Asked by Bethany

Several cows from a Caribou Herd took 4 days longer to travel 70km to Forde Lake, than it took them to travel 60km north beyond forde lake.They averaged 5km/h less before Forde lake because the foraging was better. What was their average speed for the part beyond forde lake?

Could someone please give a detailed answer solving this question with all the steps please? Thanks in advance! I made this chart, with 'distance', 'time' 'speed', and 'to forge' and 'beyond forge'. I do not know how to fill in the time and distance part and let x equal to what(I am guessing rate/speed because that's what's asked?)
Answered by Damon
speed before Forde = (s-5)
speed after Forde = s
time to Forde = (t+96) note 4 days = 96 hr
time after Forde = t
then distance = speed * time
70 = (s-5)(t+96)
60 = s t or t = 60/s

70 = (s-5)(60/s + 96)

70 = 60 + 96 s - 300/s - 480

490 = 96 s - 300/s
245 = 48 s - 150/s
or
48 s^2 -245 s -150 = 0
s = (1/96)(245 +/- 5 sqrt(3553) )
s = (1/96)(543)
s = 5.66 km/hr
Answered by Bethany
I don't understand what s = (1/96)(245 +/- 5 sqrt(3553) ) means. like is the +/- the plu/minus sign? How does 245+/-5 square root of 3553 equal 543?
Answered by Bethany
wait hold on, is it like the equation x=[-b+/- sqrt(b^2-4ac)]/2a ? Thanks so much for the help Damon!
Answered by Damon
yes, quadratic equation
Answered by Bethany
ok awesome, but I don't understand how u got 543 from 245+5sqrt(3553)- cant seem to get a whole number
Answered by Bethany
did u go 245-5 or 245+5? How did 3553 come out?
Answered by Bethany
Aha sorry, I got it now! But why did we use the positive and not the negative?
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