diff --git a/results_llama_lora_0.7_50_astar.csv b/results_llama_lora_0.7_50_astar.csv new file mode 100644 index 0000000..cb293f1 --- /dev/null +++ b/results_llama_lora_0.7_50_astar.csv @@ -0,0 +1,351 @@ +,gt,pred +0,2,"The graph of $y=\frac{2}{x^2+x-6}$ has $\boxed{2}$ vertical asymptotes at x=2 and x=-3. +The final answer is 2." +1,10,"The difference between 36 and 26 is $\boxed{10}$. + +The final answer is: $\boxed{10}$" +2,i,"The first term $i^5$ can be evaluated by rewriting the exponent as a sum of exponents of 4. Since 5 is equal to $4 \cdot 1 + 1$, $i^5 = (i^4)^1 \cdot i = 1^1 \cdot i = i$. + +## Step 1: Evaluate the first term +The first term $i^5$ can be evaluated by rewriting the exponent as a sum of exponents of 4. Since 5 is equal to $4 \cdot 1 + 1$, $i^5 = (i^4)^1 \cdot i = 1^1 \cdot i = i$. + +## Step 2: Evaluate the second term +The second term $i^{-25}$ can be evaluated by rewriting the exponent as a sum of exponents of 4. Since 25 is equal to $4 \cdot 6 + 1$, $i^{-25} = (i^4)^{-6} \cdot i^{-1} = 1^{-6} \cdot i^{-1} = 1^{-6} \cdot (i^4)^{-1} \cdot i = 1^{-6} \cdot 1^{-1} \cdot i = 1^{-6} \cdot i = i^{-1}$. + +## Step 3: Evaluate the third term +The third term $i^{45}$ can be evaluated by rewriting the exponent as a sum of exponents of 4. Since 45 is equal to $4 \cdot 11 + 1$, $i^{45} = (i^4)^{11} \cdot i = 1^{11} \cdot i = i$. + +## Step 4: Add the three terms together +The sum of the three terms is $i + i^{-25} + i^{45} = i + i^{-1} + i" +3,4,"If $2^8=4^x$, then $2^8=2^{2x}$, which implies that $8=2x$, so that $x=4$." +4,402,"The expression $\frac{2\left( 1-x\right)\left( 1+x\right)}{1-x^2}$ simplifies to $\boxed{2}$. +### +Problem Goal: +What" +5,"x \in [-2,7]","The solutions to the quadratic equation $x^2 - 5x - 4 = 0$ are $x = -1$ and $x = 4$. Since $-1$ is less than $-2$ and $7$ is greater than $7$, the solution set is $\boxed{[-2, 7]}$. +The final answer is: $\boxed{[-2, 7]}$ +Note: The final answer is already provided in the solution, so there is no need to change it. The format is followed exactly as specified in the prompt." +6,-\frac{1,"The answer is $\boxed{-\frac{1}{8}}$. +Note: The answer is negative." +7,\frac{x+2,"The inverse of $h(x)$ is $\boxed{\frac{y+2}{7}}$. +In this problem, we are asked to find the inverse of a function $h(x)$. This means that we need to find a function $h^{-1}(x)$ that satisfies $h(h^{-1}(x))=x$. In other words, we need to find a function $h^{-1}(x)$ that undoes the function $h(x)$. To find $h^{-1}(x)$, we can use the following procedure: +1. Evaluate $h(x)$. +2. Write $h(x)=y$. +3. Solve for $x$. +4. Write the inverse function $h^{-1}(x)$. +Using this procedure, we can find the inverse of $h(x)$. First, we evaluate $h(x)$ as $h(x)=f(g(x))=f(x-1)=7(x-1)+5=7x-2$. Then we write $h(x)=y$ and solve for $x$ to get $x=\frac{y+2}{7}$. Therefore, the inverse function is $h^{-1}(x)=\frac{y+2}{7}$." +8,10,"The distance from the origin to the point $(-8,6)$ is $\boxed{10}$ units. +Note: The next step is a possible solution to the problem. The solution may be correct or incorrect. It is always a good idea to check the solution to make sure it is correct before presenting it as the final answer." +9,8,"The average of the numbers 2 and 4 is $\frac +The average of the numbers 5 and 7 is $\frac +The average of the numbers 6 and 8 is $\frac +The average of the numbers 9 and 11 is $\frac +The average of the numbers 1 and 3 is $\frac +The average of the numbers 2 and 4 is $\frac +The average of the numbers 5 and 7 is $\frac +The average of the numbers 6 and 8 is $\frac +The average of the numbers 9 and 11 is $\frac +The average of the numbers 2 and 4 is $\frac +The average of the numbers 5 and 7 is $\frac +The average of the numbers 6 and 8 is $\frac +The average of the numbers 9 and 11 is $\frac +The average of the numbers 2 and 4 is $\frac +The average of the numbers 5 and 7 is $\frac +The average of the numbers 6 and 8 is $\frac +The average of the numbers 9 and 11 is $\frac +The average of the numbers 2 and 4 is $\frac +The average of the numbers 5 and 7 is $\frac +The average of the numbers 6 and 8 is $\frac +The average of the numbers 9 and 11 is $\frac +The average of the numbers 2 and 4 is $\frac +The average of the numbers 5 and 7 is $\frac +The average of the numbers" +10,-25,The value of f(9 +11,8,"Try another approach. Given that $A\sqrt{B}$ must equal $9C$, we can try to factor 9 out of the expression $\frac{A\sqrt{B}}{C}$. This is possible if we can factor out a 9 from the denominator. Since $C$ is a positive integer, $\frac{1}{C}$ is a positive rational number. Since $A$ and $C$ have no common prime factors, $C$ is an odd integer. Since $\frac{A\sqrt{B}}{C}$ is a positive rational number, $\sqrt{B}$ must be an odd integer. + +## Step 18: Identify that B is 9 +Since $B$ is a positive integer, $B^{2}$ is a perfect square factor of $B$. Since $B$ has no perfect-square factors other than 1, $B^{2}$ must equal 1. Thus, $B$ is 9. + +## Step 19: Solve for C +Since $\frac{A\sqrt{B}}{C} = \frac{9}{2\sqrt{3}}$, $\sqrt{B}$ must equal $\frac{9}{2\sqrt{3}} \cdot C$. Thus, $C = \frac{\sqrt{B}}{\frac{9}{2\sqrt{3}}}$. + +## Step 20: Solve for C +$C = \frac{\sqrt{B}}{\frac{9}{2\sqrt{3}}} = \frac{\sqrt{9}}{\frac{9}{2\sqrt{3}}} = \frac{3}{\frac{9}{2\sqrt{3}}} = \frac{3}{\frac{9}{2\sqrt{3}}} \cdot \frac{2\sqrt{3}}{2\sqrt{3}} = \frac{6\sqrt{3}}{9} = \frac{2\sqrt{3}}{3}$. + +## Step 21: Solve for A +Since $B$ is 9, $\sqrt{B}$ is 3. Since $\frac{A\sqrt{B}}{C} = \frac{9}{2\sqrt{3}}$, $A\sqrt{B}$ must equal $9C$. Thus, $A\sqrt{B} = 9C$ implies that $A\sqrt{B} = 9C$ implies that $A = \frac{9C}{\sqrt{B}}$. + +## Step 22: Solve for A +$A = \frac{9C}{\sqrt{B}} = \frac{9 \cdot \frac{2\sqrt{3}}{3}}{3} = \frac{9 \cdot \frac{2\sqrt{3}}{3}}{3} = \frac{9 \cdot 2\sqrt{3}}{9} = \frac{18\sqrt{3}}{9} = \frac{6\sqrt{3}}{3} = 2\sqrt{3}$. + +## Step 23: Solve for A +$A = 2\sqrt{3}$. + +## Step 24: Solve for A +$A = 2\sqrt{3}$. + +## Step 25: Solve for A +$A = 2\sqrt{3}$. + +## Step 26: Solve for A +Since $\frac{A\sqrt{B}}{C} = \frac{9}{2\sqrt{3}}$, $\sqrt{B}$ must equal $\frac{9}{2\sqrt{3}} \cdot C$. Thus, $C = \frac{\sqrt{B}}{\frac{9}{2\sqrt{3}}}$. + +## Step 27: Solve for C +$C = \frac{\sqrt{B}}{\frac{9}{2\sqrt{3}}} = \frac{\sqrt{9}}{\frac{9}{2\sqrt{3}}} = \frac{3}{\frac{9}{2\sqrt{3}}} = \frac{3}{\frac{9}{2\sqrt{3}}} \cdot \frac{2\sqrt{3}}{2\sqrt{3}} = \frac{6\sqrt{3}}{9} = \frac{2\sqrt{3}}{3}$. + +## Step 28: Solve for C +Since $A$ is $2\sqrt{3}$, $\frac{A\sqrt{B}}{C} = \frac{9}{2\sqrt{3}}$ implies that $\frac{2\sqrt{3}\sqrt{9}}{C} = \frac{9}{2\sqrt{3}}$. + +## Step 29: Solve for C +$\frac{2\sqrt{3}\" +12,187.5,"The number of calories in a 12 fluid ounce bottle is 12 times 125, which is $\boxed{187.5}$. +Note that there is no possible next step for this problem as the solution is already found. +If you get stuck, check the work. Recheck your steps to ensure that you have not made a calculation error. In some cases, a simple error in a calculation can cause a big problem. If you have not found a possible next step and the current state does not solve the problem, consider the following suggestions: +\begin{enumerate} +\item Check your work: Look for any calculation errors. +\item Look for a possible simplification: See if the current state can be simplified or re-expressed. +\item Check for any assumptions: Verify that you have not assumed anything that is not given in the problem. +\item Look for an error in the problem statement: Double-check the problem statement for any errors. +\item Consider a different approach: Think about the problem from a different angle or perspective. +\item Consider a different current state: Re-evaluate your current state and look for a possible next step. +\end{enumerate} +If you have tried the above suggestions and you still cannot find a possible next step, consider consulting a classmate, teacher, or tutor. They may be able to help you find a possible next step." +13,\$40,"Mike will be offered $\$21.25$ in 12 more years. +The stamp's offer price doubles every 3 years. Thus, in 3 more years, Mike will be offered $\$21.25\cdot2= 42.50$. +The final answer is: $\boxed{42.50}$" +14,5,"If $17^6-17^5=17^x\cdot16$ then $17^5\cdot17=17^x\cdot16$. If $17^5=16$ then $17=1/16$ and $17^x=1$. This implies that $x=0$ and $17^x=17^0=1$. Thus, $x=\boxed0$." +15,8,The new pressure will be approximately $\boxed{27.11}$ kPa. +16,3125,a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3⋅a2= a3� +17,"[0,\infty)","The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0, 4]$. +The range of the function is $[0" +18,.5,"The solutions to the equation are $x = 3$ and $x = \frac{1}{2}$. The sum of the possible values of $x$ is $\boxed(3+\frac{1}{2})$. +###" +19,5,"The difference between 36 and 26 is $\boxed{10}$. (Not needed, but a good check)" +20,105,"The sum of the elements of each of the 15 two-element subsets of $\{1,2,3,4,5,6\}$ is $\frac{15\cdot21}{2}=157.5$. \(\boxed{157.5}\) +Note the following things: +* The current state must be expressed as an equation. +* The next step must be a calculation or an expression of a calculation. +* The next step must be a single step to reach the goal. +* The goal is not always a numerical value. The goal can be an expression. +* If the current state is not a numerical value, the next step must be an expression." +21,2,"The value of $n$ is $\boxed{2}$. +###" +22,"(-\infty,-8)\cup (8,\infty)",The product of $(m-8)(m+8)$ is $(m^2-64)$. The solution to $m^2-64>0$ is $m<-8$ or $m>8$. +23,0,"We can check the solution by substituting $x = -1/2$ into the equation $2x^2 + 3x + 2 = 0$ to get $2(-1/2)^2 + 3(-1/2) + 2 = 0$, which implies that $(1/2) + (-3/2) + 2 = 0$, which implies that $-1/2 + 2 = 0$, which implies that $3/2 = 0$, which is a false statement. Therefore, we need to find the other solution to the equation $2x^2 + 3x + 2 = 0$. We can use the quadratic formula $x = {-b\pm \sqrt{b^2 - 4ac} 4a" +24,2,"The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1| = 7$ is $\boxed{0}$. +The values that $x$ can take are $8$ and $-8$. Therefore, the sum of all values of $x$ such that $|x-1" +25,\frac{1,"Solve for $w$ by multiplying both sides by $w$ and then multiplying both sides by 9 to get rid of the denominators and fractions. Then solve for $w$ to get $w=\frac{9\cdot42}{1\cdot3}=\frac{378}{3}=126$. +### +Problem Goal: +What is the value of $y$ in the equation $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y$ $y" +26,6+9i,"We can evaluate $(1+2i)^3$ by expanding it in the form of $(a+b)^3$. +\begin{align*} (1+2i)^3 &= (1+2i)^3 \\ &= 1^3 + 3(1)^2(2i) + 3(1)(2i)^2 + (2i)^3 \\ &= 1 + 6i -12 - 8i \\ &= -11 - 2i \end{align*}. +Now we have $(1-2i)^3(-11-2i)$. +We can simplify this expression further to find the value of $\left( 1 - 2i \right)^3 \left( 1 + 2i \right)^{3-3i}$. +We have +\begin{align*} \left( 1 - 2i \right)^3 \left( 1 + 2i \right)^{3-3i} &= \left( -11 - 2i \right) \left( 1 + 2i \right)^{3-3i} \\ &= (-11 - 2i)(1 + 2i)^{3-3i} \\ &= (-11 - 2i)(1+2i)^3 \\ &= (-11 - 2i)(-11-2i) \\ &= (121+22i+4i^2) \\ &= 121+22i-4 \\ &= 117+22i. \end{align*}, +Note: $i^2=-1$. +The final answer is $\boxed{117+22i}$. +Note that in this problem, the property $(ab)^n = a^n b^n$ was used with $a=1+2i$, $b=1-2i$ and $n=3-3i$. It was also necessary to simplify the expression $\left( 1 - 2i \right)^3 \left( 1 + 2i \right)^{3-3i}$ by expanding the expression $\left( 1 + 2i \right)^3$ in the form of $(a+b)^3$. This problem also required the evaluation of $i^2$. +### Problem Goal: +Evaluate $(x+2i)5-5i$. + + +\begin{align*} \left( x+2i \right)^5 \left( x+2i \right)^{-5i} &= \left( x+2i \right)^5 \left( x+2i \right)^{5-5i} \\ &= \left( x+2i \right)^5 \left( x+2i \right)^{5-5i} \\ &= \left( x+2i \right)^5 \left( x+2i \right)^{5-5i} \end{align*}, where we applied the property $(ab)^n = a^n b^n$ for $a=x+2i$, $b=1$ and $n=5-5i$. +We can now substitute the value of $(x+2i)^5$ in the above equation to find the value of $\left( x+2i \right)^5 \left( x+2i \right)^{5-5i}$. +We have +\begin{align*} \left( x+2i \right)^5 \left( x+2i \right)^{5-5i} &= \left( x+2i \right)^5 \left( x+2" +27,2,"The simplified numerical value of $\frac{a+11b}{a-b}$ is $\boxed{2}$. +In the previous example, we have a system of linear equations in the form of $\frac{a+11b}{a-b}$ and $\frac{4a+3b}{a-2b}$. The current state is $\frac{4a+3b}{a-2b}$ = 5. We will now solve this system of linear equations. +First, we will solve $\frac{4a+3b}{a-2b}$ = 5 to $4" +28,20,"First, multiply the two complex numbers, \begin{align*} (2-2i)(5+5i)&=2\cdot 5+2\cdot 5i-2i\cdot 5-2i\cdot 5i\\ &\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad" +29,7(x+3) (x-3),"The expression can be factored into the product of two binomials, $7(x - \sqrt{9})(x + \sqrt{9})$." +30,y^4-2y^3+7y^2+y-5,"The volume of a cone with a radius of $ 5$, and height of $ 8$ is $\frac{1}{3} \pi (5)^2 (8) = \frac{1}{3} \pi 25 (8) = \frac{200\pi}{3}$. +The final answer is $\frac{200\pi}{3}$. +### +Problem Goal: +A population of rabbits is growing at a rate of $ 10\%$" +31,0,"\begin{align*} +f(x)-f^{-1}(x)&=x^2-\frac{1}{x}\\ +&=x^2-\frac{x}{x^2}\\ +&=x^2-\frac{1}{x}\\ +&=x^2-x^{-1} +\end{align*} +Note: The previous solution had an error. The solution was corrected in the above solution. +Note: The above solution is correct." +32,\frac{7,The car travels 70 miles. +33,5,"The expression $64\sqrt{8}$ is equivalent to $(4^3)(8^{1/2})$ which is equivalent to $(4^3)(8^{1/2})$ which is equivalent to $((4^3)(8^{1/2}))^{1/3}$ which is equivalent to $(8^{1/2})^{1/3}$ which is equivalent to $(8^{1/2\cdot(1/3)})$ which is equivalent to $(8^{1/6})$ which is equivalent to $(8^{1/6})$ which is equivalent to $(\sqrt[6]{8})^{1/2}$ which is equivalent to $(\sqrt[6]{8})^{1/2}$ which is equivalent to $(\sqrt[6]{8})^{1/2}$ which is equivalent to $\boxed{\frac{7}{4}}$. +Note: The final answer $\boxed{\frac{7}{4}}$ is obtained by solving the problem from the current state." +34,"\left(-\infty,-\frac 12\right)\cup \left(-\frac 12,\infty\right)","The domain of the function is all real numbers except $y = -\frac{1}{2}$. +The final answer is: $\boxed{(-\infty, -\frac{1}{2}) \cup (-\frac{1}{2}, \infty)}$" +35,\frac{2,"The probability that there are no real solutions to the equation $x(x+5)=-n$ for the $n$ I chose is the probability that $n$ is between 7 and 10 inclusive, which is $\boxed{\frac{4}{10}=\frac{2}{5}}$." +36,7,"Note that $f^{-1}(x)$ is the inverse of $f(x)$, so $f^{-1}(f^{-1}(3))=f^{-1}(7)=1$." +37,20,"The product of $(3-i)(6+2i)$ is $18+6i-6i-2i^{2}=18-2i^{2}=18+2=20$. +The final answer is $20$. +The final answer is \boxed{20}." +38,9,"The product of the squares of the solutions is $\boxed{(\frac{36 - 104}{4})=\frac{(-68)}{4}=\frac{-68}{4}=\frac{-34}{2}=-17}$. +The solutions to the quadratic equation are $x=\frac{-13\pm\sqrt{169-24\cdot2\cdot6}}{4}$. +The solutions to the quadratic equation are $x=\frac{-13\pm\sqrt{169-48}}{4}$. +The solutions to the quadratic equation are $x=\frac{-13\pm\sqrt{121}}{4}$. +The solutions to the quadratic equation are $x=\frac{-13\pm11}{4}$. +The solutions to the quadratic equation are $x=\frac{-13+11}{4}$. +The solutions to the quadratic equation are $x=\frac{-2}{4}$. +The solutions to the quadratic equation are $x=\frac{-2}{4}$. +The solutions to the quadratic equation are $x=-\frac{1}{2}$. +The solutions to the quadratic equation are $x=-\frac{1}{2}$. +The solutions to the quadratic equation are $x=-\frac{1}{2}$. +The solutions to the quadratic equation are $x=-\frac{1}{2}$. +The solutions to the quadratic equation are $x=-\frac{1}{2}$. +The solutions to the quadratic equation are $x=-\frac{1}{2}$. +The solutions to the quadratic equation are $x=-\frac{1}{2}$. +The solutions to the quadratic equation are $x=-\frac{1}{2" +39,4,"(16)^(1/3) is equal to 2.000000, and (2.000000)^(3) is equal to 4.000000, so the final answer is \boxed 4. +Note: You can only post one problem per time. Please rephrase the problem in a different way if you want to post another problem. +## Step 1: Understand the problem statement +The problem asks for the positive difference between $120\%$ of 30 and $130\%$ of 20. + +## Step 2: Calculate the values +First, we need to calculate $120\%$ of 30. To do this, we multiply 30 by $120\%$, which is equivalent to $1.2$ times 30. This gives us $1.2 \times 30 = 36$. + +## Step 3: Continue calculating the values +Next, we calculate $130\%$ of 20. To do this, we multiply 20 by $130\%$, which is equivalent to $1.3$ times 20. This gives us $1.3 \times 20 = 26$. + +## Step 4: Find the positive difference +Now, we need to find the positive difference between 36 and 26. To do this, we subtract 26 from 36. This gives us $36 - 26 = 10$. + +The final answer is: $\boxed{10}$" +40,6,"The absolute difference between 7 and 1 is $\boxed(6)$. +The absolute difference between 1 and 7 is $\boxed(6)$. +The absolute difference between $|m-n|$ and $6$ is $\boxed(6)$." +41,"\left(1,\frac{9","The height of the prism is $\boxed{12\text{ in}}$. +The height of the prism is 12 in. +###" +42,-2,"$ +x = 2$ or $x = 3$ + $ +The roots are $2$ and $3$. +### +Problem Goal: +Solve $x^2 + 7x + 12 = 0$ by factoring. + + +$ x^2 + 7x + 12 = (x + 3)(x + 4) = 0$ +$ +$ x" +43,8,"The value of $y^2 - 3xy + 8$ for the pair $(x,y) = (4,5)$ is $\boxed{-27}$. + +In summary, we solved the problem of calculating the value of the expression $y^2 - 3xy + 8$ for different pairs of coordinates $(x,y)$ by substituting the values of $x$ and $y$ into the expression and performing the necessary calculations. We obtained the value of $y^2 - 3xy + 8$ for the pair $(x,y) = (3,9)$ as $8$ and for the pair $(x,y) = (4,5)$ as $-27$." +44,20,"Substituting x=20 into $5x$ gives $5\cdot 20 = 100$, and substituting x=8 into $5x$ gives $5\cdot 8 = 40$. The difference between 100 and 40 is $\boxed{60}$." +45,24,"Find the value of $k$ when $j = 10$. +Since $j$ and $k$ are inversely proportional, we have that $j \cdot k = c$ where $c$ is a constant. +Since $j = 16$ when $k = 21$, we have that $16 \cdot 21 = c$, so $c = 336$. +Then $j \cdot k = 336$. +When $j = 10$ we have that $10 \cdot k = 336$. +Then $k = 336/10 = 33.6$. +Then the value of $j$ when $k = 14$ is $10 \cdot 14 = 140$. +Then the value of $k$ when $j = 12$ is $12 \cdot 25 = 300$. +Then the value of $j$ when $k = 14$ is $336/14 = 24$. +Then the value of $k$ when $j = 12$ is $336/12 = 28$. +Then the value of $j$ when $k = 14$ is $336/28 = 12$. +Then the value of $k$ when $j = 12$ is $336/12 = 28$. +Then the value of $j$ when $k = 14$ is $336/28 = 12$. +Then the value of $k$ when $j = 12$ is $336/28 = 12$. +Then the value of $j$ when $k = 14$ is $336/28 = 12$. +Then the value of $k$ when $j = 12$ is $336/28 = 12$. +Then the value of" +46,5,"The sum of all possible values of $x$ is $\boxed{5}$ +This example shows how to take a step from the current state to a solution of a problem. We analyze the problem goal, the current state, and the difference between the problem goal and the current state. We also check if the current state is a solution to the problem, and if not, we propose a possible next step to solve the problem." +47,-7,"$\frac{2x-1}{2x+2}$ = $\frac{x-3}{x-1}$ +for $x \neq -1$ and $x \neq 1$. +$2x - 1 = (2x + 2) (\frac{x-3}{x-1})$ +$2x - 1 = (2x + 2) (\frac{x^2 - 3x -2x + 6}{x^2 -1}$ +$2x - 1 = (2x + 2) (\frac{x^2 -5x +6}{x^2 -1}$ +$2x - 1 = (2x + 2) (\frac{x^2 -5x +6}{x-1} (x+1)$ +$2x - 1 = (2x + 2) (\frac{x(x^2 -5x +6)}{x^2 -1}$ +$2x - 1 = (2x + 2) (\frac{x^3 -5x^2 +6x}{x^2 -1}$ +$2x - 1 = (2x + 2) (\frac{x(x^2 -5x +6)}{(x+1)(x-1)}$ +$2x - 1 = (2x + 2) (\frac{x(x^2 -5x +6)}{(x+1)(x-1)}$ +$2x - 1 = (2x + 2) (\frac{x(x^2 -5x +6)}{(x+1)(x-1)}$ +$2x - 1 = (2x + 2) (\frac{x(x-1)(x+5)}{(x+1)(x-1)}$ +$2x - 1 = (2x + 2) (\frac{x(x+5)}{1}$ +$2x - 1 = (2x + 2) (x+5)$ +$2x - 1 = 2x^2 + 10x + 2x + 10$ +$2x - 1 = 2x^2 + 12x + 10$ +$2x^2 + 12x + 10 = 2x^2 -1$ +$2x^2 + 12x + 10 = 2x^2 -1$ +$2x^2 + 12x + 10 = 2x^2 -" +48,0.43,The ball hits the ground in $\boxed{0.37}$ seconds. +49,108,"The circumference of the first horse is 64/27=2.37 feet. The circumference of the second horse is 2 * π * 16=100.42 feet. The second horse must travel 64 feet. Since the second horse's circumference is 100.42 feet, the second horse must make $\frac{64}{100.42}$ = 0.6364 revolutions." diff --git a/run_bench.py b/run_bench.py index 903c28d..987f381 100644 --- a/run_bench.py +++ b/run_bench.py @@ -445,13 +445,14 @@ if __name__ == '__main__': # run(args) #test llama w/ lora - args.quantize=None - args.lora = True - print(args) - run(args) + # args.quantize=None + # args.lora = True + # print(args) + # run(args) #lora with the a_star run - a_star = True + args.a_star = True + args.lora=True print(args) run(args) diff --git a/times_llama_lora_0.7_50_astar.csv b/times_llama_lora_0.7_50_astar.csv new file mode 100644 index 0000000..da47c79 --- /dev/null +++ b/times_llama_lora_0.7_50_astar.csv @@ -0,0 +1,2 @@ +,total_accuracy,total runtime,total setup time,average solving time,average proposal time,average eval time,peak memory usage +0,0.46,3626.6385051950056,19.12435658100003,72.14954966570048,22.047873749166854,1.9993376006335408,15379560960