Ahare and a tortoise compete in a race over a course 1.50 km long. the tortoise crawls straight and steadily at its maximum speed of 0.290 m/s toward the finish line. the hare runs at its maximum speed of 7.55 m/s toward the goal for 1.200 km and then stops to tease the tortoise. how close to the goal can the hare let the tortoise approach before resuming the race, which the tortoise wins in a photo finish? assume that, when moving, both animals move steadily at their respective maximum speeds.

The answer is Th=39s and X= 11m

Step-by-step explanation:

The necessary time for the hare to complete the 0.290 km

Th=xf-xi/vh = 1500m-1200m÷7.55m/s=39s

then the tortoise must be  x distance from finish line so that the hare continues the race

x=vt

x= 0.290m/s(39s)=11,3m

18

step-by-step explanation:

draw the shape and you will see that f and k are actually at adjacent vertices (points of the hexagon that are neighboring).   so find the distance between (1,2) and (4,2).   that is easy since (1,2) and (4,2) is just a vertical distance to count, which is 3.   it is a regular polygon which means all sides are congruent.   you have 6 sides (hexa) so multiply 6 by 3 to receive 18.

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743