d=rt
distance = speed x time
Q=9+S
155=5Q + 5S = 5(9+S) + 5S = 45 + 5S + 5S = 45 +10S
10S = 110
S = 11 mph
Q=20 mph
Patty S.
asked 12/14/20Spencer and Quinn both leave the park at the same time, but in opposite directions. If Quinn travels 9 mph faster than Spencer and after 5 hours they are 155 miles apart, how fast is each traveling?
d=rt
distance = speed x time
Q=9+S
155=5Q + 5S = 5(9+S) + 5S = 45 + 5S + 5S = 45 +10S
10S = 110
S = 11 mph
Q=20 mph
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