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Part 1

An athlete whose event is the shot put releases the shot. When the shot is released at an angle of

2525​°,

its path can be modeled by the formula

y equals negative 0.01 x squared plus 0.5 x plus 5.9y=−0.01x2+0.5x+5.9

in which x is the​ shot's horizontal​ distance, in​ feet, and y is its​ height, in feet. This formula is shown by one of the​ graphs, (a) or​ (b), in the figure. Use the formula to answer the questions below.

(a)(b)

A coordinate system has a horizontal axis labeled Horizontal Distance from 0 to 160 in increments of 20 and a vertical axis labeled Height from 0 to 20 in increments of 5. There are two curves, labeled (a) and (b) both starting from a (0, 5.9). The curve labeled (a) rises at a decreasing rate to a maximum at (25, 12.15) before falling at an increasing rate to (59.9, 0). The curve labeled (b) rises at a decreasing rate to (25, 6.28) before falling at an increasing rate to (127.3, 0). All coordinates are approximate.

Horizontal Distance[0,160,20] by [0,20,5]

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Part 1

Use the formula to determine the​ shot's maximum distance.

The maximum distance is approximately

enter your response here

feet.

​(Round to the nearest tenth as​ needed.)

To determine the shot's maximum distance using the given equation:

\[
y = -0.01x^2 + 0.5x + 5.9
\]

We know that the shot will hit the ground when \( y = 0 \) (i.e., when the height is zero). To find the maximum distance, we need to solve the equation for \( x \) when \( y = 0 \).

### Step 1: Set \( y = 0 \)

\[
0 = -0.01x^2 + 0.5x + 5.9
\]

This is a quadratic equation in the form of \( ax^2 + bx + c = 0 \), where \( a = -0.01 \), \( b = 0.5 \), and \( c = 5.9 \).

### Step 2: Use the quadratic formula

The quadratic formula is:

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

Substituting the values \( a = -0.01 \), \( b = 0.5 \), and \( c = 5.9 \):

\[
x = \frac{-0.5 \pm \sqrt{(0.5)^2 - 4(-0.01)(5.9)}}{2(-0.01)}
\]

Simplifying inside the square root:

\[
x = \frac{-0.5 \pm \sqrt{0.25 + 0.236}}{-0.02}
\]

\[
x = \frac{-0.5 \pm \sqrt{0.486}}{-0.02}
\]

\[
x = \frac{-0.5 \pm 0.697}{-0.02}
\]

### Step 3: Solve for \( x \)

There are two possible solutions:

1. \( x = \frac{-0.5 + 0.697}{-0.02} \)
2. \( x = \frac{-0.5 - 0.697}{-0.02} \)

#### For the first case:

\[
x = \frac{0.197}{-0.02} = -9.85
\]

Since this is negative, it is not a valid distance.

#### For the second case:

\[
x = \frac{-1.197}{-0.02} = 59.85
\]

### Step 4: Conclusion

The maximum horizontal distance is approximately 59.9 feet. 

If you have any questions or want more details, feel free to ask! Here are some related questions:

1. How does the angle of release affect the shot's distance?
2. What would the graph of this equation look like in a real-world scenario?
3. How can you find the maximum height of the shot using the equation?
4. What other factors influence the range of a shot put throw?
5. How would adjusting the coefficients of the quadratic equation change the shot's trajectory?

**Tip:** To solve any quadratic equation, you can always use the quadratic formula, even when coefficients seem complex.

An athlete whose event is the shot put releases the shot. When the shot is released at an angle of 25°, its path can be modeled by the formula y = -0.01x^2 + 0.5x + 5.9 in which x is the shot's horizontal distance, in feet, and y is its height, in feet. Use the formula to determine the shot's maximum distance. The maximum distance is approximately ___ feet.

Solve a real-world problem involving the trajectory of a shot put using a quadratic equation. Learn how to calculate the maximum distance using the quadratic formula and understand the factors affecting the shot's path. This detailed guide is suitable for students in Grades 9-11.

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