You have it figured out.
Rearranging and combining you had an error.
I got 5r^2-81r+235=0
Solve using the quadratic formula. One of the solutions of the quadratic will be less than 5 which would imply that the canoe would change directions for the (r-5) rate.
start with 47/r+9/r-5=5hrs OK to here. I will rewrite to see it more clearly.
[47/r] + [9/(r-5)] = 5
multiply both sides by r(r-5) to get
47(r-5) + 9r = 5(r)(r-5)
47r - 235 + 9r = 5r^2 - 25r
47r - 235 + 9r -5r^2 + 25r = 0
-5r^2 + 81r -235 = 0
I changed the sign here but it isn't necessary.
5r^2 -81r + 235 = 0
and solve by the quadratic formula. I et r = 12.4 and if I remember the problem that is mph. Then r-5 = 7.4 mph.
I hope this helps.
Then I multiply both sides by the LCD?
47r-???+9r=3r^2? I am lost here
I GOT IT!!!!! THANK YOU!!! THANK YOU!!! THANK YOU!!!
Algebra distance problem - I can't figure this one out During a trip, a canoe ...
Algebra - during the first part of a canoe trip the canoe travels 71 miles at a ...
algebra - during the first part of a trip, a conoeist travels 70 miles at a ...
algebra - Can someoe please help me figure this out? During the first part of a ...
algebra - a two part canoe trip,the first part of the trip the canoist traveled...
Algebra - During the first part of a trip, a canoeist travels 98 miles at a ...
Algebra - During the first part of a trip, a canoeist travels 57 miles at a ...
algebra - during the first trip a canoeist travels 79 miles at a certain speed. ...
algebra 2 - During the first part of a trip a canoeist travels 93 miles at a ...
Algebra 2 - During the first part of a trip a canoeist travels 93 miles at a ...