A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the
maximum height reached by the rocket, to the nearest tenth of a foot.
y= -16x² + 201x + 132

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°

m∠a is 90°

m∠b is 40°

8+10

step-by-step explanation:

you multiply 4 by 2 and 5 by 2