Question

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.?

Answer 1

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`