2007-12-27 06:48:49 · 3 answers · asked by Alexander 6 in Science & Mathematics ➔ Mathematics
they would jump one square up and another square diagonal the fact that theyre in the middle doesnt make a differance as it would start and also end in the middle so it would be the length of a square plus the the square root of the lenght of the sum of two sides squared (pythagoras)
2007-12-27 06:56:40 · answer #1 · answered by jamie 2 · 0⤊ 0⤋
are you sure you didnt forget some auxialary contraints ?
the answer as the questions stands now is infinitely many
2007-12-27 15:12:13 · answer #2 · answered by gjmb1960 7 · 1⤊ 0⤋
since the knight jumps in an L shaped pattern, assume he goes 2 spaces up and one space right. when he starts, he will be at y = 0 (the center of the 1st square). the center of the 2nd square is y = 8, the center of the 3rd is y = 16. then when you go over one it becomes x = 8 (from the center of square 3 to the center of the square one over)
so he started at (0,0) and finished at (8, 16)
now u can use the distance formula to find the distance between those 2 spots
sqrt [ (16-0)^2 + (8-0)^2) = 17.9 <-- distance the knight jumps
2007-12-27 14:58:42 · answer #3 · answered by shadowsjc 2 · 0⤊ 0⤋