Since there is no requirement for the line to be straight, the answer is infinitely many.
Otherwise, 4.
7
When two circles intersect, they can create a maximum of 2 intersection points. Each straight line can intersect with each of the two circles at a maximum of 2 points, contributing 10 points from the lines and circles. Additionally, the five straight lines can intersect each other, yielding a maximum of ( \binom{5}{2} = 10 ) intersection points. Therefore, the total maximum points of intersection are ( 2 + 10 + 10 = 22 ).
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.]
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
When two circles intersect, they can create a maximum of 2 intersection points. Each straight line can intersect with each of the two circles at a maximum of 2 points, contributing 10 points from the lines and circles. Additionally, the five straight lines can intersect each other, yielding a maximum of ( \binom{5}{2} = 10 ) intersection points. Therefore, the total maximum points of intersection are ( 2 + 10 + 10 = 22 ).
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'.
The maximum possible IQ score that a person can achieve is 200.
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.]