A cannonball is launched upwards. Its height is described by the equation h= -5t2 + 40t + 45, where
is measured in yards and t is measured in seconds.
How high is the cannonball at 0 second?

F(x) = x2 − 9, and g(x) = x − 3 f(x) = x2 − 4x + 3, and g(x) = x − 3 f(x) = x2 + 4x − 5, and g(x) = x − 1 f(x) = x2 − 16, and g(x) = x − 4 h(x) = x + 5 arrowright h(x) = x + 3 arrowright h(x) = x + 4 arrowright h(x) = x − 1 arrowright