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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) = -16t2 + 64t + 80

144 ft

Step-by-step explanation:

The object's maximum height is the same as the y-coordinate of the vertex.

To find that, first, finds the x- coordinate(t).You can find it by: first identifying a, b and c.

Multiply the b by negative 1 or simply change its sign and divide it by 2a.

Afterward, plug the answer in the equation to find y, your answer.(ax+bx²+c=-16t²+64t+80).

Therefore, a=-6, b= 64 and c=80-64/2(-16)=-64/-32=2-16(2)²+64(2)+80=144 ft

So the maximum height of the object, when launched 64ft/sec from a platform that is 80 ft high, is 144 ft

70

step-by-step explanation:

2*5=10

4*15=60

60+10=

70

answer: hey there,

the answer you are looking for is a..

hope this you!