An object is launched at 20 m/s from a height of 65 m. the equation for the height (h) in terms of time (t) is given by h(t) = -4.9t? + 20t + 65. what is the object's maximum height? the numeric answer only, rounded to the nearest meter.

85 meters

Step-by-step explanation:

The maximum height is the y-value of the vertex.

h(t) = -4.9t² + 20t + 65

      a=-4.9   b=20  c=65

h(2) = -4.9(2)² + 20(2) + 65

      = -19.6 + 40 + 65

      = 85.4

use substitution to solve each system of equations. y = x + 5 3x + y = 25 62/87,21 y = x + 5 3x + y = 25 substitute x + 5 for y in the second equation. substitute the solution for x into either equation to find y. the solution is (5, 10)

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step-by-step explanation:

umumumumum

im not sure