rp sequencelengths 0 10 | labels sequencelengths 0 10 |
|---|---|
[
"Step 1: The distance between $(2,5)$ and $(-6,-1)$ is given by the distance formula: $$d = \\sqrt{(2-(-6))^2 + (5-(-1))^2} = \\sqrt{8^2 + 6^2} = \\sqrt{64 + 36} = \\sqrt{100} = 10$$",
"Step 1: Calculate the difference between the $x$-coordinates: $-6 - 2 = -8$. Step 2: Calculate the difference between the $y$-co... | [
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"Step 1: The distance between $(2,5)$ and $(-6,-1)$ is given by the distance formula. Step 2: $\\sqrt{(2 - (-6))^2 + (5 - (-1))^2} = \\sqrt{8^2 + 6^2} = \\sqrt{64 + 36} = \\sqrt{100} = 10$ units.",
"Step 1: The distance between $(2,5)$ and $(-6,-1)$ is given by the distance formula: $$\\sqrt{(2-(-6))^2 + (5-(-1))... | [
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"Step 1: The distance between $(2,5)$ and $(-6,-1)$ is given by the distance formula: $$\\sqrt{(2-(-6))^2 + (5-(-1))^2} = \\sqrt{8^2 + 6^2} = \\sqrt{64+36} = \\sqrt{100} = 10.$$ Step 2: Therefore, the distance between $(2,5)$ and $(-6,-1)$ is $10$ units.",
"Step 1: The distance between the points $(2,5)$ and $(-6... | [
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"Step 1: To find the distance between $(2,5)$ and $(-6,-1)$, we use the distance formula, which is $\\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Step 2: Plugging in the given coordinates, we get $\\sqrt{(2 - (-6))^2 + (5 - (-1))^2} = \\sqrt{(8)^2 + (6)^2} = \\sqrt{64 + 36} = \\sqrt{100} = 10$.",
"Step 1: The distance ... | [
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"Step 1: The distance between $(2,5)$ and $(-6,-1)$ is given by the distance formula:\n\n$$d = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$$\n\nwhere $(x_1, y_1) = (2,5)$ and $(x_2, y_2) = (-6,-1)$. Step 2: Substituting these values, we get:\n\n$$d = \\sqrt{(-6 - 2)^2 + (-1 - 5)^2}$$\n\nStep 3: Simplifying, we get:\n\n$$... | [
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"Step 1: The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by the distance formula $\\sqrt{(x_1-x_2)^2 + (y_1-y_2)^2}$. Step 2: Substituting the given points gives $\\sqrt{(2-(-6))^2 + (5-(-1))^2}$, which simplifies to $\\sqrt{8^2 + 6^2}$. Step 3: Simplifying further gives $\\sqrt{64 + 36} = \\... | [
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"Step 1: The distance between the two points can be calculated using the distance formula: $d = \\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$. Step 2: Substituting the coordinates into the formula gives $d = \\sqrt{(-6-2)^2 + (-1-5)^2} = \\sqrt{8^2 + 6^2}$. Step 3: Simplifying gives $d = \\sqrt{64 + 36} = \\sqrt{100} = 10$.",... | [
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"Step 1: Use the distance formula, which is $d = \\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. Step 2: Plug in the coordinates $(2,5)$ and $(-6,-1)$ to get $d = \\sqrt{(-6-2)^2 + (-1-5)^2} = \\sqrt{(-8)^2 + (-6)^2} = \\sqrt{64 + 36} = \\sqrt{100}$.",
"Step 1: The distance between $(2,5)$ and $(-6,-1)$ is the distance b... | [
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"Step 1: Expand the right-hand side of the equation to get $\\frac{1}{2}\\left( 3x^2-1\\right) = x^4 - 75x^3 - 150x^2 + 1250x + 50$. Step 2: Combine like terms to get $\\frac{1}{2}\\left( 3x^2-1\\right) = x^4 - 75x^3 - 200x^2 + 1250x + 50$. Step 3: Subtract $\\frac{1}{2}\\left( 3x^2-1\\right)$ from both sides to ge... | [
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"Step 1: Expand the equation by multiplying the right-hand side: $\\frac{1}{2}(3x^2-1)=x^4-50x^3-10x^2+25x^3+1250x+50x^2+250$. Step 2: Combine like terms to get $\\frac{1}{2}(3x^2-1)=x^4+25x^3+25x^2+1275x+50$. Step 3: Now move all terms to one side and simplify: $\\frac{1}{2}(3x^2-1)-x^4-25x^3-25x^2-1275x-50=0$. St... | [
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"Step 1: Expand the right-hand side: $$\\frac{1}{2}\\left( 3x^2-1\\right) = \\left( x^2-50x-10\\right)\\left( x^2+25x+5\\right) = x^4 - 75x^3 - 60x^2 + 1250x + 50$$ Step 2: Group like terms: $$\\frac{1}{2}\\left( 3x^2-1\\right) = x^4 - 75x^3 - 65x^2 + 1255x + 50$$ Step 3: Since the left-hand side is a quadratic, th... | [
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"Step 1: Expand the right-hand side of the equation: $$\\frac{1}{2}\\left( 3x^2-1\\right) = \\left( x^2-50x-10\\right)\\left( x^2+25x+5\\right) = x^4-75x^3-25x^2+50x^3-1250x-10x^2+250x+50.$$Step 2: Combine like terms: $$\\frac{1}{2}\\left( 3x^2-1\\right) = x^4-125x^2+200x+50.$$Step 3: Since the equation is an equal... | [
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"Step 1: Expand the right-hand side of the equation to get $\\frac{1}{2}\\left( 3x^2-1\\right) = \\left( x^4-75x^3-10x^2+250x+50\\right)$. Step 2: Expand the left-hand side to get $\\frac{1}{2}\\left( 3x^2-1\\right) = \\frac{3x^2}{2}-\\frac{1}{2}$. Step 3: Equate the two expressions to get $\\frac{3x^2}{2}-\\frac{1... | [
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"Step 1: Expand the right-hand side to get $\\frac{1}{2}\\left( 3x^2-1\\right) = x^4-75x^3-60x^2+250x+10$. Step 2: Rewrite the left-hand side as $\\frac{3x^2-2}{2}$. Step 3: The equation now becomes $3x^2-2 = 2x^4-150x^3-120x^2+500x+20$. Step 4: Move all terms to one side and factor the left-hand side: $0 = x^4-150... | [
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"Step 1: The equation is equivalent to $\\frac{1}{2} (3x^2 - 1) = x^4 - 50x^3 - 10x^2 + 25x^3 + 125x^2 + 50x - 10x^2 - 50x - 5$ Step 2: This simplifies to $\\frac{1}{2} (3x^2 - 1) = x^4 + 75x^2 - 40x - 5$ Step 3: We can rewrite this as $0 = x^4 + 76x^2 - 40x - \\frac{1}{2}$ Step 4: The sum of the coefficients of $x... | [
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"Step 1: Expand the right-hand side of the equation:\n\n$\\frac{1}{2}\\left( 3x^2-1\\right) = \\left( x^2-50x-10\\right)\\left( x^2+25x+5\\right)$\n$\\frac{1}{2}\\left( 3x^2-1\\right) = x^4-50x^3-10x^2 + 25x^3+1250x^2+100x+5x^2-250x-50$\n$\\frac{1}{2}\\left( 3x^2-1\\right) = x^4+1075x^2+60x-50$\n\nStep 2: Write the... | [
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"Step 1: If $75$ workers can build the embankment in $4$ days, then $1$ worker can build the embankment in $4 \\cdot 75 = 300$ days. Step 2: If $50$ workers build the embankment, it will take them $\\frac{300}{50} = 6$ days.",
"Step 1: If $75$ workers take $4$ days to build an embankment, then one worker takes $4... | [
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"Step 1: If $75$ workers can build the embankment in $4$ days, then the total work done is $75 \\cdot 4 = 300$ worker-days. Step 2: If only $50$ workers were available, the total time it would take for them to do the same amount of work is $300/50 = 6$ days.",
"Step 1: If it took 75 workers 4 days to build the em... | [
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"Step 1: Let $d$ be the number of days it would take $50$ workers to build the embankment. Step 2: Since $75$ workers built the embankment in $4$ days, the total amount of work required is $4 \\cdot 75 = 300$ person-days. Step 3: The number of person-days required by $50$ workers is $50 \\cdot d$. Step 4: Equating ... | [
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"Step 1: If $75$ workers can build an embankment in $4$ days, then each worker can build $\\frac{1}{75} \\cdot 4 = \\frac{4}{75}$ of the embankment in one day. Step 2: If there are $50$ workers, they can build $\\frac{50}{75}$ of the embankment in one day.",
"Step 1: Let $d$ be the total number of days it would t... | [
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"Step 1: Since $75$ workers took $4$ days to build the embankment, the total work done is $4 \\cdot 75 = 300$ man-days. Step 2: If only $50$ workers had been available, the work done per day by one worker is $\\frac{300}{50 \\cdot 4} = \\frac{3}{2}$. Step 3: Therefore, it would take $1$ worker $\\frac{300}{\\frac{3... | [
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"Step 1: We know that if it takes $75$ workers $4$ days to build the embankment, then $75$ workers can build $1$ unit of the embankment in $4 \\cdot \\frac{1}{75} = \\frac{1}{15}$ days. Step 2: If only $50$ workers are available, it will take them $\\frac{1}{15} \\div \\frac{50}{75} = \\frac{1}{15} \\cdot \\frac{75... | [
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"Step 1: The total labor contributed by the 75 workers in one day is $75L$, where $L$ is the amount of labor contributed by one worker in one day. Step 2: The total labor required to build the embankment is $4 \\times 75L = 300L$. Step 3: Since 50 workers worked, the amount of labor contributed in one day is $50L$,... | [
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"Step 1: Let $x$ be the time it would have taken for the $50$ workers to build the embankment. Step 2: The amount of work done by each worker in one day is $\\frac{1}{4}$ (since $4$ days are required for $75$ workers to complete the task). Step 3: Since there are $50$ workers, the total amount of work done per day ... | [
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"Step 1: Let $c$ be the number of students in the chemistry class and $b$ be the number of students in the biology class. Step 2: We are given that $c = 3b$ and that $c + b - 5 = 43$. Step 3: Substitute $c = 3b$ into the second equation to get $3b + b - 5 = 43$. Step 4: This simplifies to $4b = 48$, so $b = 12$. St... | [
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"Step 1: Let the number of students taking chemistry be $C$ and the number of students taking biology be $B$. Step 2: We know $C = 3B$, and $C + B - 5 = 43$. Step 3: Substituting $C = 3B$ into the second equation gives $3B + B - 5 = 43$. Step 4: This simplifies to $4B = 48$, so $B = 12$. Step 5: We know $C = 3B$, s... | [
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"Step 1: Let $c$ be the number of students in the chemistry class, and $b$ be the number of students in the biology class. Step 2: We know that $c + b = 43$ and $c = 3b$. Step 3: Substituting the second equation into the first gives $3b + b = 43$, so $4b = 43$. Step 4: Solving for $b$ gives $b = \\frac{43}{4} = 10\... | [
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"Step 1: Let $b$ be the number of students in the biology class. Step 2: The number of students in the chemistry class is $3b$, since it is three times as large. Step 3: The total number of students in both classes is the sum of the number of students taking biology and chemistry, minus the 5 students taking both: ... | [
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"Step 1: Let $c$ be the number of students taking chemistry and $b$ be the number of students taking biology. Step 2: The equation $c + b - 5 = 43$ describes the total number of students, where we subtract 5 to account for the 5 students taking both classes. Step 3: Since the chemistry class is three times as large... | [
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"Step 1: Let the number of students in the biology class be $b$. Step 2: Then, the number of students in the chemistry class is $3b$. Step 3: The number of students taking only chemistry is $3b - 5$, since 5 students are taking both classes. Step 4: The number of students taking only biology is $b - 5$, since 5 stu... | [
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"Step 1: Let $c$ be the number of students in the chemistry class, and $b$ be the number of students in the biology class. Step 2: The number of students taking chemistry is $c$, and the number of students taking biology is $b$. Step 3: The number of students taking both classes is 5. Step 4: This gives the equatio... | [
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"Step 1: Let $B$ be the number of students taking biology, and let $C$ be the number of students taking chemistry. Step 2: The number of students taking biology or chemistry is $B + C - 5$, since the 5 students taking both are counted twice. Step 3: Since each student takes at least one class, this number must equa... | [
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"Step 1: Let $x$ be the number of students in each of the three rows with the same number of students. Step 2: Then the fourth row has $x + 1$ students. Step 3: The total number of students is $3x + (x + 1) = 4x + 1$. Step 4: Since there are over 30 students, $4x + 1 > 30$. Step 5: The smallest integer value of $x$... | [
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"Step 1: Let $x$ be the number of students in each of the three rows with the same number of students. Step 2: The total number of students is $x + x + x + (x + 1) = 4x + 1$. Step 3: Since there are more than 30 students, we need $4x + 1 > 30$. Step 4: This implies $x > 7.25$. Step 5: Since $x$ is an integer, we ne... | [
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"Step 1: Let the number of students in each of the three rows with equal numbers be $x$. Step 2: Then, the number of students in the fourth row is $x+1$. Step 3: The total number of students in the class is $4x+x+1=5x+1$. Step 4: Since the class has more than 30 students, $5x+1>30$. Step 5: Subtracting 1 from both ... | [
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"Step 1: Let $x$ be the number of students in each of the three rows that have the same number of students. Step 2: The number of students in the fourth row is $x + 1$. Step 3: The total number of students is greater than 30, so we can write $3x + (x + 1) > 30$. Step 4: Simplifying, we get $4x > 29$. Step 5: The sm... | [
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0.11367728292009859,
0.11367728292009859,
0.09307108745303111,
0.11367728292009859
] |
[
"Step 1: Let the number of students in each of the three rows be $x$. Step 2: Since one row has one more student than the other three rows, the total number of students is $3x + (x + 1) = 4x + 1$. Step 3: Since the class has more than 30 students, $4x + 1 > 30$. Step 4: We can subtract 30 from both sides to get $4x... | [
0.08688170329867509,
0.09844986769270467,
0.09844986769270467,
0.1432437122631831,
0.1432437122631831,
0.1432437122631831,
0.1432437122631831,
0.1432437122631831
] |
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