a circle 9 cm from point b
I was co fused by this but you just do a diagram and write this
The locus point is the perpendicular bisector of AB. The locus point is the perpendicular bisector of AB.
That's a circle, centered at 'a', with a radius of 2 cm.
The locus of all points such that the sum of the distances from the point to two fixed points is a constant (in this case, 6 cm) is an ellipse. The two fixed points are called the foci of the ellipse. The total distance of 6 cm is the major axis length of the ellipse, indicating that the foci are separated by a distance less than 6 cm, ensuring that the ellipse is defined.
Axiom from 1952: "A point has no magnitude". This implies an infinite number of points in any distance (or area or volume).
A concave lens will appear!
The locus point is the perpendicular bisector of AB. The locus point is the perpendicular bisector of AB.
The locus of points (or collection of all points) that are 10 centimeters from a given point would be a circle (of radius 10 cm) in two dimensions, and a sphere (of radius 10 cm) in three dimensions.
That's a circle, centered at 'a', with a radius of 2 cm.
A pair of parallel lines at a distance of 1 cm from the line Q.
5-3/x2-0x
It is the set of all points that are at a distance of 1 cm from a fixed point (called the centre).
Axiom from 1952: "A point has no magnitude". This implies an infinite number of points in any distance (or area or volume).
A circle with a radius of 2 units has the following properties: Diameter: 2 × 2 = 4 2×2=4 units Circumference: 2 𝜋 × 2 = 4 𝜋 ≈ 12.57 2π×2=4π≈12.57 units Area: 𝜋 × 2 2 = 4 𝜋 ≈ 12.57 π×2 2 =4π≈12.57 square units
A concave lens will appear!
It's 3. I got it right on the quiz.
A sphere with radius 4/pi = 0.6366 cm will fit the bill - after a fashion. Four points, equidistant on the equator, along with a point at each pole. Of the 15 possible pairs, all will be 1 cm apart along the great circle apart from the 3 pairs of diametrically opposite points, which will be 2 cm apart. There is no way in which 6 points can be marked on any sphere so that each of the 15 pairs is the same distance apart.
There are an infinity of points 4cm from a given line. These points form 2 lines parallel to and either side of the original line. Equally, there are an infinite number of points 4cm from a given point on the original line. These points lie on the circumference of a circle radius 4cm with its centre at the given point. There are only 2 points that fulfil both conditions. These points are found on the circumference of the circle where a diameter perpendicular to the original line and passing through the given point meets the circumference of the circle. These two points are also where the two parallel lines form tangents with the circle.