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Sagot :
To solve this problem, we will analyze the provided function [tex]\( D(t) \)[/tex] and determine the distances at specified times. We are given the piecewise function:
[tex]\[ D(t) = \left\{ \begin{array}{cl} 300t + 125, & 0 \leq t < 2.5 \\ 875, & 2.5 \leq t \leq 3.5 \\ 75t + 612.5, & 3.5 < t \leq 6 \end{array} \right. \][/tex]
We need to determine the distances at specific times using this function:
1. Starting distance, at 0 hours
For [tex]\( t = 0 \)[/tex]:
[tex]\[ D(0) = 300(0) + 125 = 125 \text{ miles} \][/tex]
Therefore, the starting distance at 0 hours is 125 miles.
2. Distance at 2 hours
For [tex]\( t = 2 \)[/tex]:
[tex]\[ D(2) = 300(2) + 125 = 600 + 125 = 725 \text{ miles} \][/tex]
Therefore, at 2 hours, the traveler is 725 miles from home.
3. Distance at 2.5 hours
For [tex]\( t = 2.5 \)[/tex]:
Using the second part of the piecewise function:
[tex]\[ D(2.5) = 875 \text{ miles} \][/tex]
Therefore, at 2.5 hours, the distance is 875 miles, and the traveler is not moving but staying constant at this distance within the range [tex]\( 2.5 \leq t \leq 3.5 \)[/tex].
4. Distance at 3 hours
For [tex]\( t = 3 \)[/tex]:
Using the same part of the piecewise function as before:
[tex]\[ D(3) = 875 \text{ miles} \][/tex]
Therefore, at 3 hours, the distance is still 875 miles.
5. Total distance at 6 hours
For [tex]\( t = 6 \)[/tex]:
Using the third part of the piecewise function:
[tex]\[ D(6) = 75(6) + 612.5 = 450 + 612.5 = 1062.5 \text{ miles} \][/tex]
Therefore, the total distance from home after 6 hours is 1062.5 miles.
Given these calculations, we can now select the true statements:
- The starting distance at 0 hours is 125 miles (not 300 miles).
- At 2 hours, the traveler is 725 miles from home.
- At 2.5 hours, the traveler is not moving and the distance is 875 miles.
- At 3 hours, the distance is constant, at 875 miles.
- The total distance from home after 6 hours is 1062.5 miles.
Thus, the correct options are:
- At 2 hours, the traveler is 725 miles from home.
- At 3 hours, the distance is constant, at 875 miles.
- The total distance from home after 6 hours is 1,062.5 miles.
[tex]\[ D(t) = \left\{ \begin{array}{cl} 300t + 125, & 0 \leq t < 2.5 \\ 875, & 2.5 \leq t \leq 3.5 \\ 75t + 612.5, & 3.5 < t \leq 6 \end{array} \right. \][/tex]
We need to determine the distances at specific times using this function:
1. Starting distance, at 0 hours
For [tex]\( t = 0 \)[/tex]:
[tex]\[ D(0) = 300(0) + 125 = 125 \text{ miles} \][/tex]
Therefore, the starting distance at 0 hours is 125 miles.
2. Distance at 2 hours
For [tex]\( t = 2 \)[/tex]:
[tex]\[ D(2) = 300(2) + 125 = 600 + 125 = 725 \text{ miles} \][/tex]
Therefore, at 2 hours, the traveler is 725 miles from home.
3. Distance at 2.5 hours
For [tex]\( t = 2.5 \)[/tex]:
Using the second part of the piecewise function:
[tex]\[ D(2.5) = 875 \text{ miles} \][/tex]
Therefore, at 2.5 hours, the distance is 875 miles, and the traveler is not moving but staying constant at this distance within the range [tex]\( 2.5 \leq t \leq 3.5 \)[/tex].
4. Distance at 3 hours
For [tex]\( t = 3 \)[/tex]:
Using the same part of the piecewise function as before:
[tex]\[ D(3) = 875 \text{ miles} \][/tex]
Therefore, at 3 hours, the distance is still 875 miles.
5. Total distance at 6 hours
For [tex]\( t = 6 \)[/tex]:
Using the third part of the piecewise function:
[tex]\[ D(6) = 75(6) + 612.5 = 450 + 612.5 = 1062.5 \text{ miles} \][/tex]
Therefore, the total distance from home after 6 hours is 1062.5 miles.
Given these calculations, we can now select the true statements:
- The starting distance at 0 hours is 125 miles (not 300 miles).
- At 2 hours, the traveler is 725 miles from home.
- At 2.5 hours, the traveler is not moving and the distance is 875 miles.
- At 3 hours, the distance is constant, at 875 miles.
- The total distance from home after 6 hours is 1062.5 miles.
Thus, the correct options are:
- At 2 hours, the traveler is 725 miles from home.
- At 3 hours, the distance is constant, at 875 miles.
- The total distance from home after 6 hours is 1,062.5 miles.
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