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\boxed{(3,\frac{\pi}{2})} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{(3,\ \dfrac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates corresponds to (3, π/2) in polar coordinates.
**Step-by-step reasoning:**
1. **Understand the conversion formulas:**
- The formulas to convert from rectangular coordinates (x, y) to polar coordinates (r, θ) are:
- \( r = \sqrt{x^2 + y^2} \)
- \( \theta... |
The point (0,3) in rectangular coordinates has a distance r from the origin calculated as r = √(0² + 3²) = √9 = 3.
The angle θ is found using the arctangent function, but since the point is on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right... |
\boxed{(3,\dfrac{\pi}{2})} |
\[ \boxed{(3, \frac{\pi}{2})} \] |
\[ \boxed{(3, \frac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates is located on the positive y-axis. In polar coordinates, this corresponds to a radius r of 3, since the distance from the origin to (0,3) is √(0² + 3²) = 3. The angle θ is π/2 radians, as it lies on the positive y-axis. Therefore, the polar coordinates are (3, π/2).
\boxed... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3, \frac{\pi}{2})} \] |
\boxed{(3,\frac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{(3,\ \dfrac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates corresponds to (3, π/2) in polar coordinates.
\boxed{\left(3,\dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = 3, and the angle θ is π/2 radians, which is 90 degrees. Therefore, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{(3,\dfrac{\pi}{2})} |
\[ \boxed{(3, \frac{\pi}{2})} \] |
\[ \boxed{(3, \frac{\pi}{2})} \]
The point (0, 3) is on the positive y-axis. The distance from the origin is 3, so \( r = 3 \). The angle from the positive x-axis is \( \frac{\pi}{2} \) radians, which is 90 degrees. This satisfies the condition \( 0 \le \theta < 2\pi \). |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = 3. The angle θ is π/2 radians, as it is measured from the positive x-axis to the positive y-axis. Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{(3,\frac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \]
This is the polar coordinate representation of the point (0,3) in rectangular coordinates. |
\boxed{(3,\frac{\pi}{2})}
The point (0,3) in rectangular coordinates corresponds to the polar coordinates (3, π/2), since the distance from the origin is 3 and the angle is π/2 radians (90 degrees) from the positive x-axis. This satisfies r > 0 and 0 ≤ θ < 2π.
\boxed{(3,\frac{\pi}{2})}
\boxed{(3,\frac{\pi}{2})}... |
\[ \boxed{(3,\ \frac{\pi}{2})} \]
The point (0,3) is on the positive y-axis. The distance from the origin is 3, and the angle is π/2 radians. This satisfies r > 0 and 0 ≤ θ < 2π. |
\boxed{(3,\ \dfrac{\pi}{2})}
The point (0,3) in rectangular coordinates is on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is π/2 radians, as it is 90 degrees from the positive x-axis. Thus, the polar coordinates are (3, π/2). |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \]
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is 3, and the angle from the positive x-axis is π/2 radians. Thus, the polar coordinates are (3, π/2). |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates corresponds to (3, π/2) in polar coordinates.
**Step-by-step reasoning:**
1. Identify the rectangular coordinates: x = 0, y = 3.
2. Calculate the radius r using the formula r = √(x² + y²):
r = √(0² + 3²) = √9 = 3.
3. Calculate the angle θ using the formula θ = t... |
The point (0,3) in rectangular coordinates has x = 0 and y = 3.
To find r, use the formula r = √(x² + y²) = √(0² + 3²) = √9 = 3.
To find θ, use the formula θ = tan⁻¹(y/x). Since x = 0 and y = 3, the point is on the positive y-axis.
The angle θ is π/2 radians, which is equivalent to 90 degrees, and it lies with... |
\boxed{(3,\dfrac{\pi}{2})}
The point (0,3) in rectangular coordinates has a distance from the origin of 3 and is located on the positive y-axis. In polar coordinates, this corresponds to r = 3 and θ = π/2 radians. The angle θ is measured from the positive x-axis, and since the point is on the y-axis, θ is π/2, whi... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3, \frac{\pi}{2})} \]
The point (0, 3) is on the positive y-axis. The distance from the origin is 3, so r = 3. The angle θ is π/2 radians, as it is 90 degrees from the positive x-axis.
\[ \boxed{(3, \frac{\pi}{2})} \]
**Reasoning:**
- Calculate r using the formula r = √(x² + y²).
r = √(0² + 3²... |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The distance r is calculated as √(0² + 3²) = √9 = 3.
- The angle θ is found using the arctangent function. Since x = 0 and y = 3, the point is on the positive y-axis, so θ = π/2.
- Thus, the polar coordinates are (3, π/2... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is π/2 radians, as it lies on the positive y-axis. Therefore, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance r is calculated as r = √(0² + 3²) = √9 = 3.
The angle θ is determined from the rectangular coordinates. Since x = 0 and y = 3, the point lies on the positive y-axis. Therefore, θ = π/2.
The polar coordinates a... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \]
The point (0,3) is on the positive y-axis. The distance from the origin is 3, and the angle is π/2 radians. This satisfies r > 0 and 0 ≤ θ < 2π. |
\[ \boxed{(3, \frac{\pi}{2})} \]
The point (0,3) in rectangular coordinates is at (r, θ) = (3, π/2) in polar coordinates. Here, r = 3 is the distance from the origin, and θ = π/2 is the angle from the positive x-axis. |
\boxed{(3,\frac{\pi}{2})}
The point (0,3) in rectangular coordinates corresponds to the polar coordinates (3, π/2), since the distance from the origin is 3 and the angle is π/2 radians (90 degrees) along the positive y-axis. This satisfies r > 0 and 0 ≤ θ < 2π. |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{(3,\ \dfrac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The polar coordinates are (3, π/2).
**Step-by-Step Reasoning:**
- Rectangular coordinates: (x, y) = (0, 3)
- Radius r: r = √(x² + y²) = √(0² + 3²) = √9 = 3
- Angle θ: Since x = 0 and y > 0, the point is on the positive y-axis. Thus, θ = π/2 (90 degrees)
- Polar coordinates: (r, θ) = (3, π/2)
\boxed{\left... |
The point (0,3) in rectangular coordinates has x = 0 and y = 3.
To find r, use the formula r = √(x² + y²) = √(0² + 3²) = √9 = 3.
To find θ, use the formula θ = tan⁻¹(y/x). Since x = 0 and y = 3, the point is on the positive y-axis.
The angle θ is π/2 radians, which is equivalent to 90 degrees, and it lies with... |
\boxed{(3,\dfrac{\pi}{2})}
The point (0,3) in rectangular coordinates has a distance from the origin of 3 and is located on the positive y-axis. In polar coordinates, this corresponds to r = 3 and θ = π/2 radians. The angle θ is measured from the positive x-axis, and since the point is on the positive y-axis, θ = ... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3, \frac{\pi}{2})} \]
The point (0, 3) is on the positive y-axis. The distance from the origin is 3, so r = 3. The angle θ is π/2 radians, as it is 90 degrees from the positive x-axis. Thus, the polar coordinates are (3, π/2). |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radius r is the distance from the origin: r = √(0² + 3²) = √9 = 3.
- The angle θ is found using the arctangent function. Since x = 0 and y = 3, the point is on the positive y-axis. The angle is θ = arctan(y/x), but sin... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{(3,\frac{\pi}{2})}
The point (0,3) in rectangular coordinates has x = 0 and y = 3. The distance from the origin is r = √(0² + 3²) = √9 = 3. The angle θ is found using the arctangent function, but since x = 0 and y > 0, the point lies on the positive y-axis, so θ = π/2. Therefore, the polar coordinates are (... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \]
The point (0,3) is on the positive y-axis. The distance from the origin is 3, and the angle from the positive x-axis is π/2 radians. This satisfies the conditions r > 0 and 0 ≤ θ < 2π. |
\[ \boxed{(3, \frac{\pi}{2})} \]
The point (0,3) in rectangular coordinates is at (r, θ) = (3, π/2) in polar coordinates. Here, r = 3 is the distance from the origin, and θ = π/2 is the angle from the positive x-axis. |
\boxed{(3,\frac{\pi}{2})} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{(3,\ \dfrac{\pi}{2})} |
\boxed{(3, \frac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
The angle θ is found using the arctangent function, but since the x-coordinate is 0 and y is positive, the point lies on the positive y-axis. Thus, θ = π/2 radians.
Therefore, the polar coordinates are (3, π/2).... |
The point (0,3) in rectangular coordinates has x = 0 and y = 3.
To find r, use the formula r = √(x² + y²) = √(0² + 3²) = √9 = 3.
Since x = 0 and y > 0, the point is on the positive y-axis.
The angle θ is π/2 radians, which is between 0 and 2π.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \df... |
\boxed{(3,\dfrac{\pi}{2})} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3, \frac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The distance r is calculated as r = √(0² + 3²) = √9 = 3.
- The angle θ is determined from the coordinates. Since x = 0 and y = 3, the point lies on the positive y-axis. Therefore, θ = π/2 radians (or 90 degrees).
- Both ... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is π/2 radians, as it lies on the positive y-axis. Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance r is calculated as r = √(0² + 3²) = √9 = 3.
The angle θ is determined using the arctangent function. Since x = 0 and y = 3, the point is on the positive y-axis, so θ = π/2.
Thus, the polar coordinates are (3, ... |
The point (0,3) in rectangular coordinates has x = 0 and y = 3.
The radius r is calculated as:
r = √(0² + 3²) = √9 = 3.
The angle θ is found using the arctangent function. Since x = 0 and y = 3, the point is on the positive y-axis. Therefore, θ = π/2 radians.
The polar coordinates are (3, π/2).
\boxed{\left(3,... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{(3,\frac{\pi}{2})} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{(3,\ \dfrac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates corresponds to the polar coordinates (3, π/2).
**Step-by-step reasoning:**
1. **Calculate the radius \(r\):**
The formula for \(r\) is \(r = \sqrt{x^2 + y^2}\).
Substituting \(x = 0\) and \(y = 3\):
\(r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
2. **Determ... |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{(3,\dfrac{\pi}{2})} |
\[ \boxed{(3, \frac{\pi}{2})} \] |
\[ \boxed{(3, \frac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates is on the positive y-axis. The distance from the origin is r = √(0² + 3²) = √9 = 3. Since the point is on the positive y-axis, the angle θ is π/2 radians. Therefore, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. Since the point is on the positive y-axis, the angle θ is π/2 radians. Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = √9 = 3. Since the point is on the positive y-axis, the angle θ is π/2 radians, which is equivalent to 90 degrees and falls within the range [0, 2π).
Thus, the polar coordinates are (3, π/2)... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3, \frac{\pi}{2})} \]
The point (0,3) is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is π/2 radians, as it is 90 degrees from the positive x-axis. Thus, the polar coordinates are (3, π/2). |
\boxed{(3,\dfrac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{(3,\ \dfrac{\pi}{2})} |
The point (0,3) in rectangular coordinates has a distance from the origin given by \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
Since the point lies on the positive y-axis, the angle \( \theta \) is \( \frac{\pi}{2} \) radians.
Thus, the polar coordinates are \( (3, \frac{\pi}{2}) \).
\boxed{\left(3,\ \dfrac{\pi}... |
The point (0, 3) in rectangular coordinates is on the positive y-axis.
- The distance from the origin, \( r \), is \( \sqrt{0^2 + 3^2} = 3 \).
- The angle \( \theta \) is \( \frac{\pi}{2} \), as it lies on the positive y-axis.
Thus, the polar coordinates are \( \left(3, \frac{\pi}{2}\right) \).
\boxed{\left... |
The point (0,3) in rectangular coordinates corresponds to (3, π/2) in polar coordinates.
**Step-by-step reasoning:**
1. The formulas for conversion are:
- \( r = \sqrt{x^2 + y^2} \)
- \( \theta = \tan^{-1}\left(\frac{y}{x}\right) \), with adjustments for the quadrant.
2. For (0,3):
- \( x = 0 \)... |
The point (0,3) in rectangular coordinates has x = 0 and y = 3.
The distance from the origin is r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, θ = π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates has x = 0 and y = 3.
The distance r is given by r = √(x² + y²) = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\dfrac{\pi}{2}\right)} |
The point (0,3) is located on the positive y-axis. In polar coordinates, this corresponds to a radius \(r\) equal to the distance from the origin, which is \(r = \sqrt{0^2 + 3^2} = 3\). The angle \(\theta\) is \(\frac{\pi}{2}\) radians, as it lies on the positive y-axis and satisfies \(0 \leq \theta < 2\pi\).
\boxe... |
The point (0,3) in rectangular coordinates has a distance r from the origin calculated as r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is on the positive y-axis.
The distance from the origin is r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = √9 = 3. Since the point is on the positive y-axis, the angle θ is π/2 radians, as this is the standard angle for the positive y-axis in the range [0, 2π).
Thus, the polar coordinates are (3... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is located on the positive y-axis.
The distance from the origin is r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates has x = 0 and y = 3.
The distance from the origin is r = √(0² + 3²) = √9 = 3.
Since x = 0 and y = 3, the point lies on the positive y-axis. The angle θ is π/2 radians, which is 90 degrees.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \]
\[ \boxed{(3,\ \dfrac{3\pi}{2})} \]
\[ \boxed{(3,\ \dfrac{5\pi}{2})} \]
\[ \boxed{(3,\ \dfrac{7\pi}{2})} \]
\[ \boxed{(3,\ \dfrac{9\pi}{2})} \]
\[ \boxed{(3,\ \dfrac{11\pi}{2})} \]
\[ \boxed{(3,\ \dfrac{13\pi}{2})} \]
\[ \boxed{(3,\ \dfrac{15\pi}{2})} \]
\[ \boxed{(... |
\boxed{(3,\ \dfrac{\pi}{2})} |
\[ \boxed{(3,\ \pi/2)} \] |
\boxed{(3,\ \pi/2)} |
\[ \boxed{(3,\ \pi/2)} \] |
\[ \boxed{(3,\ \pi/2)} \] |
The polar coordinates are (3, π/2).
**Step-by-Step Reasoning:**
- Rectangular coordinates: (x, y) = (0, 3)
- Radius r: r = √(x² + y²) = √(0² + 3²) = √9 = 3
- Angle θ: Since x = 0 and y > 0, the point is on the positive y-axis. Thus, θ = π/2 (90 degrees) in the standard range [0, 2π).
- Polar coordinates: (r... |
The point (0,3) in rectangular coordinates has a distance r from the origin calculated as r = √(0² + 3²) = √9 = 3.
Since the point lies on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{(3,\dfrac{\pi}{2})} |
\[ \boxed{(3, \frac{\pi}{2})} \] |
\[ \boxed{(3,\ \pi/2)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- Since \(x = 0\) and \(y = 3\), the point lies on the positive y-axis. The angle \(\theta\) is \(\frac{\pi}{2}\) radi... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. Since the point is on the positive y-axis, the angle θ is π/2 radians. Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{(3,\ \pi/2)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, r, is calculated as:
r = √(0² + 3²) = √9 = 3.
The angle θ is determined using the arctangent function, but since x = 0, the point lies on the positive y-axis. Therefore, θ = π/2.
Thus, th... |
\[ \boxed{(3,\ \pi/2)} \] |
\[ \boxed{(3,\ \pi/2)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The polar coordinates are (3, π/2).
**Step-by-step reasoning:**
- The rectangular coordinates are (0, 3).
- The distance from the origin, r, is calculated as √(0² + 3²) = √9 = 3.
- The angle θ is determined based on the position on the y-axis. Since x = 0 and y > 0, θ = π/2 radians.
- The point (0, 3) is in... |
The point (0,3) in rectangular coordinates has a distance r from the origin calculated as r = √(0² + 3²) = √9 = 3.
The angle θ is located on the positive y-axis, so θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined based on the position. Since the point lies on the positive y-axis, \(\theta = \f... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and since the point lies on the positive y-axis, θ = π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\r... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \(r\) from the origin is calculated as:
\(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is determined based on the position of the point, which lies on the positive y-axis. T... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) convert to polar coordinates (r, θ) as follows:
- r is the distance from the origin: √(0² + 3²) = √9 = 3.
- θ is the angle from the positive x-axis. Since the point is on the positive y-axis, θ = π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{... |
The point (0,3) in rectangular coordinates has a distance r from the origin calculated as r = √(0² + 3²) = √9 = 3.
The angle θ is found using the position on the positive y-axis, which corresponds to θ = π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined based on the position. Since the point lies on the positive y-axis, \(\theta = \f... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2. Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates (0,3) correspond to polar coordinates (3, π/2).
- **r** is the distance from the origin: √(0² + 3²) = √9 = 3.
- **θ** is the angle from the positive x-axis. Since the point is on the positive y-axis, θ = π/2.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The polar coordinates are (3, π/2).
**Step-by-step reasoning:**
- The rectangular coordinates are (0, 3).
- The distance from the origin, r, is calculated as √(0² + 3²) = √9 = 3.
- The angle θ is determined based on the position on the y-axis. Since x = 0 and y > 0, θ = π/2 radians.
- The point (0, 3) is on... |
The point (0,3) in rectangular coordinates has a distance r from the origin calculated as r = √(0² + 3²) = √9 = 3.
The angle θ is found using the standard position, where a point on the positive y-axis corresponds to θ = π/2.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined based on the position of the point. Since the point is on the positive y-axis, \(... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and since the point is on the positive y-axis, θ = π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\r... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, r, is calculated as:
r = √(0² + 3²) = √9 = 3.
The angle θ is determined based on the position of the point on the positive y-axis, which corresponds to θ = π/2 radians.
Thus, the polar co... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The polar coordinates are (3, π/2).
**Step-by-step reasoning:**
- The rectangular coordinates are (0, 3).
- The distance from the origin, r, is calculated as √(0² + 3²) = √9 = 3.
- The angle θ is determined based on the position on the y-axis. Since x = 0 and y > 0, θ = π/2 radians.
- The point (0, 3) is on... |
The point (0,3) in rectangular coordinates has a distance r from the origin given by r = √(0² + 3²) = √9 = 3.
The angle θ is measured from the positive x-axis. Since the point is on the positive y-axis, θ = π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined based on the position of the point. Since the point is on the positive y-axis, \(... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2. Therefore, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \(r\) from the origin is calculated as:
\(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is determined by the position on the y-axis. Since the point is on the positive y-axis... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The polar coordinates are (3, π/2).
**Step-by-step reasoning:**
- The rectangular coordinates are (0, 3).
- The distance from the origin, \(r\), is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined by considering the position on the positive y-axis.... |
The point (0,3) in rectangular coordinates has a distance r from the origin calculated as r = √(0² + 3²) = √9 = 3.
The angle θ is found by considering the position on the positive y-axis, which corresponds to θ = π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates has a distance r from the origin calculated as r = √(0² + 3²) = √9 = 3.
The angle θ is determined to be π/2 radians, as the point lies on the positive y-axis.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined by the position on the positive y-axis, which corresponds to \(\theta = \frac{\pi... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2. Therefore, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \(r\) from the origin is calculated as:
\(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is the direction from the positive x-axis. Since the point is on the positive y-axis, ... |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The polar coordinates are (3, π/2).
**Step-by-step reasoning:**
- The rectangular coordinates are (0, 3).
- The distance from the origin, r, is calculated as √(0² + 3²) = √9 = 3.
- The angle θ is determined by the direction from the origin. Since the point is on the positive y-axis, θ = π/2 radians.
- The r... |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point lies on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the direction from the positive ... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, r, is calculated as:
r = √(0² + 3²) = √9 = 3.
The angle θ is determined based on the position of the point. Since the point lies on the positive y-axis, θ is π/2 radians.
Thus, the polar ... |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(x² + y²) = √(0² + 3²) = √9 = 3
- θ = π/2 radians, since the point lies on the positive y-axis.
The point (0,3) is on the positive y-axis, so θ = π... |
The point (0,3) in rectangular coordinates has a distance r from the origin given by r = √(0² + 3²) = √9 = 3.
The angle θ is measured from the positive x-axis. Since the point is on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{(3,\ \dfrac{\pi}{2})} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined based on the position. Since the point lies on the positive y-axis, \(\theta = \f... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates (0, 3) correspond to polar coordinates (3, π/2).
- The radius r is the distance from the origin, given by r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- The angle θ is found using the position. Since the point is on the positive y-axis, θ = π/2 radians, which is equivalent to 90 degrees and sat... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) = (3, π/2).
- The distance r is calculated as r = √(0² + 3²) = √9 = 3.
- The angle θ is determined since the point lies on the positive y-axis, so θ = π/2 radians.
This satisfies the conditions r > 0 and 0 ≤ θ < 2π. |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point lies on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the direction from the positive ... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates (0, 3) correspond to polar coordinates (3, π/2).
- The radius \( r \) is the distance from the origin: \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
- The angle \( \theta \) is found using the position on the coordinate plane. Since the point is on the positive y-axis, \( \theta = \pi/2 \) ... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) = (3, π/2).
- The distance r is calculated as r = √(0² + 3²) = √9 = 3.
- The angle θ is determined since the point lies on the positive y-axis, so θ = π/2 radians.
This satisfies the conditions r > 0 and 0 ≤ θ < 2π. |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = 3.
Since the point lies on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{(3,\ \dfrac{\pi}{2})} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined based on the position. Since the point lies on the positive y-axis, \(\theta = \f... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
Given x = 0 and y = 3:
- r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- θ = π/2 radians, since the point lies on the positive y-axis.
The point (0,3) is on ... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The radial distance \(r\) is given by \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is determined by the direction from the origin. Since the point is on the positive y-axis, \(\theta = \frac{\pi}{2}\) radians.
Thus, the polar coordinate... |
The point (0,3) in rectangular coordinates has a distance r from the origin given by r = √(0² + 3²) = √9 = 3.
The angle θ is measured from the positive x-axis. Since the point is on the positive y-axis, θ = π/2 radians.
The polar coordinates are (3, π/2).
\boxed{(3,\ \dfrac{\pi}{2})} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the direction from the positive ... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{(3,\ \dfrac{\pi}{2})} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates are (0, 3).
The distance from the origin, r, is calculated as √(0² + 3²) = √9 = 3.
The angle θ is determined to be π/2 radians, as the point lies on the positive y-axis.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point lies on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = 3.
The angle θ is determined to be π/2 radians, as the point lies on the positive y-axis.
Thus, the polar coordinates are (3, π/2).
\boxed{(3,\ \dfrac{\pi}{2})} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined based on the position of the point. Since the point is on the positive y-axis, \(... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \(r\) from the origin is calculated as:
\(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is determined based on the position of the point. Since the point is on the positive y... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point lies on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point $(0,3)$ in rectangular coordinates has $r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
Since the point lies on the positive y-axis, $\theta = \frac{\pi}{2}$.
Thus, the polar coordinates are $(3, \frac{\pi}{2})$.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the direction from the positive ... |
The point \((0, 3)\) in rectangular coordinates corresponds to the polar coordinates \((3, \frac{\pi}{2})\).
- \(r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\)
- \(\theta = \frac{\pi}{2}\) radians, as the point lies on the positive y-axis.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \(r\) from the origin is calculated as:
\(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is determined based on the position of the point. Since the point is on the positive y... |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point is on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
The angle θ is the direction from the positive x-axis. Since the point is on the positive y-axis, θ = π/2.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = 3.
The angle θ is determined to be π/2 radians, as the point lies on the positive y-axis.
Thus, the polar coordinates are (3, π/2).
\boxed{(3,\ \dfrac{\pi}{2})} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the direction from the positive ... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \(r\) from the origin is calculated as:
\(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is determined based on the position of the point. Since the point is on the positive y... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
The angle θ is the direction from the positive x-axis. Since the point is on the positive y-axis, θ = π/2 radians.
The polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = 3.
The angle θ is found to be π/2 radians, as the point lies on the positive y-axis.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the direction from the positive ... |
The point (0,3) in rectangular coordinates is located on the positive y-axis. The distance from the origin is r = √(0² + 3²) = 3. The angle θ is measured from the positive x-axis, and for a point on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \(r\) from the origin is calculated as:
\(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is determined based on the position of the point. Since the point is on the positive y... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(0² + 3²) = √9 = 3
- θ = π/2 radians (since the point is on the positive y-axis)
The point (0,3) is not the same as (0,-3), which would correspond t... |
The point $(0,3)$ in rectangular coordinates has $r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
Since the point lies on the positive y-axis, $\theta = \frac{\pi}{2}$.
Thus, the polar coordinates are $(3, \frac{\pi}{2})$.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the direction from the positive ... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
Given x = 0 and y = 3:
- r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- θ = π/2 radians, since the point lies on the positive y-axis.
The angle θ is defined... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(0² + 3²) = √9 = 3
- θ = π/2 radians (since the point is on the positive y-axis)
The polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2... |
The point $(0,3)$ in rectangular coordinates has $r = \sqrt{0^2 + 3^2} = 3$ and $\theta = \pi/2$ since it lies on the positive y-axis. Thus, the polar coordinates are $(3, \pi/2)$.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-ax... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \).
Thus, the polar coordinates are \( (3, \frac{\pi}{2}) \).
\... |
The point $(0,3)$ in rectangular coordinates has $r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
Since the point lies on the positive y-axis, $\theta = \frac{\pi}{2}$.
Thus, the polar coordinates are $(3, \frac{\pi}{2})$.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, but ... |
The point \((0, 3)\) in rectangular coordinates is converted to polar coordinates as \((r, \theta) = (3, \frac{\pi}{2})\).
- \(r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\), which is positive.
- \(\theta = \tan^{-1}\left(\frac{3}{0}\right)\) is undefined, but since the point is on the positive y-axis, \(\theta = \frac{\pi}... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates are (0,3).
The distance from the origin is r = √(0² + 3²) = √9 = 3.
The angle θ is the angle from the positive x-axis. Since the point is on the positive y-axis, θ = π/2 radians.
This satisfies 0 ≤ θ < 2π and r > 0.
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(x² + y²) = √(0² + 3²) = √9 = 3
- θ = tan⁻¹(y/x) is undefined when x = 0, but since y = 3 > 0, θ = π/2 radians.
The angle θ = π/2 is in the range ... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is the direction from the positive x-axis. Since the point is on the positive y-axis, $\theta = \pi/2$.
T... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(0² + 3²) = √9 = 3
- θ = π/2 radians (since the point is on the positive y-axis)
The angle θ is given in the range [0, 2π), and r is positive.
\box... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is the direction from the positive x-axis. Since the point is on the positive y-axis, $\theta = \pi/2$.
T... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
Given x = 0 and y = 3:
- r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- θ = π/2 radians, since the point lies on the positive y-axis.
The value of θ is with... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \).
Both \( r > 0 \) and \( 0 \leq \theta < 2\pi \) are satisfied.... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is the direction from the positive x-axis. Since the point is on the positive y-axis, $\theta = \pi/2$.
T... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \).
Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is determined using the arctangent functi... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
The rectangular coordinates are (0,3).
The distance from the origin is r = √(0² + 3²) = √9 = 3.
The angle θ is the angle from the positive x-axis. Since the point is on the positive y-axis, θ = π/2 radians.
This satisfies 0 ≤ θ < 2π, and r > 0.
The polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{(3,\ \dfrac{\pi}{2})} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) = (3, π/2).
- The distance r is calculated as r = √(0² + 3²) = √9 = 3.
- The angle θ is determined since the point lies on the positive y-axis, so θ = π/2 radians.
This satisfies the conditions r > 0 and 0 ≤ θ < 2π.
\boxed{\left(3,\ \dfrac{\... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The distance $r$ from the origin is given by $r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is determined by the direction from the positive x-axis. Since the point is on the positive y-axis, $\theta = \pi/2$ radian... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, \(r\), is calculated as:
\[ r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \(\theta\) is determined based on the position of the point. Since the point lies on the posit... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Determine θ: Since the point (0,3) lies on the positive y-axis, θ = π/2 radians.
The point (0,3) is i... |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = 3.
The angle θ is found to be π/2 radians, as the point lies on the positive y-axis.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, \(r\), is calculated as:
\[ r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \(\theta\) is determined based on the position of the point. Since the point lies on the posit... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Determine θ: Since x = 0 and y = 3 (positive), the point lies on the positive y-axis, so θ = π/2 radia... |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = 3.
The angle θ is found to be π/2 radians, as the point lies on the positive y-axis.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, but ... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies \( r > 0 \) and \( 0 \leq \theta < 2\pi... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, r, is given by:
r = √(0² + 3²) = √9 = 3.
The angle θ is determined using the position of the point. Since the point lies on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinat... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies \( r > 0 \) and \( 0 \leq \theta < 2\pi... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, r, is given by:
r = √(0² + 3²) = √9 = 3.
The angle θ is determined using the position of the point. Since the point lies on the positive y-axis, θ = π/2 radians.
Thus, the polar coordinat... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies \( r > 0 \) and \( 0 \leq \theta < 2\pi... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, \(r\), is given by \(r = \sqrt{x^2 + y^2}\). Substituting \(x = 0\) and \(y = 3\):
\[r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3.\]
The angle \(\theta\) is the direction from the positive x-axis. Si... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies \( r > 0 \) and \( 0 \leq \theta < 2\pi... |
The point (0,3) in rectangular coordinates has a distance from the origin of r = √(0² + 3²) = √9 = 3.
Since the point lies on the positive y-axis, the angle θ is π/2 radians.
Thus, the polar coordinates are (3, π/2).
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, \(r\), is given by \(r = \sqrt{x^2 + y^2}\). Substituting \(x = 0\) and \(y = 3\):
\[r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3.\]
The angle \(\theta\) is the direction from the positive x-axis. Si... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(x² + y²) = √(0² + 3²) = √9 = 3
- θ = tan⁻¹(y/x) is undefined when x = 0, but since y = 3 > 0, θ = π/2 radians.
The point (0,3) lies on the positi... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, \(r\), is given by \(r = \sqrt{x^2 + y^2}\). Substituting \(x = 0\) and \(y = 3\):
\[r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3.\]
The angle \(\theta\) is the direction from the positive x-axis. Si... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, \(r\), is calculated as:
\[ r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \(\theta\) is determined based on the position of the point. Since the point (0,3) lies on the... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \frac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \).
Verification: \( x = 3 \cos\left(\frac{\pi}{2}\right) = 3 \cdo... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is the direction from the positive x-axis. Since the point is on the positive y-axis, $\theta = \frac{\pi}{2... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, but ... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Determine θ: Since the point (0,3) is on the positive y-axis, θ = π/2 radians.
Verification:
- x = r... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, r, is calculated as:
r = √(0² + 3²) = √9 = 3.
The angle θ is determined based on the position of the point. Since the point lies on the positive y-axis, θ = π/2 radians.
Verification:
x... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r:
r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Calculate θ:
The point (0,3) lies on the positive y-axis. The angle for the positive y-axis is... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, but ... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Determine θ: Since x = 0 and y = 3 (positive y-axis), θ = π/2.
The point (0,3) is not the origin, so t... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is determined using the arctangent function, considering the quadrant. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) rad... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies \( r > 0 \) and \( 0 \leq \theta < 2\pi... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(0² + 3²) = √9 = 3.
- Determine θ: Since the point is on the positive y-axis, θ = π/2 radians.
Verify:
- x = r cos θ = 3 * cos(π/2) = 3... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Determine θ: Since x = 0 and y = 3 (positive), the point lies on the positive y-axis, so θ = π/2 radian... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using $\theta = \tan^{-1}(y/x)$, but since $x=0$ and $y>0$, $\theta = \pi/2$.
Thus, the polar co... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies the conditions \( r > 0 \) and \( 0 \le... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(0² + 3²) = √9 = 3
- θ = π/2 radians (since the point is on the positive y-axis)
The point (0,3) is not the origin, so the representation is unique ... |
\[ \boxed{(3,\ \dfrac{\pi}{2})} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies \( r > 0 \) and \( 0 \leq \theta < 2\pi... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using $\theta = \tan^{-1}(y/x)$, but since $x=0$ and $y>0$, $\theta = \pi/2$.
Thus, the polar co... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-ax... |
The point \((0, 3)\) in rectangular coordinates is converted to polar coordinates using the formulas \(r = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}(y/x)\), but with adjustments for the quadrant and axes.
Here, \(x = 0\) and \(y = 3\).
- \(r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- \(\theta\) is the angle from ... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r:
r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Calculate θ:
The point (0,3) lies on the positive y-axis. The angle θ is π/2 radians, as this... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using the arctangent function, considering the quadrant. Since $x=0$ and $y=3>0$, the point lies on... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-ax... |
The point \((0, 3)\) in rectangular coordinates is converted to polar coordinates as follows:
- The distance \(r\) from the origin is \(\sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is \(\frac{\pi}{2}\) radians, as the point lies on the positive y-axis.
Thus, the polar coordinates are \(\left(3, \f... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Determine θ: Since the point (0,3) lies on the positive y-axis, θ = π/2 radians.
Verification:
- x ... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The polar coordinate $\theta$ is the angle from the positive x-axis. Since the point is on the positive y-axis, $\theta = \pi/2... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is determined to be \( \theta = \frac{\pi}{2} \) radians, as the point lies on the positiv... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(x² + y²) = √(0² + 3²) = √9 = 3
- θ = tan⁻¹(y/x) is undefined when x = 0, but since y = 3 > 0, θ = π/2 radians.
The point (0,3) lies on the positi... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The polar coordinate $\theta$ is the angle from the positive x-axis. Since the point is on the positive y-axis, $\theta = \pi/2... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle θ is found using the arctangent function, considering the quadrant. Since x = 0 and y = 3, the point lies on th... |
The point \((0, 3)\) in rectangular coordinates is converted to polar coordinates as follows:
- The distance \(r\) from the origin is \(\sqrt{0^2 + 3^2} = 3\).
- The angle \(\theta\) is \(\frac{\pi}{2}\) radians, as the point lies on the positive y-axis.
Verification:
- \(x = r \cos \theta = 3 \cdot \cos\le... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies \( r > 0 \) and \( 0 \leq \theta < 2\pi... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using $\theta = \tan^{-1}(y/x)$, but since $x=0$ and $y>0$, $\theta = \pi/2$.
Thus, the polar co... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-ax... |
The point \((0, 3)\) in rectangular coordinates is converted to polar coordinates using the formulas \(r = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}(y/x)\), but with adjustments for the quadrant and axes.
Here, \(x = 0\) and \(y = 3\).
First, calculate \(r\):
\(r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
Next, ... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \(r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
The angle \(\theta\) is found using the arctangent function, considering the position on the positive y-axis. Since \(x = 0\) and \(y > 0\), \(\theta = \frac{\pi}{2}\) radians.
This satisfies \(r > 0... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The polar coordinate $\theta$ is the angle from the positive x-axis. Since the point is on the positive y-axis, $\theta = \pi/2... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance \( r \) from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-ax... |
The point \((0, 3)\) in rectangular coordinates is converted to polar coordinates using the formulas \(r = \sqrt{x^2 + y^2}\) and \(\theta = \tan^{-1}(y/x)\), but with adjustments for the quadrant and axes.
Here, \(x = 0\) and \(y = 3\).
First, calculate \(r\):
\(r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
Next, ... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r:
r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Calculate θ:
Since the point is on the positive y-axis, θ = π/2 radians.
- Verify:
x = r ... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using $\theta = \tan^{-1}(y/x)$, but since $x=0$ and $y>0$, $\theta = \pi/2$ radians.
Thus, the ... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle θ is found using the arctangent function, considering the position on the positive y-axis. Since x = 0 and y = 3, θ = π/2 radians.
Thus, the polar coordinates are (3, π/2).
Verification... |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates are (0,3).
The distance from the origin is \( r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \).
The angle \( \theta \) is the angle from the positive x-axis. Since the point is on the positive y-axis, \( \theta = \frac{\pi}{2} \) radians.
This satisfies \( r > 0 \) and \( 0 \leq \theta < 2\pi... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using the arctangent function, considering the quadrant. Since $x=0$ and $y=3>0$, the point lies on... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, \(r\), is calculated as:
\[ r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \(\theta\) is determined using the arctangent function, considering the position on the positi... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ), where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Determine θ: Since the point is on the positive y-axis, θ = π/2 radians. This satisfies 0 ≤ θ < 2π and... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is the direction from the positive x-axis. Since the point is on the positive y-axis, $\theta = \frac{\pi}{2... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(0² + 3²) = √9 = 3.
- Determine θ: Since the point is on the positive y-axis, θ = π/2 radians. This is confirmed using the atan2 function,... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is the direction from the positive x-axis. Since the point is on the positive y-axis, $\theta = \frac{\pi}{2... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- r = √(0² + 3²) = √9 = 3
- θ = π/2 radians (since the point is on the positive y-axis)
The point (0,3) is not the origin, so the representation is unique ... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using the arctangent function, considering the quadrant. Since $x=0$ and $y=3>0$, the point lies on... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
- The radial distance \(r\) is calculated as \(r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3\).
- The angle \(\theta\) is determined using the arctangent function, considering the position on the positive y-axis. Si... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r: r = √(0² + 3²) = √9 = 3.
- Determine θ: Since the point is on the positive y-axis, θ = π/2 radians.
Verify:
- x = r cos θ = 3 * cos(π/2) = 3... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using the arctangent function, considering the quadrant. Since $x=0$ and $y=3>0$, the point lies on... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r:
r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Calculate θ:
The point (0,3) lies on the positive y-axis. The angle for the positive y-axis is... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using the arctangent function, considering the quadrant. Since $x=0$ and $y=3>0$, the point lies on... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin, \(r\), is given by \(r = \sqrt{x^2 + y^2}\). Substituting \(x = 0\) and \(y = 3\):
\[r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3.\]
The angle \(\theta\) is found using the arctangent function, consi... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r:
r = √(x² + y²) = √(0² + 3²) = √9 = 3.
- Calculate θ:
The point (0,3) lies on the positive y-axis. The angle for the positive y-axis is... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using the arctangent function, considering the quadrant. Since $x=0$ and $y=3>0$, the point lies on... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r:
r = √(0² + 3²) = √9 = 3
- Calculate θ:
Since the point is on the positive y-axis, θ = π/2 radians.
The point (0,3) satisfies x = r co... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using the arctangent function, considering the quadrant. Since $x=0$ and $y=3>0$, the point lies on... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The rectangular coordinates (0,3) correspond to polar coordinates (r, θ) where r is the distance from the origin and θ is the angle from the positive x-axis.
- Calculate r:
r = √(0² + 3²) = √9 = 3
- Calculate θ:
Since the point is on the positive y-axis, θ = π/2 radians.
The point (0,3) satisfies x = r co... |
The point $(0,3)$ in rectangular coordinates has $x=0$ and $y=3$.
The polar coordinate $r$ is the distance from the origin:
$r = \sqrt{x^2 + y^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.
The angle $\theta$ is found using the arctangent function, considering the quadrant. Since $x=0$ and $y=3>0$, the point lies on... |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left(3,\ \dfrac{\pi}{2}\right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
The point (0,3) in rectangular coordinates is converted to polar coordinates as follows.
The distance from the origin is given by \( r = \sqrt{x^2 + y^2} \). Substituting \( x = 0 \) and \( y = 3 \):
\[ r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3 \]
The angle \( \theta \) is found using the arctangent function, cons... |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\boxed{\left(3,\ \dfrac{\pi}{2}\right)} |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
\[ \boxed{\left( 3,\ \dfrac{\pi}{2} \right)} \] |
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