College Algebra Formula Chart
I’m soooooooooo lost in my college intermediate algebra class…can someone help me please?
I’m really lost on the “quadratic equations, inequalities, and functions” chapter.
Solving a motion problem.
- On a windy day Eduardo found that he could go 16 mi downstream and then 4 mi back upsteam at top speed in a total of 48 min. What was the top speed of Eduardo’s book if the current was 15 mph?
it has something to do with the distance formula d = rt and then using that solve for time I think, t = d/r
they have a chart in the book with 4 sections
the 1st row showed : upstream / (d) 4 / (r) x – 5/ (t) blank
the 2nd row showed: downsteam / (d) 16) / (r) blank / (t) blank
how do I solve it?
thanks!
yes, it has something to do with d = rt
let x be the speed at which Eduarado travel
when he travels upstream, his resutant speed is x – 15.
when he travels downstream, his resultant speed is x + 15
now you need to convert minute to hours
48min = 4/5hrs
given that the total time is 4/5hrs. That means the time it takes to travel upstream plus the time it takes to travel downstream will equal to 4/5
recall that d = rt
then t = d/r
upstream: given the distance is 4 min
speed: x – 15
so t = 4/(x-15)
downstream: given the distance is 16 mile
speed: x + 15
so = t = 16/(x + 15)
4/(x – 15) + 16/(x+15) = 4/5
now just solve
4x + 60 + 16x – 240 = (4x^2 – 900)/5
20x – 180 = (4x^2 – 900) / 5
100x – 900 = 4x^2 – 900
100x = 4x^2
0 = 4x^2 – 100x
0 = 4x (x – 25)
x = 0 or 25
take 25 as the answer
so the maximun speed is 25 mi/hr.
hope it helps.
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