An object is launched directly in the air at a speed of 48 feet per second from a platform located 12 feet above the ground. The position of the object can be modeled using the function f(t)=−16t2+48t+12, where t is the time in seconds and f(t) is the height, in feet, of the object. What is the maximum height, in feet, that the object will reach?

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  48 ft

Step-by-step explanation:

For the quadratic ax^2 +bx +c, the axis of symmetry is x = -b/(2a). For the given quadratic, which defines a parabola opening downward, the axis of symmetry defines the time at which the maximum height is reached.

  t = -48/(2(-16)) = 1.5

Then the maximum height is ...

  f(1.5) = (-16·1.5 +48)1.5 +12 = (24·1.5) +12

  f(1.5) = 48

The maximum height the object will reach is 48 feet.

answer: howdy

step-by-step explanation:

8(4)^8=72^feet to say that the total can conclude to 8(4) which as you know would mean 8 x 4. that equals 32w. 32^ could make a 72 answer as well. but your coefficent will remain w so your answer is 72w.

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