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| 1 | gt | pred | |
|---|---|---|---|
| 2 | 0 | 2 | The equation has only one solution if and only if $x>0$. |
| 3 | 1 | 10 | The difference between 36 and 26 is $\boxed{10}$. ### |
| 4 | 2 | i | Since $i^2=-1$, we have $$i^5=i^4\cdot i=i\cdot i=-1\cdot i$$ $$ |
| 5 | 3 | 4 | The value of $x$ is $\boxed{4}$. The difference between 36 and 26 is $\boxed{10}$ The value of $x$ is $\boxed{4}$ The difference between $120\%$ of 30 and $130\%$ of 20 is $\boxed{10}$ The value of $x$ is $\boxed{4}$ The difference between $120\%$ of 30 and $130\%$ of 20 is $\boxed{10}$ The final answer is: $\boxed{4}$ |
| 6 | 4 | 402 | The difference between 36 and 26 is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. The final answer for the first problem is $\boxed{10}$. |
| 7 | 5 | x \in [-2,7] | The next step is to write the final answer in interval notation. $x \ge 1/2$ and $x \le 14$ becomes $\boxed{[1/2,14]}$. Note that this solution does not include the value $x=5/2$ because the absolute value is not defined at $x=5/2$. |
| 8 | 6 | -\frac{1 | The solution is $x=-\frac{1}{4}$. The final answer is: $\boxed{-\frac{1}{4}}$ |
| 9 | 7 | \frac{x+2 | The inverse of h(x) is $x=\frac{y+2}{7}$. The final answer is: \boxed{\frac{y+2}{7}} |
| 10 | 8 | 10 | The distance from the origin to the point $(-8,6)$ in a coordinate system is $\boxed{10}$ units. |
| 11 | 9 | 8 | Subtract $\frac{4}{3}\pi$ from both sides of the equation to get $1000-\frac{4}{3}\pi=\frac{4}{3}\pi r^3$. ### Here is an example to help: Problem Goal: A rectangular prism has a base area of 20 square units and a height of 5 units. What is the |
| 12 | 10 | -25 | f(-10)= 3(-10)+ 5 = -30 + 5 = -25. |
| 13 | 11 | 8 | The sum of $3 + 3 + 2$ is $\boxed{8}$. The final answer is: $\boxed{8}$ |
| 14 | 12 | 187.5 | The solutions to the equation are $x=\frac{-2\pm\sqrt{11}}$, which are two possible values for x. The solutions to the equation are $x=\frac{-2\pm\sqrt{11}} |
| 15 | 13 | \$40 | The new price of the vitamin C is $\boxed{4.0959}$. ### |
| 16 | 14 | 5 | The cost per day is \boxed{\1.25}. |
| 17 | 15 | 8 | The pressure in the 1.835 liter container will be \[ P=\frac{ 2\cdot 1.835\cdot 100 }{ 3.67\cdot 2} \] \[ P=\frac{ 2\cdot 1.835\cdot 100 }{ 7.34\cdot 2} \] \[ P=\frac{ 2\cdot 1.835\cdot 100 }{ 14.68 } \] \[ P=\frac{ 2\cdot 1.835\cdot 100 }{ 14.68 } \] \[ P=\frac{ 2\cdot 183.5 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ 14.68 } \] \[ P=\frac{ 367 }{ |
| 18 | 16 | 3125 | The final answer is: $\boxed{3125}$ This is a general example of the problem solving strategy. Note: In a real problem, there may be multiple steps to solve the problem, and we should check the result of each step to ensure the solution is correct. |
| 19 | 17 | [0,\infty) | Let $y = (x^2 + 1)/(x^2 - 1)$. Then $y^2 = ((x^2 + 1)/(x^2 - 1))^2 = (x^2 + 1)^2/((x^2 - 1)^2)$. Then $y^2(x^2 - 1) = (x^2 + 1)^2/((x^2 - 1)^2) = (x^2 + 1)^2/(x^2 - 1)^2$. Then $(y^2)(x^2 - 1) = ( |
| 20 | 18 | .5 | |
| 21 | 19 | 5 | The final answer is: $\boxed5$ The final answer is: $\boxed5$ |
| 22 | 20 | 105 | In order to find the sum of these 15 sums of the elements of the subsets of {1,2,3,4,5,6}, we will add up the 15 sums. |
| 23 | 21 | 2 | The value of n is $\boxed{2}$. |
| 24 | 22 | (-\infty,-8)\cup (8,\infty) | The quadratic formula can be used to solve the quadratic equation. Since the equation has two distinct real roots, the discriminant is positive. The quadratic formula is \(x=\frac{-b±\sqrt{b^2-4ac}}{2a}\), where a = 2, b = m, and c = 8. The discriminant is \(b^2-4ac=m^2-4\cdot2\cdot8=m^2-64\). Since the discriminant is positive, we have \(m^2-64>0\). Since m^2 > 64, we can take the square root of both sides to get m > ±√64 or m < ±√64. Simplifying √64 to 8, we have m > ±8 or m < ±8. Expressing the possible values of m in interval notation, we have m ∈ (−∞,−8) ∪ (8,∞). |
| 25 | 23 | 0 | Therefore, $x + 2 = 0 \Rightarrow x = -2$. ### Problem Goal: Find the value of $x$ in the equation $x^2 - 4x + 4 = 0$. The equation $x^2 - 4x + 4 = 0$ can be rewritten as $( |
| 26 | 24 | 2 | The final answer is 2. |
| 27 | 25 | \frac{1 | The simplified fraction $\frac{1}{84}$ is already in simplest form. The final answer is: $\boxed{\frac{1}{84}}$ |
| 28 | 26 | 6+9i | Use the Rational Root Theorem to find possible roots of the equation $x^3 + 2x^2 - 7x - 12 = 0$. The possible roots are $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 12$. Test the possible roots $\pm 1$, $\pm 2$, $\pm 3$, $\pm 4$, $\pm 6$, $\pm 12$ to find the actual roots. The rational roots of the equation are $x= -1$, $x= -2$, $x= 3$. Since the rational roots are not solutions to the inequality $x^3 + 2x^2 - 7x - 12 > 0$, the inequality is true for all values of $x$. The inequality $x^3 + 2x^2 - 7x - 12 > 0$ is true for all values of x. The inequality $x^3 + 2x^2 - 7x - 12 > 0$ is true for all values of x. The inequality $x^3 + 2x^2 - 7x - 12 > 0$ is true for all values of x. The inequality $x |
| 29 | 27 | 2 | The length ### Problem Goal: The length |
| 30 | 28 | 20 | The result of the multiplication is $\boxed{20 |
| 31 | 29 | 7(x+3) (x-3) | \begin{aligned}7x^2-63 &= 7x^2-9\cdot7x \\&= 7x^2-9\cdot7x \cdot\(1/7) \\ &= 7x(x-3)\end{aligned} \begin{aligned}7x^2-63 &= 7x^2-9\cdot7x \cdot\(1/7) \\ &= 7x(x-3)\end{aligned} Note: The solution is not provided as the answer is already in the simplified form. ### |
| 32 | 30 | y^4-2y^3+7y^2+y-5 | This is the correct answer. |
| 33 | 31 | 0 | The expression $f(x)-f^{-1}(x)$ simplifies to $\boxed{0}$. |
| 34 | 32 | \frac{7 | The value of $\log_28\sqrt{2}$ is $\boxed{2}$. Note: $\log_28\sqrt{2}$ is equal to 2. ### Step 10: We can use the property $\log_ab = \frac{\log_ca}{\log_cb}$, where we choose $c=2$, to simplify $\log_2\sqrt{2}$. This gives us $\log_2\sqrt{2} = \frac{\log_2 2}{\log_2\sqrt{2}} = \frac{1}{\log_2\sqrt{2}}$. ## Step 11: We can use the property $\log_ab = \frac{\log_ca}{\log_cb}$, where we choose $c=2$, to simplify $\log_2\sqrt{2}$. This gives us $\log_2\sqrt{2} = \frac{\log_2 2}{\log_2\sqrt{2}} = \frac{1}{\log_2\sqrt{2}}$. ## Step 12: We can use the property $\log_ab = \frac{\log_ca}{\log_cb}$, where we choose $c=2$, to simplify $\log_2\sqrt{2}$. This gives us $\log_2\sqrt{2} = \frac{\log_2 2}{\log_2\sqrt{2}} = \frac{1}{\log_2\sqrt{2}}$. ## Step 13: We can use the property $\log_ab = \frac{\log_ca}{\log_cb}$, where we choose $c=2$, to simplify $\log_2\sqrt{2}$. This gives us $\log_2\sqrt{2} = \frac{\log_2 2}{\log_2\sqrt{2}} = \frac{1}{\log_2\sqrt{2}}$. ## Step 14: We can use the property $\log_ab = \frac{\log_ca}{\log_cb}$, where we choose $c=2$, to simplify $\log_2\sqrt{2}$. This gives us $\log_2\sqrt{2} = \frac{\log_2 2}{\log_2\sqrt{2}} = \frac{1}{\log_2\sqrt{2}}$. ## Step 15: We can use the property $\log_ab = \frac{\log_ca}{\log_cb}$, where we choose $c=2$, to simplify $\log_2\sqrt{2}$. This gives us $\log_2\sqrt{2} = \frac{\log_2 2}{\log_2\sqrt{2}} = \frac{1}{\log_2 |
| 35 | 33 | 5 | The final answer is $\boxed{\ |
| 36 | 34 | \left(-\infty,-\frac 12\right)\cup \left(-\frac 12,\infty\right) | Now we can see that the function can be rewritten as $$k(y) = \frac{1}{2y+1} = \frac{2y+1}{2y+1} = 1 = \frac{2y+1}{2y+1}$$ This means that $y=-\frac{1}{2}$ is a root of the function, and we can use this root to find the other root. The other root can be found by multiplying both sides by the denominator, $2y+1$, to get $$1 = \frac{2y+1}{2y+1}$$ $$1(2y+1) = 2y+1$$ $$2y+1 = 2y+1$$ This means that the function can be rewritten as $$k(y) = \frac{1 |
| 37 | 35 | \frac{2 | The number of integers between $1$ and $1000$ inclusive that are divisible by $2$, $3$, or $5$ is $58$. The final answer is: $\boxed{58}$$ ### Problem Goal: A sequence $a_n$ is defined by the recurrence relation $a_{n+1} = 4a_n + a_{n-1}$ for $n \geq 1$, with initial conditions $a_0 = 1$ and $a_1 = 3$. What is the value of $a_{10}$? $a_2 = |
| 38 | 36 | 7 | To find $f^{-1}(f^{-1}(3))$, we first find the value of $f^{-1}(3)$, which is 4, and then find the value of $f^{-1}(4)$, which is 7. The final answer is $\boxed{7}$. ### |
| 39 | 37 | 20 | Note: I corrected the simplification process, following the proper steps to evaluate the expression $(3-i)(6+2i)$. The correct simplification is $20+6i$, not $20-6i$. The final answer is: $\boxed{20+6i}$ |
| 40 | 38 | 9 | Let $x$ = $(-5)$ and $y$ = $(-4)$. $x^2$ = 25 and $y^2$ = 16. The product of the squares of the solutions is $25\cdot 16$ = $\boxed{400}$. Note that the current state was incorrect, but the process of proposing a next step and correcting the error is shown. |
| 41 | 39 | 4 | If $a=8$, what is the value |
| 42 | 40 | 6 | The absolute difference between 8 and 7 is $\boxed{1}$. The final answer is: $\boxed{1}$ |
| 43 | 41 | \left(1,\frac{9 | Note that the midpoint of $\overline{PQ}$ is $(1,4.5)$, which is $(5-3)/2, (3+6)/2)$. Note that $PQ$ is the hypotenuse of a right triangle with legs of length 8 and 3. Note that the area of $\overline{PQ}$ is $\sqrt{8^{2}+3^{2}}=\sqrt{73}$. Note that the midpoint of $\overline{PQ}$ is $(\frac{5-3}{2}, \frac{3+6}{2})$. Note that $\sqrt{73}$ is the area of $\overline{PQ}$, and the midpoint of $\overline{PQ}$ is $(\frac{5-3}{2}, \frac{3+6}{2})$. Note that $\overline{PQ}$ is the hypotenuse of a right triangle with legs of length 8 and 3. Note that the area of $\overline{PQ}$ is $\sqrt{73}$, and the midpoint of $\overline{PQ}$ is $(\frac{5-3}{2}, \frac{3+6}{2})$. Note that $\sqrt{73}$ is the area of $\overline{PQ}$, and the midpoint of $\overline{PQ}$ is $(\frac{5-3}{2}, \frac{3+6}{2})$. Note that the midpoint of $\overline |
| 44 | 42 | -2 | (There is no next step to take in this case.) |
| 45 | 43 | 8 | $n^2 + n - 126 = (n - 7)(n + 18) = 0$. Note that in the example above, the goal is to find the difference between two given values, and the current state is to evaluate the difference. In general, the next step is to evaluate the expression and see if it can be simplified to find the goal. In the example above, the next step is to find the difference between the two values. In general, the next step is to evaluate the expression and see if it can be simplified to find the goal. The final answer is the difference between the two values. |
| 46 | 44 | 20 | If the difference between the two numbers is $\frac{8}{2}=4$, then the greater of the two numbers is $5\cdot4=20$. |
| 47 | 45 | 24 | We can check that $j$ and $k$ are inversely proportional by solving for $k$ in the equation $jk = c$. Substituting $j=24$ into the equation, we have $24k=c$, which gives $c=24k$. Substituting $k=14$ into the equation, we have $24\cdot 14=c$, which gives $c=336$. Solving for $k$ in the equation $c=24k$, we find $k=\frac{336}{24}$, which gives $k=\boxed{14}$. The final answer is: $\boxed{24}$ |
| 48 | 46 | 5 | (You can add more possible next steps here, but this is not necessary.) |
| 49 | 47 | -7 | The value of $x$ that makes the fractions equal is $\boxed{5}$. The final answer is: $\boxed{5}$ |
| 50 | 48 | 0.43 | The time it takes for the ball to hit the ground is $\boxed{0.04}$ second(s). |
| 51 | 49 | 108 | The number of revolutions the horse 16 feet from the center must make to travel the same distance is $\boxed(108)$. The next step is to verify the answer to see if it is correct or not. The answer 108 is among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet from the center must make 27 revolutions. (4 x 27 = 108 revolutions. This is not among the choices, so you can eliminate a. The answer is b. 64 feet is the distance around the entire merry-go-round. So to travel the same distance, the horse 16 feet |
| 52 | 50 | 2 | The problem is asking for vertical asymptotes, which are the values of x that make the denominator equal to zero. The denominator $x^2+x-6$ is zero when $x=2$ and $x=-3$, so there are $\boxed{2}$ vertical asymptotes. Note: The final answer is the number of vertical asymptotes, not the values themselves. |
| 53 | 51 | 10 | The difference between 36 and 26 is $\boxed(10)$. The final answer is: $\boxed{10}$ |
| 54 | 52 | i | Expand the factored form of the quadratic equation $(x-a)(x-b)=x^2-4x+3$ to get $x^2-ax-bx+ab=x^2-(a+b)x+ab$. The coefficients of the quadratic equation $x^2-ax-bx+ab=x^2-(a+b)x+ab$ are $a+b$ and $ab$, and these coefficients must be $-4$ and $3$, respectively. Solve for $a+b$ and $ab$ by setting $a+b$ equal to $-4$ and $ab$ equal to $3$. $ab=3$ implies that $a$ and $b$ are factors of $3$. The factors of $3$ are $-1$ and $3$ and $1$ and $3$. Since the |
| 55 | 53 | 4 | $1=2^{x^2}$ and $16=2^{x^2}$. ## Step 15: Solve for $x$ Since $1=2^{x^2}$, it follows that $x^2=0$, so $x=0$. Since $16=2^{x^2}$, it follows that $x^2=4$, so $x=\pm 2$. The final answer is: $\boxed{0}$ |
| 56 | 54 | 402 | The final answer is: $\boxed{402}$. |
| 57 | 55 | x \in [-2,7] | The inequality $x^2 - 5x - 4 \le 10$ is equivalent to $x^2 - 5x - 14 \ge 0$. Factoring, we get $(x - 7)(x + 2) \ge 0$. This is true when $x \le -2$ or $x \ge 7$. Note: The first case is true since $-2$ is less than $7$. Note: The second case is true since $7$ is greater than $-2$. Note: The interval $\boxed{(-\infty, -2) \cup (7, \infty)}$ is the solution set. Note: $x \in \boxed{(-\infty, -2) \cup (7, \infty)}$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2) \cup (7, \infty)$. Note: $(-\infty, -2 |
| 58 | 56 | -\frac{1 | This is a good solution, but the problem asks for the smallest value of x. The current solution is $x = 1$, but the problem statement also includes $x = \frac{3}{2}$ as a possible answer. We can see that $\frac{3}{2}$ is indeed smaller than 1. Therefore, the current state is not the final answer. We can use the problem statement to guide us to propose a new next step. ## Step 6: Re-evaluate the current state The current state states that $x = 1$. However, this is not the smallest value of $x$. We need to re-evaluate the current state. ## Step 7: Propose a new next step Since $x = 1$ is not the smallest value of $x$, we need to find another solution. The problem statement also includes $x = \frac{3}{2}$ as a possible answer. We can see that $\frac{3}{2}$ is indeed smaller than 1. Therefore, we propose a new next step. ## Step 8: Use the new next step to propose a new final answer The difference between 36 and 26 is $\boxed(10)$. The final answer is: $\boxed{\frac{3}{2}}$ |
| 59 | 57 | \frac{x+2 | The inverse of h(x) is \boxed{x= \( \frac{y+2}{7} \). Note: The solution requires the calculation of the inverse of the function $h(x)=f(g(x))$. The inverse of h(x) is obtained by substituting g(x) into f(x) and then finding the inverse. The correct solution is the inverse of h(x) is x= \( \frac{y+2}{7} \). |
| 60 | 58 | 10 | Find the distance between the city center and the outskirts of the city by using the distance formula: \[d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\] The final answer is: $\boxed{75}$. The final answer is: $\boxed{9000}$. The final answer is: $\boxed{9}$. The final answer is: $\boxed{10}$. The final answer is: $\boxed{120}$. The final answer is: $\boxed{75}$. The final answer is: $\boxed{9000}$. The final answer is: $\boxed{9}$. The final answer is: $\boxed{10}$. The final answer is: $\boxed{120}$. The final answer is: $\boxed{75}$. The The final answer is: $\boxed{9000}$. The final answer is: $\boxed{9}$. The final answer is: $\boxed{10}$. The final answer is: $\boxed{120}$. The final answer is: $\boxed{75}$. The final answer is: $\boxed{9000}$. The final answer is: $\boxed{9}$. The final answer is: $\boxed{10}$. The final answer is: $\boxed{120}$. The final answer is: $\boxed{75}$. The The final answer is: $\boxed{9000}$. The final answer is: $\boxed{9}$. The final answer is: $\boxed{10}$. The final answer is: $\boxed{120}$. The final answer is: $\boxed{75}$. The final answer is: $\boxed{9000}$. The final answer is: $\boxed{9}$. The final answer is: $\boxed{10}$. The final answer is: $\boxed{120}$. The final answer is: $\boxed{75}$. The The final answer is: $\boxed{9000}$. The final answer is: $\boxed{9}$. The final answer is: $\boxed{10}$. The final answer is: $\boxed{120}$. The final answer is: $\boxed{75}$. The final answer is: $\boxed{9000}$. The final answer is: $\boxed{9}$. The final answer is: $\boxed{10}$. The final answer is: $\boxed{120}$. The final answer |
| 61 | 59 | 8 | The midpoint of the segment is $\boxed{\left(\frac{1+1}{2},\frac{4+10}{2}\right)=\left(1,7\right)}$, so the sum of the coordinates is $\boxed{1+7=8}$. Note: A different solution is possible. The final answer is: $\boxed{8}$ |
| 62 | 60 | -25 | Subtract 6 from both sides to get $2=0$, then subtract 2a from both sides to get $8=6$. Subtract 2a from both sides to get $2a=0$, then divide both sides by 2 to get $a=0$. The difference between the current state and the proposed next step is that the current state does not solve for $a$, but it solves for the constant term. The proposed next step solves for $a$. ### Problem Goal: Find the value |
| 63 | 61 | 8 | No further steps are needed, as the answer is found from the current state. The final answer is: $\boxed{15}$ |
| 64 | 62 | 187.5 | The sum of 8 and 4 is \(\boxed{12}\). \(\Box\) **16.**- 5.4 fluid ounces = 5.4 fluid ounces \(\times\) 125 calories/fluid ounce = 687 There are 687 calories in a 5.4 fluid ounce bottle of Vitamin Water. \(\Box\) **17.**- 2.1 fluid ounces = 2.1 fluid ounces \(\times\) 125 calories/fluid ounce = 262.5 There are 262.5 calories in a 2.1 fluid ounce bottle of Vitamin Water. \(\Box\) **18.**- 4.6 fluid ounces = 4.6 fluid ounces \(\times\) 125 calories/fluid ounce = 575 There are 575 calories in a 4.6 fluid ounce bottle of Vitamin Water. \(\Box\) **19.**- 3.1 fluid ounces = 3.1 fluid ounces \(\times\) 125 calories/fluid ounce = 382.5 There are 382.5 calories in a 3.1 fluid ounce bottle of Vitamin Water. \(\Box\) **20.**- 1.5 fluid ounces = 1.5 fluid ounces \(\times\) 125 calories/fluid ounce = 187.5 There are 187.5 calories in a 1.5 fluid ounce bottle of Vitamin Water. \(\Box\) **21.**- 2.9 fluid ounces = 2.9 fluid ounces \(\times\) 125 calories/fluid ounce = 363.25 There are 363.25 calories in a 2.9 fluid ounce bottle of Vitamin Water. \(\Box\) **22.**- 4.3 fluid ounces = 4.3 fluid ounces \(\times\) 125 calories/fluid ounce = 546.25 There are 546.25 calories in a 4.3 fluid ounce bottle of Vitamin Water. \(\Box\) **23.**- 3.5 fluid ounces = 3.5 fluid ounces \(\times\) 125 calories/fluid ounce = 437.5 There are 437.5 calories in a 3.5 fluid ounce bottle of Vitamin Water. \(\Box\) **24.**- 1.4 fluid ounces = 1.4 fluid ounces \(\times\) 125 calories/fluid ounce = 175 There are 175 calories in a 1.4 fluid ounce bottle of Vitamin Water. \(\Box\) **25.**- 3.9 fluid ounces = 3.9 fluid ounces \(\times\) 125 calories/fluid ounce = 482.5 There are 482.5 calories in a 3.9 fluid ounce bottle of Vitamin Water. \(\Box\) **26.**- 1.3 fluid ounces = 1.3 fluid ounces \(\times\) 125 calories/fluid ounce = 162.5 There are 162.5 calories in a 1.3 fluid ounce bottle of Vitamin Water. \(\Box\) **27.**- 2.5 fluid ounces = 2.5 fluid ounces \(\times\) 125 calories/fluid ounce = 312.5 There are 312.5 calories in a 2.5 fluid ounce bottle of Vitamin Water. \(\Box\) **28.**- 5.7 fluid ounces = 5.7 fluid ounces \(\times\) 125 calories/fluid ounce = 713.25 There are 713.25 calories in a 5.7 fluid ounce bottle of Vitamin Water. \(\Box\) **29.**- 2.2 fluid ounces = 2.2 fluid ounces \(\times\) 125 calories/fluid ounce = 275 There are 275 calories in a 2.2 fluid ounce bottle of Vitamin Water. \(\Box\) **30.**- 4.8 fluid ounces = 4.8 fluid ounces \(\times\) 125 calories/fluid ounce = 600 There are 600 calories in a 4.8 fluid ounce bottle of Vitamin Water. |
| 65 | 63 | \$40 | \begin{align*} x^{10}-x^{9}+x^{7}-x^{6}+x^{4}-x^{3}+x^{2}-x^{0}+x^{1} &=(x^{10}+x^{9})-(x^{9}+x^{8}+x^{7})+(x^{7}+x^{6})-(x^{6}+x^{5}+x^{4})\\ & +(x^{4}+x^{3}+x^{2})-(x^{3}+x^{2}+x^{1})+x^{2}-x^{0}+x^{1}\\ & =x^{10}+x^{9}-x^{9}-x^{8}-x^{7}+x |
| 66 | 64 | 5 | The value of $x$ in the equation $(17^6-17^5)\div16=17^x$ is $x=\boxed5$. The current state is correct, so no change is needed. Note that the next step is the current state, so the answer is given without a box. |
| 67 | 65 | 8 | The pressure of the hydrogen is $\boxed{7.00}$ kPa. The pressure of the hydrogen is 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = 4(3.67) / 1.835 p2 = 7.00 kPa. The pressure of the hydrogen is 7.00 kPa. To find the new pressure of the hydrogen, p2, we need to find p2. To find p2, we can use the proportion p1/v1 = p2/v2. p2 = p1 v1 / v2. p2 = |
| 68 | 66 | 3125 | The value of $1/1/1/1/1$ is $\boxed(1)$. ### Problem Goal: What is |
| 69 | 67 | [0,\infty) | The square root of any nonnegative real number is a real number, so the range is all real numbers. |
| 70 | 68 | .5 | The sum of the possible values of $x$ is $\boxed(1.5+2.5+3.5)$. |
| 71 | 69 | 5 | \begin{align*} x^2 + 4x + y^2 - 8y & = 0\\ 2^2 + 4(2) + y^2 - 8y & = 0\\ 4 + 8 + y^2 - 8y & = 0\\ y^2 - 8y + 12 & = -12\\ y^2 - 8y + 12 + 12 & = -12 + 12\\ y^2 - 8y + 24 & = 0\\ (y - 2)(y - 6) & = 0\\ \end{align*} Thus, $y = 2$ or $y = 6$. The value of $y$ when $x = 2$ is $\boxed{6}$. ### Problem Goal: Find the value of $ |
| 72 | 70 | 105 | The 15 sums are: $(1+2)$, $(1+3)$, $(1+4)$, $(1+5)$, $(1+6)$, $(2+3)$, $(2+4)$, $(2+5)$, $(2+6)$, $(3+4)$, $(3+5)$, $(3+6)$, $(4+5)$, $(4+6)$, $(5+6)$. The sum of these 15 sums is $\boxed{150}$. The final answer is: $\boxed{150}$ |
| 73 | 71 | 2 | We find that the given equation can be factored as $(x-1)(x^{2}-x-18)=0$, and we can then factor $ x^{2}-x-18$ as $(x-6)(x+3)$. We can set the factors equal to zero and solve for x: \begin{align*} (x-1)(x-6)(x+3)&=0\\ x-1&=0 \quad \text{or} \quad x-6=0 \quad \text{or} \quad x+3=0\\ x&=1 \quad \text{or} \quad x=6 \quad \text{or} \quad x=-3 \end{align*} The solutions are $x=1$, $x=6$, and $x=-3$. We can now plug these solutions into the original equation $x^{3}-2x^{2}-9x+18=0$ to see if any of them are true. We find that $1^{3}-2\cdot1^{2}-9\cdot1+18=1-2-9+18=8$, $6^{3}-2\cdot6^{2}-9\cdot6+18=216-72-54+18=108$, and $(-3)^{3}-2\cdot(-3)^{2}-9\cdot(-3)+18=-27-18+27+18=-0$. Since $-0=0$ is true, $x=-3$ is a solution to the original equation. We can |
| 74 | 72 | (-\infty,-8)\cup (8,\infty) | The value of $m$ can be any real number greater than 8 or less than 8, so $m$ is equal to $\boxed{(-\infty,-8) \cup (8,\infty)}$. The final answer is: $(-\infty,-8) \cup (8,\infty)$ The difference between 36 and 26 is $\boxed(10)$. The final answer is: $\boxed(10)$ Note that this is not a full solution to a typical math problem. A full solution would include an explanation of the reasoning, the identification of the variables and constants, and the solution steps. This is a very short example of how to propose a next step for a given problem. To solve a typical math problem, you need to: 1. Read and understand the problem goal. 2. Read and understand the current state. 3. Identify the variables and constants. 4. Propose a next step to solve the problem. 5. Solve the proposed next step. 6. Check the proposed solution to see if it solves the problem. If it does, reiterate the current state. If it does not, try again. |
| 75 | 73 | 0 | The distance between the city and the next town is x, and we have 20 = 0.2x, so x = 20/0.2 = 100. The distance between the city and the next town is $\boxed{100}$. ### Problem Goal: Solve for the variable $x$ in the equation $x^2 + 2x + 2 = 0$. We have $x^2 + 2x + 2 = 0$, which is quadratic. We could attempt to factor it, but it does not factor. We can use the quadratic formula instead. ### Step 1: Write down the quadratic formula We have $x = \( rac{x1 + x2 - rac{2} \cdot 1} {2a} \cdot 1}$, where a = 1, b = 2, and c = 2. This simplifies to $x = \( rac{-2 + \sqrt{2^2 - 4 \cdot 1 \cdot 2} \cdot 1} {2 \cdot 1}$. ### Step |
| 76 | 74 | 2 | The equation can be split into two cases: 1. $x-1=7$ 2. $x-1=-7$ Case 1: $x-1=7$ $x=8$ The second value is $x=8-7=1$ Case 2: $x-1=-7$ $x=4$ The first value is $x=4+7=11$ The sum of the two values is $\boxed{2\cdot 11}$ The final answer is: $\boxed{22}$ |
| 77 | 75 | \frac{1 | We have found that w is equal to $\frac{1}{12}$, which is the solution to the problem. The final answer is: $\boxed{\frac{1}{12}}$ |
| 78 | 76 | 6+9i | \begin{align*} 62 - 3i \end{align*} Note: We simplified the expression and obtained a new expression, which is the current state. |
| 79 | 77 | 2 | \[\frac{4a+3b}{a-2b}=5\implies4a+3b=5a-10b\impliesa=13b\implies\frac{a+11b}{a-b}=\frac{13b+11b}{13b-b}=\frac{24b}{12b}=\frac{2}{1}=2.\] |
| 80 | 78 | 20 | (2-2i)(5+5i) = 10+10i-10i-10i^2 = 10+0i+10 = 20 The final answer is 20. |
| 81 | 79 | 7(x+3) (x-3) | The expression $7x^2-63$ factors as $(7x-9)(x+7)$. The final answer is: $\boxed{7x^2-63}$ |
| 82 | 80 | y^4-2y^3+7y^2+y-5 | Find the derivative of f(y) + g(y) with respect to y, which is 4y^3 - 6y^2 + 14y + 1. The final answer is: $\boxed{4y^3 - 6y^2 + 14y + 1}$ |
| 83 | 81 | 0 | Note that if we had used the definition of inverse function, we would have also had to use the fact that $f^{-1}(f(x))=x$ to simplify the expression. Since $f^{-1}(f(x))=x$, we can simplify the expression as $f(x)-f^{-1}(x)=f(x)-f^{-1}(f^{-1}(f(x)))$. We can now substitute $f(f(x))=x$ into the expression. ## Step 8: Substitute $f(f(x))=x$ into the expression $f(x)-f^{-1}(x)=f(x)-f^{-1}(x)=f(x)-x$. ## Step 9: Solve for $f(x)-f^{-1}(x)$ Since $f(x)-x$ is a constant, we have $f(x)-f^{-1}(x)=\boxed{0}$. The final answer is: $\boxed{0}$ |
| 84 | 82 | \frac{7 | The value of $\log |
| 85 | 83 | 5 | The final answer is $\boxed{5}$. |
| 86 | 84 | \left(-\infty,-\frac 12\right)\cup \left(-\frac 12,\infty\right) | The domain of the function $$k(y) = \frac{1}{2y+1}~$$ is $$\left(-\infty,-\frac{1}{2}\right)\cup\left(-\frac{1}{2},\infty\right)~.$$ Note: Note that the domain of a rational expression, such as $$k(y) = \frac{1}{2y+1}~$$ is the set of values of $$y$ that do not make the denominator equal to zero. In this case, the denominator is zero when $$2y+1=0$$, which happens when $$y=-\frac{1}{2}~.$$ So the domain of $$k(y) = \frac{1}{2y+1}~$$ is the set of all values of $$y$$ except $$y=-\frac{1}{2}~.$$ The domain of the function $$k(y) = \frac{1}{2y+1}~$$ is $$\left(-\infty,-\frac{1}{2}\right)\cup\left(-\frac{1}{2},\infty\right)~.$$ If the current state is a solution to the problem, the next step is to check if the proposed next step is a valid solution. In this case, the current state is not a solution to the problem. So the next step is to find a new current state that is a solution to the problem. If the current state is not a solution to the problem, the next step is to find a new current state that is a solution to the problem. In this case, the new current state is $$\ |
| 87 | 85 | \frac{2 | The probability that it will rain tomorrow and not today is $\frac{1}{5} \cdot \frac{4}{5} = \frac{4}{25}$. The final answer is: $\boxed{\frac{4}{25}}$ |
| 88 | 86 | 7 | The inverse of $f^{-1}(4)$ is $f^{-1}(3)$ which is $\boxed{7}$. |
| 89 | 87 | 20 | Multiply the numerators and denominators and combine like terms. \begin{align*} \left(3-i\right)\left(6+2i\right)&=3\left(6+2i\right)-i\left(6+2i\right)\\ &=18+6i-6i-2i^2\\ &=18+6i-6i+2\\ &=\boxed{20+0i} \end{align*} \begin{align*} \left(3-i\right)\left(6+2i\right)&=3\left(6+2i\right)-i\left(6+2i\right)\\ &=18+6i-6i-2i^2\\ &=18+6i-6i+2\\ &=\boxed{20+0i} \end{align*} \begin{align*} \left(3-i\right)\left(6+2i\right)&=3\left(6+2i\right)-i\left(6+2i\right)\\ &=18+6i-6i-2i^2\\ &=18+6i-6i+2\\ &=\boxed{20+0i} \end{align*} \begin{align*} \left(3-i\right)\left(6+2i\right)&=3\left(6+2i\right)-i\left(6+2i\right)\\ &=18+6i-6i-2i^2\\ &=18+6i-6i+2\\ &=\boxed{20+0i} \end{align*} \begin{align*} \left(3-i\right)\left(6+2i\right)&=3\left(6+2i\right)-i\left(6+2i\right)\\ &=18+6i-6i-2i^2\\ &=18+6i-6i+2\\ &=\boxed{20+0i} \end{align*} \begin{align*} \left(3-i\right)\left(6+2i\right)&=3\left(6+2i\right)-i\left(6+2i\right)\\ &=18+6i-6i-2i^2\\ &=18+6i-6i+2\\ &=\boxed{20+0i} \end{align*} \begin{align*} \left(3-i\right)\left(6+2i\right)&=3\left(6+2i\right)-i\left(6+2i\right)\\ &=18+6i-6i-2i^2\\ &=18+6i-6i+2\\ &=\boxed{20+0i} \end{align*} \begin{align*} \left(3-i\right)\left(6+2i\right)&=3\left(6+2i\right)-i\left(6+2i\right)\\ &=18+6i-6i-2i^2\\ &=18+6i-6i+2\\ &=\boxed{20+0i} \end{align*} \begin{align*} \left(3-i\right)\left( |
| 90 | 88 | 9 | The product of the squares of the solutions is $\boxed{9}$ . |
| 91 | 89 | 4 | The value of $\left(16\sqrt[3]{a^2}\right)^{\frac 13}$ when $a = 8$ is $\boxed 4$. ### Problem Goal: Solve for $x$ in the equation $2^x = 64$. None ## Step 1: We need to solve for $x$ in the equation $2^x = 64$. ## Step 2: Since $64$ is equal to $2^6$, we can rewrite the equation as $2^x = 2^6$. ## Step 3: This implies that $x = 6$. The final answer is: $\boxed 6$ |
| 92 | 90 | 6 | The difference between 0.5 and 0.5 is $\boxed{0}$. Note that the absolute value of a number is the distance from the number to 0, on the number line. Therefore, the absolute value of any number $a$ is $|a|=\begin{cases}a&\text{if }a\ge0\\ -a&\text{if }a<0\end{cases}$. Therefore, the absolute value of 0.5 is $\boxed{0.5}$. In a similar way, the absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0.5}$. The absolute value of 0.5 is $\boxed{0. |
| 93 | 91 | \left(1,\frac{9 | The midpoint of $\overline{PQ}$ is $\left(1,4.5\right)$. The midpoint formula is a shortcut for finding the midpoint of $\overline{PQ}$. We can also find the midpoint by finding the average of the $x$-coordinates of $P$ and $Q$ and the average of the $y$-coordinates of $P$ and $Q$ separately. The midpoint of $\overline{PQ}$ is $\left(1,4.5\right)$. The midpoint formula is a shortcut for finding the midpoint of $\overline{PQ}$. We can also find the midpoint by finding the average of the $x$-coordinates of $P$ and $Q$ and the average of the $y$-coordinates of $P$ and $Q$ separately. The midpoint of $\overline{PQ}$ is $\left(1,4.5\right)$. The midpoint formula is a shortcut for finding the midpoint of $\overline{PQ}$. We can also find the midpoint by finding the average of the $x$-coordinates of $P$ and $Q$ and the average of the $y$-coordinates of $P$ and $Q$ separately. The midpoint of $\overline{PQ}$ is $\left(1,4.5\right)$. The midpoint formula is a shortcut for finding the midpoint of $\overline{PQ}$. We can also find the midpoint by finding the average of the $x$-coordinates of $P$ and $Q$ and the average of the $y$-coordinates of $P$ and $Q$ separately. The midpoint of $\overline{PQ}$ is $\left(1,4.5\right)$. The midpoint formula is a shortcut for finding the midpoint of $\overline{PQ}$. We can also find the midpoint by finding the average of the $x$-coordinates of $P$ and $Q$ and the average of the $y$-coordinates of $P$ and $Q$ separately. The midpoint of $\overline{PQ}$ is $\left(1,4.5\right)$. The midpoint formula is a shortcut for finding the midpoint of $\overline{PQ}$. We can also find the midpoint by finding the average of the $x$-coordinates of $P$ and $Q$ and the average of the $y$-coordinates of $P$ and $Q$ separately. The midpoint of $\overline{PQ}$ is $\left(1,4.5\right)$. The midpoint formula is a shortcut for finding the midpoint of $\overline{PQ}$. We can also find the midpoint by finding the average of the |
| 94 | 92 | -2 | The value of h is $\boxed(-2)$. The expression $x^2 + 4x + 5$ can be written in the form $a(x - h)^2 + k$ as $x^2 + 4x + 5=(x^2+4x+4)+1=(x+2)^2+1$. The value of h is -2. |
| 95 | 93 | 8 | The distance between $(5,8)$ and $(2,4)$ is $\boxed(5)$. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is 5. The distance between $(5,8)$ and $(2,4)$ is |
| 96 | 94 | 20 | The new ratio is 7:2, so the two numbers are 7x and 2x. The value of the lesser number is 2x = 2(4) = 8. The value of the larger number is 7x = 7(4) = 28. The problem asks for the greater of the two numbers before 8 is added to it. The greater number is 28. |
| 97 | 95 | 24 | Subtracting $2x$ from both sides gives $5=10x-3$. Adding 3 to both sides gives $8=10x$. Dividing both sides by 10 gives $x=\frac{8}{10}=\frac{4}{5}$. The final answer is: $\boxed{\frac{4}{5}}$ Note that for this example, the current state is the solution to the problem, not a current state of the world. This example is more of a simple arithmetic problem. Note that this is not the only possible next step, there are many possible next steps. |
| 98 | 96 | 5 | The sum of 5 and -1 is $\boxed(4)$. ### |
| 99 | 97 | -7 | The equation $x^2 - x + \frac{5}{2} = 0$ is in the form $ax^2 + bx + c = 0$. We can see that $a=1$, $b=-1$, and $c=\frac{5}{2}$. We can try to use the quadratic formula, $x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$, to solve for $x$. The quadratic formula will give us two possible solutions for $x$. We can check to see if both solutions satisfy the original equation. The solutions to the equation $x^2 - x + \frac{5}{2} = 0$ are $\frac{1}{2}$ and $\frac{1}{2}$. Both solutions satisfy the original equation. The final answer is: $\boxed{\frac{1}{2}}$ |
| 100 | 98 | 0.43 | The ball hit the ground in $\boxed{0.37}$ seconds. |
| 101 | 99 | 108 | This is the correct number of revolutions for the second horse, which is a whole number. |