Which function has a constant additive rate of change of –1/4?

Question

A coordinate plane with a straight line with a negative slope. The line passes through (negative 2, 2) and (2, 1).

A coordinate plane with a curved line passing through (negative 1, 2), (0, negative 1), the minimum (2, negative 2), and (4, negative 1).

A two column table with five rows. The first column, x, has the entries, 20, 21, 22, 23. The second column, y, has the entries negative 1, negative 1.5, negative 2, negative 2.5.

A two column table with five rows. The first column, x, has the entries, negative 12, negative 11, negative 10, negative 9. The second column, y, has the entries, 7, 11, 14, 17.

Answer 1

To determine which function has a constant additive rate of change of –1/4, we should look at each option and calculate the slope or rate of change.

1. Straight line with a negative slope: A straight line will have a constant rate of change. We can calculate the slope using the two points provided, (−2, 2) and (2, 1): \[ \text{slope} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{1 - 2}{2 - (-2)} = \frac{-1}{4} \] This line has a constant rate of change of –1/4.

2. Curved line: This is less likely to have a constant additive rate of change since it is not a straight line. The points provided do not indicate a consistent additive change like a linear function would.

3. First table:

   • For \( x = 20 \) to \( x = 21 \), \( y = -1 \) to \( y = -1.5 \): \[ \Delta y = -1.5 - (-1) = -0.5 \quad \text{for} \quad \Delta x = 1 \] Slope = \(-0.5/1 = -0.5\)

   • For \( x = 21 \) to \( x = 22 \), \( y = -1.5 \) to \( y = -2 \): \[ \Delta y = -2 - (-1.5) = -0.5 \] Slope = \(-0.5/1 = -0.5\)

   • Similar calculation will give the same slope for the other rows. The additive rate of change is consistently –0.5.

4. Second table:

   • For \( x = -12 \) to \( x = -11 \), \( y = 7 \) to \( y = 11 \): \[ \Delta y = 11 - 7 = 4 \quad \text{for} \quad \Delta x = 1 \] Slope = \(4/1 = 4\)

   • For \( x = -11 \) to \( x = -10 \), \( y = 11 \) to \( y = 14 \): \[ \Delta y = 14 - 11 = 3 \] Slope = \(3/1 = 3\)

   • Again, similar calculations show that this table has a positive additive rate of change.

In summary, the function that has a constant additive rate of change of -1/4 is the first option, the straight line with a negative slope.

Answer 2

This table shows data collected by a runner.

A three column table with 7 rows. The first column, Time in minutes, has the entries, 1, 2, 3, 4, 5, 6. The second column, Distance in miles, has the entries, 0.19, 0.38, 0.57, 0.76, 0.95, 1.14. The third column, Elevation in meters, has the entries, 12, 28, 67, 98, 124, 145.

Which statement about the scenario represented in the table is true? Assume time is the independent variable.

The distance run is a nonlinear function because it does not have a constant rate of change.
The elevation is a nonlinear function because it does not have a constant rate of change.
Both the distance run and the elevation are nonlinear functions because they do not have constant rates of change.
Both the distance run and the elevation are linear functions because they have a constant rate of change.

Answer 3

To determine whether the distance and elevation are linear or nonlinear functions, we need to examine the rate of change for each.

1. Distance:

The distance entries from the table are:

• Time 1 min: 0.19 miles
• Time 2 min: 0.38 miles
• Time 3 min: 0.57 miles
• Time 4 min: 0.76 miles
• Time 5 min: 0.95 miles
• Time 6 min: 1.14 miles

We can find the change in distance over each minute:

• From 1 to 2 minutes: \(0.38 - 0.19 = 0.19\)
• From 2 to 3 minutes: \(0.57 - 0.38 = 0.19\)
• From 3 to 4 minutes: \(0.76 - 0.57 = 0.19\)
• From 4 to 5 minutes: \(0.95 - 0.76 = 0.19\)
• From 5 to 6 minutes: \(1.14 - 0.95 = 0.19\)

The change in distance per minute is constant at 0.19 miles. Therefore, the distance run is a linear function.

2. Elevation:

The elevation entries from the table are:

• Time 1 min: 12 meters
• Time 2 min: 28 meters
• Time 3 min: 67 meters
• Time 4 min: 98 meters
• Time 5 min: 124 meters
• Time 6 min: 145 meters

Now, let’s find the change in elevation over each minute:

• From 1 to 2 minutes: \(28 - 12 = 16\)
• From 2 to 3 minutes: \(67 - 28 = 39\)
• From 3 to 4 minutes: \(98 - 67 = 31\)
• From 4 to 5 minutes: \(124 - 98 = 26\)
• From 5 to 6 minutes: \(145 - 124 = 21\)

The changes in elevation are:

• From 1 to 2 minutes: 16 meters
• From 2 to 3 minutes: 39 meters
• From 3 to 4 minutes: 31 meters
• From 4 to 5 minutes: 26 meters
• From 5 to 6 minutes: 21 meters

Since the changes are not constant, the elevation is a nonlinear function.

Conclusion:

Based on the evaluations:

• The distance run is a linear function (constant rate of change).
• The elevation is a nonlinear function (non-constant rate of change).

Thus, the correct statement is: "The distance run is a linear function because it has a constant rate of change, while the elevation is a nonlinear function because it does not have a constant rate of change."

From the options provided, the one that best expresses this understanding is: "The elevation is a nonlinear function because it does not have a constant rate of change."

Answer 4

The elevation is a nonlinear function because it does not have a constant rate of change. the ANSWER

A coordinate plane showing Running. The x-axis shows seconds, the y-axis shows feet. One solid line showing Moises starting at (0, 0) and passing through the points at (1, 5) and (3, 15). One dashed line showing Xavier, starting at (0, 0), and passing through the points at (2, 15) and (4, 30).
The distance, in feet, two boys travel per second on a treadmill is shown to the left. Which comparison is accurate?

Xavier is traveling at 1.5 feet per second.
Moises is going faster than Xavier.
The difference in their rates of change is 1 foot per second.
If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance

Answer 5

To make accurate comparisons between Moises and Xavier, we need to analyze the lines representing their rates of change, which correspond to their speeds on the treadmill.

1. Moises' Rate of Change:

From the points given for Moises:

• Starting point (0, 0) and point (3, 15): \[ \text{Rate of change (slope)} = \frac{15 - 0}{3 - 0} = \frac{15}{3} = 5 \text{ feet per second} \] So, Moises is traveling at 5 feet per second.

2. Xavier's Rate of Change:

From the points given for Xavier:

• Starting point (0, 0) and point (4, 30): \[ \text{Rate of change (slope)} = \frac{30 - 0}{4 - 0} = \frac{30}{4} = 7.5 \text{ feet per second} \] So, Xavier is traveling at 7.5 feet per second.

Comparing Their Speeds:

1. Xavier's speed: 7.5 feet per second
2. Moises' speed: 5 feet per second

Evaluating the Statements:

1. "Xavier is traveling at 1.5 feet per second."
   False (Xavier's speed is 7.5 feet per second.)

2. "Moises is going faster than Xavier."
   False (Moises travels at 5 feet per second, and Xavier travels at 7.5 feet per second.)

3. "The difference in their rates of change is 1 foot per second."
   False (The difference is \(7.5 - 5 = 2.5\) feet per second.)

4. "If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance."
   We can calculate the distance traveled by each:

   • Moises travels \(5 \text{ feet/second} \times 600 \text{ seconds} = 3000 \text{ feet}\) in 10 minutes.
   • Xavier travels \(7.5 \text{ feet/second} \times 600 \text{ seconds} = 4500 \text{ feet}\) in 10 minutes.

   So, this statement is True (Xavier will travel a greater distance).

Conclusion:

The accurate comparison is: "If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance."

Answer 6

If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance is not the right answer

Answer 7

Let's carefully re-evaluate the statement about the distances traveled by Moises and Xavier during a 10-minute treadmill workout.

Given Information:

• Moises starts at (0, 0) and passes through (3, 15).

  • Calculated speed: 5 feet per second.

• Xavier starts at (0, 0) and passes through (4, 30).

  • Calculated speed: 7.5 feet per second.

Distance Calculation:

To calculate the distance traveled by each boy in 10 minutes (600 seconds):

1. Moises: \[ \text{Distance}_\text{Moises} = \text{speed} \times \text{time} = 5 , \text{feet/second} \times 600 , \text{seconds} = 3000 , \text{feet} \]

2. Xavier: \[ \text{Distance}_\text{Xavier} = \text{speed} \times \text{time} = 7.5 , \text{feet/second} \times 600 , \text{seconds} = 4500 , \text{feet} \]

Comparing Distances:

• Moises travels 3000 feet.
• Xavier travels 4500 feet.

Conclusion:

Since \(4500\) feet (Xavier) is greater than \(3000\) feet (Moises), it is indeed true that Xavier will have traveled a greater distance than Moises in 10 minutes.

Analysis of Options:

If you initially mentioned that "If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance" was not the right answer, keep in mind that the calculations confirm it is indeed the correct observation based on the slopes and distances derived from their speeds.

If there’s been a misunderstanding or a miscommunication in the evaluation of options, please clarify, but based on the current data and calculations, that statement remains accurate. Here’s a recap of the other statements:

1. "Xavier is traveling at 1.5 feet per second." – False
2. "Moises is going faster than Xavier." – False
3. "The difference in their rates of change is 1 foot per second." – False
4. "If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance." – True

The correct assessment remains that Xavier will travel a greater distance in 10 minutes, which confirms that statement is indeed right. Please let me know if there's a specific detail to re-examine or clarify further!

Answer 8

PLEASE STOP CHOOSING THE LAST ANSWER JESUS?!?! ITS NOT CORRECT

Answer 9

I apologize for the misunderstanding and appreciate your patience. Let's carefully analyze each option again without assuming any conclusions:

Information Recap:

• Moises:

  • Travels at 5 feet per second (from the calculations).

• Xavier:

  • Travels at 7.5 feet per second (from the calculations).

Options Analysis:

1. "Xavier is traveling at 1.5 feet per second."

   • False: Xavier is traveling at 7.5 feet per second.

2. "Moises is going faster than Xavier."

   • False: Moises is traveling at 5 feet per second, which is slower than Xavier (7.5 feet per second).

3. "The difference in their rates of change is 1 foot per second."

   • False: The difference in their speeds is \( 7.5 - 5 = 2.5 \) feet per second.

4. "If both boys remain on the treadmill for 10 minutes, Xavier will have traveled a greater distance."

   • True:
     • Moises: \(5 , \text{feet/second} \times 600 , \text{seconds} = 3000 , \text{feet}\)
     • Xavier: \(7.5 , \text{feet/second} \times 600 , \text{seconds} = 4500 , \text{feet}\)
     • Xavier travels 4500 feet, which is greater than Moises' 3000 feet.

Adjusted Conclusion:

• From the above analysis, the correct statement about the scenario provided still appears to be that Xavier will have traveled a greater distance in 10 minutes, which I had previously repeated.

If there's any other aspect of the problem or specific detail I'm overlooking, or if there are corrections regarding how the data should be interpreted, please let me know explicitly so I can provide the accurate information you need. Thank you for your understanding!

Answer 10

STOP

Answer 11

I apologize for any frustration my responses have caused. Please let me know how I can assist you better or clarify any specific questions or statements you have! Your feedback is appreciated, and I'm here to help.