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☪Λaгѳԓ☪ wrote:
Yeah but if you can't give a formula and TTK can then you both get a 135ωⅇɢṡɪⅇṣ wrote:
If its .333 then I win 😄ThatTallKid_ wrote:
.83 is too fast but .33 may be correct. I'll check itYOU wrote:
2.5/3= .83mph. I dunno that's what I got.YOU wrote:
The guy is going 1mph. He goes 3 miles back to his starting point. The stick covers 1 mile in 3 hrs so .3mph..3mph
That might be wrong so 2.5m/3hrs➡ Is your answer.
Either .3mph or .83mph is your answer. I don't remember Algebra 2 freshman year with those kind of problems😁 -
♛Çąŋժूƴɱąŋ♛ wrote:
Shit I just reread it. Give this man your loot.Doc Xray wrote:
👆👆👆👆thisAssuming the rower travels at a constant rate of speed (both upstream and downstream, which is bogus), then the stick is traveling at 1/3 the speed of the rower (moves one "segment" while the rower moves 3 segments). We're told that the rower is traveling at 1 mile per hour, thus the current (stick) is moving at 1/3 mph. Reality would require the rower to travel downstream faster than upstream. If you assume constant effort by the rower (without tiring), and also assume the rower moves twice as fast downstream as upstream, then the stick moves one segment in the same amount of time as the rower moving two segments (1 at speed x, and 2 at speed 2x). Therefore, the current is moving at 0.5 mph. Too many assumptions to resolve exactly!
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Oh duh my bad but we still don't know that he was going 1 mph so we have the stick going one mile at s speed with the rower going one hour at r speed and one hour and one mile at s+r speed the hours can be converted into miles by multiplying them by the speed so we have 1s=s+r+r^2+[s+r]^2 I think correct me if I am wrong
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Lady Katsa wrote:
There's no way to solve for r or s with that thoughOh duh my bad but we still don't know that he was going 1 mph so we have the stick going one mile at s speed with the rower going one hour at r speed and one hour and one mile at s+r speed the hours can be converted into miles by multiplying them by the speed so we have 1s=s+r+r^2+[s+r]^2 I think correct me if I am wrong
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ωⅇɢṡɪⅇṣ wrote:
I know I'm sorry I'm working on that but does anyone see anything I missed on that?Lady Katsa wrote:
There's no way to solve for r or s with that thoughOh duh my bad but we still don't know that he was going 1 mph so we have the stick going one mile at s speed with the rower going one hour at r speed and one hour and one mile at s+r speed the hours can be converted into miles by multiplying them by the speed so we have 1s=s+r+r^2+[s+r]^2 I think correct me if I am wrong
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Lady Katsa wrote:
I tried the same thing you're doing and I just ended up with a hell of a lot of useless variables😔ωⅇɢṡɪⅇṣ wrote:
I know I'm sorry I'm working on that but does anyone see anything I missed on that?Lady Katsa wrote:
There's no way to solve for r or s with that thoughOh duh my bad but we still don't know that he was going 1 mph so we have the stick going one mile at s speed with the rower going one hour at r speed and one hour and one mile at s+r speed the hours can be converted into miles by multiplying them by the speed so we have 1s=s+r+r^2+[s+r]^2 I think correct me if I am wrong
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K how about using this too rowers total distance is d so d=1+s+2r not sure if that helps cause it throws in d but just an idea
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Argh... So much math!
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Im gonna say its .5 mph. If she is going 1mph then it takes her 1.5 mph to get to the spot she turned at cause she is going upstream so you add the currents speed to her speed. The stick would arrive in 2 hours. When she turns around she is going .5 mph cause you subtract it cause she is going with the current. That means she must travel 1 and a half miles in .5 mph and it would take her 1.5 hours to do it. Add the 1.5 hours and .5 and get 2 hours. The stick arrives at 2 hours cause it is traveling .5 mph so it takes 2 hours. I don't see a problem with this and don't know how people are getting .33 but I'm assuming you must apply the current.
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Vince Carolli wrote:
The problem is you are assuming that the rower is going 1mph while that may be true and perhaps I am just over complicating things the question doesn't say thatIm gonna say its .5 mph. If she is going 1mph then it takes her 1.5 mph to get to the spot she turned at cause she is going upstream so you add the currents speed to her speed. The stick would arrive in 2 hours. When she turns around she is going .5 mph cause you subtract it cause she is going with the current. That means she must travel 1 and a half miles in .5 mph and it would take her 1.5 hours to do it. Add the 1.5 hours and .5 and get 2 hours. The stick arrives at 2 hours cause it is traveling .5 mph so it takes 2 hours. I don't see a problem with this and don't know how people are getting .33 but I'm assuming you must apply the current.
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Lady Katsa wrote:
In assuming he traveled 3 miles if he is going at a constant rate. And the stick travels same amount of time, but a 3rd of the distance. Asumptions here.Vince Carolli wrote:
The problem is you are assuming that the rower is going 1mph while that may be true and perhaps I am just over complicating things the question doesn't say thatMath Post here
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ThatTallKid_ wrote:
Constant speed yea but its never stated what that speed is and the current affects that speedLady Katsa wrote:
In assuming he traveled 3 miles if he is going at a constant rate. And the stick travels same amount of time, but a 3rd of the distance. Asumptions here.Vince Carolli wrote:
The problem is you are assuming that the rower is going 1mph while that may be true and perhaps I am just over complicating things the question doesn't say thatMath Post here
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YOU wrote:
Amendment to the above equation s=r^2+r(s+r)+s+rOh duh my bad but we still don't know that he was going 1 mph so we have the stick going one mile at s speed with the rower going one hour at r speed and one hour and one mile at s+r speed the hours can be converted into miles by multiplying them by the speed so we have 1s=s+r+r^2+[s+r]^2 I think correct me if I am wrong
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Chacha failed me, I gotta put my thinking cap on here
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Lady Katsa wrote:
The only units given were 1 mile. Then 1 hour so I'm assuming 1mph. Or the answer is a ratio since its a word problem answer would be something along the lines of "1/3 of rowers speed" something like that, I dunnoThatTallKid_ wrote:
Constant speed yea but its never stated what that speed is and the current affects that speedLady Katsa wrote:
In assuming he traveled 3 miles if he is going at a constant rate. And the stick travels same amount of time, but a 3rd of the distance. Asumptions here.Vince Carolli wrote:
The problem is you are assuming that the rower is going 1mph while that may be true and perhaps I am just over complicating things the question doesn't say thatMath Post here
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Yeah I think it has to be in terms of the rowers speed because and again I could b wrong but the question only says that the rower went one mile then another hour then back but never that the mile took exactly one hour
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Lady Katsa wrote:
Yes same conclusion must be some sort of ratio. Case Closed!📁Yeah I think it has to be in terms of the rowers speed because and again I could b wrong but the question only says that the rower went one mile then another hour then back but never that the mile took exactly one hour
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Lawl I'm in 8th grade don't ask me
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ThatTallKid_ wrote:
It cannot be a ratio, there is only one correct answer because the moment the current is over 1 mph it beats the rower regardlessLady Katsa wrote:
Yes same conclusion must be some sort of ratio. Case Closed!📁Yeah I think it has to be in terms of the rowers speed because and again I could b wrong but the question only says that the rower went one mile then another hour then back but never that the mile took exactly one hour
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jcisabeast wrote:
Precisely there is only one answer but it can't be found until the rowers speed is found which needs more information to be determinedThatTallKid_ wrote:
It cannot be a ratio, there is only one correct answer because the moment the current is over 1 mph it beats the rower regardlessLady Katsa wrote:
Yes same conclusion must be some sort of ratio. Case Closed!📁Yeah I think it has to be in terms of the rowers speed because and again I could b wrong but the question only says that the rower went one mile then another hour then back but never that the mile took exactly one hour
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Lady Katsa wrote:
This is a quite possible question, I could get it by guess and check eventually but the formula is stumping mejcisabeast wrote:
Precisely there is only one answer but it can't be found until the rowers speed is found which needs more information to be determinedThatTallKid_ wrote:
It cannot be a ratio, there is only one correct answer because the moment the current is over 1 mph it beats the rower regardlessLady Katsa wrote:
Yes same conclusion must be some sort of ratio. Case Closed!📁Yeah I think it has to be in terms of the rowers speed because and again I could b wrong but the question only says that the rower went one mile then another hour then back but never that the mile took exactly one hour
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jcisabeast wrote:
You can guess, but no real way of checking since its mostly assumptions of the info given. CLady Katsa wrote:
This is a quite possible question, I could get it by guess and check eventually but the formula is stumping mejcisabeast wrote:
Precisely there is only one answer but it can't be found until the rowers speed is found which needs more information to be determinedThatTallKid_ wrote:
It cannot be a ratio, there is only one correct answer because the moment the current is over 1 mph it beats the rower regardlessLady Katsa wrote:
Yes same conclusion must be some sort of ratio. Case Closed!📁Good reasoning
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I've got it, websites do you have pal so I can send ya a pic?
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*wegsies
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Doc Xray wrote:
Damn, this answer makes me wanna give him a 135 loot turf.Assuming the rower travels at a constant rate of speed (both upstream and downstream, which is bogus), then the stick is traveling at 1/3 the speed of the rower (moves one "segment" while the rower moves 3 segments). We're told that the rower is traveling at 1 mile per hour, thus the current (stick) is moving at 1/3 mph. Reality would require the rower to travel downstream faster than upstream. If you assume constant effort by the rower (without tiring), and also assume the rower moves twice as fast downstream as upstream, then the stick moves one segment in the same amount of time as the rower moving two segments (1 at speed x, and 2 at speed 2x). Therefore, the current is moving at 0.5 mph. Too many assumptions to resolve exactly!
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If you account for the speed of the river when the rower goes upstream as well as when they go downstream then you can solve this with no more variables. The answer does boil down to .5 mph since it ends up taking the stick 2 hrs to go one mile. And no you don't have to assume that the rower was going 1 mph.
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Hint...independent of the rower speed of say X mph, after one hour they are exactly X miles apart since the rower will be slowed down as much as the stick was aided by the current.
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I have the answer and I will include the work in the next message... The answer is that the current is flowing at a rate of 0.5 miles per hour
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Let's set the variables:
x = rower's speed independent of current
y = rate of current
z = time for rower to row back downstream after rowing upstream for an hour after traveling 1 mile from starting point -
When traveling upstream for an hour, the rower covers a distance of:
1 hr * (x - y) mi/hr = x - y miWhen traveling back downstream, the rower covers a distance of (x - y) + 1 miles. During this time z, the rower is traveling at a rate of (x + y) mi/hr. The equation to represent distance traveled by speed*time is:
z hr * (x + y) mi/hr = (x - y + 1) mi
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