2 hours traveling+1 hour waiting=3 hours, 6hours total-3= 3hours return trip. X= mph on first leg 2hours X=3hours (X-20) 2x=3x-60 -1x=-60 x=60 60mph first leg, 40mph second leg
Mr. Derbyshire traveled at 60 mph on the trip to Loganville and 40 mph on the return trip home.
To solve the problem of determining Mr. Derbyshire's speeds for the two legs of his trip, follow these steps:
Let x be the speed in miles per hour (mph) on the trip from Mr. Derbyshire’s house to Loganville.
Speed going to Loganville: x mph[/tex]
Speed returning home: x - 20 \) m
Trip Details:
Travel Time to Loganville:
Time to Loganville = Speed Distance = x d
Since this time is 2 hours:
x d = 2 ⟹ d = 2 x
Travel Time Returning Home:
Time to return home = Speed Distance = x − 20 d
Substitute d = 2x \
Time to return home = x − 20 2 x
Total Time Away from Home:
Mr. Derbyshire is gone for a total of 6 hours, including the 1 hour break in Loganville:
Total time = Time to Loganville + Time to return home + 1
2 + x − 20 2 x + 1 = 6
Solve for x \
2 + x − 20 2 x + 1 = 6
3 + x − 20 2 x = 6
x − 20 2 x = 6 − 3
x − 20 2 x = 3
Cross-multiply to solve for x \
2 x = 3 ( x − 20 )
2 x = 3 x − 60
2 x − 3 x = − 60
− x = − 60
x = 60
Speed to Loganville is [tex] x = 60 mph
Speed returning home:
x - 20 = 60 - 20 = 40 \) m
Mr. Derbyshire traveled at 60 mph to Loganville and returned at 40 mph. This was determined by calculating the total time away and the different speeds for each leg of the trip. The setup involved equating distances traveled during both legs of the journey to find the speeds.
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