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The equation provided in your homework is:

\[
h = 112 + 128t - 16t^2
\]

You are asked to find \( h \) for two different values of \( t \):

1. \( t = 2 \)
2. \( t = 6 \)

Let's solve for both cases.

### a. For \( t = 2 \):
Substitute \( t = 2 \) into the equation for \( h \):

\[
h = 112 + 128(2) - 16(2^2)
\]
\[
h = 112 + 256 - 16(4)
\]
\[
h = 112 + 256 - 64
\]
\[
h = 304
\]

### b. For \( t = 6 \):
Substitute \( t = 6 \) into the equation for \( h \):

\[
h = 112 + 128(6) - 16(6^2)
\]
\[
h = 112 + 768 - 16(36)
\]
\[
h = 112 + 768 - 576
\]
\[
h = 304
\]

So for both values \( t = 2 \) and \( t = 6 \), \( h = 304 \).

Would you like more details on this? Here are five related questions you could explore:

1. How can you find the vertex of the quadratic equation \( h = 112 + 128t - 16t^2 \)?
2. What is the significance of the \( t^2 \) term in the equation?
3. How can you determine when \( h \) reaches its maximum value?
4. Can you find when \( h = 0 \) by solving the quadratic equation?
5. How would you graph this equation and identify its key features?

**Tip**: When dealing with quadratic equations, remember that they often represent parabolas, and the term with \( t^2 \) controls the curvature.

Find h if h = 112 + 128t - 16t^2 and for a. t = 2, b. t = 6

Learn how to solve the quadratic equation h = 112 + 128t - 16t² for two values of t (t = 2 and t = 6). This detailed explanation covers substitution and quadratic concepts, suitable for students in Grades 9-12.

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