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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

1.22 seconds
1.22 seconds

1.03 seconds
1.03 seconds

−1.03 seconds
negative 1.03 seconds

23.01 feet

11 answers

To find the time when the volleyball reaches its maximum height, we can use the vertex formula for a quadratic equation in the form \( h(t) = at^2 + bt + c \). The time \( t \) when the maximum height occurs is given by the formula:

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

In your equation \( h = -16t^2 + 33t + 6 \), the coefficients are:

  • \( a = -16 \)
  • \( b = 33 \)

Now, we can substitute these values into the formula:

\[ t = -\frac{33}{2(-16)} = -\frac{33}{-32} = \frac{33}{32} \approx 1.03125 \text{ seconds} \]

Rounding it gives approx \( 1.03 \) seconds.

The question asks for the domain of the ball when it reaches its maximum height. The closest option from the responses provided is:

1.03 seconds.

This indicates that approximately at this time, the ball reaches its maximum height.

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
, where h is the height in feet and t is the time in seconds. Using this equation, what is the maximum range the ball will reach? Round to the nearest hundredth.(1 point)
feet

To find the maximum height (range) the football will reach, we need to determine the time at which this occurs using the vertex formula for the quadratic equation in the form:

\[ h(t) = at^2 + bt + c \]

In your equation \( h = -16t^2 + 60t + 0 \), the coefficients are:

  • \( a = -16 \)
  • \( b = 60 \)

The time \( t \) when the maximum height occurs can be found using:

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

Substituting the values:

\[ t = -\frac{60}{2(-16)} = \frac{60}{32} = \frac{15}{8} = 1.875 \text{ seconds} \]

Next, we substitute this time back into the height equation to find the maximum height:

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

Calculating \( (1.875)^2 \):

\[ (1.875)^2 = 3.515625 \]

Now substituting this into the height equation:

\[ h(1.875) = -16(3.515625) + 60(1.875) \]

Calculating each term:

\[ -16(3.515625) = -56.25 \] \[ 60(1.875) = 112.5 \]

Combining these:

\[ h(1.875) = -56.25 + 112.5 = 56.25 \]

Thus, the maximum range (height) the ball will reach is approximately 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 \( A \) that Nora can achieve for her dog run, given the equation \( A = -w^2 + 60w \), we can use the vertex formula for a quadratic function.

In the quadratic equation \( A = -w^2 + 60w \):

  • The coefficient \( a = -1 \)
  • The coefficient \( b = 60 \)

The maximum value of a quadratic function occurs at the vertex, and the width \( w \) at the vertex can be calculated using the formula:

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

Substituting in the values:

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

Now we can substitute \( w = 30 \) back into the area equation to find the maximum area:

\[ A = -w^2 + 60w \] \[ A = -(30)^2 + 60(30) \] \[ A = -900 + 1800 \] \[ A = 900 \]

Thus, the maximum area of the dog run is 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 that corresponds to the x-value of \(-1\) for the function \( f(x) = 2x^2 - 2x \), we need to substitute \(-1\) into the function.

  1. Start with the function: \[ f(x) = 2x^2 - 2x \]

  2. Substitute \(x = -1\): \[ f(-1) = 2(-1)^2 - 2(-1) \]

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

  4. Now substitute this back into the equation: \[ f(-1) = 2(1) - 2(-1) \] \[ f(-1) = 2 - (-2) \] \[ f(-1) = 2 + 2 = 4 \]

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

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 will use the formula for the vertex of a parabola, since the profit function is a quadratic equation.

The given profit function is:

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

In this equation, the coefficients are:

  • \(a = -5\)
  • \(b = 1000\)

The x-value that gives the maximum profit can be found using the vertex formula:

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

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

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

This means that the company should spend $100,000 on marketing to achieve the maximum profit.

Thus, the spending in thousands directed toward marketing for maximum profit is $100.

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

5
5

110
1 tenth

10
10

2

To determine when the rock hits the water, we need to find the time \( t \) when the height \( h \) equals 0 (which represents the height of the water level).

The equation given is:

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

Setting \( h = 0 \) to find when the rock hits the water:

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

Rearranging gives:

\[ 4.9t^2 = 10 \]

Now, divide both sides by 4.9:

\[ t^2 = \frac{10}{4.9} \approx 2.0408 \]

Taking the square root:

\[ t \approx \sqrt{2.0408} \approx 1.43 \text{ seconds} \]

Now, regarding a reasonable scale for the range (time in seconds), the time from 0 seconds to about 1.5 seconds would be appropriate for the context of how long it takes for the rock to hit the water.

Looking at the provided responses:

  • 5 seconds
  • 110 (presumably 1 tenth)
  • 10 seconds
  • 2 seconds

A reasonable scale for the time range would be 2 seconds given that it is just slightly over the computed time of approximately 1.43 seconds, allowing for an easier table setup and including the possibility of some measured intervals. Thus, the answer is 2 seconds.