anonymous
  • anonymous
Kevin can jog to work in 7/12 of an hour. When he rides his bike, it takes him 1/6 of an hour. If he rides 12 miles per hour faster than he jogs, how far away is his work?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
It's the same idea as before. The distance to work doesn't change.
anonymous
  • anonymous
Remember the formula \(d = vt\).
anonymous
  • anonymous
Since \(d\) is the same no matter how he gets there, \((vt)_{jog} = (vt)_{bike}\).

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anonymous
  • anonymous
but I don't know why in sometimes it becomes d1+d2
anonymous
  • anonymous
All I can think for that would be \(d_{1} + d_{2} = d_{total}\)
anonymous
  • anonymous
But that formula wouldn't apply for this question.
anonymous
  • anonymous
\[vt 1=vt2\] \[\frac{ 1 }{ 7 }t=\frac{ 1 }{ 6 } (12-t) \]
anonymous
  • anonymous
?
anonymous
  • anonymous
Once you've determined what to equate (in this case, \(d\)), it's best to write out everything you need and see what you already know. That way, you know what to solve for. You're looking for \((vt)_{jog} = (vt)_{bike}\). There are 4 variables there: \[\begin{align*} v_{jog} &= \\ v_{bike} &= \\t_{jog} &= \\t_{bike} &= \end{align*}\]
anonymous
  • anonymous
So V for both will be 12 ?
anonymous
  • anonymous
You've used \(12 - t\), which is not correct because the 12 relates to his velocity, not time. Go back to your question and see what you can plug in. You should end up with three answers (one will be relative), leaving one variable to solve for.
anonymous
  • anonymous
No, V cannot be 12 for both.
anonymous
  • anonymous
Okay so v bike = 12 and v jop= 12-t
anonymous
  • anonymous
No.
anonymous
  • anonymous
He doesn't bike at a velocity of 12.
anonymous
  • anonymous
he rides 12 miles per hour faster than he jogs
anonymous
  • anonymous
Yes! There you go.
anonymous
  • anonymous
He will ride a bike 12 miles per hour and he will jog 12-t
anonymous
  • anonymous
Nope. That's the same thing you said before. he does not bike at a speed of 12 mph. He bikes 12 mph *faster* than he jogs. You need to know how fast he jogs to know how fast he bikes.
anonymous
  • anonymous
" rides 12 miles per hour FASTER THAN" he jogs
anonymous
  • anonymous
Have you plugged in your values for these? \[\begin{align*} v_{jog} &= \\ v_{bike} &= \\t_{jog} &= \\t_{bike} &= \end{align*}\]
anonymous
  • anonymous
nope
anonymous
  • anonymous
If he bikes 12 mph faster than he jogs, \(v_{bike} = v_{jog} + 12\), right?
anonymous
  • anonymous
That should get you started. There are two more values you can plug in.
anonymous
  • anonymous
@zello then you'd better fill it in, keep in mind the unit match with variable !
anonymous
  • anonymous
Yes! Keeping track of units is crucial, as is remembering what your variable means when you're done.
anonymous
  • anonymous
all i can get is he jogs for 35 minutes, and rides for 10 minutes. He rides 2 miles faster per 10 minutes than he jogs O.O
anonymous
  • anonymous
V bike= V jog+12 V jog = V bike-12 T jog =7/12 T bike = 1/6
anonymous
  • anonymous
Your times are right, but not your velocities.
anonymous
  • anonymous
V-jog is wrong.
anonymous
  • anonymous
It's just V-jog. Don't equate it to anything. It will be the variable you solve for.
anonymous
  • anonymous
Okay
anonymous
  • anonymous
\[vt Bike = vt Jog\] \[\frac{ 1 }{ 6 }(vJog+12) =\frac{ 7 }{ 12 } v\]
anonymous
  • anonymous
Yes! Good.
anonymous
  • anonymous
now what ?
anonymous
  • anonymous
Now solve for \(v_{jog}.\)
anonymous
  • anonymous
And remember—the v on the right side is also \(v_{jog}\).
anonymous
  • anonymous
Call it \(v\), or \(x\), or whatever you want. It's the same no matter how you approach it. If the subscript "jog" confuses you, omit it, but remember what \(v\) represents.
anonymous
  • anonymous
Ok
anonymous
  • anonymous
\[\frac{ 1 }{ 6 }(v+12)=\frac{ 7 }{ 12 }v\] \[\frac{ 1 }{ 6 }v+2=\frac{ 7 }{ 12 }\] \[\frac{ 1 }{ 6}v-\frac{ 7 }{ 12 }v=-2\] \[0.16v-0.58=-2\] \[-0.42 v=-2\] \[v=4.76\]
anonymous
  • anonymous
I left my answers in fractions, so I'm not sure. What do you get for your final answer?
anonymous
  • anonymous
I just solve V jog
anonymous
  • anonymous
@zello something isn't right with your calculation !
anonymous
  • anonymous
No, that's not what the question wants. Go back and read the last sentence.
anonymous
  • anonymous
@Chlorophyll It's just rounding error. I got \(v_{jog} = \frac{24}{5}\), so I think he's okay.
anonymous
  • anonymous
Although I only looked at his answer, not his work.
anonymous
  • anonymous
I'm a girl -_-
anonymous
  • anonymous
Oh, I'm sorry. I judge too quickly by avatars.
anonymous
  • anonymous
Although that's obviously not you... :S
anonymous
  • anonymous
Yes, it's very neat V = 24/5 ( 4.8 mi/hr) @zello Better keep it in fraction form!
anonymous
  • anonymous
@zello When you get to this stage: \[\frac{ 1 }{ 6}v-\frac{ 7 }{ 12 }v=-2\] Multiply everything by 12 and simplify to get rid of the fractions.
anonymous
  • anonymous
I have a headache because from 9 A.M until now, I still working on my math >.<
anonymous
  • anonymous
Well, you've been working hard, and definitely showing growth. You're welcome to take a break, but you're almost done this question. Remember, we needed to find \(v_{jog}\) to solve the equation, but \(v_{jog}\) is not what the question asks for. It asks how far away his work is.
anonymous
  • anonymous
how can i found the distance then
anonymous
  • anonymous
find *
anonymous
  • anonymous
You know your distance formula. We've used it a lot.
anonymous
  • anonymous
d= vt d jog= d bike
anonymous
  • anonymous
Top one.
anonymous
  • anonymous
vt jog = vt bike
anonymous
  • anonymous
No, just \(d = vt\). You know \(v\) now, and you already knew \(t\), so solve for \(d\).
anonymous
  • anonymous
d bike = v bike * t bike d= 4.76 * 0.58 d= 2.7608
anonymous
  • anonymous
Good. Don't forget your units. It's technically 2.8, so you're a bit off due to rounding errors, but that's right.
anonymous
  • anonymous
Excellent job working through this.
anonymous
  • anonymous
@zello I wonder if you know how to say thanks by distributing the medal (?)
anonymous
  • anonymous
Thank you so much and I'm sorry that I took a lot of your time. I really appreciate that
Hero
  • Hero
I got 2.8 exactly with my methods.
anonymous
  • anonymous
Not a problem at all. Glad I could help. Thank you for making the effort to solve it on your own and learn.
anonymous
  • anonymous
Yup, that's why I say zello's calculation is incorrect!
anonymous
  • anonymous
It' not incorrect, it's just because I didn't round my answer
anonymous
  • anonymous
It's exact, so there's no need to round!
anonymous
  • anonymous
it's not exact in my calculator
anonymous
  • anonymous
So you're unable to calculate the fraction?
anonymous
  • anonymous
I know how to calculate the fraction but the question says that they want the answer rounded to the nearest tenth
anonymous
  • anonymous
First, you should keep the number in fraction if it isn't exact result Second, in this case it's exact! 7v / 12 = ( v + 12 ) /6 (7/ 12 - 1/6)v = 2 -> v = 2 * 12/5 = 24/5 = 2.8 mi/ hr
anonymous
  • anonymous
There's nothing to round off :)
anonymous
  • anonymous
when I calculate it, I didn't use the fraction
anonymous
  • anonymous
However, thank you

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