Since there is no requirement for the line to be straight, the answer is infinitely many.
Otherwise, 4.
7
21
12
Maximum 12 intersections are there! You can simply use the formula: No. of intersections = n(n-1)/2 [where 'n' is the number of lines] This is derived as each new line can intersect (at most) all the previously drawn lines. There if there is: 1 line => 0 intersections 2 lines => 1 intersection 3 lines => 3 intersections [1 that was there + 2 by the new line can intersect both the previous lines.] 4 lines => 6 intersections [3 that were already there + 3 because the new line can intersect all the 3 lines that were present previously.]
If x is the unknown or variable in an equation it can have many possible maximum or minimum values
4
No two circles can intersect more than twice. Each circle can intersect with each other circle. Thus there ought to be 2 × 30 × (30 - 1) intersections. However, this counts each intersection twice: once for each circle. Thus the answer is half this, giving: maximum_number_of_intersections = ½ × 2 × 30 × (30 - 1) = 30 × 29 = 870.
7
There can be a maximum of 15 intersections. With two non-parallel lines, there will be one intersection, a third (non-parallel) line can be drawn to cut the other two, and that makes 2 more intersections for a total of 3. You can actually draw this out, and with a fourth, fifth and sixth line, you will create a maximum of 3, 4 and 5 more intersections (respectively), and this will bring your total to fifteentotal intersections for the six lines. You can get each successive line to cut all of the other existing lines if you draw them in a prejudicial (maximized) way.
Six.
discuss the possible number of points of interscetion of two distinct circle
Maximum possible vacuum is -0.1 Mpa
21
12
Greatest possible number of intersections for following lines:1(line(s)) = 0 (intersections)2= 13= 1+24= 1+2+35= 1+2+3+4....8=1+2+3+4+5+6+7= 28Every time you add a line, you are able to create 1 more maximum intersection than the previous state. You can add up until 'number of lines - (minus) 1'.
Maximum 12 intersections are there! You can simply use the formula: No. of intersections = n(n-1)/2 [where 'n' is the number of lines] This is derived as each new line can intersect (at most) all the previously drawn lines. There if there is: 1 line => 0 intersections 2 lines => 1 intersection 3 lines => 3 intersections [1 that was there + 2 by the new line can intersect both the previous lines.] 4 lines => 6 intersections [3 that were already there + 3 because the new line can intersect all the 3 lines that were present previously.]
If x is the unknown or variable in an equation it can have many possible maximum or minimum values