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 1 185k | 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 | 580 |
precalculus__8ee539 | 1 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) |
Okay, so I need to convert the rectangular coordinates (0, 3) to polar coordinates. Hmm, let me remember how that works. Polar coordinates are (r, θ), right? Where r is the distance from the origin to the point, and θ is the angle from the positive x-axis to the point.
First, let me find r. The formula for r is the... | true | true | stop | 545 |
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's find r. The formula for r is the square root of (x squared plus y square... | true | true | stop | 438 |
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 r is the distance from the origin to the point, and θ is the angle made with the positive x-axis.
Alright, starting with r. The formula for r is the square... | true | true | stop | 533 |
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 (r, θ), right? So, first, I remember that r is the distance from the origin to the point, and θ is the angle made with the positive x-axis.
Starting with r. The formula for r is the squar... | true | true | stop | 515 |
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 the formulas for conversion.
Right, the formulas are r = sqrt(x² + y²) an... | true | true | stop | 521 |
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 t... | true | true | stop | 492 |
precalculus__8ee539 | 7 | 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), right? Where r is the distance from the origin and θ is the angle from the positive x-axis. Let me think about how to do this.
First, the point given is (0, 3). In rectangular coordinates, that's on... | true | true | stop | 733 |
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 from the positive x-axis.
First, let's find r. The formula for r is the square root of (x squared plus ... | true | true | stop | 464 |
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 sq... | true | true | stop | 503 |
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 | 448 |
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) where r is positive and θ is between 0 and 2π. Hmm, let me remember how polar coordinates work.
First, polar coordinates are based on the distance from the origin (r) and the angle (θ) from the positive x-axis. To convert from... | true | true | stop | 606 |
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 wai... | true | true | stop | 647 |
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 r is the distance from the origin to the point, and θ is the angle made with the positive x-axis.
Starting with the point (0, 3). Since the x-co... | true | true | stop | 521 |
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). But wa... | true | true | stop | 410 |
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. Let me remember how polar coordinates work. In polar coordinates, a point is represented as (r, θ), where r is the distance from the origin to the point, and θ is the angle measured from the positive x-axis to the line conne... | true | true | stop | 505 |
precalculus__8ee539 | 16 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) |
Okay, so I need to convert the rectangular coordinates (0, 3) to polar coordinates. Hmm, let's see. Polar coordinates are (r, θ), right? Where r is the distance from the origin, and θ is the angle from the positive x-axis.
First, I should find r. The formula for r is the square root of (x squared plus y squared). So... | true | true | stop | 488 |
precalculus__8ee539 | 17 | 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 usually 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 the formulas for converting between rectangular and polar coordinates... | true | true | stop | 707 |
precalculus__8ee539 | 18 | 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? Where r is the distance from the origin and θ is the angle from the positive x-axis. Let me recall the formulas. I think r is calculated using the Pythagorean theorem, so r = sqrt(x... | true | true | stop | 630 |
precalculus__8ee539 | 19 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) |
Okay, so I need to convert the point (0, 3) from rectangular coordinates to polar coordinates. Let me recall how to do that. Polar coordinates are given as (r, θ), where r is the distance from the origin to the point, and θ is the angle made with the positive x-axis.
First, I need to find r. The formula for r is th... | true | true | stop | 548 |
precalculus__8ee539 | 20 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) |
Okay, so I need to convert the point (0, 3) from rectangular coordinates to polar coordinates. Let me remember how to do this. Polar coordinates are (r, θ), 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 (x² + y²)... | true | true | stop | 457 |
precalculus__8ee539 | 21 | 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 given as (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.
First, the radial component r is calculated using t... | true | true | stop | 704 |
precalculus__8ee539 | 22 | 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 r is the distance from the origin, and θ is the angle from the positive x-axis.
First, let me remember the formulas. To convert from rectangular (x, y) to polar (r, θ), the formul... | true | true | stop | 548 |
precalculus__8ee539 | 23 | 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 the formulas for converting between rectangular and polar coordinates.
The formulas are... | true | true | stop | 471 |
precalculus__8ee539 | 24 | 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) 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, the formula for r is the square root of (x squared plus... | true | true | stop | 674 |
precalculus__8ee539 | 25 | 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, let me remember how polar coordinates work. Polar coordinates are usually given as (r, θ), where r is the distance from the origin to the point, and θ is the angle made with the positive x-axis.
First, I should find ... | true | true | stop | 480 |
precalculus__8ee539 | 26 | 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 made with the positive x-axis.
First, to find r, I think the formula is r = sqrt(x² + y²). Here, x is 0 and y i... | true | true | stop | 532 |
precalculus__8ee539 | 27 | 2 | precalculus | \left( 3, \frac{\pi}{2} \right) |
Okay, so I need to convert the point (0, 3) from rectangular coordinates to polar coordinates. Let me recall how polar coordinates work. In polar coordinates, a point is represented as (r, θ), where r is the distance from the origin to the point, and θ is the angle made with the positive x-axis.
First, let's find r... | true | true | stop | 556 |
precalculus__8ee539 | 28 | 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 that works.
First, the formula for r is the square root of (x squared plus y squared). Since the point is (0, 3), x is 0 and y is 3. So plugging thos... | true | true | stop | 553 |
precalculus__8ee539 | 29 | 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 conversion.
First, I remember that r is calculated using the formula r = √(x² + y²). The x-coordinate here is 0, and the y-coordinate is 3... | true | true | stop | 502 |
precalculus__8ee539 | 30 | 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 usually given as (r, θ), right? Where r is the distance from the origin and θ is the angle made with the positive x-axis. Let me recall the formulas. I think r is calculated using the Pythagorean theorem: r ... | true | true | stop | 545 |
precalculus__8ee539 | 31 | 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, let me remember how polar coordinates work. Polar coordinates are usually given as (r, θ), where r is the distance from the origin to the point, and θ is the angle made with the positive x-axis.
First, let's recall t... | true | true | stop | 652 |
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 right, the distance formula is derived from the Pythagorean theorem. It's sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Let me make sure I have ... | true | true | stop | 558 |
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 | 529 |
algebra__e16674 | 2 | 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. I think it's the square root of [(x2 - x1)^2 + (y2 - y1)^2]. Yeah, that sounds right. Let me write that down.
First, label the points. Let's say (x1, y1) is (2, -6) and (x2, y2) is (-4, 3). S... | true | true | stop | 505 |
algebra__e16674 | 3 | 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 plane. Let me think. Oh right, the distance formula is derived from the Pythagorean theorem. It's sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Let me make sure I got that right. ... | true | true | stop | 710 |
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 sqrt[(x2 - x1)^2 + (y2 - y1)^2].
Alright... | true | true | stop | 609 |
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 | 586 |
algebra__e16674 | 6 | 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. 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 need to plug in the coordinates into that formula.
Let me wri... | true | true | stop | 620 |
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 | 626 |
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[(x2 - x1)^2 + (y2 - y1)^2].
Alright... | true | true | stop | 566 |
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].
Alr... | 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. The formula is the square root of [(x2 - x1) squared plus (y2 - y1) squared]. So first, I should identify the coordinates of the two points. Let me label them as (x1, y1) and (x... | true | true | stop | 621 |
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), the... | true | true | stop | 557 |
algebra__e16674 | 12 | 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? So if you have two points, (x1, y1) and (x2, y2), th... | true | true | stop | 577 |
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. So I should plug in the coordinates from the given points into this formula.
Fi... | true | true | stop | 537 |
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].
Alright, l... | true | true | stop | 571 |
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 sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Yeah, that's right. So let me plug in the values.
First, x1 is 2 and y1 is -6. Then x2 is -4 and y2 is 3.... | true | true | stop | 457 |
algebra__e16674 | 16 | 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 the square root of [(x2 - x1)^2 + (y2 - y1)^2... | true | true | stop | 683 |
algebra__e16674 | 17 | 3 | algebra | 3\sqrt{13} |
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, right, distance between two points in a coordinate plane. I remember there's a formula for that. Let me think... Oh yeah, the distance formula. It's derived from the Pythagorean theorem. The formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2].
... | true | true | stop | 581 |
algebra__e16674 | 18 | 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 is derived from the Pythagorean theorem. So, if I have two points (x1, y1) and (x2, y2), the di... | true | true | stop | 562 |
algebra__e16674 | 19 | 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 the formula for that is the square root of [(x2 - x1) squared plus (y2 - y1) squared]. Let me make sure I remember that correctly. Yeah, the distance formula is derived from the... | true | true | stop | 612 |
algebra__e16674 | 20 | 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 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]. Let me make sure I g... | true | true | stop | 609 |
algebra__e16674 | 21 | 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? So if I have two points, (x1, y1) and (x2, y2), the ... | true | true | stop | 552 |
algebra__e16674 | 22 | 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]. Let me ... | true | true | stop | 597 |
algebra__e16674 | 23 | 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? So, if I have two points (x1, y1) and (x2, y2), the ... | true | true | stop | 559 |
algebra__e16674 | 24 | 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 the formula for that is the distance formula, right? Let me recall. Oh yeah, it's derived from the Pythagorean theorem. The formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Let me ma... | true | true | stop | 613 |
algebra__e16674 | 25 | 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 the square root of [ (x2 - x1)^2 + (y2 -... | true | true | stop | 538 |
algebra__e16674 | 26 | 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 something like the square root of [(x2 - x... | true | true | stop | 591 |
algebra__e16674 | 27 | 3 | algebra | 3\sqrt{13} |
Okay, so I need to find the distance between the points (2, -6) and (-4, 3). Hmm, I remember that the distance formula is derived from the Pythagorean theorem. Let me recall the formula. I think it's something like the square root of [(x2 - x1)^2 + (y2 - y1)^2]. Let me make sure. Yeah, that sounds right. So the formu... | true | true | stop | 527 |
algebra__e16674 | 28 | 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. It's sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Yeah, that sounds righ... | true | true | stop | 515 |
algebra__e16674 | 29 | 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 between two points in a plane is something like the square root of [(x2 - x1) squared plus (y2 - y1) squared]. Let me recall the exact formula.
Yes, the formula is distance = √[(x2 - x1)² + (y2 - y1)²]. So I... | true | true | stop | 608 |
algebra__e16674 | 30 | 3 | algebra | 3\sqrt{13} |
Okay, let's see. I need to find the distance between the points (2, -6) and (-4, 3). Hmm, distance formula, right? The formula is sqrt[(x2 - x1)^2 + (y2 - y1)^2]. Let me make sure I remember that correctly. Yeah, that's the distance between two points in a plane.
So first, let me label the points. Let's call the fir... | true | true | stop | 516 |
algebra__e16674 | 31 | 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 remember there's a formula for that. 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]. Let me ... | true | true | stop | 564 |
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 wo... | true | true | stop | 571 |
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, sub... | true | true | stop | 604 |
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 | 754 |
intermediate_algebra__687b74 | 3 | 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 the signs alternate between positive and negative. 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. E... | true | true | stop | 564 |
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 | 805 |
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, the second negative, and so on, ending with 100 being subtracted.
First, I should figure out how many ... | true | true | stop | 1,442 |
intermediate_algebra__687b74 | 6 | 1 | intermediate_algebra | -50 |
Okay, let's see. I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, so it's alternating between positive and negative numbers, right? The pattern starts with positive 1, then subtracts 2, adds 3, subtracts 4, and so on until 99 minus 100.
First, let me check how many terms there are. The sequence st... | true | true | stop | 681 |
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.
First, maybe I can group the terms in pairs. Let's see: (1 - 2) + (3 - 4) + (5 - 6) + ... + (99 ... | true | true | stop | 783 |
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 | 834 |
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 which... | true | true | stop | 759 |
intermediate_algebra__687b74 | 10 | 1 | intermediate_algebra | -50 |
Okay, let's see. I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, so it's alternating signs, starting with positive, then negative, 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 would be ... | true | true | stop | 524 |
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 pairs... | true | true | stop | 511 |
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, so it's alternating between adding and subtracting consecutive numbers. Let me try to figure out a pattern here.
First, let's break it down. The series starts at 1, subtracts 2, adds 3, subtracts 4, and so on until it ... | true | true | stop | 752 |
intermediate_algebra__687b74 | 13 | 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 is starting with 1, then subtract 2, add 3, subtract 4, and so on, ending with subtracting 100.
First, I should figure out... | true | true | stop | 772 |
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, subtracts 2, adds 3, subtracts 4, and so on until 99 is subtracted and 100 is added? Wait, no, the problem s... | true | true | stop | 772 |
intermediate_algebra__687b74 | 15 | 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, so it's alternating signs, starting with positive. The pattern is +, -, +, -, and so on. Each pair of numbers is subtracted from the previous one. Let me think about how to approach this.
First, maybe I can group the t... | true | true | stop | 780 |
intermediate_algebra__687b74 | 16 | 1 | intermediate_algebra | -50 |
Okay, let's see. I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, alternating signs starting with positive. Let me think about how to approach this.
First, maybe group the terms in pairs. Like (1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100). Each pair would be (-1) because 1 - 2 is -1, 3 - 4 is -1, ... | true | true | stop | 522 |
intermediate_algebra__687b74 | 17 | 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, right? Starting with positive, then negative, positive, and so on. The last term is -100. Let me figure out how many terms there are here. From 1 to 100, that's 100 numbers. Since the signs a... | true | true | stop | 511 |
intermediate_algebra__687b74 | 18 | 1 | intermediate_algebra | -50 |
Okay, let's see. I need to compute the sum: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm. So it's an alternating series where the signs go positive, negative, positive, etc. Let me think about how to approach this.
First, maybe I can group the terms in pairs. Like, pair the first two terms, then the next two, and so on. ... | true | true | stop | 960 |
intermediate_algebra__687b74 | 19 | 1 | intermediate_algebra | -50 |
Okay, so I need to compute this alternating sum: 1 - 2 + 3 - 4 + 5 - ... - 100. Hmm, let me think. First, let's try to understand the pattern here. The signs alternate between positive and negative. The first term is positive, then negative, positive, and so on. The last term is -100, which is the 100th term. Since 1... | true | true | stop | 631 |
intermediate_algebra__687b74 | 20 | 1 | intermediate_algebra | -50 |
Okay, let's see. I need to compute the sum: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, so it's an alternating series where each pair of numbers is subtracted. Let me try to break it down.
First, let's see how many terms there are. The series goes from 1 to 100, so there are 100 terms. Since it's alternating, each pair... | true | true | stop | 606 |
intermediate_algebra__687b74 | 21 | 1 | intermediate_algebra | -50 |
Okay, let's see. I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, so it's an alternating series where the signs switch every term. Let me think about how to approach this.
First, maybe I can group the terms in pairs. Like, (1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100). That might simplify things. ... | true | true | stop | 522 |
intermediate_algebra__687b74 | 22 | 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 first term is positive, the second negative, third positive, and so on. The last term is -100.
First, let me figure out how many t... | true | true | stop | 1,117 |
intermediate_algebra__687b74 | 23 | 1 | intermediate_algebra | -50 |
Okay, let's see. I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, this is an alternating series where the signs alternate between positive and negative. 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... | true | true | stop | 854 |
intermediate_algebra__687b74 | 24 | 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 how many terms there are in this series. The series starts at 1 and goes up to 100, with each term alternating between positive and negative. Since the last term is -100, which is the 100th ... | true | true | stop | 829 |
intermediate_algebra__687b74 | 25 | 1 | intermediate_algebra | -50 |
Okay, so I need to compute this series: 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let me think. It's an alternating series where the signs switch each term. The pattern is positive, negative, positive, and so on. Let me check if I can find a pattern or a formula for such a series.
First, let me write out the first fe... | true | true | stop | 766 |
intermediate_algebra__687b74 | 26 | 1 | intermediate_algebra | -50 |
Okay, so I need to compute the sum 1 - 2 + 3 - 4 + 5 - ... + 99 - 100. Hmm, let me think about how to approach this. The pattern here is alternating between adding and subtracting consecutive integers. Let me see if there's a way to group these terms to make the calculation easier.
First, let's consider how the sequ... | true | true | stop | 867 |
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