HELP ASAP ILL GIVE U BRAINIEST
An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t^2 + 64t + 80?

Step-by-step explanation: The max. height of the object is the same as it's partner, the y-coordinate of the vertex. In order to find that, we first have to find the X-Coordinate, which is T. We can find it by...

1. Identifying A,B,and C.

2. Multiply B by -1 or simplify and change its sign. Then divide it by 2A.

After this is all completed, plug in your equation in order to seek Y, your answer.

(AX+BX^2+C=-16T+64T+80)...In other words, A=-6, B=64 and C=80!

-64/2(-16)=-64/-32=2

-16(2)^2+64(2)+80=144 feet.

I hope this helps you!

HELP ASAP ILL GIVE U BRAINIEST An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t^2 + 64t + 80?

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HELP ASAP ILL GIVE U BRAINIEST
An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the objects maximum height if the equation of height (h) in terms of time (t) of the object is given by h(t) = -16t^2 + 64t + 80?