# The temperate at noon was 8° F. By 6:00 p.m. the temperate had dropped 10°. what was the temperate at 6:00 p.m.?

Because the temperature had dropped ${10}^{o}$ we need to subtract the ${10}^{o}$ from the starting temperature of ${8}^{o}$ giving
${8}^{o} - {10}^{0} = - {2}^{0}$
At 6:00 PM the temperature was a chilly $- {2}^{o}$