To determine which girl arrives at the activity center first, we need to find out when each girl's distance function reaches 6 miles.
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Bethany's distance function: \( y = 4x \)
- To find when she reaches 6 miles, set \( y = 6 \): \[ 6 = 4x \implies x = \frac{6}{4} = 1.5 \text{ hours} \]
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Morgan's distance function: \( y = 8x - 4 \)
- Morgan leaves half an hour after Bethany, so we need to consider this delay and set \( y = 6 \): \[ 6 = 8x - 4 \implies 10 = 8x \implies x = \frac{10}{8} = 1.25 \text{ hours} \]
- Since she leaves 0.5 hours after Bethany, we add 0.5 hours: \[ 1.25 + 0.5 = 1.75 \text{ hours after school} \]
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Taylor's distance function: \( y = 20x - 20 \)
- Taylor leaves one hour after Bethany, so we set \( y = 6 \): \[ 6 = 20x - 20 \implies 26 = 20x \implies x = \frac{26}{20} = 1.3 \text{ hours} \]
- Since she leaves 1 hour after Bethany, we add 1 hour: \[ 1.3 + 1 = 2.3 \text{ hours after school} \]
Now we summarize the arrival times after school:
- Bethany arrives at 1.5 hours
- Morgan arrives at 1.75 hours
- Taylor arrives at 2.3 hours
From these calculations, we can conclude that Bethany arrives at the activity center first.
Final answer: Bethany