A long distance trucker traveled a 180 miles during a snow storm in one direction. The return trip on a rainy day at double the speed it took 3 hours less. Find the speed going

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A long distance trucker traveled a 180 miles during a snow storm in one direction. The return trip on a rainy day at double the speed it took 3 hours less. Find the speed going

Mathematics
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Did you try to figure out how to solve it?
say the speed going is v. say it too time t to reach the destination while going. express v and t in terms of the distance.
took* time t

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nikki, we need you to give us some feedback in order to help you.
yeah, solving your math homework is not the point of Openstudy
I did try...I ended up with 2 different variables and cannot solve.
Write down what you got. We can correct you
I got for the return trip 180=2r(t-3)
r is the speed and t is time
right, what is the going trip formula?
180=r(t+3)
I do not know the time to figure out the speed
If im not mistaken the original should be 180 = r* (t)
ok but how do you figure out the time to get the rate?
why did you write 180 = r(t+3)? Does the trucker take 6 extra hours on the way to the destination?
well simplify the second equation
first define what t is.
because I added three hours to the time of the return trip
t is time
define what t is. what do you mean by t?
t is time taken to do what?
the time for the total distance of 180
time for total distance of 180? what does that mean? how can time be for distance?
Nikki, t represents the time of the going trip
What it the time of the going back trip?
it takes t amount of time to go 180 miles on the going trip, and it takes t-3hours on the return trip. im confused on how to find t because i need t to find r
I think this is a classic mechanic problem. So you can use the velocity formula (V=d/t)
right, so it takes t amount of time, going at speed r to reach 180 miles.
so 180 = t*r
That is what I am using i learned it as d=rt which the same thing with only the distance in a different spot
okay while returning, the speed is 2r, and the time take to travel back the same distance reduces by 3 hours. so 180 = 2r(t-3)
yes i have that. but how do you find the speed from that with another variable
so, lets call 180 = tr ------------------------- equation 1 180 = 2r(t-3) ------------------------- equation 2. from equation 1 we have t = 180/r, correct?
yes!!! thank you i got it from here!!! much thanks
so substitute t = 180/r in equation 2 and post what you got.
alright, you are welcome.
30=r

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