You can
put this solution on YOUR website!(79/v)+2/(v-5)=2
Multiply out gives
(79v-395+2v)/v^2-5v = 2
81v-395 = 2v^2-10V
-395 = 2v^2-91v
divide each side by 2
-395/2 = v^2-91v/2
Complete the square
-395/2 = (v-91/4)^2 - (91/4)^2
(-395/2 + (91/4)^2)^1/2 = v - 91/4
v = 17.89 + 91/4 = 40.64
OR
v = -17.89 + 91/4 = 4.86
since you can not have a negative speed (4.86 - 5) then we can ignore the second solution for v.
so the speed on the first part of the trip was 40.64 mph and 35.64mph on the second part.
You can
put this solution on YOUR website! During the first part of a trip, a canoeist travels 79 miles at a certain speed. The canoeist travels 2 miles on the second part of the trip at a speed 5mph slower. The total time for the trip is 2 hrs. What was the speed on each part of the trip?
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1st part DATA:
distance = 79 miles ; rate = x mph ; time = d/r = 79/x hrs.
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2nd part DATA:
distance = 2 miles ; rate = x-5 mph ; time = d/r = 2/(x-5) hrs
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Equation:
time + time = 2 hrs
79/x + 2/(x-5) = 2
Multiply thru by x(x-5) to get:
79(x-5) + 2x = 2x(x-5)
79x - 395 + 2x = 2x^2 - 10x
2x^2 -91x + 395 = 0
Use Quadratic Formula to
get a reasonable value of "x":
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The answers are 4.86 and 40.64
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Neither answer is reasonable for canoeing speed.
It can't be 4.86 because x-5 would be negative.
It can't be 40.64 because canoes don't go that fast.
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Cheers,
stan H.
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