Yes.
figure
Each point can be connected to every other point. That gives you 10 x 9. However, since this counts each line segment twice, this result should be divided by 2.
In the English alphabet, the uppercase letters that have perpendicular line segments are A, B, D, E, F, H, K, L, T, and Z. This totals to 10 uppercase letters. Each of these letters includes at least one pair of line segments that meet at a right angle.
It depends on the orange. The number of segments usually range from 9-11. So if there are 10, then, yes, it would be an even number.
Cut each apple into fifths (creating 90 1/5 segments). Then, give each kid 9 segments.
11
10 collinear points form one set of overlapping line segments, of which there are 45.
figure
Each point can be connected to every other point. That gives you 10 x 9. However, since this counts each line segment twice, this result should be divided by 2.
In the English alphabet, the uppercase letters that have perpendicular line segments are A, B, D, E, F, H, K, L, T, and Z. This totals to 10 uppercase letters. Each of these letters includes at least one pair of line segments that meet at a right angle.
The sum is 6.
You need any number between 10 to 20 distinct end points.
There are 10 segments of 0.3 centimeters each in 3 centimeters. You can divide 3 by 0.3 to get the answer.
You need any number between 10 to 20 distinct end points.
Points:(4, 3) and (10, -5) Midpoint: (4+10)/2, (3-5)/2 = (7, -1)
8:58:58 - i think so.
All crustaceans have 10 legs, 2 body segments, and 1 pair of antennae.