June rowed 45 upstream.
The next day she rowed the same distance Downstream taking her 4hrs less.
If she can row 7mph in still water,
find the speed of the current river.
6 answers
you need the time upstream
45 miles** Upstream
upstream
(7-current)*time=45
downstream:
(7+current)*(time-4)=45
solve for time in the first equation
time=45/(7-current)
put it in the second equation
(7+current)*(45/(7-crrent) -4)=45
(7+c)(45-4(7-c)=45*(7-c)
well, it is messy, and I dont see a simple trick
(7+c)(45-28+4c)=315-45c
(7+c)(17+4c)=315-45c
Can you FOIL the left side out, then combine units to get a quadratic equation, and use the quadratic formula?
(7-current)*time=45
downstream:
(7+current)*(time-4)=45
solve for time in the first equation
time=45/(7-current)
put it in the second equation
(7+current)*(45/(7-crrent) -4)=45
(7+c)(45-4(7-c)=45*(7-c)
well, it is messy, and I dont see a simple trick
(7+c)(45-28+4c)=315-45c
(7+c)(17+4c)=315-45c
Can you FOIL the left side out, then combine units to get a quadratic equation, and use the quadratic formula?
ignore the first post, you dont need that. See second post.
Yes I can.
I tried doing just that but I don't think it's right.
I ended up getting:
119+28c+17c+4c(2*) = 315-45c
0=4c(2*)+196-90c
0=2(2c[2*]+98-45c)
But now I can't find a possible solution?
Q: Am I doing something wrong?
I tried doing just that but I don't think it's right.
I ended up getting:
119+28c+17c+4c(2*) = 315-45c
0=4c(2*)+196-90c
0=2(2c[2*]+98-45c)
But now I can't find a possible solution?
Q: Am I doing something wrong?
(2c[2*] -45c+98)=0
quadratic equation
c= (45+-sqrt(45^2-8*98))/4
c= 2.44
check: upstream: 7-2.44=5.56
45=5.56*t
donwtream
45=9.44*t-4
subtract the first from second
0=4.88t-4*9.44
t= 8 hrs going up, 4 hours coming back
quadratic equation
c= (45+-sqrt(45^2-8*98))/4
c= 2.44
check: upstream: 7-2.44=5.56
45=5.56*t
donwtream
45=9.44*t-4
subtract the first from second
0=4.88t-4*9.44
t= 8 hrs going up, 4 hours coming back