Renate launched an object vertically from a point that is 58.9 meters above ground level with an initial velocity of 21.6 meters per second. This situation can be represented by the equation h=−4.9t^2+21.6t+58.9, where h is the height of the object in meters and t is the time in seconds after the object is launched. What is the maximum height of the object?

answer: standard form is ax^2 + bx + c = 0

step-by-step explanation:

standard form is ax^2 + bx + c = 0

(2x + 1)(x - 2) = 0

distribute to get into standard form

2x^2 - 4x + x - 2 = 0

2x^2 - 3x - 2 = 0

a = 2, b= -3, c = -2

a is your answer

answer: i don't understand. what are the options

step-by-step explanation:

tangent

step-by-step explanation:

step-by-step explanation:

no solution.