Mark estimates that the distance from his house to to his school is 15.5 mi the actual distance is 14.5 mi to the nearest tenth of a percent what is the percent error in mark's estimate?
anwser only if u know the anwser got !

step-by-step explanation:

there is a formula i can give you:

it is r^2*n*sin(360/n) divided by 2 where r represents the "radius" and n is the number of sides.

plug them in:

(10sqrt(3))^2*6*sin(360/6) divided by 2

100*3*6*sin(60) divided by 2

900*sin(60)

900*sqrt(3)/2

450*sqrt(3)

approx is 779.4 cm^2