\n\\cline{2-6}\nx^2 - 2x + k & x^4 & -6x^3 & +16x^2 & -25x & +10 \\\\\n\\multicolumn{2}{r}{x^2} & -2x^3 & + kx^2 \\\\ \n\\cline{2-4}\n\\multicolumn{2}{r}{0} & -4x^3 & +(16 - k)x^2 \\\\\n\\multicolumn{2}{r}{} &- 4x^3 & +8x^2 & - 4kx \\\\ \n\\cline{3-5}\n\\multicolumn{2}{r}{} & 0 & +(8 - k)x^2 & +(4k - 25)x \\\\\n\\multicolumn{2}{r}{} & & +(8 - k)x^2 & +(2k - 16)x & +k(8 - k) \\\\\n\\cline{4-6}\n\\multicolumn{2}{r}{} & & 0 & +(2k - 9)x & +(k^2 - 8k + 10) \\\\\n\\end{array}\n\\]Thus, the remainder is $(2k - 9)x + (k^2 - 8k + 10).$ We want this to be $x + a,$ so $2k - 9 = 1$ and $k^2 - 8k + 10 = a.$ Solving, we find $(k,a) = \\boxed{(5,-5)}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The real numbers $a,$ $b,$ $c,$ and $d$ satisfy\n\\[a^2 + b^2 + c^2 + 1 = d + \\sqrt{a + b + c - d}.\\]Find $d.$", "role": "user" }, { "content": "Let $x = \\sqrt{a + b + c - d}.$ Then $x^2 = a + b + c - d,$ so $d = a + b + c - x^2,$ and we can write\n\\[a^2 + b^2 + c^2 + 1 = a + b + c - x^2 + x.\\]Then\n\\[a^2 - a + b^2 - b + c^2 - c + x^2 - x + 1 = 0.\\]Completing the square in $a,$ $b,$ $c,$ and $x,$ we get\n\\[\\left( a - \\frac{1}{2} \\right)^2 + \\left( b - \\frac{1}{2} \\right)^2 + \\left( c - \\frac{1}{2} \\right)^2 + \\left( x - \\frac{1}{2} \\right)^2 = 0.\\]Hence, $a = b = c = x = \\frac{1}{2},$ so\n\\[d = a + b + c - x^2 = \\frac{1}{2} + \\frac{1}{2} + \\frac{1}{2} - \\frac{1}{4} = \\boxed{\\frac{5}{4}}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find the last three digits of $9^{105}.$", "role": "user" }, { "content": "We can write $9^{105} = (10 - 1)^{105}.$ Then by the Binomial Theorem,\n\\[(10 - 1)^{105} = 10^{105} - \\binom{105}{1} 10^{104} + \\binom{105}{2} 10^{103} - \\dots + \\binom{105}{102} 10^3 - \\binom{105}{103} 10^2 + \\binom{105}{104} 10 - 1.\\]All the terms up to $\\binom{105}{102} 10^3$ are divisible by $10^3,$ so for the purpose of finding the last three digits, we can ignore them. We are left with\n\\begin{align}\n-\\binom{105}{103} 10^2 + \\binom{105}{104} 10 - 1 &= -\\binom{105}{2} 10^2 + \\binom{105}{1} 10 - 1 \\\\\n&= -\\frac{105 \\cdot 104}{2} \\cdot 10^2 + 105 \\cdot 10 - 1 \\\\\n&= -546000 + 1050 - 1 \\\\\n&= -546000 + 1049.\n\\end{align}Hence, the last three digits are $\\boxed{049}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Compute the value of the infinite series \\[\n\\sum_{n=2}^{\\infty} \\frac{n^4+3n^2+10n+10}{2^n \\cdot \\left(n^4+4\\right)}\n\\]", "role": "user" }, { "content": "We factor the denominator: \\[n^4+4 = (n^2+2)^2-(2n)^2 = (n^2-2n+2)(n^2+2n+2).\\]Now,\n\n\\begin{eqnarray}\n\\frac{n^4+3n^2+10n+10}{n^4+4} & = & 1 + \\frac{3n^2+10n+6}{n^4+4} \\\\\n& = & 1 + \\frac{4}{n^2-2n+2} - \\frac{1}{n^2+2n+2} \\\\\n\\Longrightarrow \\sum_{n=2}^{\\infty} \\frac{n^4+3n^2+10n+10}{2^n \\cdot \\left(n^4+4\\right)} & = & \\sum_{n=2}^{\\infty} \\frac{1}{2^n} + \\frac{4}{2^n\\cdot(n^2-2n+2)} - \\frac{1}{2^n\\cdot(n^2+2n+2)} \\\\\n& = & \\frac{1}{2} + \\sum_{n=2}^{\\infty} \\frac{1}{2^{n-2}\\cdot\\left((n-1)^2+1\\right)} - \\frac{1}{2^n\\cdot\\left((n+1)^2+1\\right)}\n\\end{eqnarray}The last series telescopes to $\\frac{1}{2} + \\frac{1}{10}$; thus, our our desired answer is $\\frac{1}{2} + \\frac{1}{2} + \\frac{1}{10} = \\boxed{\\frac{11}{10}}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $f : \\mathbb{R} \\to \\mathbb{R}$ be a function such that\n\\[f(f(x) - y) = f(x) + f(f(y) - f(-x)) + x\\]for all real numbers $x$ and $y.$\n\nLet $n$ be the number of possible values of $f(3),$ and let $s$ be the sum of all possible values of $f(3).$ Find $n \\times s.$", "role": "user" }, { "content": "Setting $x = y = 0,$ we get\n\\[f(f(0)) = 2f(0).\\]Let $c = f(0),$ so $f(c) = 2c.$\n\nSetting $x = 0$ and $y = c,$ we get\n\\[f(0) = f(0) + f(f(c) - c).\\]Then $f(c) = 0,$ so $c = 0.$\n\nSetting $x = 0,$ we get\n\\[f(-y) = f(f(y))\\]for all $y.$\n\nSetting $y = f(x),$ we get\n\\[0 = f(x) + f(f(f(x)) - f(-x)) + x.\\]Since $f(f(x)) = f(-x),$ this becomes $f(x) = -x$ for all $x.$ We can check that this function works.\n\nThus, $n = 1$ and $s = -3,$ so $n \\times s = \\boxed{-3}.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Find all solutions to the inequality \\[\\frac{x^2}{(x-3)^2} \\ge 0.\\](Give your answer in interval notation.)", "role": "user" }, { "content": "As the quotient of two perfect squares, the left-hand side is always nonnegative when it is defined. The left-hand side is defined whenever $x \\neq 3,$ so the solution set is $\\boxed{ (-\\infty, 3) \\cup (3, \\infty) }.$", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $F(x)$ be a polynomial such that $F(6) = 15$ and\\[\\frac{F(3x)}{F(x+3)} = 9-\\frac{48x+54}{x^2+5x+6}\\]for $x \\in \\mathbb{R}$ such that both sides are defined. Find $F(12)$.\n", "role": "user" }, { "content": "Combining denominators and simplifying,\\[\\frac{F(3x)}{F(x+3)} = \\frac{9(x^2+5x+6)-48x-54}{x^2+5x+6} = \\frac{9x^2 - 3x}{x^2 + 5x + 6}= \\frac{3x(3x-1)}{(x+3)(x+2)}\\]It becomes obvious that $F(x) = ax(x-1)$, for some constant $a$, matches the definition of the polynomial. To prove that $F(x)$ must have this form, note that\\[(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)\\]\nSince $3x$ and $3x-1$ divides the right side of the equation, $3x$ and $3x-1$ divides the left side of the equation. Thus $3x(3x-1)$ divides $F(3x)$, so $x(x-1)$ divides $F(x)$.\nIt is easy to see that $F(x)$ is a quadratic, thus $F(x)=ax(x-1)$ as desired.\nBy the given, $F(6) = a(6)(5) = 15 \\Longrightarrow a = \\frac 12$. Thus, $F(12) = \\frac{1}{2}(12)(11) = \\boxed{66}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: The expansion of $(x+1)^n$ has 3 consecutive terms with coefficients in the ratio $1:2:3$ that can be written in the form\\[{n\\choose k} : {n\\choose k+1} : {n \\choose k+2}\\]Find the sum of all possible values of $n+k$.\n", "role": "user" }, { "content": "By definition, ${n\\choose k} = \\frac{n!}{k!(n-k)!}$. The ratio of the first two terms give us that\\begin{align}\\frac{1}{2} &= \\frac{\\frac{n!}{k!(n-k)!}}{\\frac{n!}{(k+1)!(n-k-1)!}} = \\frac{k+1}{n-k}\\\\ 2&=n-3k\\end{align}The ratio of the second and third terms give us that\\begin{align}\\frac{2}{3} &= \\frac{\\frac{n!}{(k+1)!(n-k-1)!}}{\\frac{n!}{(k+2)!(n-k-2)!}} = \\frac{k+2}{n-k-1}\\\\ 8&=2n-5k\\end{align}This is a linear system of two equations with two unknowns, indicating that there is a unique solution. Solving by substitution or multiplying the top equation and subtracting, we find $k = 4, n = 14$. Thus, $n+k=\\boxed{18}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $S = (1+i)^{17} - (1-i)^{17}$, where $i=\\sqrt{-1}$. Find $\|S\|$.\n", "role": "user" }, { "content": "Rewriting the complex numbers in polar notation form, $1+i = \\sqrt{2}\\,\\text{cis}\\,\\frac{\\pi}{4}$ and $1-i = \\sqrt{2}\\,\\text{cis}\\,-\\frac{\\pi}{4}$, where $\\text{cis}\\,\\theta = \\cos \\theta + i\\sin \\theta$. By De Moivre's Theorem,\\begin{align} \\left(\\sqrt{2}\\,\\text{cis}\\,\\frac{\\pi}{4}\\right)^{17} - \\left(\\sqrt{2}\\,\\text{cis}\\,-\\frac{\\pi}{4}\\right)^{17} &= 2^{17/2}\\,\\left(\\text{cis}\\,\\frac{17\\pi}{4}\\right) - 2^{17/2}\\,\\left(\\text{cis}\\,-\\frac{17\\pi}{4}\\right) \\\\ &= 2^{17/2}\\left[\\text{cis}\\left(\\frac{\\pi}{4}\\right) - \\text{cis}\\left(-\\frac{\\pi}{4}\\right)\\right] \\\\ &= 2^{17/2}\\left(2i\\sin \\frac{\\pi}{4}\\right) \\\\ &= 2^{17/2} \\cdot 2 \\cdot 2^{-1/2}i = 2^9i = \\boxed{512}\\,i \\end{align}", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: A $\\frac 1p$ -array is a structured, infinite, collection of numbers. For example, a $\\frac 13$ -array is constructed as follows:\n\\begin{align} 1 \\qquad \\frac 13\\,\\ \\qquad \\frac 19\\,\\ \\qquad \\frac 1{27} \\qquad &\\cdots\\\\ \\frac 16 \\qquad \\frac 1{18}\\,\\ \\qquad \\frac{1}{54} \\qquad &\\cdots\\\\ \\frac 1{36} \\qquad \\frac 1{108} \\qquad &\\cdots\\\\ \\frac 1{216} \\qquad &\\cdots\\\\ &\\ddots \\end{align}\nIn general, the first entry of each row is $\\frac{1}{2p}$ times the first entry of the previous row. Then, each succeeding term in a row is $\\frac 1p$ times the previous term in the same row. If the sum of all the terms in a $\\frac{1}{2008}$ -array can be written in the form $\\frac mn$, where $m$ and $n$ are relatively prime positive integers, find the remainder when $m+n$ is divided by $2008$.\n", "role": "user" }, { "content": "Note that the value in the $r$th row and the $c$th column is given by $\\left(\\frac{1}{(2p)^r}\\right)\\left(\\frac{1}{p^c}\\right)$. We wish to evaluate the summation over all $r,c$, and so the summation will be, using the formula for an infinite geometric series:\\begin{align}\\sum_{r=1}^{\\infty}\\sum_{c=1}^{\\infty} \\left(\\frac{1}{(2p)^r}\\right)\\left(\\frac{1}{p^c}\\right) &= \\left(\\sum_{r=1}^{\\infty} \\frac{1}{(2p)^r}\\right)\\left(\\sum_{c=1}^{\\infty} \\frac{1}{p^c}\\right)\\\\ &= \\left(\\frac{1}{1-\\frac{1}{2p}}\\right)\\left(\\frac{1}{1-\\frac{1}{p}}\\right)\\\\ &= \\frac{2p^2}{(2p-1)(p-1)}\\end{align}Taking the denominator with $p=2008$ (indeed, the answer is independent of the value of $p$), we have $m+n \\equiv 2008^2 + (2008-1)(2\\cdot 2008 - 1) \\equiv (-1)(-1) \\equiv 1 \\pmod{2008}$ (or consider FOILing). The answer is $\\boxed{1}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Consider the following function $g(x)$ defined as\\[(x^{2^{2008}-1}-1)g(x) = (x+1)(x^2+1)(x^4+1)\\cdots (x^{2^{2007}}+1) - 1\\]Find $g(2)$.\n", "role": "user" }, { "content": "Multiply both sides by $x-1$; the right hand side collapses by the reverse of the difference of squares.\n\\begin{align}(x-1)(x^{2^{2008}-1}-1)g(x) &= (x-1)(x+1)(x^2+1)(x^4+1)\\cdots (x^{2^{2007}}+1) - (x-1)\\\\ &= (x^2-1) (x^2+1)(x^4+1)\\cdots (x^{2^{2007}}+1) - (x-1)\\\\ &= \\cdots\\\\ &= \\left(x^{2^{2008}}-1\\right) - (x-1) = x^{2^{2008}} - x \\end{align}Substituting $x = 2$, we have\\[\\left(2^{2^{2008}-1}-1\\right) \\cdot g(2) = 2^{2^{2008}}-2 = 2\\left(2^{2^{2008}-1}-1\\right)\\]Dividing both sides by $2^{2^{2008}-1}$, we find $g(2) = \\boxed{2}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $S$ denote the value of the sum\\[\\sum_{n=0}^{668} (-1)^{n} {2004 \\choose 3n}\\]Determine the remainder obtained when $S$ is divided by $1000$.\n", "role": "user" }, { "content": "Consider the polynomial\\[f(x)=(x-1)^{2004}=\\sum_{n=0}^{2004}\\binom{2004}{n}\\cdot(-1)^n x^{2004-n}.\\]\nLet $\\omega^3=1$ with $\\omega\\neq 1$. We have\n\\begin{align} \\frac{f(1)+f(\\omega)+f(\\omega^2)}{3} &= \\frac{(1-1)^{2004}+(\\omega-1)^{2004}+(\\omega^2-1)^{2004}}{3} \\\\ &= \\frac{1}{3}\\sum_{n=0}^{2004}\\binom{2004}{n}\\cdot(-1)^n\\cdot(1^{2004-n}+\\omega^{2004-n}+(\\omega^2)^{2004-n}) \\\\ &= \\sum_{n=0}^{668}(-1)^n \\binom{2004}{3n}. \\end{align}\nwhere the last step follows because $1^k+\\omega^k+\\omega^{2k}$ is 0 when $k$ is not divisible by 3, and $3$ when $k$ is divisible by 3.\nWe now compute $\\frac{(1-1)^{2004}+(\\omega-1)^{2004}+(\\omega^2-1)^{2004}}{3}$. WLOG, let $\\omega = \\frac{-1+\\sqrt{3}i}{2}, \\omega^2=\\frac{-1-\\sqrt{3}i}{2}$. Then $\\omega-1=\\frac{-3+\\sqrt{3}i}{2} = \\sqrt{3}\\cdot \\frac{-\\sqrt{3}+i}{2}$, and $\\omega^2-1=\\sqrt{3}\\cdot\\frac{-\\sqrt{3}-i}{2}$. These numbers are both of the form $\\sqrt{3}\\cdot\\varphi$, where $\\varphi$ is a 12th root of unity, so both of these, when raised to the 2004-th power, become $3^{1002}$. Thus, our desired sum becomes $2\\cdot3^{1001}$.\nTo find $2\\cdot3^{1001} \\pmod{1000}$, we notice that $3^{\\phi{500}}\\equiv 3^{200}\\equiv 1 \\pmod{500}$ so that $3^{1001}\\equiv 3 \\pmod{500}$. Then $2\\cdot3^{1001}=2(500k+3)=1000k+6$. Thus, our answer is $\\boxed{6}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: Let $a$ and $b$ be the two real values of $x$ for which\\[\\sqrt[3]{x} + \\sqrt[3]{20 - x} = 2\\]The smaller of the two values can be expressed as $p - \\sqrt{q}$, where $p$ and $q$ are integers. Compute $p + q$.\n", "role": "user" }, { "content": "Let $a=\\sqrt[3]{x}, b = \\sqrt[3]{20-x}$. Then $a+b = 2$ and $a^3 + b^3 = 20$. Factoring,\\[a^3 + b^3 = (a+b)((a+b)^2-3ab) = 2(4-3ab)= 8-6ab=20 \\Longrightarrow ab = -2\\]\nSolving $a+b=2, ab=-2$ gives us the quadratic $a^2 - 2a - 2 = 0$. The quadratic formula yields $a = \\frac{2 - \\sqrt{12}}{2} = 1 - \\sqrt{3}$, and $x = a^3 = (1-\\sqrt{3})^3 = 1 - 3\\sqrt{3} + 9 - 3\\sqrt{3} = 10 - \\sqrt{108}$. Therefore, $p+q=\\boxed{118}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: If $p, q,$ and $r$ are three non-zero integers such that $p + q + r = 26$ and\\[\\frac{1}{p} + \\frac{1}{q} + \\frac{1}{r} + \\frac{360}{pqr} = 1,\\] compute $pqr$.\n", "role": "user" }, { "content": "\\begin{align} \\frac {1}{p} + \\frac {1}{q} + \\frac {1}{r} + \\frac {360}{pqr} & = 1 \\\\ pq + pr + qr + 360 & = pqr \\\\ 360 & = pqr - pq - pr - qr \\\\ & = (p - 1)(q - 1)(r - 1) - (p + q + r) + 1 \\\\ & = (p - 1)(q - 1)(r - 1) - 25 \\\\ 385 & = (p - 1)(q - 1)(r - 1) \\\\ \\end{align}\nFrom here, you can factor $385$ as $5 \\cdot 7 \\cdot 11$, giving corresponding values of $6, 8,$ and $12$. The answer is $6 \\cdot 8 \\cdot 12=\\boxed{576}$.", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: $x$ is a real number with the property that $x+\\tfrac1x = 3$. Let $S_m = x^m + \\tfrac{1}{x^m}$. Determine the value of $S_7$.\n", "role": "user" }, { "content": "We can calculate\\[x^2 + \\dfrac{1}{x^2} = \\left(x + \\dfrac{1}{x}\\right)^2 - 2 = 3^2 -2 = 7.\\]Similarly,\\[x^3 + \\dfrac{1}{x^3} = \\left(x + \\dfrac{1}{x}\\right) \\left(x^2 + \\dfrac{1}{x^2}\\right) - \\left(x + \\dfrac{1}{x}\\right) = 3 \\cdot 7 - 3 = 18\\]and\\[x^4 + \\dfrac{1}{x^4} = \\left(x^2 + \\dfrac{1}{x^2}\\right)^2 - 2 = 7^2 - 2 = 47.\\]Finally,\\[x^7 + \\dfrac{1}{x^7} = \\left(x^3 + \\dfrac{1}{x^3}\\right) \\left(x^4 + \\dfrac{1}{x^4}\\right) - \\left(x + \\dfrac{1}{x}\\right) = 18 \\cdot 47 - 3 = \\boxed{843}.\\]", "role": "assistant" } ]
[ { "content": "Question: Find the domain of the expression $\\frac{\\sqrt{x-2}}{\\sqrt{5-x}}$.}\nAnswer:The expressions inside each square root must be non-negative.\nTherefore, $x-2 \\ge 0$, so $x\\ge2$, and $5 - x \\ge 0$, so $x \\le 5$.\nAlso, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$.\nTherefore, the domain of the expression is $\\boxed{[2,5)}$.\n\nQuestion: If $\\det \\mathbf{A} = 2$ and $\\det \\mathbf{B} = 12,$ then find $\\det (\\mathbf{A} \\mathbf{B}).$\nAnswer:We have that $\\det (\\mathbf{A} \\mathbf{B}) = (\\det \\mathbf{A})(\\det \\mathbf{B}) = (2)(12) = \\boxed{24}.$\n\nQuestion: Terrell usually lifts two 20-pound weights 12 times. If he uses two 15-pound weights instead, how many times must Terrell lift them in order to lift the same total weight?\nAnswer:If Terrell lifts two 20-pound weights 12 times, he lifts a total of $2\\cdot 12\\cdot20=480$ pounds of weight. If he lifts two 15-pound weights instead for $n$ times, he will lift a total of $2\\cdot15\\cdot n=30n$ pounds of weight. Equating this to 480 pounds, we can solve for $n$: \\begin{align}\n30n&=480\\\\\n\\Rightarrow\\qquad n&=480/30=\\boxed{16}\n\\end{align}\n\nQuestion: If the system of equations\n\n\\begin{align}\n6x-4y&=a,\\\\\n6y-9x &=b.\n\\end{align}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{a}{b},$ assuming $b$ is nonzero.\nAnswer:If we multiply the first equation by $-\\frac{3}{2}$, we obtain\n\n$$6y-9x=-\\frac{3}{2}a.$$Since we also know that $6y-9x=b$, we have\n\n$$-\\frac{3}{2}a=b\\Rightarrow\\frac{a}{b}=\\boxed{-\\frac{2}{3}}.$$\n\nQuestion: A function $f(x)$ is defined for all real numbers $x$. For all non-zero values $x$, we have\n\\[2f\\left(x\\right) + f\\left(\\frac{1}{x}\\right) = 5x + 4\\]\nLet $S$ denote the sum of all of the values of $x$ for which $f(x) = 2004$. Compute the integer nearest to $S$.\n", "role": "user" }, { "content": "Substituting $\\frac{1}{x}$, we have\n\\[2f\\left(\\frac 1x\\right) + f\\left(x\\right) = \\frac{5}{x} + 4\\]\nThis gives us two equations, which we can eliminate $f\\left(\\frac 1x\\right)$ from (the first equation multiplied by two, subtracting the second):\n\\begin{align} 3f(x) &= 10x + 4 - \\frac 5x \\\\ 0 &= x^2 - \\frac{3 \\times 2004 - 4}{10}x + \\frac 52\\end{align}\nClearly, the discriminant of the quadratic equation $\\Delta > 0$, so both roots are real. By Vieta's formulas, the sum of the roots is the coefficient of the $x$ term, so our answer is $\\left[\\frac{3 \\times 2004 - 4}{10}\\right] = \\boxed{601}$.", "role": "assistant" } ]

Subsets and Splits

No community queries yet

The top public SQL queries from the community will appear here once available.