SOLUTION: Cross-country cycling. Erin was traveling across the
desert on her bicycle. Before lunch she traveled 60 miles
(mi); after lunch she traveled 46 mi. She put in 1 hour
more after
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-> SOLUTION: Cross-country cycling. Erin was traveling across the
desert on her bicycle. Before lunch she traveled 60 miles
(mi); after lunch she traveled 46 mi. She put in 1 hour
more after
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Question 191981This question is from textbook
: Cross-country cycling. Erin was traveling across the
desert on her bicycle. Before lunch she traveled 60 miles
(mi); after lunch she traveled 46 mi. She put in 1 hour
more after lunch than before lunch, but her speed was
4 mph slower than before. What was her speed before
lunch and after lunch?
I tried solving this problem by using the d=r(t) equation.
x=rate y=time
before lunch= 60/x=y
after lunch= 48/x-4=y+1
so I substituted 60/x for y
46/x-4=60/x +1
46x=60(x-4) +x(x-4)
46x= 60x-240+x^2-4x
46x=x^2+56x-240
0=x^2+10x-240
I plug this into the quadratic formula, and it doesn't work, any suggestions?
This question is from textbook