Datasets:
problem_key stringclasses 560
values | rollout_index int64 0 31 | level int64 1 5 | capability stringclasses 9
values | gt stringclasses 345
values | solution stringlengths 9 410k | correct bool 2
classes | committed bool 2
classes | finish_reason stringclasses 2
values | output_tokens int64 4 31.7k |
|---|---|---|---|---|---|---|---|---|---|
precalculus__8ee539 | 0 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the point (0, 3) from rectangular coordinates to polar coordinates. Hmm, polar coordinates are in the form (r, θ), right? Let me recall how to do this.
First, I remember that to convert from rectangular (x, y) to polar (r, θ), the formulas are r = sqrt(x² + y²) and θ = arctan(y/x). But wai... | true | true | stop | 682 |
precalculus__8ee539 | 1 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, let's see. I need to convert the rectangular coordinates (0, 3) to polar coordinates. Hmm, polar coordinates are in the form (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis. Alright, so first, I should find r.
The formula for r is the square root of (x squared plus y ... | true | true | stop | 714 |
precalculus__8ee539 | 2 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the rectangular coordinates (0, 3) to polar coordinates. Let me remember how that works. Polar coordinates are (r, θ), where r is the distance from the origin, and θ is the angle from the positive x-axis.
First, let me recall the formulas. To convert from rectangular (x, y) to polar (r, ... | true | true | stop | 888 |
precalculus__8ee539 | 3 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, let's see. I need to convert the rectangular coordinates (0, 3) to polar coordinates. Hmm, polar coordinates are (r, θ), right? So, first, I remember that to convert from rectangular (x, y) to polar, the formulas are r = sqrt(x² + y²) and θ = arctan(y/x). But wait, I have to be careful with the quadrant where ... | true | true | stop | 750 |
precalculus__8ee539 | 4 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, let's see. I need to convert the point (0, 3) from rectangular coordinates to polar coordinates. Hmm, polar coordinates are usually written as (r, θ), right? So I need to find r and θ.
First, I remember that in polar coordinates, r is the distance from the origin to the point. So for the point (0, 3), that's 3... | true | true | stop | 573 |
precalculus__8ee539 | 5 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the point (0, 3) from rectangular coordinates to polar coordinates. Hmm, polar coordinates are in the form (r, θ), where r is the distance from the origin, and θ is the angle from the positive x-axis. Let me remember how to do this conversion.
First, for r. I think r is calculated using t... | true | true | stop | 500 |
precalculus__8ee539 | 6 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, let's see. I need to convert the point (0, 3) from rectangular coordinates to polar coordinates. Hmm, polar coordinates are in the form (r, θ), right? So first, I remember that r is the distance from the origin to the point, calculated using the Pythagorean theorem. The formula for r is sqrt(x² + y²).
Given ... | true | true | stop | 551 |
precalculus__8ee539 | 7 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
[asy]import graph; size(6cm); real r = 1.5; real s = 0.2; real t = 0.8; real R = 1; real x = 0; real y = 0; real y1 = 0; real y2 = 0; real y3 = 0; real x1 = 0; real x2 = 0; real x3 = 0; real y4 = 0; real y5 = 0; real x4 = 0; real x5 = 0; real y6 = 0; real x6 = 0; draw((0,0)--(0,-3),Arrows); draw((0,-3)--(0,-6),Arrows... | false | false | length | 31,744 |
precalculus__8ee539 | 8 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the rectangular coordinates (0, 3) to polar coordinates. Let me remember how that works. Polar coordinates are (r, θ), right? Where r is the distance from the origin, and θ is the angle made with the positive x-axis.
First, let's find r. The formula for r is the square root of (x squared ... | true | true | stop | 639 |
precalculus__8ee539 | 9 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the rectangular coordinates (0, 3) to polar coordinates. Hmm, polar coordinates are in the form (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis. Let me recall how to do this conversion.
First, I think the formula for r is the square root of (x s... | true | true | stop | 507 |
precalculus__8ee539 | 10 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the point (0,3) from rectangular coordinates to polar coordinates. Hmm, polar coordinates are (r, θ), right? Let me recall the formulas for converting.
I think the formula for r is the square root of (x squared plus y squared). So in this case, x is 0 and y is 3. Let me calculate that. r... | true | true | stop | 776 |
precalculus__8ee539 | 11 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the rectangular coordinates (0, 3) to polar coordinates (r, θ). Hmm, polar coordinates are based on the radius r and the angle θ from the positive x-axis. Let me recall the formulas for conversion.
I think the formula for r is the square root of (x squared plus y squared). So, r = sqrt(x... | true | true | stop | 686 |
precalculus__8ee539 | 12 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the rectangular coordinates (0, 3) to polar coordinates. Hmm, polar coordinates are (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis. Let me recall the formulas for conversion.
Right, the formulas are r = sqrt(x² + y²) and θ = arctan(y/x). But wa... | true | true | stop | 514 |
precalculus__8ee539 | 13 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, let's see. I need to convert the rectangular coordinates (0, 3) to polar coordinates. Hmm, polar coordinates are in the form (r, θ), right? So, first, I remember that to convert from rectangular to polar, there are some formulas. Let me recall them.
The formulas are r squared equals x squared plus y squared, ... | true | true | stop | 833 |
precalculus__8ee539 | 14 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the rectangular coordinates (0, 3) to polar coordinates (r, θ). Hmm, polar coordinates are based on the distance from the origin and the angle from the positive x-axis, right? Let me recall the formulas for conversion.
I think the formulas are r = sqrt(x² + y²) and θ = arctan(y/x). Wait,... | true | true | stop | 653 |
precalculus__8ee539 | 15 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) | .
Okay, so I need to convert the point (0,3) from rectangular coordinates to polar coordinates. Hmm, polar coordinates are in the form (r, θ), right? Where r is the distance from the origin and θ is the angle from the positive x-axis. Let me recall the formulas for converting between rectangular and polar coordinates.... | true | true | stop | 637 |
algebra__e16674 | 0 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think. Oh yeah, the distance formula is derived from the Pythagorean theorem. Right, so if you have two points, (x1, y1) and (x2, y2), the di... | true | true | stop | 691 |
algebra__e16674 | 1 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, distance between two points in a coordinate plane... I think that's the distance formula. Let me recall. The distance formula is derived from the Pythagorean theorem, right? So if I have two points, (x1, y1) and (x2, y2), the distance b... | true | true | stop | 708 |
algebra__e16674 | 2 | 3 | algebra | 3\sqrt{13} | .
Okay, let's see. I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think. Oh right, the distance formula! It's derived from the Pythagorean theorem. So, the formula is the square root of [(x2 - x1) sq... | true | true | stop | 570 |
algebra__e16674 | 3 | 3 | algebra | 3\sqrt{13} | .
Okay, let's see. I need to find the distance between the points (2, -6) and (-4, 3). Hmm, how do I do that again? Oh right, the distance formula! It's like the Pythagorean theorem in two dimensions. So the formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Let me write that down.
First, I should identify the coordinates. T... | true | true | stop | 511 |
algebra__e16674 | 4 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think... Oh right, the distance formula! It's derived from the Pythagorean theorem. The formula is the square root of [(x2 - x1) squared plus... | true | true | stop | 853 |
algebra__e16674 | 5 | 3 | algebra | 3\sqrt{13} | .
Okay, let's see. I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me recall... Oh right, the distance formula. It's derived from the Pythagorean theorem, right? So if you have two points, (x1, y1) and ... | true | true | stop | 671 |
algebra__e16674 | 6 | 3 | algebra | 3\sqrt{13} | .
Okay, let's see. I need to find the distance between the two points (2, -6) and (-4, 3). Hmm, distance formula, right? The formula is something like the square root of [(x2 - x1)^2 + (y2 - y1)^2]. Let me recall. Yeah, that's right. So first, I should identify the coordinates. The first point is (2, -6), so x1 is 2 an... | true | true | stop | 647 |
algebra__e16674 | 7 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think... Oh right, the distance formula! It's derived from the Pythagorean theorem, right?
The formula is something like the square root of... | true | true | stop | 625 |
algebra__e16674 | 8 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think... Oh right, the distance formula! It's derived from the Pythagorean theorem. The formula is \( \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} \)... | true | true | stop | 661 |
algebra__e16674 | 9 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the two points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think... Oh right, the distance formula! It's derived from the Pythagorean theorem. The formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2].
Al... | true | true | stop | 589 |
algebra__e16674 | 10 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, right, the distance formula. Let me recall how that works. I think it's based on the Pythagorean theorem. The formula is the square root of [(x2 - x1)^2 + (y2 - y1)^2]. Let me make sure I remember that correctly.
So, first, I should l... | true | true | stop | 721 |
algebra__e16674 | 11 | 3 | algebra | 3\sqrt{13} | .
Okay, let's see. I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me recall. Oh right, the distance formula is derived from the Pythagorean theorem. So, if you have two points (x1, y1) and (x2, y2), th... | true | true | stop | 637 |
algebra__e16674 | 12 | 3 | algebra | 3\sqrt{13} | and to only include the answer, not any other text.
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think... Oh right, the distance formula! It's derived from the Pythagorean theorem. The f... | true | true | stop | 653 |
algebra__e16674 | 13 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, distance formula, right? Let me recall. The distance between two points (x1, y1) and (x2, y2) is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Yeah, that's the formula. Alright, let's plug in the values.
First, let's label the points. Let me call t... | true | true | stop | 581 |
algebra__e16674 | 14 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember there's a formula for the distance between two points in a coordinate plane. Let me think... Oh right, the distance formula is derived from the Pythagorean theorem. The formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Let me make... | true | true | stop | 572 |
algebra__e16674 | 15 | 3 | algebra | 3\sqrt{13} | .
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, right, distance formula. Let me recall. The distance between two points (x1, y1) and (x2, y2) is the square root of [(x2 - x1)^2 + (y2 - y1)^2]. Yeah, that's right. So I should apply that here.
First, let's identify the coordinates. Th... | true | true | stop | 596 |
intermediate_algebra__687b74 | 0 | 1 | intermediate_algebra | -50 | .
Okay, let me see. I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, alternating signs, starting with positive. So the pattern is +, -, +, -, and so on. Let me think about how to approach this.
First, maybe I can group the terms in pairs. Like (1 - 2) + (3 - 4) + (5 - 6) + ... and so on. Each pair w... | true | true | stop | 1,290 |
intermediate_algebra__687b74 | 1 | 1 | intermediate_algebra | -50 | .
Okay, let me try to figure out this problem. So, the question is to compute the sum: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, that's an alternating series where the signs switch between positive and negative. Let me see how to approach this.
First, let me write out the series to see the pattern. Starting from 1, su... | true | true | stop | 966 |
intermediate_algebra__687b74 | 2 | 1 | intermediate_algebra | -50 | .
Okay, so I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let's see. This looks like an alternating series where the signs alternate between positive and negative. Let me try to figure out a pattern here.
First, let me write out the first few terms to see if I can spot a pattern. The series starts... | true | true | stop | 906 |
intermediate_algebra__687b74 | 3 | 1 | intermediate_algebra | -50 | .
Okay, so I need to compute the sum of the series 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let's see. First, I should figure out if there's a pattern here. The signs are alternating, starting with positive for the first term. So the first term is positive, the second negative, third positive, and so on. The last term... | true | true | stop | 1,150 |
intermediate_algebra__687b74 | 4 | 1 | intermediate_algebra | -50 | .
Okay, let's see. I need to compute the sum of this series: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, so it's alternating signs, starting with positive, then negative, positive, and so on. Let me try to figure out a pattern here.
First, maybe I can group the terms in pairs. Let's see, starting from the beginning:
1 -... | true | true | stop | 750 |
intermediate_algebra__687b74 | 5 | 1 | intermediate_algebra | -50 | .
Okay, so I need to compute the sum of the series 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let's see. The pattern here is alternating between adding and subtracting consecutive integers. The first term is positive 1, then subtract 2, add 3, subtract 4, and so on until we reach 99 - 100.
First, I should figure out h... | true | true | stop | 1,387 |
intermediate_algebra__687b74 | 6 | 1 | intermediate_algebra | -50 | .
Okay, let's see. I need to compute the sum $1 - 2 + 3 - 4 + 5 - \dots + 99 - 100$. Hmm. So it's alternating addition and subtraction of consecutive numbers starting from 1. Let me try to figure out a pattern here.
First, maybe I should write out the first few terms to see if there's a pattern. Let's see:
1 - 2 = -1... | true | true | stop | 1,928 |
intermediate_algebra__687b74 | 7 | 1 | intermediate_algebra | -50 | .
Okay, so I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let's see. This looks like an alternating series where each odd number is added and each even number is subtracted. Let me try to break it down step by step.
First, maybe I can group the terms in pairs. Like, (1 - 2) + (3 - 4) + (5 - 6) + .... | true | true | stop | 959 |
intermediate_algebra__687b74 | 8 | 1 | intermediate_algebra | -50 | .
Okay, so I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let's see. The pattern here is alternating between adding an odd number and subtracting an even number. Let me write out the first few terms to see if there's a pattern.
Starting with 1, then subtract 2, add 3, subtract 4, add 5, subtract 6... | true | true | stop | 1,029 |
intermediate_algebra__687b74 | 9 | 1 | intermediate_algebra | -50 | .
Okay, so I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let's see. This looks like an alternating series where each pair of numbers is subtracted. Let me break it down.
First, maybe I can group the terms into pairs. Each pair would be (1 - 2), then (3 - 4), and so on, up until the last pair whic... | true | true | stop | 2,069 |
intermediate_algebra__687b74 | 10 | 1 | intermediate_algebra | -50 | .
Okay, let's see. I need to compute this alternating sum: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, so it's adding and subtracting consecutive integers starting from 1, alternating each time. Let me think about how to approach this.
First, maybe I can group the terms in pairs. Like, (1 - 2) + (3 - 4) + (5 - 6) + ... a... | true | true | stop | 1,024 |
intermediate_algebra__687b74 | 11 | 1 | intermediate_algebra | -50 | .
Okay, let's see. I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, this looks like an alternating series where each pair of numbers is subtracted and added. Let me try to break it down.
First, maybe I can group the terms in pairs. Like (1 - 2) + (3 - 4) + (5 - 6) + ... and so on. Each of these pair... | true | true | stop | 1,199 |
intermediate_algebra__687b74 | 12 | 1 | intermediate_algebra | -50 | .
Okay, let's see. I need to compute the sum of the series: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, alternating addition and subtraction. So each pair is like (1 - 2), then (3 - 4), and so on. Let me check how many terms there are. The series goes from 1 to 100, so that's 100 terms. Since each pair is two terms, that... | true | true | stop | 504 |
intermediate_algebra__687b74 | 13 | 1 | intermediate_algebra | -50 | .
Okay, let's see. I need to compute the sum: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, alternating signs, right? So it starts with positive 1, then negative 2, positive 3, and so on, ending with negative 100. Let me think about how to approach this.
First, maybe I should group the terms in pairs. Like, (1 - 2) + (3 - ... | true | true | stop | 859 |
intermediate_algebra__687b74 | 14 | 1 | intermediate_algebra | -50 | .
Okay, so I need to compute the sum: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let's see. This looks like an alternating series where the signs alternate between positive and negative. The pattern starts with 1, then subtracts 2, adds 3, subtracts 4, and so on up to 100.
First, I should figure out how many terms the... | true | true | stop | 1,020 |
End of preview. Expand in Data Studio
- Downloads last month
- -