P(t) = 0.5t^2 - 9.65t + 100 pt as 200 used in a different form

To transform the function \( P(t) = 0.5t^2 - 9.65t + 100 \) into a different form with \( P(t) = 200 \), we can set the equation equal to 200:

\[
0.5t^2 - 9.65t + 100 = 200
\]

Subtract 200 from both sides:

\[
0.5t^2 - 9.65t + 100 - 200 = 0
\]

This simplifies to:

\[
0.5t^2 - 9.65t - 100 = 0
\]

Now, to solve for \( t \), we can multiply the entire equation by 2 to eliminate the decimal:

\[
t^2 - 19.3t - 200 = 0
\]

Next, we can apply the quadratic formula:

\[
t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -19.3 \), and \( c = -200 \):

1. Calculate the discriminant:

\[
b^2 - 4ac = (-19.3)^2 - 4(1)(-200) = 372.49 + 800 = 1172.49
\]

2. Substitute into the quadratic formula:

\[
t = \frac{19.3 \pm \sqrt{1172.49}}{2}
\]

3. Calculate the square root and the two potential values for \( t \).

Would you like to see the calculations for \( t \) or have any specific questions about this transformation? Here are some related questions you might consider:

1. What are the exact solutions for \( t \)?
2. How can we interpret the physical meaning of \( P(t) \) in this context?
3. Can we graph this quadratic function to visualize the changes?
4. What happens to \( P(t) \) if the coefficients change?
5. How do you find the vertex of the quadratic function? 

**Tip:** Always check your discriminant to determine the nature of the roots in a quadratic equation!

Explore the transformation of the quadratic function P(t) = 0.5t^2 - 9.65t + 100 to find t when P(t) = 200. This detailed solution involves the quadratic formula and discriminant calculations, suitable for Grades 9-11.

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