It depends very much on what a, b and c are.
No. In plane geometry, two lines can intersect at at most one point. This means that for n points, the maximum number of intersections is limited by the number of pairs of lines. For lines a,b,c,d, there are six pairs: (a,b), (a,c), (a,d), (b,c), (b,d), (c,d). So four lines can have at most six intersections.
Three noncollinear points A, B, and C determine exactly three lines. Each pair of points can be connected to form a line: line AB between points A and B, line AC between points A and C, and line BC between points B and C. Thus, the total number of lines determined by points A, B, and C is three.
y = mx + c y = mx + b We see that the lines have the same slope m. If the lines have slopes with the same sign, then the lines are parallel. Soppose that the slope is positive, m > 0. Case 1: If c > 0, b> 0, and c > b, or c < 0, b < 0, and c < b, then we have to work in the same way to find the distance between lines, which is also the same distance. Let work for the first possibility where c > 0, b > 0, and c > b. If we draw the lines, we see that from the interception of the lines with x and y-coordinate axis, are formed two similar right triangles on the second quadrant. The bigger triangle vertices are (0, c), (0, 0), and (-c, 0) The midpoint of the hypotenuse is (-c/2, c/2). The distance between the origin and the midpoint is (c√2)/2. √[(-c/2)2 + (c/2)2] = √(c2/4 + c2/4) = √ (2c2/4) = (c√2)/2 The smaller triangle vertices are (0, b), (0, 0), and (-b, 0) The midpoint of the hypotenuse is (-b/2, b/2). The distance between the origin and the midpoint is (b√2)/2. √[(-b/2)2 + (b/2)2] = √(b2/4 + b2/4) = √ (2b2/4) = (b√2)/2 The distance between the lines is (√2)/2)(c - b). (c√2)/2 - (b√2)/2 = (√2)/2)(c - b) Case 2: If c > 0, b< 0 or c < 0, b > 0. If we draw the lines, we see that from the interception of the lines with x and y-coordinate axis are formed two similar right triangles on the second and fourth quadrant. In this case the distance between lines is (√2)/2)(c + b). If the slope is negative for both lines, then we find the same result as above, but the formed right triangles are in the first and third quadrant. If the lines have slopes with different sign, then the lines intersect.
4*3/2 = 6 lines.
If A ~ B and B ~ C then A ~ C. The above statement is true is you substitute "is parallel to" for ~ or if you substitute "is congruent to" for ~.
A b c b d b a b c b a b c b a b c b a b c b a b c b a b c b
b & c
# 1
No. In plane geometry, two lines can intersect at at most one point. This means that for n points, the maximum number of intersections is limited by the number of pairs of lines. For lines a,b,c,d, there are six pairs: (a,b), (a,c), (a,d), (b,c), (b,d), (c,d). So four lines can have at most six intersections.
5 its 4
C
y = mx + c y = mx + b We see that the lines have the same slope m. If the lines have slopes with the same sign, then the lines are parallel. Soppose that the slope is positive, m > 0. Case 1: If c > 0, b> 0, and c > b, or c < 0, b < 0, and c < b, then we have to work in the same way to find the distance between lines, which is also the same distance. Let work for the first possibility where c > 0, b > 0, and c > b. If we draw the lines, we see that from the interception of the lines with x and y-coordinate axis, are formed two similar right triangles on the second quadrant. The bigger triangle vertices are (0, c), (0, 0), and (-c, 0) The midpoint of the hypotenuse is (-c/2, c/2). The distance between the origin and the midpoint is (c√2)/2. √[(-c/2)2 + (c/2)2] = √(c2/4 + c2/4) = √ (2c2/4) = (c√2)/2 The smaller triangle vertices are (0, b), (0, 0), and (-b, 0) The midpoint of the hypotenuse is (-b/2, b/2). The distance between the origin and the midpoint is (b√2)/2. √[(-b/2)2 + (b/2)2] = √(b2/4 + b2/4) = √ (2b2/4) = (b√2)/2 The distance between the lines is (√2)/2)(c - b). (c√2)/2 - (b√2)/2 = (√2)/2)(c - b) Case 2: If c > 0, b< 0 or c < 0, b > 0. If we draw the lines, we see that from the interception of the lines with x and y-coordinate axis are formed two similar right triangles on the second and fourth quadrant. In this case the distance between lines is (√2)/2)(c + b). If the slope is negative for both lines, then we find the same result as above, but the formed right triangles are in the first and third quadrant. If the lines have slopes with different sign, then the lines intersect.
10 Nodes A B C D E A-B B-C C-D D-E A-C A-D E-B E-C E-A B-D
The RER or reseau express regional has5 lines, A, B, C, D and E. They plan to expand E westward by 2020.
a...................b . . . . . c...................d a to c = vertical line a to b = horizontal line a to d = diagonal line
C = B * log2(1 + SNR) C= channel capacity B= Bandwidth , telephone lines have a usable range of around 3400Hz = =
4*3/2 = 6 lines.