Question

pronto ta ai me ajuda agora pf

Answer 1

A) d²=Δx²+Δy²
$d^{2}=(3-1)^{2} + (7-4)^{2} \\ d ^{2} = 2^{2}+3^{2} \\ d^{2}=4+9 \\ \sqrt{d^{2}}= \sqrt{13} \\ d= \sqrt{13}$

D) $d^2=(0- \sqrt{5})^2+(-2-(-2))^2 \\ d^2=(- \sqrt{5} )^2+(-2+2)^2 \\ d^2=5+0 \\ \sqrt{d^2} = \sqrt{5} \\ d= \sqrt{5}$