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Sara went to meet her friend Rianna in Rome, but Rianna left Rome on a train 15 minutes earlier. The distance traveled by the t...Question
Sara went to meet her friend Rianna in Rome, but Rianna left Rome on a train 15 minutes earlier. The distance traveled by the train, in miles, with respect to time can be modeled by the equation y=40x+10
. Sara can buy a ticket on a faster train that will leave in 10 minutes and follow the same route. If the distance of the faster train with respect to time can be modeled by the equation y=60x−10
, which two lines on the graph show the solution to when Sara’s train catches up to Rianna’s train?
(1 point)
Responses
Line 3 and Line 4
Line 3 and Line 4
Line 2 and Line 3
Line 2 and Line 3
Line 1 and Line 4
Line 1 and Line 4
Line 1 and Line 2
. Sara can buy a ticket on a faster train that will leave in 10 minutes and follow the same route. If the distance of the faster train with respect to time can be modeled by the equation y=60x−10
, which two lines on the graph show the solution to when Sara’s train catches up to Rianna’s train?
(1 point)
Responses
Line 3 and Line 4
Line 3 and Line 4
Line 2 and Line 3
Line 2 and Line 3
Line 1 and Line 4
Line 1 and Line 4
Line 1 and Line 2
Answers
There are no human answers yet.
Answered by
GPT-4o mini
To determine when Sara’s train catches up to Rianna's train, we need to find the point where the two distance equations are equal.
Rianna's train's distance can be modeled by the equation:
\[ y = 40x + 10 \]
Sara's train's distance can be modeled by the equation:
\[ y = 60x - 10 \]
Here, \( y \) represents the distance in miles, and \( x \) represents the time in hours from when Rianna left the station. However, since Sara leaves 10 minutes (or \(\frac{1}{6}\) hours) later, we need to adjust her equation by accounting for this delay.
Let's express \( y \) for Sara's position in terms of the same time reference as Rianna's:
- When Sara starts, Rianna has already traveled for \( x - \frac{1}{6} \) hours. Thus:
\[ \text{For Sara: } y = 60\left(x - \frac{1}{6}\right) - 10 = 60x - 10 - 10 = 60x - 20 \]
Now we have:
1. Rianna's distance: \( y = 40x + 10 \)
2. Sara's distance: \( y = 60x - 20 \)
To find when they meet, we set the two equations equal to each other:
\[ 40x + 10 = 60x - 20 \]
Now, let's solved for \( x \):
1. Subtract \( 40x \) from both sides:
\[ 10 = 20x - 20 \]
2. Add \( 20 \) to both sides:
\[ 30 = 20x \]
3. Finally, divide by \( 20 \):
\[ x = \frac{3}{2} \text{ hours = 1.5 hours} \]
Now, substitute \( x = 1.5 \) back into either equation to find the distance:
Using Rianna's equation:
\[ y = 40(1.5) + 10 = 60 + 10 = 70 \text{ miles} \]
Using Sara's equation:
\[ y = 60(1.5) - 20 = 90 - 20 = 70 \text{ miles} \]
Both trains meet at 70 miles after 1.5 hours.
Now, in terms of interpreting the question regarding which lines represent this equation on a graph, we must identify which lines correspond to the given equations:
1. **Rianna's Train \(y = 40x + 10\)**: This is a straight line with a slope of 40 and y-intercept of 10.
2. **Sara's Train \(y = 60x - 20\)**: This is a straight line with a slope of 60 and y-intercept of -20.
To find the lines on the graph:
- **Line 1**: can represent \( y = 40x + 10 \)
- **Line 2**: could represent \( y = 60x - 20 \)
- **Line 3** and **Line 4** would represent other equations.
In this case, **the correct lines that show the solution to when Sara’s train catches up with Rianna are Line 1 (Rianna's train) and Line 2 (Sara's train)**.
So the correct response would be:
**Line 1 and Line 2**
Rianna's train's distance can be modeled by the equation:
\[ y = 40x + 10 \]
Sara's train's distance can be modeled by the equation:
\[ y = 60x - 10 \]
Here, \( y \) represents the distance in miles, and \( x \) represents the time in hours from when Rianna left the station. However, since Sara leaves 10 minutes (or \(\frac{1}{6}\) hours) later, we need to adjust her equation by accounting for this delay.
Let's express \( y \) for Sara's position in terms of the same time reference as Rianna's:
- When Sara starts, Rianna has already traveled for \( x - \frac{1}{6} \) hours. Thus:
\[ \text{For Sara: } y = 60\left(x - \frac{1}{6}\right) - 10 = 60x - 10 - 10 = 60x - 20 \]
Now we have:
1. Rianna's distance: \( y = 40x + 10 \)
2. Sara's distance: \( y = 60x - 20 \)
To find when they meet, we set the two equations equal to each other:
\[ 40x + 10 = 60x - 20 \]
Now, let's solved for \( x \):
1. Subtract \( 40x \) from both sides:
\[ 10 = 20x - 20 \]
2. Add \( 20 \) to both sides:
\[ 30 = 20x \]
3. Finally, divide by \( 20 \):
\[ x = \frac{3}{2} \text{ hours = 1.5 hours} \]
Now, substitute \( x = 1.5 \) back into either equation to find the distance:
Using Rianna's equation:
\[ y = 40(1.5) + 10 = 60 + 10 = 70 \text{ miles} \]
Using Sara's equation:
\[ y = 60(1.5) - 20 = 90 - 20 = 70 \text{ miles} \]
Both trains meet at 70 miles after 1.5 hours.
Now, in terms of interpreting the question regarding which lines represent this equation on a graph, we must identify which lines correspond to the given equations:
1. **Rianna's Train \(y = 40x + 10\)**: This is a straight line with a slope of 40 and y-intercept of 10.
2. **Sara's Train \(y = 60x - 20\)**: This is a straight line with a slope of 60 and y-intercept of -20.
To find the lines on the graph:
- **Line 1**: can represent \( y = 40x + 10 \)
- **Line 2**: could represent \( y = 60x - 20 \)
- **Line 3** and **Line 4** would represent other equations.
In this case, **the correct lines that show the solution to when Sara’s train catches up with Rianna are Line 1 (Rianna's train) and Line 2 (Sara's train)**.
So the correct response would be:
**Line 1 and Line 2**
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