"Problem:\nFind the domain of the expression $\frac{{\sqrt{{x-2}}}}{{\sqrt{{5-x}}}}$.}}\nSolution:"
"Problem:\nFind the domain of the expression $\\frac{{\sqrt{{x-2}}}}{{\sqrt{{5-x}}}}$.}}\nSolution:"
),
dict(
role="BOT",
prompt=
"The expressions inside each square root must be non-negative. Therefore, $x-2 \ge 0$, so $x\ge2$, and $5 - x \ge 0$, so $x \le 5$. Also, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$. Therefore, the domain of the expression is $\boxed{{[2,5)}}$.\nFinal Answer: The final answer is $[2,5)$. I hope it is correct."
"The expressions inside each square root must be non-negative. Therefore, $x-2 \ge 0$, so $x\ge2$, and $5 - x \ge 0$, so $x \le 5$. Also, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$. Therefore, the domain of the expression is $\\boxed{{[2,5)}}$.\nFinal Answer: The final answer is $[2,5)$. I hope it is correct.\n"
),
dict(
role="HUMAN",
...
...
@@ -27,7 +27,7 @@ math_infer_cfg = dict(
dict(
role="BOT",
prompt=
"We have that $\det (\mathbf{{A}} \mathbf{{B}}) = (\det \mathbf{{A}})(\det \mathbf{{B}}) = (2)(12) = \boxed{{24}}.$\nFinal Answer: The final answer is $24$. I hope it is correct."
"We have that $\det (\mathbf{{A}} \mathbf{{B}}) = (\det \mathbf{{A}})(\det \mathbf{{B}}) = (2)(12) = \\boxed{{24}}.$\nFinal Answer: The final answer is $24$. I hope it is correct.\n"
),
dict(
role="HUMAN",
...
...
@@ -37,17 +37,17 @@ math_infer_cfg = dict(
dict(
role="BOT",
prompt=
"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*}} 30n&=480\\ \Rightarrow\qquad n&=480/30=\boxed{{16}} \end{{align*}}\nFinal Answer: The final answer is $16$. I hope it is correct."
"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*}} 30n&=480\\\\ \Rightarrow\qquad n&=480/30=\\boxed{{16}} \end{{align*}}\nFinal Answer: The final answer is $16$. I hope it is correct.\n"
),
dict(
role="HUMAN",
prompt=
"Problem:\nIf the system of equations: \begin{{align*}} 6x-4y&=a,\\ 6y-9x &=b. \end{{align*}}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\frac{{a}}{{b}},$ assuming $b$ is nonzero.\nSolution:"
"Problem:\nIf the system of equations: \\begin{{align*}} 6x-4y&=a,\\\\ 6y-9x &=b. \end{{align*}}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{{a}}{{b}},$ assuming $b$ is nonzero.\nSolution:"
),
dict(
role="BOT",
prompt=
"If we multiply the first equation by $-\frac{{3}}{{2}}$, we obtain $$6y-9x=-\frac{{3}}{{2}}a.$$Since we also know that $6y-9x=b$, we have $$-\frac{{3}}{{2}}a=b\Rightarrow\frac{{a}}{{b}}=\boxed{{-\frac{{2}}{{3}}}}.$$\nFinal Answer: The final answer is $-\frac{{2}}{{3}}$. I hope it is correct."
"If we multiply the first equation by $-\\frac{{3}}{{2}}$, we obtain $$6y-9x=-\\frac{{3}}{{2}}a.$$Since we also know that $6y-9x=b$, we have $$-\\frac{{3}}{{2}}a=b\Rightarrow\\frac{{a}}{{b}}=\\boxed{{-\\frac{{2}}{{3}}}}.$$\nFinal Answer: The final answer is $-\\frac{{2}}{{3}}$. I hope it is correct.\n"
Find the domain of the expression $\frac{{\sqrt{{x-2}}}}{{\sqrt{{5-x}}}}$.}}
Find the domain of the expression $\\frac{{\sqrt{{x-2}}}}{{\sqrt{{5-x}}}}$.}}
Solution:
The expressions inside each square root must be non-negative. Therefore, $x-2 \ge 0$, so $x\ge2$, and $5 - x \ge 0$, so $x \le 5$. Also, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$. Therefore, the domain of the expression is $\boxed{{[2,5)}}$.
The expressions inside each square root must be non-negative. Therefore, $x-2 \ge 0$, so $x\ge2$, and $5 - x \ge 0$, so $x \le 5$. Also, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$. Therefore, the domain of the expression is $\\boxed{{[2,5)}}$.
Final Answer: The final answer is $[2,5)$. I hope it is correct.
Problem:
If $\det \mathbf{{A}} = 2$ and $\det \mathbf{{B}} = 12,$ then find $\det (\mathbf{{A}} \mathbf{{B}}).$
Solution:
We have that $\det (\mathbf{{A}} \mathbf{{B}}) = (\det \mathbf{{A}})(\det \mathbf{{B}}) = (2)(12) = \boxed{{24}}.$
We have that $\det (\mathbf{{A}} \mathbf{{B}}) = (\det \mathbf{{A}})(\det \mathbf{{B}}) = (2)(12) = \\boxed{{24}}.$
Final Answer: The final answer is $24$. I hope it is correct.
Problem:
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?
Solution:
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*}} 30n&=480\\ \Rightarrow\qquad n&=480/30=\boxed{{16}} \end{{align*}}
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*}} 30n&=480\\\\ \Rightarrow\qquad n&=480/30=\\boxed{{16}} \end{{align*}}
Final Answer: The final answer is $16$. I hope it is correct.
Problem:
If the system of equations: \begin{{align*}} 6x-4y&=a,\\ 6y-9x &=b. \end{{align*}}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\frac{{a}}{{b}},$ assuming $b$ is nonzero.
If the system of equations: \\begin{{align*}} 6x-4y&=a,\\\\ 6y-9x &=b. \end{{align*}}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{{a}}{{b}},$ assuming $b$ is nonzero.
Solution:
If we multiply the first equation by $-\frac{{3}}{{2}}$, we obtain $$6y-9x=-\frac{{3}}{{2}}a.$$Since we also know that $6y-9x=b$, we have $$-\frac{{3}}{{2}}a=b\Rightarrow\frac{{a}}{{b}}=\boxed{{-\frac{{2}}{{3}}}}.$$
Final Answer: The final answer is $-\frac{{2}}{{3}}$. I hope it is correct.
If we multiply the first equation by $-\\frac{{3}}{{2}}$, we obtain $$6y-9x=-\\frac{{3}}{{2}}a.$$Since we also know that $6y-9x=b$, we have $$-\\frac{{3}}{{2}}a=b\Rightarrow\\frac{{a}}{{b}}=\\boxed{{-\\frac{{2}}{{3}}}}.$$
Final Answer: The final answer is $-\\frac{{2}}{{3}}$. I hope it is correct.
Find the domain of the expression $\frac{{\sqrt{{x-2}}}}{{\sqrt{{5-x}}}}$.}}
Find the domain of the expression $\\frac{{\sqrt{{x-2}}}}{{\sqrt{{5-x}}}}$.}}
Solution:
The expressions inside each square root must be non-negative. Therefore, $x-2 \ge 0$, so $x\ge2$, and $5 - x \ge 0$, so $x \le 5$. Also, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$. Therefore, the domain of the expression is $\boxed{{[2,5)}}$.
The expressions inside each square root must be non-negative. Therefore, $x-2 \ge 0$, so $x\ge2$, and $5 - x \ge 0$, so $x \le 5$. Also, the denominator cannot be equal to zero, so $5-x>0$, which gives $x<5$. Therefore, the domain of the expression is $\\boxed{{[2,5)}}$.
Final Answer: The final answer is $[2,5)$. I hope it is correct.
Problem:
If $\det \mathbf{{A}} = 2$ and $\det \mathbf{{B}} = 12,$ then find $\det (\mathbf{{A}} \mathbf{{B}}).$
Solution:
We have that $\det (\mathbf{{A}} \mathbf{{B}}) = (\det \mathbf{{A}})(\det \mathbf{{B}}) = (2)(12) = \boxed{{24}}.$
We have that $\det (\mathbf{{A}} \mathbf{{B}}) = (\det \mathbf{{A}})(\det \mathbf{{B}}) = (2)(12) = \\boxed{{24}}.$
Final Answer: The final answer is $24$. I hope it is correct.
Problem:
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?
Solution:
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*}} 30n&=480\\ \Rightarrow\qquad n&=480/30=\boxed{{16}} \end{{align*}}
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*}} 30n&=480\\\\ \Rightarrow\qquad n&=480/30=\\boxed{{16}} \end{{align*}}
Final Answer: The final answer is $16$. I hope it is correct.
Problem:
If the system of equations: \begin{{align*}} 6x-4y&=a,\\ 6y-9x &=b. \end{{align*}}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\frac{{a}}{{b}},$ assuming $b$ is nonzero.
If the system of equations: \\begin{{align*}} 6x-4y&=a,\\\\ 6y-9x &=b. \end{{align*}}has a solution $(x, y)$ where $x$ and $y$ are both nonzero, find $\\frac{{a}}{{b}},$ assuming $b$ is nonzero.
Solution:
If we multiply the first equation by $-\frac{{3}}{{2}}$, we obtain $$6y-9x=-\frac{{3}}{{2}}a.$$Since we also know that $6y-9x=b$, we have $$-\frac{{3}}{{2}}a=b\Rightarrow\frac{{a}}{{b}}=\boxed{{-\frac{{2}}{{3}}}}.$$
Final Answer: The final answer is $-\frac{{2}}{{3}}$. I hope it is correct.
If we multiply the first equation by $-\\frac{{3}}{{2}}$, we obtain $$6y-9x=-\\frac{{3}}{{2}}a.$$Since we also know that $6y-9x=b$, we have $$-\\frac{{3}}{{2}}a=b\Rightarrow\\frac{{a}}{{b}}=\\boxed{{-\\frac{{2}}{{3}}}}.$$
Final Answer: The final answer is $-\\frac{{2}}{{3}}$. I hope it is correct.
