t=-0.021t^2 + .504t + 98.1 what was the max at what time 

The equation given is:

\[
h(t) = -0.021t^2 + 0.504t + 98.1
\]

This is a quadratic function that models a parabola opening downwards (since the coefficient of \( t^2 \) is negative). The maximum value of this function occurs at the vertex of the parabola.

To find the time \( t \) at which the maximum occurs, we use the vertex formula:

\[
t_{\text{max}} = \frac{-b}{2a}
\]

Here, \( a = -0.021 \) and \( b = 0.504 \).

Let's calculate \( t_{\text{max}} \):

\[
t_{\text{max}} = \frac{-0.504}{2 \times (-0.021)} = \frac{-0.504}{-0.042} 
\]

Let me calculate that for you.The maximum value occurs at \( t = 12 \) seconds.

Now, to find the maximum height at this time, we substitute \( t = 12 \) back into the original equation:

\[
h(12) = -0.021(12)^2 + 0.504(12) + 98.1
\]

I'll calculate this for you.

t=-0.021t^2 + .504t + 98.1 what was the max at what time

Learn how to find the maximum value and the time at which it occurs for the quadratic function h(t) = -0.021t^2 + 0.504t + 98.1. This solution explains key algebraic concepts like the vertex formula and solving quadratic functions. Ideal for students in Grades 9-11.

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