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The problem asks to simplify \( H(-t) \), where \( H(t) = 6t^2 - 3t + 3 \).

To find \( H(-t) \), we substitute \( -t \) for \( t \) in the given expression for \( H(t) \):

\[
H(-t) = 6(-t)^2 - 3(-t) + 3
\]

Step-by-step simplification:

1. \( (-t)^2 = t^2 \), so \( 6(-t)^2 = 6t^2 \)
2. \( -3(-t) = 3t \)
3. The constant \( 3 \) remains unchanged.

Thus, \( H(-t) = 6t^2 + 3t + 3 \).

This is the simplified expression for \( H(-t) \).

Would you like more details or have any questions? 

Here are 5 related questions:
1. How would the simplification change if \( H(t) \) included higher powers of \( t \)?
2. What is the significance of even and odd functions in relation to this problem?
3. How would \( H(-t) \) change if the linear term had a higher coefficient?
4. Could \( H(t) \) represent a real-world situation, and how would changing \( t \) to \( -t \) affect it?
5. How does substituting \( -t \) affect the symmetry of a function?

**Tip:** Even powers of \( t \) remain unchanged when substituting \( -t \), while odd powers change sign.

Simplify H(-t), where H(t) = 6t^2 - 3t + 3.

This problem demonstrates how to simplify H(-t), given the quadratic function H(t) = 6t^2 - 3t + 3. Follow a step-by-step guide to substitute -t into the equation and simplify it. Suitable for students in Grades 9-10 learning substitution in quadratic functions.

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