To determine which girl arrives at the activity center first, we can find the point at which each girl's distance (in miles) reaches 6 miles, as this indicates they have arrived.
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Bethany's equation: \( y = 4x \)
- Set \( y = 6 \): \[ 6 = 4x \implies x = \frac{6}{4} = 1.5 \text{ hours} \]
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Morgan's equation: \( y = 8x - 4 \)
- Morgan leaves half an hour after Bethany, so we need to account for that by letting \( x \) start at 0.5 instead. Hence, we rewrite Morgan's equation:
- For Morgan, if we set \( d \) (distance) to 6: \[ 6 = 8x - 4 \implies 8x = 10 \implies x = \frac{10}{8} = 1.25 \text{ hours} \]
- However, since she leaves half an hour later, the total time from when Bethany leaves is \( 1.25 + 0.5 = 1.75 \text{ hours} \).
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Taylor's equation: \( y = 20x - 20 \)
- Taylor leaves one hour after Bethany, so we need to account for that. Therefore, we set \( x \) starting from 1 hour:
- Set \( d = 6 \): \[ 6 = 20x - 20 \implies 20x = 26 \implies x = \frac{26}{20} = 1.3 \text{ hours} \]
- Adding the one hour delay, the total time from when Bethany leaves is \( 1.3 + 1 = 2.3 \text{ hours} \).
Now we summarize the arrival times:
- Bethany arrives at 1.5 hours after leaving.
- Morgan arrives at 1.75 hours after leaving (1.25 + 0.5).
- Taylor arrives at 2.3 hours after leaving (1.3 + 1).
Conclusion: Bethany arrives first at the activity center at 1.5 hours.
Answer: Bethany