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+ [
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+ {
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+ "instruction": "Let \\[f(x) = \\left\\{\n\\begin{array}{cl} ax+3, &\\text{ if }x>2, \\\\\nx-5 &\\text{ if } -2 \\le x \\le 2, \\\\\n2x-b &\\text{ if } x <-2.\n\\end{array}\n\\right.\\]Find $a+b$ if the piecewise function is continuous (which means that its graph can be drawn without lifting your pencil from the paper).",
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+ "input": "input 5",
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+ "type": "Algebra",
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+ "output": "For the piecewise function to be continuous, the cases must \"meet\" at $2$ and $-2$. For example, $ax+3$ and $x-5$ must be equal when $x=2$. This implies $a(2)+3=2-5$, which we solve to get $2a=-6 \\Rightarrow a=-3$. Similarly, $x-5$ and $2x-b$ must be equal when $x=-2$. Substituting, we get $-2-5=2(-2)-b$, which implies $b=3$. So $a+b=-3+3=\\boxed{0}$."
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+ },
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+ {
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+ "instruction": "If $A=2+i$, $O=-4$, $P=-i$, and $S=2+4i$, find $A-O+P+S$.",
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+ "input": "input 3",
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+ "type": "Algebra",
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+ "output": "Adding real parts and imaginary parts separately, we have $(2-(-4)+0+2)+(1+0-1+4)i=\\boxed{8+4i}$."
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+ },
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+ {
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+ "instruction": "The perimeter of a rectangle is 24 inches. What is the number of square inches in the maximum possible area for this rectangle?",
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+ "input": "input 3",
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+ "type": "Algebra",
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+ "output": "Let one pair of parallel sides have length $x$ and the other pair of parallel sides have length $12-x$. This means that the perimeter of the rectangle is $x+x+12-x+12-x=24$ as the instruction states. The area of this rectangle is $12x-x^2$. Completing the square results in $-(x-6)^2+36\\le 36$ since $(x-6)^2\\ge 0$, so the maximum area of $\\boxed{36}$ is obtained when the rectangle is a square of side length 6 inches."
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+ },
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+ {
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+ "instruction": "What is the sum of all values of $y$ for which the expression $\\frac{y+6}{y^2-5y+4}$ is undefined?",
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+ "input": "input 3",
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+ "type": "Algebra",
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+ "output": "The given expression is undefined when the denominator is zero. Thus, we want to find the sum of the zeros $y$ to the quadratic $y^2-5y+4$. Since for a quadratic with the equation $ax^2+bx+c=0$, the sum of the outputs is $-b/a$, the sum of the zeros of the quadratic $y^2-5y+4$ is $5/1=\\boxed{5}$."
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+ },
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+ {
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+ "instruction": "If $3p+4q=8$ and $4p+3q=13$, what is $q$ equal to?",
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+ "input": "input 3",
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+ "type": "Algebra",
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+ "output": "Because the question only asks for the value of $q$, we can begin by eliminating $p$. In order to do this, we multiply the first equation by 4 and the second equation by 3, giving us a system of two equations that both have 12 as the coefficient of $p$ \\begin{align*} 12p+16q&=32\n\\\\ 12p+9q&=39\n\\end{align*}From here, we can just subtract the second equation from the first. This gives us $(12p+16q)-(12p+9q)=32-(39)$, which simplifies to $7q=-7$ or $q=\\boxed{-1}$."
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+ },
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+ {
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+ "instruction": "Find the distance between the points $(-5,-2)$ and $(7,3)$.",
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+ "input": "input 2",
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+ "type": "Algebra",
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+ "output": "We use the distance formula to find that the distance is \\[\\sqrt{(-5 -7)^2 + (-2-3)^2} = \\!\\sqrt{144 + 25} = \\boxed{13}.\\]"
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+ },
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+ {
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+ "instruction": "For what values of $b$ is $-2$ not in the range of the function $f(x)=x^2+bx+2$? Express your answer in interval notation.",
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+ "input": "input 5",
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+ "type": "Algebra",
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+ "output": "We see that $-2$ is not in the range of $f(x) = x^2 + bx + 2$ if and only if the equation $x^2 + bx + 2 = -2$ has no real roots. We can re-write this equation as $x^2 + bx + 4 = 0$. The discriminant of this quadratic is $b^2 - 4 \\cdot 4 = b^2 - 16$. The quadratic has no real roots if and only if the discriminant is negative, so $b^2 - 16 < 0$, or $b^2 < 16$. The set of values of $b$ that satisfy this inequality is $b \\in \\boxed{(-4,4)}$."
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+ },
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+ {
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+ "instruction": "After traveling 50 miles by taxi, Ann is charged a fare of $\\$120$. Assuming the taxi fare is directly proportional to distance traveled, how much would Ann be charged (in dollars) if she had traveled 70 miles?",
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+ "input": "input 1",
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+ "type": "Algebra",
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+ "output": "Let $d$ equal the cost of a 70 mile taxi ride. Since we know that Ann was charged $120 dollars a 50 mile taxi ride, we can set up the proportion $\\frac{120}{50}=\\frac{d}{70}$. If we solve for $d$ by multiplying both sides by 70, we find that $d=\\left(\\frac{120}{50}\\right)(70)=\\boxed{168}$ dollars."
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+ },
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+ {
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+ "instruction": "Sixteen is 64$\\%$ of what number?",
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+ "input": "input 1",
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+ "type": "Algebra",
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+ "output": "If the number is $x$, we can set up the equation $\\frac{16}{x}=\\frac{64}{100}$. We divide both sides by $4$ to get $\\frac{1}{x}=\\frac{4}{100}=\\frac{1}{25}$, so $x=\\boxed{25}$."
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+ },
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+ {
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+ "instruction": "Let \\[f(x) =\n\\begin{cases}\n2x + 9 &\\text{if }x<-2, \\\\\n5-2x&\\text{if }x\\ge -2.\n\\end{cases}\n\\]Find $f(3).$",
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+ "input": "input 2",
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+ "type": "Algebra",
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+ "output": "Because $3\\ge -2,$ we use the second case to determine that $f(3) = 5-2(3) = \\boxed{-1}.$"
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+ }
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+ ]