A football player kicks the ball with an initial upward velocity of 60 feet per second. This situation is modeled using the equation h=−16t2+60t+0

To find the maximum height (or range) the football will reach, we can determine the vertex of the quadratic equation given, which is in the form \( h = -16t^2 + 60t + 0 \).

The formula for the time \( t \) at which the maximum height occurs is given by:

\[ t = -\frac{b}{2a} \]

where \( a \) and \( b \) are the coefficients from the equation \( h = at^2 + bt + c \).

From the given equation \( h = -16t^2 + 60t + 0 \), we have:

• \( a = -16 \)
• \( b = 60 \)

Now, substituting these values into the formula for \( t \):

\[ t = -\frac{60}{2 \times -16} = \frac{60}{32} = 1.875 \text{ seconds} \]

Next, we will substitute \( t = 1.875 \) back into the original equation to find the maximum height \( h \):

\[ h = -16(1.875)^2 + 60(1.875) + 0 \]

Calculating \( (1.875)^2 \):

\[ (1.875)^2 = 3.515625 \]

Now plug it back into the height equation:

\[ h = -16(3.515625) + 60(1.875) \] \[ h = -56.25 + 112.5 \] \[ h = 56.25 \text{ feet} \]

Thus, the maximum height (range) the ball will reach is 56.25 feet.

Final Answer (rounded to the nearest hundredth):

56.25 feet.

Nora is creating a dog run for her dog, Mr. Darcey. She has enough fencing material for a fence with a perimeter of 120 feet. This situation is modeled using the equation A=−w2+60w
where A
is the area and w is the width. Using this equation, identify the maximum range or area the fence can have.(1 point)
square feet

To find the maximum area of the dog run that Nora can create, we will analyze the given equation for area:

\[ A = -w^2 + 60w \]

This is a quadratic equation of the form \( A = ax^2 + bx + c \), where \( a = -1 \) and \( b = 60 \).

The maximum value of a quadratic equation occurs at its vertex, and the \( w \)-coordinate of the vertex can be calculated using the formula:

\[ w = -\frac{b}{2a} \]

Substituting the values of \( a \) and \( b \):

\[ w = -\frac{60}{2 \times -1} = \frac{60}{2} = 30 \]

Now that we have the width \( w \), we will substitute it back into the area equation to find the maximum area \( A \):

\[ A = -w^2 + 60w \] \[ A = -(30)^2 + 60(30) \] \[ A = -900 + 1800 \] \[ A = 900 \text{ square feet} \]

Thus, the maximum area that the fence can have is 900 square feet.

Final Answer:

900 square feet.

Graph the function f(x)=2x2−2x
. What is the y-value that corresponds to the x-value of −1
?(1 point)

To find the \( y \)-value corresponding to \( x = -1 \) for the function \( f(x) = 2x^2 - 2x \), we need to substitute \( -1 \) into the function:

\[ f(-1) = 2(-1)^2 - 2(-1) \]

Calculating this step-by-step:

1. Calculate \( (-1)^2 \): \[ (-1)^2 = 1 \]

2. Multiply by 2: \[ 2 \cdot 1 = 2 \]

3. Calculate \( -2(-1) \): \[ -2(-1) = 2 \]

4. Now combine the results: \[ f(-1) = 2 + 2 = 4 \]

Thus, the \( y \)-value that corresponds to the \( x \)-value of \( -1 \) is 4.

Final Answer:

4.

Graph the function f(x)=2x2−2x
. True or false: The x-intercepts of this graph are (0,0)
and (1,0)
.

Type 1 for true.

Type 2 for false.

(1 point)

To determine the x-intercepts of the function \( f(x) = 2x^2 - 2x \), we need to set \( f(x) \) equal to zero and solve for \( x \):

\[ 2x^2 - 2x = 0 \]

We can factor out \( 2x \):

\[ 2x(x - 1) = 0 \]

Setting each factor equal to zero gives us the possible solutions:

1. \( 2x = 0 \) which simplifies to \( x = 0 \)
2. \( x - 1 = 0 \) which simplifies to \( x = 1 \)

Thus, the x-intercepts are at:

• \( (0, 0) \)
• \( (1, 0) \)

Since the statement claims that the x-intercepts of the graph are \( (0, 0) \) and \( (1, 0) \), this statement is true.

Final answer: 1 (true).

The profit (in thousands of dollars) of a company is represented as P=−5x2+1,000x+5,000
, where P
represents the profit and x represents the amount spent on marketing (in thousands of dollars). How much spending in the thousands will be directed toward marketing to achieve the maximum profit?(1 point)

To find the amount spent on marketing that will achieve the maximum profit, we need to determine the vertex of the quadratic function given by:

\[ P = -5x^2 + 1000x + 5000 \]

The formula for the \( x \)-coordinate of the vertex of a parabola represented by the equation \( P = ax^2 + bx + c \) is given by:

\[ x = -\frac{b}{2a} \]

In this case, the coefficients are:

• \( a = -5 \)
• \( b = 1000 \)

Now we can substitute these values into the vertex formula:

\[ x = -\frac{1000}{2 \cdot -5} = \frac{1000}{10} = 100 \]

Thus, the amount spent on marketing to achieve the maximum profit is 100 thousand dollars.

Final Answer:

100 (in thousands of dollars).

A volleyball is served by a 6-foot player at an initial upward velocity of 33 feet per second. The situation is modeled by the equation h=−16t2+33t+6
h representing the height in feet and t representing the time in seconds. Using this equation, define the domain of the ball when it reaches its maximum height.(1 point)
Responses

−1.03 seconds
negative 1.03 seconds

1.22 seconds
1.22 seconds

1.03 seconds
1.03 seconds

23.01 feet

While hiking, Marek throws a rock off a cliff that is roughly 10 meters above a lake. This situation is modeled by the equation h=−4.9t2+10
, with h representing the height in meters and t the time in seconds. If you were to set up a table to determine when the rock hits the water, what would be a reasonable scale for the range?(1 point)
Responses

2
2

5
5

10
10

1/10

An owl swoops down from its perch in a tree at 20 feet to catch field mice below. It follows the function h=−6t2−24t+20
. Plot points to graph the equation. Which ordered pair represents the vertex?(1 point)
Responses

(44,−2)
left parenthesis 44 comma negative 2 right parenthesis

(1,0)
left parenthesis 1 comma 0 right parenthesis

(−2,44)
left parenthesis negative 2 comma 44 right parenthesis

(−5,0)
left parenthesis negative 5 comma 0 right parenthesis

Luca is building a garden in his backyard. He needs to erect a fence to keep his dog Luna from digging in it. He has enough materials for a fence with 120 feet around the perimeter. This function is represented by the formula A=−w2+60w
, where A
is the area and w is the width. Which of the following correctly identifies the x-intercept(s)?(1 point)
Responses

(0,0)
and (60,0)
left parenthesis 0 comma 0 right parenthesis and left parenthesis 60 comma 0 right parenthesis

(60,0)
left parenthesis 60 comma 0 right parenthesis

(0,0)
left parenthesis 0 comma 0 right parenthesis

(60,0)
and (30,900)

An owl swoops down from its perch in a tree at 30 feet to catch field mice. It follows the function h=6t2 −24t+30
, where t is the time in seconds and h is the height in feet. Which of the following best interprets the vertex?(1 point)
Responses

(6,2)
left parenthesis 6 comma 2 right parenthesis

(2,6)
left parenthesis 2 comma 6 right parenthesis

(30,0)
left parenthesis 30 comma 0 right parenthesis

(0,30)

Let's analyze each question one by one:

1. Volleyball Height Equation

Equation:
\( h = -16t^2 + 33t + 6 \)

To find the time at which the maximum height occurs, we use the vertex formula. The vertex \( t \) can be found using:

\[ t = -\frac{b}{2a} \]

In this case:

• \( a = -16 \)
• \( b = 33 \)

Calculating:

\[ t = -\frac{33}{2 \cdot -16} = \frac{33}{32} \approx 1.03 \text{ seconds} \]

The domain of the volleyball when it reaches its maximum height is generally from the time it is served until it reaches its peak height (1.03 seconds) and then it will fall back down. However, the maximum height occurs specifically at 1.03 seconds.

Response:
1.03 seconds (or the option that correctly states this).

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2. Marek's Rock Throwing Equation

Equation:
\( h = -4.9t^2 + 10 \)

To determine when the rock hits the water, we need to set \( h = 0 \):

\[ 0 = -4.9t^2 + 10 \]

Solving for \( t \):

\[ 4.9t^2 = 10 \ t^2 = \frac{10}{4.9} \ t \approx \sqrt{2.04} \approx 1.43 \text{ seconds} \]

When setting up a table, a reasonable scale for determining when the rock hits the water could be in 2 second intervals, as it provides a clear measure of time to reach just over 1.43 seconds.

Response:
2

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3. Owl Swooping Equation

Equation:
\( h = -6t^2 - 24t + 20 \)

To find the vertex of this parabola (which indicates the maximum height reached), we again use:

\[ t = -\frac{b}{2a} \]

Where:

• \( a = -6 \)
• \( b = -24 \)

Calculating:

\[ t = -\frac{-24}{2 \cdot -6} = \frac{24}{-12} = -2 \]

Now substituting \( t = -2 \) back into the equation:

\[ h = -6(-2)^2 - 24(-2) + 20 \] \[ h = -6(4) + 48 + 20 \] \[ h = -24 + 48 + 20 = 44 \]

Thus, the vertex is at \( (-2, 44) \).

Response:
(−2, 44)

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4. Luca's Garden Fence Equation

Equation:
\( A = -w^2 + 60w \)

To find the x-intercepts, we set \( A = 0 \):

\[ -w^2 + 60w = 0 \ w(w - 60) = 0 \]

The solutions are:

• \( w = 0 \) (no width = area of 0)
• \( w = 60 \)

Thus, the x-intercepts are \( (0, 0) \) and \( (60, 0) \).

Response:
(0, 0) and (60, 0)

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5. Owl Swooping Interpretation

Equation:
\( h = 6t^2 - 24t + 30 \)

To find the vertex:

\[ t = -\frac{b}{2a} = -\frac{-24}{2 \cdot 6} = \frac{24}{12} = 2 \]

Now substituting \( t = 2 \):

\[ h = 6(2)^2 - 24(2) + 30 \ h = 6(4) - 48 + 30 \ h = 24 - 48 + 30 = 6 \]

The vertex is \( (2, 6) \), which represents the time it takes (2 seconds) to reach the maximum height (6 feet).

Response:
(2, 6)

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Please let me know if you need further clarification or additional assistance with these concepts!