question stringlengths 45 247 | ground_truth stringlengths 1 4 | full_response stringlengths 410 3.25k | extracted_answer stringlengths 1 9 | extraction_method stringclasses 4
values | extraction_confidence float64 0.3 1 | has_boxed bool 2
classes | has_placeholder bool 1
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classes | generation_time float64 6.15 40.6 | response_length int64 410 3.25k | problem_id int64 0 99 | category stringclasses 10
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Find the area of the region bounded by the curves f(x) = -1e^(1(1x^2 + 4x + 4)) + 3, g(x) = (2x + 1)/(-2x + 3), x = -3.1, and x = -0.9. Round your answer to 1 decimal place. | 4.0 | es, we need to integrate the difference between the two functions over the interval from \( x = -3.1 \) to \( x = -0.9 \).
First, let's define the functions clearly:
\[ f(x) = -e^{(1(1x^2 + 4x + 4))} + 3 \]
\[ g(x) = \frac{2x + 1}{-2x + 3} \]
The area \( A \) is given by the integral:
\[ A = \int_{-3.1}^{-0.9} [f(x) ... | 2.5 | boxed | 1 | true | false | false | 20.224603 | 1,187 | 0 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = (1(3x^2 - 5x + 2) + 2)/(-2(3x^2 - 5x + 2) - 3) and g(x) = 4x^2 - 5x - 1 from x = -0.1 to x = 1.3. Round to 1 decimal place. | 1.8 | f(x) = \frac{1(3x^2 - 5x + 2) + 2}{-2(3x^2 - 5x + 2) - 3} \) and \( g(x) = 4x^2 - 5x - 1 \) from \( x = -0.1 \) to \( x = 1.3 \), we need to follow these steps:
1. Simplify the function \( f(x) \).
2. Determine the points of intersection between \( f(x) \) and \( g(x) \).
3. Set up and evaluate the integral for the ar... | 2.5 | boxed | 1 | true | false | false | 21.20116 | 1,536 | 1 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -2sin(3π(-1|x - 3| - 1) - 2) + 2, g(x) = 4x, x = 0.6, and x = 0.8. Round your answer to 1 decimal place. | 0.2 | es, we need to integrate the difference between the upper and lower functions over the specified interval. Here's the step-by-step process:
1. **Identify the functions and the interval:**
- \( f(x) = -2\sin(3\pi(-1|x - 3| - 1)) - 2 + 2 = -2\sin(3\pi(-1|x - 3| - 1)) \)
- \( g(x) = 4x \)
- The interval is from ... | 0.6 | boxed | 1 | true | false | true | 29.269932 | 1,705 | 2 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = 2|(3sin(2πx + 2) - 1) + 2| - 1 and g(x) = 1|x + 2| - 2 from x = 1.3 to x = 1.5. Round to 1 decimal place. | 0.2 | f(x) = 2|(3\sin(2\pi x + 2) - 1) + 2| - 1 \) and \( g(x) = 1|x + 2| - 2 \) from \( x = 1.3 \) to \( x = 1.5 \), we need to follow these steps:
### Step 1: Simplify the Functions
First, simplify \( f(x) \):
\[ f(x) = 2|(3\sin(2\pi x + 2) - 1) + 2| - 1 \]
\[ f(x) = 2|3\sin(2\pi x + 2) + 1| - 1 \]
Next, simplify \( g(x... | 0.3 | boxed | 1 | true | false | true | 21.247809 | 1,486 | 3 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 1cos(3π(3x) - 3) + 3, g(x) = 1x^2 - 1x + 3, x = -0.1, and x = 0.0. Round your answer to 1 decimal place. | 0.1 | es \( f(x) = 1\cos(3\pi(3x) - 3) + 3 \), \( g(x) = x^2 - x + 3 \), \( x = -0.1 \), and \( x = 0.0 \), we need to follow these steps:
### Step 1: Determine the points of intersection
First, we need to check if there are any points of intersection between \( f(x) \) and \( g(x) \) in the interval \([-0.1, 0.0]\).
Set \... | 3 | last_number | 0.3 | false | false | false | 40.27407 | 2,148 | 4 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -3(-3x^2 + 1x + 2)^2 + 3(-3x^2 + 1x + 2) - 2, g(x) = 3x^2 + 1x - 2, x = -0.6, and x = -0.4. Round your answer to 1 decimal place. | 0.1 | es, we need to follow these steps:
1. **Identify the curves and the interval**: We have \( f(x) = -3(-3x^2 + x + 2)^2 + 3(-3x^2 + x + 2) - 2 \) and \( g(x) = 3x^2 + x - 2 \). The interval is from \( x = -0.6 \) to \( x = -0.4 \).
2. **Set up the integral**: The area between two curves \( y = f(x) \) and \( y = g(x) \... | 0.0 | boxed | 1 | true | false | true | 29.30612 | 1,786 | 5 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -2sin(2π(-1x^2 - 3x + 2) + 2) - 1 and g(x) = (-2x)/(1x) from x = 4.0 to x = 4.1. Round to 1 decimal place. | 0.1 | f(x) = -2\sin(2\pi(-1x^2 - 3x + 2)) + 2 \) and \( g(x) = -2x \) from \( x = 4.0 \) to \( x = 4.1 \), we need to follow these steps:
1. **Define the functions:**
\[
f(x) = -2\sin(2\pi(-1x^2 - 3x + 2)) + 2
\]
\[
g(x) = -2x
\]
2. **Set up the integral for the area between the curves:**
The area \( A... | 0.1 | boxed | 1 | true | false | true | 23.456345 | 1,455 | 6 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = (-1((-1x + 1)/(-2x + 1)) + 2)/(-3((-1x + 1)/(-2x + 1))) and g(x) = 5x from x = -0.1 to x = 0.9. Round to 1 decimal place. | 1.4 | between the two given functions \( f(x) \) and \( g(x) \) over the interval \( x = -0.1 \) to \( x = 0.9 \).
First, let's simplify the function \( f(x) \):
\[ f(x) = \frac{-1\left(\frac{-1x + 1}{-2x + 1}\right) + 2}{-3\left(\frac{-1x + 1}{-2x + 1}\right)} \]
Simplify the numerator:
\[ -1\left(\frac{-1x + 1}{-2x + 1... | 1 | last_number | 0.3 | false | false | true | 40.233303 | 2,285 | 7 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 3(2x^2 + 1x - 2) + 1 and g(x) = (2x - 1)/(2x) between the vertical lines x = -1.3 and x = 0.1. Round to 1 decimal place. | 29.1 | hs of \( f(x) = 3(2x^2 + x - 2) + 1 \) and \( g(x) = \frac{2x - 1}{2x} \) between \( x = -1.3 \) and \( x = 0.1 \), we need to follow these steps:
### Step 1: Simplify the functions
First, simplify \( f(x) \):
\[ f(x) = 3(2x^2 + x - 2) + 1 \]
\[ f(x) = 6x^2 + 3x - 6 + 1 \]
\[ f(x) = 6x^2 + 3x - 5 \]
Next, simplify \(... | 1 | last_number | 0.3 | false | false | false | 39.957415 | 1,864 | 8 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = 2cos(2π(-4x^2 - 5x - 5) - 2) - 2 and g(x) = -4x - 4 from x = -0.2 to x = -0.1. Round to 1 decimal place. | 0.1 | f(x) = 2\cos(2\pi(-4x^2 - 5x - 5)) - 2 \) and \( g(x) = -4x - 4 \) from \( x = -0.2 \) to \( x = -0.1 \), we need to follow these steps:
### Step 1: Define the functions
\[ f(x) = 2\cos(2\pi(-4x^2 - 5x - 5)) - 2 \]
\[ g(x) = -4x - 4 \]
### Step 2: Set up the integral for the area
The area \( A \) between two curves \... | 0.1 | boxed | 1 | true | false | true | 40.141133 | 2,288 | 9 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -2e^(-2(4x^2 - 1x + 4)), g(x) = 5x^2 - 3x, x = 0.0, and x = 0.6. Round your answer to 1 decimal place. | 0.2 | f(x) = -2e^{-2(4x^2 - x + 4)} \), \( g(x) = 5x^2 - 3x \), \( x = 0.0 \), and \( x = 0.6 \), we need to integrate the difference between the two functions over the interval from \( x = 0 \) to \( x = 0.6 \).
Step 1: Set up the integral.
The area \( A \) is given by:
\[ A = \int_{0}^{0.6} [g(x) - f(x)] \, dx \]
\[ A = \... | 0.2 | boxed | 1 | true | false | true | 14.166476 | 939 | 10 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -1(-2cos(1πx + 1) + 3) + 3 and g(x) = (1x - 3)/(-1x - 2) between the vertical lines x = 3.2 and x = 4.2. Round to 1 decimal place. | 1.4 | hs of \( f(x) = -1(-2\cos(\pi x + 1) + 3) + 3 \) and \( g(x) = \frac{x - 3}{-x - 2} \) between \( x = 3.2 \) and \( x = 4.2 \), we need to follow these steps:
### Step 1: Simplify the functions
First, simplify \( f(x) \):
\[ f(x) = -1(-2\cos(\pi x + 1) + 3) + 3 \]
\[ f(x) = 2\cos(\pi x + 1) - 3 + 3 \]
\[ f(x) = 2\cos... | 0.5 | boxed | 1 | true | false | false | 26.440694 | 1,753 | 11 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 2|((3x + 1)/(3x + 2)) + 3| - 2 and g(x) = (-3x - 3)/(3x - 1) between the vertical lines x = -0.6 and x = 0.1. Round to 1 decimal place. | 1.6 | hs of \( f(x) = 2\left|\frac{3x + 1}{3x + 2}\right| + 3 - 2 \) and \( g(x) = \frac{-3x - 3}{3x - 1} \) between \( x = -0.6 \) and \( x = 0.1 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the expression for \( f(x) \):
\[ f(x) = 2\left|\frac{3x + 1}{3x + 2}\right| + 3 - 2 \]
\[ f(x)... | 1.2 | boxed | 1 | true | false | true | 23.603862 | 1,823 | 12 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = (2(-3|x - 1| + 1) - 2)/(-3(-3|x - 1| + 1) - 1) and g(x) = -5x^2 - 2 from x = 0.4 to x = 1.5. Round to 1 decimal place. | 19.3 | f(x) = \frac{2(-3|x - 1| + 1) - 2}{-3(-3|x - 1| + 1) - 1} \) and \( g(x) = -5x^2 - 2 \) from \( x = 0.4 \) to \( x = 1.5 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the expression for \( f(x) \):
\[ f(x) = \frac{2(-3|x - 1| + 1) - 2}{-3(-3|x - 1| + 1) - 1} \]
\[ = \frac{-6|x - 1|... | 1 | last_number | 0.3 | false | false | false | 40.073969 | 2,100 | 13 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -4(-1sin(1πx - 1))^2 - 1(-1sin(1πx - 1)) + 4 and g(x) = -1x + 5 from x = 1.4 to x = 2.2. Round to 1 decimal place. | 1.9 | f(x) = -4(-1\sin(\pi x - 1))^2 - 1(-1\sin(\pi x - 1)) + 4 \) and \( g(x) = -x + 5 \) from \( x = 1.4 \) to \( x = 2.2 \), we need to follow these steps:
### Step 1: Simplify the Functions
First, let's simplify \( f(x) \):
\[ f(x) = -4(-\sin(\pi x - 1))^2 - (-\sin(\pi x - 1)) + 4 \]
\[ f(x) = -4\sin^2(\pi x - 1) + \sin... | 2 | last_number | 0.3 | false | false | false | 40.024501 | 1,981 | 14 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = 1(4x^2 - 5x - 2)^2 - 4(4x^2 - 5x - 2) + 3 and g(x) = (-1x + 1)/(-2x) from x = -0.7 to x = -0.4. Round to 1 decimal place. | 0.5 | f(x) = (4x^2 - 5x - 2)^2 - 4(4x^2 - 5x - 2) + 3 \) and \( g(x) = \frac{-x + 1}{-2x} \) from \( x = -0.7 \) to \( x = -0.4 \), we need to follow these steps:
### Step 1: Simplify \( g(x) \)
First, simplify \( g(x) \):
\[ g(x) = \frac{-x + 1}{-2x} = \frac{x - 1}{2x} = \frac{1}{2} - \frac{1}{2x} \]
### Step 2: Set up th... | 0.1 | boxed | 1 | true | false | true | 27.511445 | 1,722 | 15 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = (-2(3x + 3) - 2)/(-3(3x + 3) - 1), g(x) = -2|x + 2| + 1, x = -1.7, and x = -1.2. Round your answer to 1 decimal place. | 0.1 | es, we need to follow these steps:
1. **Understand the functions and boundaries:**
- \( f(x) = \frac{-2(3x + 3) - 2}{-3(3x + 3) - 1} \)
- \( g(x) = -2|x + 2| + 1 \)
- The boundaries are \( x = -1.7 \) and \( x = -1.2 \).
2. **Simplify \( f(x) \):**
\[
f(x) = \frac{-2(3x + 3) - 2}{-3(3x + 3) - 1}
\]
... | 0.5 | boxed | 1 | true | false | true | 31.45381 | 1,793 | 16 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -5(1x^2 + 3x - 1)^2 + 3(1x^2 + 3x - 1) + 1, g(x) = -3|x - 2| + 3, x = -3.8, and x = -2.9. Round your answer to 1 decimal place. | 8.9 | es, we need to follow these steps:
1. **Identify the curves and their intersections:**
- \( f(x) = -5(1x^2 + 3x - 1)^2 + 3(1x^2 + 3x - 1) + 1 \)
- \( g(x) = -3|x - 2| + 3 \)
- The vertical lines \( x = -3.8 \) and \( x = -2.9 \)
2. **Simplify the functions if possible:**
- \( f(x) \) is already in its sim... | 0.1 | boxed | 1 | true | false | false | 28.568785 | 1,776 | 17 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 2sin(2π(-4x - 5) - 1) + 3 and g(x) = -2x - 2 between the vertical lines x = -2.9 and x = -2.8. Round to 1 decimal place. | 0.1 | hs of \( f(x) = 2\sin(2\pi(-4x - 5)) - 1 + 3 \) and \( g(x) = -2x - 2 \) between the vertical lines \( x = -2.9 \) and \( x = -2.8 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, simplify the expression for \( f(x) \):
\[ f(x) = 2\sin(2\pi(-4x - 5)) - 1 + 3 \]
\[ f(x) = 2\sin(-8\... | 2 | last_number | 0.3 | false | false | false | 39.897982 | 1,995 | 18 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -1(-4x^2 - 4x - 3) - 2 and g(x) = 3|x| between the vertical lines x = -1.6 and x = -0.2. Round to 1 decimal place. | 2.0 | hs of \( f(x) = -1(-4x^2 - 4x - 3) - 2 \) and \( g(x) = 3|x| \) between the vertical lines \( x = -1.6 \) and \( x = -0.2 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the expression for \( f(x) \):
\[
f(x) = -1(-4x^2 - 4x - 3) - 2 = 4x^2 + 4x + 3 - 2 = 4x^2 + 4x + 1
\]
So, \( f(x)... | 0 | equals_end | 0.8 | false | false | false | 40.216918 | 2,121 | 19 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = (2(-3cos(2πx - 1)) - 2)/(-3(-3cos(2πx - 1)) + 4) and g(x) = (1x + 2)/(-3x + 2) from x = 0.5 to x = 0.8. Round to 1 decimal place. | 41.3 | f(x) \) and \( g(x) \) from \( x = 0.5 \) to \( x = 0.8 \), we need to follow these steps:
1. **Define the functions:**
\[
f(x) = \frac{2(-3\cos(2\pi x - 1)) - 2}{-3(-3\cos(2\pi x - 1)) + 4}
\]
\[
g(x) = \frac{x + 2}{-3x + 2}
\]
2. **Set up the integral for the area between the curves:**
The area... | 0.1 | boxed | 1 | true | false | false | 26.262132 | 1,824 | 20 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -1cos(1π(-1|x|) + 1) - 2 and g(x) = 1x^2 + 3x - 3 between the vertical lines x = -3.5 and x = 0.1. Round to 1 decimal place. | 7.5 | hs of \( f(x) = -1\cos(\pi(-1|x|) + 1) - 2 \) and \( g(x) = x^2 + 3x - 3 \) between the vertical lines \( x = -3.5 \) and \( x = 0.1 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
The function \( f(x) = -1\cos(\pi(-1|x|) + 1) - 2 \) can be simplified as follows:
- The term \( -1|x| \) ... | 10.2 | boxed | 1 | true | false | false | 39.184051 | 2,575 | 21 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -3(-1|x - 1| + 1) - 1 and g(x) = -2x^2 + 5x - 1 between the vertical lines x = 0.0 and x = 2.3. Round to 1 decimal place. | 8.0 | hs of \( f(x) = -3(-1|x - 1| + 1) - 1 \) and \( g(x) = -2x^2 + 5x - 1 \) between \( x = 0.0 \) and \( x = 2.3 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the expression for \( f(x) \):
\[ f(x) = -3(-1|x - 1| + 1) - 1 \]
\[ f(x) = -3(-|x - 1| - 1) - 1 \]
\[ f(x) = 3|x - 1| + 3 - 1... | 3 | last_number | 0.3 | false | false | false | 39.958593 | 1,969 | 22 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -2cos(3π(2x^2 - 2x - 1) - 2) - 3 and g(x) = 1x + 2 between the vertical lines x = -3.5 and x = -3.4. Round to 1 decimal place. | 0.2 | hs of \( f(x) = -2\cos(3\pi(2x^2 - 2x - 1)) - 2 \) and \( g(x) = x + 2 \) between the vertical lines \( x = -3.5 \) and \( x = -3.4 \), we need to follow these steps:
1. **Identify the points of intersection**: First, find where the two functions intersect in the interval \([-3.5, -3.4]\). This involves solving \( f(x... | 0.1 | boxed | 1 | true | false | true | 38.389262 | 2,384 | 23 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = (3(3e^(2x) + 1) - 3)/(-1(3e^(2x) + 1) + 2) and g(x) = 2|x + 3| - 2 from x = -1.9 to x = -1.0. Round to 1 decimal place. | 0.3 | f(x) = \frac{3(3e^{2x} + 1) - 3}{-1(3e^{2x} + 1) + 2} \) and \( g(x) = 2|x + 3| - 2 \) from \( x = -1.9 \) to \( x = -1.0 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the expression for \( f(x) \):
\[ f(x) = \frac{3(3e^{2x} + 1) - 3}{-1(3e^{2x} + 1) + 2} \]
\[ f(x) = \frac{9e^{2x}... | 0.5 | boxed | 1 | true | false | true | 31.05171 | 1,912 | 24 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 1cos(3π(-3x^2 + 5x + 1) + 1) and g(x) = -2x + 5 between the vertical lines x = 2.2 and x = 2.3000000000000003. Round to 1 decimal place. | 0.1 | hs of \( f(x) = 1\cos(3\pi(-3x^2 + 5x + 1)) + 1 \) and \( g(x) = -2x + 5 \) between \( x = 2.2 \) and \( x = 2.3000000000000003 \), we need to follow these steps:
1. **Identify the functions and the interval:**
- \( f(x) = 1\cos(3\pi(-3x^2 + 5x + 1)) + 1 \)
- \( g(x) = -2x + 5 \)
- Interval: \( [2.2, 2.3] \)
... | 0.1 | boxed | 1 | true | false | true | 29.535349 | 1,740 | 25 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = (-3(3cos(1πx + 2) + 1) - 2)/(1(3cos(1πx + 2) + 1) + 3) and g(x) = 1x + 1 from x = -2.9 to x = -2.1. Round to 1 decimal place. | 0.2 | f(x) = \frac{-3(3\cos(\pi x + 2) + 1) - 2}{1(3\cos(\pi x + 2) + 1) + 3} \) and \( g(x) = x + 1 \) from \( x = -2.9 \) to \( x = -2.1 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, let's simplify \( f(x) \):
\[ f(x) = \frac{-3(3\cos(\pi x + 2) + 1) - 2}{1(3\cos(\pi x + 2) + 1) + ... | 5 | last_number | 0.3 | false | false | false | 40.166621 | 1,964 | 26 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 5(1x^2 - 3x + 2)^2 - 2(1x^2 - 3x + 2) - 2 and g(x) = -1x + 1 between the vertical lines x = 0.4 and x = 2.4. Round to 1 decimal place. | 2.5 | hs of \( f(x) = 5(1x^2 - 3x + 2)^2 - 2(1x^2 - 3x + 2) - 2 \) and \( g(x) = -1x + 1 \) between \( x = 0.4 \) and \( x = 2.4 \), we need to follow these steps:
1. **Identify the functions and the interval:**
- \( f(x) = 5(1x^2 - 3x + 2)^2 - 2(1x^2 - 3x + 2) - 2 \)
- \( g(x) = -1x + 1 \)
- Interval: \( [0.4, 2.4... | 12.3 | boxed | 1 | true | false | false | 29.143909 | 1,673 | 27 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 1e^(-1(-1|x + 3| - 1)) + 3 and g(x) = 1x^2 + 1x between the vertical lines x = -4.6 and x = -3.0. Round to 1 decimal place. | 1.8 | hs of \( f(x) = e^{-1(-1|x + 3| - 1)} + 3 \) and \( g(x) = x^2 + x \) between \( x = -4.6 \) and \( x = -3.0 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, simplify the expression for \( f(x) \):
\[ f(x) = e^{-1(-1|x + 3| - 1)} + 3 \]
\[ f(x) = e^{x + 3 - 1} + 3 \]
\[ f(x) = e^{... | 4 | last_number | 0.3 | false | false | false | 40.000273 | 2,036 | 28 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -2((-1x - 2)/(-1x - 2))^2 - 2 and g(x) = -2x^2 - 5x + 3 between the vertical lines x = -3.5 and x = 1.0. Round to 1 decimal place. | 30.4 | hs of \( f(x) = -2\left(\frac{-x-2}{-x-2}\right)^2 - 2 \) and \( g(x) = -2x^2 - 5x + 3 \) between the vertical lines \( x = -3.5 \) and \( x = 1.0 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the function \( f(x) \):
\[ f(x) = -2\left(\frac{-x-2}{-x-2}\right)^2 - 2 \]
Since \(\fra... | 5416667 | last_number | 0.3 | false | false | false | 39.834801 | 1,976 | 29 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 3cos(3π(4x^2 - 4x - 4) - 1), g(x) = 1x^2 - 5x - 4, x = -1.1, and x = -1.0. Round your answer to 1 decimal place. | 0.3 | es, we need to integrate the difference between the upper and lower functions over the specified interval. Let's break this down step by step.
### Step 1: Identify the Functions and Interval
- \( f(x) = 3\cos(3\pi(4x^2 - 4x - 4)) - 1 \)
- \( g(x) = x^2 - 5x - 4 \)
- The interval is from \( x = -1.1 \) to \( x = -1.0 \... | 0.1 | boxed | 1 | true | false | true | 32.859293 | 1,823 | 30 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 3(-3x + 2) + 1, g(x) = -5x^2 + 4, x = 0.4, and x = 1.4. Round your answer to 1 decimal place. | 0.6 | es, we need to integrate the difference between the upper and lower functions over the interval from \(x = 0.4\) to \(x = 1.4\).
First, let's simplify the function \(f(x)\):
\[ f(x) = 3(-3x + 2) + 1 = -9x + 6 + 1 = -9x + 7 \]
So, we have:
\[ f(x) = -9x + 7 \]
\[ g(x) = -5x^2 + 4 \]
Next, we determine which function ... | 0.6 | boxed | 1 | true | false | true | 36.949917 | 1,879 | 31 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = 3|(3cos(2πx - 1) - 3) - 2| - 3 and g(x) = -1x from x = -3.9 to x = -3.8. Round to 1 decimal place. | 0.1 | f(x) = 3|(3\cos(2\pi x - 1) - 3) - 2| - 3 \) and \( g(x) = -1x \) from \( x = -3.9 \) to \( x = -3.8 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, simplify the expression inside the absolute value:
\[ f(x) = 3|3\cos(2\pi x - 1) - 5| - 3 \]
### Step 2: Evaluate \( f(x) \) and \... | 1 | last_number | 0.3 | false | false | false | 40.262231 | 2,049 | 32 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -2(1x^2 + 4x - 5) - 1 and g(x) = -2x from x = -4.1 to x = 1.1. Round to 1 decimal place. | 46.8 | f(x) = -2(1x^2 + 4x - 5) - 1 \) and \( g(x) = -2x \) from \( x = -4.1 \) to \( x = 1.1 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the expression for \( f(x) \):
\[ f(x) = -2(1x^2 + 4x - 5) - 1 \]
\[ f(x) = -2x^2 - 8x + 10 - 1 \]
\[ f(x) = -2x^2 - 8x + 9 \]
### Step 2: Set up th... | 46.9 | boxed | 1 | true | false | true | 37.281798 | 1,880 | 33 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 1e^(-1(-2|x + 3| + 2)) - 1, g(x) = (-2x + 1)/(-1x + 3), x = -4.4, and x = -1.7. Round your answer to 1 decimal place. | 3.9 | es, we need to integrate the difference between the upper and lower functions over the specified interval. Let's break this down step by step.
### Step 1: Identify the Functions and Interval
The functions given are:
\[ f(x) = e^{-1(-2|x + 3| + 2)} - 1 \]
\[ g(x) = \frac{-2x + 1}{-x + 3} \]
The interval is from \( x =... | 1.9 | boxed | 1 | true | false | false | 26.359193 | 1,757 | 34 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -3cos(3π(-1x - 4) - 2) + 2, g(x) = 5x - 4, x = 1.3, and x = 1.6. Round your answer to 1 decimal place. | 0.3 | es, we need to integrate the difference between the upper and lower functions over the specified interval. Here's the step-by-step process:
1. **Identify the functions and the interval:**
- \( f(x) = -3\cos(3\pi(-1x - 4)) - 2 \)
- \( g(x) = 5x - 4 \)
- The interval is from \( x = 1.3 \) to \( x = 1.6 \).
2. ... | 4 | equals_end | 0.8 | false | false | false | 39.981126 | 1,982 | 35 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = 4(-1sin(1πx) - 3) + 5 and g(x) = -1x^2 - 4x - 3 from x = -4.3 to x = 1.0. Round to 1 decimal place. | 31.4 | f(x) = 4(-\sin(\pi x) - 3) + 5 \) and \( g(x) = -x^2 - 4x - 3 \) from \( x = -4.3 \) to \( x = 1.0 \), we need to follow these steps:
### Step 1: Simplify the Functions
First, simplify \( f(x) \):
\[ f(x) = 4(-\sin(\pi x) - 3) + 5 \]
\[ f(x) = -4\sin(\pi x) - 12 + 5 \]
\[ f(x) = -4\sin(\pi x) - 7 \]
So, the functions... | 123.4 | boxed | 1 | true | false | false | 24.16632 | 1,596 | 36 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -1(3|x| + 1)^2 + 2(3|x| + 1) + 2 and g(x) = 1x^2 + 5x + 2 from x = -0.6 to x = 0.2. Round to 1 decimal place. | 0.9 | f(x) = -1(3|x| + 1)^2 + 2(3|x| + 1) + 2 \) and \( g(x) = x^2 + 5x + 2 \) from \( x = -0.6 \) to \( x = 0.2 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, let's simplify \( f(x) \):
\[ f(x) = -1(3|x| + 1)^2 + 2(3|x| + 1) + 2 \]
Since \( |x| \) is symmetric around zero, we can co... | 2 | last_number | 0.3 | false | false | false | 39.711444 | 1,974 | 37 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 3(-5x^2 - 2x)^2 + 3(-5x^2 - 2x) - 4, g(x) = -3x + 2, x = -0.9, and x = 0.4. Round your answer to 1 decimal place. | 8.1 | es, we need to integrate the difference between the upper and lower functions over the specified interval. Here's the step-by-step process:
1. **Identify the upper and lower functions:**
- The function \( f(x) = 3(-5x^2 - 2x)^2 + 3(-5x^2 - 2x) - 4 \)
- The line \( g(x) = -3x + 2 \)
2. **Determine the interval:*... | 12.3 | boxed | 1 | true | false | false | 22.981201 | 1,702 | 38 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -3cos(3π(-1x^2 - 4x + 3)) + 2 and g(x) = 2x^2 - 4x between the vertical lines x = 2.8 and x = 2.9. Round to 1 decimal place. | 0.3 | hs of \( f(x) = -3\cos(3\pi(-x^2 - 4x + 3)) + 2 \) and \( g(x) = 2x^2 - 4x \) between \( x = 2.8 \) and \( x = 2.9 \), we need to follow these steps:
1. **Identify the functions and the interval:**
- \( f(x) = -3\cos(3\pi(-x^2 - 4x + 3)) + 2 \)
- \( g(x) = 2x^2 - 4x \)
- Interval: \( [2.8, 2.9] \)
2. **Set u... | 0.0 | boxed | 1 | true | false | true | 26.264673 | 1,662 | 39 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 2sin(3π(2|x - 1| - 1) - 3) - 3, g(x) = 3x^2 - 2, x = -0.4, and x = -0.30000000000000004. Round your answer to 1 decimal place. | 0.1 | es, we need to integrate the difference between the upper and lower functions over the interval from \( x = -0.4 \) to \( x = -0.30000000000000004 \).
First, let's identify which function is above the other in the interval \([-0.4, -0.30000000000000004]\). We need to evaluate both functions at a point in this interval... | 0.1 | boxed | 1 | true | false | true | 30.757658 | 1,750 | 40 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -1(-2sin(1πx - 1) - 2) - 1, g(x) = 3|x|, x = 0.3, and x = 1.0. Round your answer to 1 decimal place. | 0.3 | es, we need to integrate the difference between the upper and lower functions over the specified interval. Let's break this down step by step.
### Step 1: Simplify the Functions
First, let's simplify the function \( f(x) \):
\[ f(x) = -1(-2\sin(1\pi x - 1) - 2) - 1 \]
\[ f(x) = 2\sin(1\pi x - 1) + 2 - 1 \]
\[ f(x) = ... | 0 | last_number | 0.3 | false | false | true | 39.844346 | 2,260 | 41 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = (-2(-3sin(3πx - 3)) - 2)/(-3(-3sin(3πx - 3))) and g(x) = 3x^2 - 5x + 3 between the vertical lines x = -0.4 and x = -0.0. Round to 1 decimal place. | 1.9 | hs of \( f(x) = \frac{-2(-3\sin(3\pi x - 3)) - 2}{-3(-3\sin(3\pi x - 3))} \) and \( g(x) = 3x^2 - 5x + 3 \) between the vertical lines \( x = -0.4 \) and \( x = -0.0 \), we need to follow these steps:
1. Simplify the function \( f(x) \).
2. Set up the integral for the area between the curves.
3. Evaluate the integral ... | 0.5 | boxed | 1 | true | false | false | 39.186219 | 2,138 | 42 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 5(-2sin(3πx - 1) + 1) + 5, g(x) = (-2x + 3)/(-2x), x = -0.3, and x = -0.1. Round your answer to 1 decimal place. | 0.7 | es, we need to follow these steps:
1. **Identify the functions and the interval:**
- \( f(x) = 5(-2\sin(3\pi x - 1) + 1) + 5 \)
- \( g(x) = \frac{-2x + 3}{-2x} \)
- The interval is from \( x = -0.3 \) to \( x = -0.1 \).
2. **Simplify the functions if possible:**
- For \( f(x) \):
\[
f(x) = 5(-2\... | 6 | equals_end | 0.8 | false | false | false | 39.861679 | 2,062 | 43 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 3cos(1π(-1|x + 1| + 3) - 1) and g(x) = -2|x - 1| + 3 between the vertical lines x = 1.4 and x = 2.1. Round to 1 decimal place. | 0.6 | hs of \( f(x) = 3\cos(\pi(-x + 3) - 1) \) and \( g(x) = -2|x - 1| + 3 \) between \( x = 1.4 \) and \( x = 2.1 \), we need to follow these steps:
### Step 1: Simplify the functions
First, let's simplify \( f(x) \):
\[ f(x) = 3\cos(\pi(-x + 3) - 1) \]
\[ f(x) = 3\cos(-\pi x + 3\pi - 1) \]
Since cosine is an even functi... | 1.0 | boxed | 1 | true | false | false | 37.307393 | 2,348 | 44 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -5(2x^2 - 4) - 2, g(x) = 2|x + 1| + 2, x = -1.2, and x = 1.1. Round your answer to 1 decimal place. | 22.2 | es, we need to integrate the difference between the upper and lower functions over the specified interval. Let's break this down step by step.
### Step 1: Identify the Functions and Interval
- \( f(x) = -5(2x^2 - 4) - 2 \)
- \( g(x) = 2|x + 1| + 2 \)
- The interval is from \( x = -1.2 \) to \( x = 1.1 \).
### Step 2:... | 3 | last_number | 0.3 | false | false | false | 39.886408 | 2,074 | 45 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = 2|(5x^2 + 1x - 5) + 3| + 2 and g(x) = 1x^2 - 4x from x = -0.9 to x = -0.6. Round to 1 decimal place. | 0.2 | f(x) = 2|(5x^2 + x - 5) + 3| + 2 \) and \( g(x) = x^2 - 4x \) from \( x = -0.9 \) to \( x = -0.6 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the expression inside the absolute value for \( f(x) \):
\[ 5x^2 + x - 5 + 3 = 5x^2 + x - 2 \]
Thus,
\[ f(x) = 2|5x^2 + x - 2| + 2 \]
### ... | 227 | last_number | 0.3 | false | false | false | 40.4229 | 2,091 | 46 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = 1|(-3x + 5)| - 2 and g(x) = 1|x - 3| - 2 from x = 1.0 to x = 2.0. Round to 1 decimal place. | 0.7 | f(x) = 1|(-3x + 5)| - 2 \) and \( g(x) = 1|x - 3| - 2 \) from \( x = 1.0 \) to \( x = 2.0 \), we need to follow these steps:
### Step 1: Simplify the Functions
First, let's simplify the expressions for \( f(x) \) and \( g(x) \).
For \( f(x) = 1|(-3x + 5)| - 2 \):
- The expression inside the absolute value, \(-3x + 5... | 2 | last_number | 0.3 | false | false | false | 40.154479 | 2,232 | 47 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = (1(-1x - 1) + 2)/(2(-1x - 1) - 1) and g(x) = 5x - 3 between the vertical lines x = -1.4 and x = 0.6. Round to 1 decimal place. | 7.2 | hs of \( f(x) = \frac{(1(-x - 1) + 2)}{(2(-x - 1) - 1)} \) and \( g(x) = 5x - 3 \) between the vertical lines \( x = -1.4 \) and \( x = 0.6 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, simplify the expression for \( f(x) \):
\[ f(x) = \frac{(1(-x - 1) + 2)}{(2(-x - 1) - 1)} \]... | 1 | last_number | 0.3 | false | false | false | 40.135361 | 2,064 | 48 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -2|(3x^2 - 5x - 5) + 1| + 2 and g(x) = -1x - 4 from x = 1.7 to x = 2.7. Round to 1 decimal place. | 4.0 | f(x) = -2|(3x^2 - 5x - 5) + 1| + 2 \) and \( g(x) = -1x - 4 \) from \( x = 1.7 \) to \( x = 2.7 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, simplify the expression inside the absolute value:
\[ 3x^2 - 5x - 5 + 1 = 3x^2 - 5x - 4 \]
Thus,
\[ f(x) = -2|3x^2 - 5x - 4| + 2 \]
### Step 2: Dete... | 7 | last_number | 0.3 | false | false | false | 39.855406 | 1,950 | 49 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -2(3|x - 2|)^2 - 2(3|x - 2|) + 3 and g(x) = -2x^2 - 2x from x = 1.4 to x = 3.1. Round to 1 decimal place. | 16.8 | f(x) = -2(3|x - 2|)^2 - 2(3|x - 2|) + 3 \) and \( g(x) = -2x^2 - 2x \) from \( x = 1.4 \) to \( x = 3.1 \), we need to follow these steps:
### Step 1: Simplify \( f(x) \)
First, let's simplify \( f(x) \):
\[ f(x) = -2(3|x - 2|)^2 - 2(3|x - 2|) + 3 \]
Since \( |x - 2| \) is piecewise defined, we need to consider two ca... | 2 | therefore | 0.8 | false | false | false | 40.15568 | 1,954 | 50 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 5(-2x + 3)^2 + 5(-2x + 3) - 4 and g(x) = (-2x + 3)/(1x) between the vertical lines x = 1.2 and x = 2.2. Round to 1 decimal place. | 3.4 | hs of \( f(x) = 5(-2x + 3)^2 + 5(-2x + 3) - 4 \) and \( g(x) = \frac{-2x + 3}{x} \) between \( x = 1.2 \) and \( x = 2.2 \), we need to follow these steps:
1. **Find the points of intersection of \( f(x) \) and \( g(x) \):**
Set \( f(x) = g(x) \):
\[
5(-2x + 3)^2 + 5(-2x + 3) - 4 = \frac{-2x + 3}{x}
\]
... | 0 | equals_end | 0.8 | false | false | false | 39.914388 | 2,042 | 51 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = 1cos(1π((1x - 1)/(2x + 3))) + 2 and g(x) = 2|x + 1| from x = -1.7 to x = -1.5999999999999999. Round to 1 decimal place. | 0.1 | f(x) = 1\cos\left(\frac{\pi((1x - 1)}{2x + 3}\right) + 2 \) and \( g(x) = 2|x + 1| \) from \( x = -1.7 \) to \( x = -1.5999999999999999 \), we need to follow these steps:
### Step 1: Simplify the Functions
First, let's simplify the given functions for clarity:
- \( f(x) = 1\cos\left(\frac{\pi((1x - 1)}{2x + 3}\right) ... | 0.0 | boxed | 1 | true | false | true | 33.545778 | 2,116 | 52 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -5(-1|x + 2| + 2) - 3 and g(x) = 4x - 2 between the vertical lines x = -2.3 and x = 1.0. Round to 1 decimal place. | 5.0 | hs of \( f(x) = -5(-1|x + 2| + 2) - 3 \) and \( g(x) = 4x - 2 \) between the vertical lines \( x = -2.3 \) and \( x = 1.0 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, simplify \( f(x) \):
\[ f(x) = -5(-1|x + 2| + 2) - 3 \]
\[ f(x) = -5(-|x + 2| - 2) - 3 \]
\[ f(x) = 5|x + 2| +... | 23 | last_number | 0.3 | false | false | false | 40.0753 | 2,070 | 53 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -1e^(2(-2|x| - 1)), g(x) = -3x^2 + 2x + 2, x = -0.6, and x = 1.2. Round your answer to 1 decimal place. | 2.8 | es, we need to integrate the difference between the upper and lower functions over the specified interval. The curves are \( f(x) = -e^{2(-2|x| - 1)} \) and \( g(x) = -3x^2 + 2x + 2 \), and the bounds are \( x = -0.6 \) and \( x = 1.2 \).
First, let's analyze the function \( f(x) = -e^{2(-2|x| - 1)} \). This function ... | 8 | last_number | 0.3 | false | false | false | 39.733454 | 2,163 | 54 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 2(-2|x + 1| + 3) - 4 and g(x) = 1x - 5 between the vertical lines x = -3.7 and x = 0.6. Round to 1 decimal place. | 17.1 | hs of \( f(x) = 2(-2|x + 1| + 3) - 4 \) and \( g(x) = x - 5 \) between the vertical lines \( x = -3.7 \) and \( x = 0.6 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, simplify \( f(x) \):
\[ f(x) = 2(-2|x + 1| + 3) - 4 \]
\[ f(x) = -4|x + 1| + 6 - 4 \]
\[ f(x) = -4|x + 1| + 2 \]... | 1 | last_number | 0.3 | false | false | false | 40.070093 | 2,031 | 55 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = 3(-5x)^2 + 2 and g(x) = 1x + 5 between the vertical lines x = -0.2 and x = 0.2. Round to 1 decimal place. | 0.8 | hs of \( f(x) = 3(-5x)^2 + 2 \) and \( g(x) = x + 5 \) between the vertical lines \( x = -0.2 \) and \( x = 0.2 \), we need to follow these steps:
1. **Simplify the function \( f(x) \):**
\[
f(x) = 3(-5x)^2 + 2 = 3(25x^2) + 2 = 75x^2 + 2
\]
2. **Set up the integral for the area between the curves:**
The a... | 0.8 | boxed | 1 | true | false | true | 37.278445 | 1,903 | 56 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = -3sin(1π(-3x - 1) - 3) + 1, g(x) = 4x^2 - 5x, x = -0.0, and x = 0.4. Round your answer to 1 decimal place. | 1.3 | es, we need to integrate the difference between the upper and lower functions over the interval from \( x = -0.0 \) to \( x = 0.4 \).
First, let's identify the upper and lower functions in the interval \([-0.0, 0.4]\):
- The function \( g(x) = 4x^2 - 5x \)
- The function \( f(x) = -3\sin(\pi(-3x - 1)) - 3 \)
We need ... | 0 | last_number | 0.3 | false | false | false | 40.10126 | 2,230 | 57 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -1(-2sin(1πx - 2) - 2)^2 - 1 and g(x) = -3x^2 + 5x + 1 from x = 2.7 to x = 3.3. Round to 1 decimal place. | 1.7 | f(x) = -1(-2\sin(1\pi x - 2) - 2)^2 - 1 \) and \( g(x) = -3x^2 + 5x + 1 \) from \( x = 2.7 \) to \( x = 3.3 \), we need to follow these steps:
1. **Determine the points of intersection (if any) within the interval [2.7, 3.3]**:
Since the functions are given and we are only asked to find the area between them over a... | 0.5 | boxed | 1 | true | false | false | 23.166026 | 1,575 | 58 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -3cos(2π(1|x - 1| + 2) - 1) + 3 and g(x) = -5x - 3 from x = -1.5 to x = -1.0. Round to 1 decimal place. | 0.9 | f(x) = -3\cos(2\pi(1|x - 1| + 2)) - 1 + 3 \) and \( g(x) = -5x - 3 \) from \( x = -1.5 \) to \( x = -1.0 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, simplify \( f(x) \):
\[ f(x) = -3\cos(2\pi(1|x - 1| + 2)) - 1 + 3 \]
\[ f(x) = -3\cos(2\pi|x - 1| + 4\pi) - 1 + 3 \]
Since \(\c... | 5 | equals_end | 0.8 | false | false | false | 39.9179 | 1,984 | 59 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 2(1sin(2πx - 1) + 3)^2 - 5(1sin(2πx - 1) + 3) - 4, g(x) = 2|x + 1| + 2, x = -1.8, and x = -1.4. Round your answer to 1 decimal place. | 1.1 | es, we need to follow these steps:
1. **Identify the curves and the bounds:**
- \( f(x) = 2(1\sin(2\pi x - 1) + 3)^2 - 5(1\sin(2\pi x - 1) + 3) - 4 \)
- \( g(x) = 2|x + 1| + 2 \)
- The bounds are \( x = -1.8 \) and \( x = -1.4 \).
2. **Determine which curve is above the other in the interval \([-1.8, -1.4]\)... | 0.8 | boxed | 1 | true | false | false | 31.477499 | 1,973 | 60 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = (-2(-1x + 2) + 1)/(3(-1x + 2) + 1), g(x) = -1x^2 + 4x, x = 2.2, and x = 4.2. Round your answer to 1 decimal place. | 33.5 | es, we need to follow these steps:
1. **Identify the functions and the bounds:**
- \( f(x) = \frac{-2(-1x + 2) + 1}{3(-1x + 2) + 1} \)
- \( g(x) = -x^2 + 4x \)
- The bounds are \( x = 2.2 \) and \( x = 4.2 \).
2. **Simplify the function \( f(x) \):**
\[
f(x) = \frac{-2(-1x + 2) + 1}{3(-1x + 2) + 1}
... | 13 | last_number | 0.3 | false | false | false | 40.239856 | 2,325 | 61 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 1e^(-1(-1|x| - 1)) - 3, g(x) = 1|x + 3| - 2, x = -0.3, and x = 0.5. Round your answer to 1 decimal place. | 0.6 | es, we need to integrate the difference between the upper and lower functions over the interval from \(x = -0.3\) to \(x = 0.5\).
First, let's identify the upper and lower functions in the interval \([-0.3, 0.5]\).
The function \(f(x) = e^{-1(-1|x| - 1)} - 3\) simplifies to:
\[ f(x) = e^{x - 1} - 3 \]
This is an expo... | 3.0 | boxed | 1 | true | false | false | 39.095515 | 2,077 | 62 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = (3(-2cos(3πx - 2) - 3) + 3)/(-2(-2cos(3πx - 2) - 3) - 1) and g(x) = 4x^2 - 4x between the vertical lines x = 0.4 and x = 0.6. Round to 1 decimal place. | 0.1 | hs of \( f(x) \) and \( g(x) \) between \( x = 0.4 \) and \( x = 0.6 \), we need to follow these steps:
1. Simplify the function \( f(x) \).
2. Set up the integral for the area between the curves.
3. Evaluate the integral numerically.
### Step 1: Simplify \( f(x) \)
Given:
\[ f(x) = \frac{3(-2\cos(3\pi x - 2) - 3) +... | 0.0 | boxed | 1 | true | false | true | 29.124318 | 1,759 | 63 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -5(-2sin(3πx)) + 2 and g(x) = -1|x - 1| - 2 between the vertical lines x = -3.4 and x = -2.9. Round to 1 decimal place. | 5.6 | hs of \( f(x) = -5(-2\sin(3\pi x)) + 2 \) and \( g(x) = -1|x - 1| - 2 \) between the vertical lines \( x = -3.4 \) and \( x = -2.9 \), we need to follow these steps:
### Step 1: Simplify the functions
First, simplify \( f(x) \):
\[ f(x) = -5(-2\sin(3\pi x)) + 2 = 10\sin(3\pi x) + 2 \]
Next, consider \( g(x) \). The a... | 4 | last_number | 0.3 | false | false | false | 39.878125 | 2,133 | 64 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 1(-3x - 4) - 5, g(x) = -1x^2 - 3x - 5, x = -2.0, and x = 2.0. Round your answer to 1 decimal place. | 10.7 | es, we need to follow these steps:
1. **Identify the functions and the bounds:**
- \( f(x) = -3x - 9 \)
- \( g(x) = -x^2 - 3x - 5 \)
- The bounds are \( x = -2.0 \) and \( x = 2.0 \).
2. **Set up the integral for the area between the curves:**
The area \( A \) between two curves \( y = f(x) \) and \( y = ... | 10.7 | boxed | 1 | true | false | true | 28.572093 | 1,573 | 65 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -2sin(3π(5x + 2)) - 2 and g(x) = -2x - 2 between the vertical lines x = -0.1 and x = 0.0. Round to 1 decimal place. | 0.1 | hs of \( f(x) = -2\sin(3\pi(5x + 2)) - 2 \) and \( g(x) = -2x - 2 \) between \( x = -0.1 \) and \( x = 0.0 \), we need to follow these steps:
1. **Identify the functions and the interval:**
- \( f(x) = -2\sin(3\pi(5x + 2)) - 2 \)
- \( g(x) = -2x - 2 \)
- Interval: \( x = -0.1 \) to \( x = 0.0 \)
2. **Set up ... | 0.1 | boxed | 1 | true | false | true | 29.233677 | 1,884 | 66 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -2(-3sin(1πx + 2) + 1) - 2 and g(x) = (-2x - 3)/(-3x - 1) between the vertical lines x = -3.9 and x = -2.4. Round to 1 decimal place. | 9.3 | hs of \( f(x) = -2(-3\sin(1\pi x + 2) + 1) - 2 \) and \( g(x) = \frac{-2x - 3}{-3x - 1} \) between the vertical lines \( x = -3.9 \) and \( x = -2.4 \), we need to follow these steps:
1. **Simplify the functions**:
- For \( f(x) \):
\[
f(x) = -2(-3\sin(1\pi x + 2) + 1) - 2
\]
Simplify inside the... | 1.8 | boxed | 1 | true | false | false | 25.906767 | 1,633 | 67 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -3cos(1π((3x + 3)/(2x - 1)) - 1) - 2 and g(x) = -4x - 3 from x = 0.4 to x = 0.5. Round to 1 decimal place. | 0.3 | f(x) = -3\cos\left(\pi\left(\frac{3x + 3}{2x - 1}\right)\right) - 1 \) and \( g(x) = -4x - 3 \) from \( x = 0.4 \) to \( x = 0.5 \), we need to follow these steps:
### Step 1: Define the integrand
The area between the curves is given by the integral of the absolute difference between the two functions over the interva... | 0.1 | boxed | 1 | true | false | true | 25.133008 | 1,703 | 68 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -3|((2x - 1)/(3x - 2))| - 3 and g(x) = -1x^2 + 1x - 4 from x = -0.5 to x = 0.4. Round to 1 decimal place. | 0.3 | losed between the two given functions \( f(x) = -3\left|\frac{2x - 1}{3x - 2}\right| - 3 \) and \( g(x) = -x^2 + x - 4 \) from \( x = -0.5 \) to \( x = 0.4 \).
### Step 1: Understand the Functions
First, let's understand the behavior of each function:
- \( f(x) = -3\left|\frac{2x - 1}{3x - 2}\right| - 3 \)
- \( g(x)... | 1.0 | boxed | 1 | true | false | false | 19.96752 | 1,576 | 69 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -1sin(2π(1x^2 - 1x) - 2) - 2 and g(x) = (-1x - 1)/(1x + 3) between the vertical lines x = -5.0 and x = -4.9. Round to 1 decimal place. | 0.1 | hs of \( f(x) = -1\sin(2\pi(1x^2 - 1x) - 2) - 2 \) and \( g(x) = \frac{-1x - 1}{1x + 3} \) between the vertical lines \( x = -5.0 \) and \( x = -4.9 \), we need to follow these steps:
### Step 1: Understand the Functions
- \( f(x) = -1\sin(2\pi(1x^2 - 1x) - 2) - 2 \)
- \( g(x) = \frac{-1x - 1}{1x + 3} \)
### Step 2: ... | 0.5 | boxed | 1 | true | false | true | 19.852794 | 1,644 | 70 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -2sin(1π(1|x + 1| - 3) + 2) - 3 and g(x) = -1|x + 1| - 1 from x = -3.4 to x = -2.4. Round to 1 decimal place. | 1.2 | f(x) = -2\sin(\pi(1|x + 1| - 3)) + 2 \) and \( g(x) = -1|x + 1| - 1 \) from \( x = -3.4 \) to \( x = -2.4 \), we need to follow these steps:
### Step 1: Simplify the Functions
First, let's simplify the expressions for \( f(x) \) and \( g(x) \).
For \( f(x) \):
\[ f(x) = -2\sin(\pi(1|x + 1| - 3)) + 2 \]
Since \( |x +... | 2 | last_number | 0.3 | false | false | false | 39.817409 | 2,249 | 71 | algebra_func_area |
Determine the area of the region enclosed by the graphs of f(x) = -3sin(2π((-1x + 1)/(-1x)) + 1) - 1 and g(x) = 4x^2 - 2x - 3 between the vertical lines x = -0.2 and x = -0.1. Round to 1 decimal place. | 0.2 | hs of \( f(x) = -3\sin\left(2\pi\left(\frac{-1x + 1}{-1x}\right)\right) + 1 \) and \( g(x) = 4x^2 - 2x - 3 \) between \( x = -0.2 \) and \( x = -0.1 \), we need to follow these steps:
1. Simplify the function \( f(x) \).
2. Set up the integral for the area between the curves.
3. Evaluate the integral numerically.
###... | 0.0 | boxed | 1 | true | false | true | 31.844204 | 1,950 | 72 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 3|(-3x + 4) - 1| - 3, g(x) = 2x + 5, x = 0.1, and x = 2.4. Round your answer to 1 decimal place. | 11.7 | es, we need to follow these steps:
1. **Understand the functions and their intersections:**
- \( f(x) = 3|(-3x + 4) - 1| - 3 \)
- \( g(x) = 2x + 5 \)
- The bounds are \( x = 0.1 \) and \( x = 2.4 \).
2. **Simplify \( f(x) \):**
\[
f(x) = 3|(-3x + 4) - 1| - 3 = 3|-3x + 3| - 3
\]
\[
f(x) = 3| -3... | 2 | last_number | 0.3 | false | false | false | 40.089194 | 2,015 | 73 | algebra_func_area |
Calculate the area enclosed between the functions f(x) = -4(3x^2 - 2x - 5) - 5 and g(x) = -5x^2 + 2x from x = -1.1 to x = 2.0. Round to 1 decimal place. | 33.1 | f(x) = -4(3x^2 - 2x - 5) - 5 \) and \( g(x) = -5x^2 + 2x \) from \( x = -1.1 \) to \( x = 2.0 \), we need to follow these steps:
### Step 1: Simplify the function \( f(x) \)
First, simplify \( f(x) \):
\[
f(x) = -4(3x^2 - 2x - 5) - 5
\]
\[
= -12x^2 + 8x + 20 - 5
\]
\[
= -12x^2 + 8x + 15
\]
### Step 2: Set up the inte... | 33.1 | boxed | 1 | true | false | true | 38.920322 | 1,889 | 74 | algebra_func_area |
Find the area of the region bounded by the curves f(x) = 2|(3x^2 - 2x - 5) - 1| - 3, g(x) = (-2x + 2)/(-3x + 2), x = -1.3, and x = -0.9. Round your answer to 1 decimal place. | 0.8 | es, we need to follow these steps:
1. **Simplify the function \( f(x) \):**
\[
f(x) = 2 \left| (3x^2 - 2x - 5) - 1 \right| - 3
\]
Simplify inside the absolute value:
\[
f(x) = 2 \left| 3x^2 - 2x - 6 \right| - 3
\]
2. **Analyze the behavior of \( f(x) \):**
The expression inside the absolute va... | 0.4 | boxed | 1 | true | false | true | 31.888529 | 1,998 | 75 | algebra_func_area |
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