Respuesta :
The distance between any two points is:
d^2=(x2-x1)^2+(y2-y1)^2 so in this case we have:
d^2=(5-1)^2+(1-3)^2, d^2=(1-5)^2+(1-1)^2, d^2=(1-3)^2+(1-1)^2
d^2=(16+4), d^2=16, d^2=4
So the total distance or perimeter, p, is:
p=√20+√16+√4
p=√20+4+2
p=6+√20 km (exact)
p≈10.47 km (to nearest hundredth of a kilometer)
d^2=(x2-x1)^2+(y2-y1)^2 so in this case we have:
d^2=(5-1)^2+(1-3)^2, d^2=(1-5)^2+(1-1)^2, d^2=(1-3)^2+(1-1)^2
d^2=(16+4), d^2=16, d^2=4
So the total distance or perimeter, p, is:
p=√20+√16+√4
p=√20+4+2
p=6+√20 km (exact)
p≈10.47 km (to nearest hundredth of a kilometer)