Answer :
Let's analyze the function [tex]\( D(t) \)[/tex] that defines a traveler's distance from home in miles over time in hours. We will evaluate it at specific times to determine the appropriate options.
1. At 0 hours:
The formula for the distance [tex]\( D(t) \)[/tex] when [tex]\( 0 \leq t < 2.5 \)[/tex] is given by:
[tex]\[ D(t) = 300t + 125 \][/tex]
Plugging in [tex]\( t = 0 \)[/tex]:
[tex]\[ D(0) = 300(0) + 125 = 125 \text{ miles} \][/tex]
So, the starting distance at 0 hours is 125 miles, not 300 miles.
2. At 2 hours:
Using the same formula as for [tex]\( 0 \leq t < 2.5 \)[/tex]:
[tex]\[ D(t) = 300t + 125 \][/tex]
Plugging in [tex]\( t = 2 \)[/tex]:
[tex]\[ D(2) = 300(2) + 125 = 600 + 125 = 725 \text{ miles} \][/tex]
So, at 2 hours, the traveler is 725 miles from home. This statement is correct.
3. At 2.5 hours:
The distance function for [tex]\( 2.5 \leq t \leq 3.5 \)[/tex] is constant:
[tex]\[ D(t) = 875 \][/tex]
Thus, at [tex]\( t = 2.5 \)[/tex]:
[tex]\[ D(2.5) = 875 \text{ miles} \][/tex]
So the traveler is not moving farther from home at 2.5 hours. The distance is constant 875 miles. This statement is incorrect.
4. At 3 hours:
The distance function for [tex]\( 2.5 \leq t \leq 3.5 \)[/tex] remains:
[tex]\[ D(t) = 875 \][/tex]
Thus, at [tex]\( t = 3 \)[/tex]:
[tex]\[ D(3) = 875 \text{ miles} \][/tex]
So, at 3 hours, the distance is indeed constant at 875 miles. This statement is correct.
5. At 6 hours:
The formula for the distance [tex]\( D(t) \)[/tex] when [tex]\( 3.5 < t \leq 6 \)[/tex] is:
[tex]\[ D(t) = 75t + 612.5 \][/tex]
Plugging in [tex]\( t = 6 \)[/tex]:
[tex]\[ D(6) = 75(6) + 612.5 = 450 + 612.5 = 1062.5 \text{ miles} \][/tex]
So, the total distance from home after 6 hours is 1062.5 miles. This statement is correct.
### Summary:
The correct statements are:
1. At 2 hours, the traveler is 725 miles from home.
2. At 3 hours, the distance is constant, at 875 miles.
3. The total distance from home after 6 hours is 1062.5 miles.
1. At 0 hours:
The formula for the distance [tex]\( D(t) \)[/tex] when [tex]\( 0 \leq t < 2.5 \)[/tex] is given by:
[tex]\[ D(t) = 300t + 125 \][/tex]
Plugging in [tex]\( t = 0 \)[/tex]:
[tex]\[ D(0) = 300(0) + 125 = 125 \text{ miles} \][/tex]
So, the starting distance at 0 hours is 125 miles, not 300 miles.
2. At 2 hours:
Using the same formula as for [tex]\( 0 \leq t < 2.5 \)[/tex]:
[tex]\[ D(t) = 300t + 125 \][/tex]
Plugging in [tex]\( t = 2 \)[/tex]:
[tex]\[ D(2) = 300(2) + 125 = 600 + 125 = 725 \text{ miles} \][/tex]
So, at 2 hours, the traveler is 725 miles from home. This statement is correct.
3. At 2.5 hours:
The distance function for [tex]\( 2.5 \leq t \leq 3.5 \)[/tex] is constant:
[tex]\[ D(t) = 875 \][/tex]
Thus, at [tex]\( t = 2.5 \)[/tex]:
[tex]\[ D(2.5) = 875 \text{ miles} \][/tex]
So the traveler is not moving farther from home at 2.5 hours. The distance is constant 875 miles. This statement is incorrect.
4. At 3 hours:
The distance function for [tex]\( 2.5 \leq t \leq 3.5 \)[/tex] remains:
[tex]\[ D(t) = 875 \][/tex]
Thus, at [tex]\( t = 3 \)[/tex]:
[tex]\[ D(3) = 875 \text{ miles} \][/tex]
So, at 3 hours, the distance is indeed constant at 875 miles. This statement is correct.
5. At 6 hours:
The formula for the distance [tex]\( D(t) \)[/tex] when [tex]\( 3.5 < t \leq 6 \)[/tex] is:
[tex]\[ D(t) = 75t + 612.5 \][/tex]
Plugging in [tex]\( t = 6 \)[/tex]:
[tex]\[ D(6) = 75(6) + 612.5 = 450 + 612.5 = 1062.5 \text{ miles} \][/tex]
So, the total distance from home after 6 hours is 1062.5 miles. This statement is correct.
### Summary:
The correct statements are:
1. At 2 hours, the traveler is 725 miles from home.
2. At 3 hours, the distance is constant, at 875 miles.
3. The total distance from home after 6 hours is 1062.5 miles.