Here's the question you clicked on:
zello
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?
It's the same idea as before. The distance to work doesn't change.
Remember the formula \(d = vt\).
Since \(d\) is the same no matter how he gets there, \((vt)_{jog} = (vt)_{bike}\).
but I don't know why in sometimes it becomes d1+d2
All I can think for that would be \(d_{1} + d_{2} = d_{total}\)
But that formula wouldn't apply for this question.
\[vt 1=vt2\] \[\frac{ 1 }{ 7 }t=\frac{ 1 }{ 6 } (12-t) \]
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: \[\begin{align*} v_{jog} &= \\ v_{bike} &= \\t_{jog} &= \\t_{bike} &= \end{align*}\]
So V for both will be 12 ?
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.
No, V cannot be 12 for both.
Okay so v bike = 12 and v jop= 12-t
He doesn't bike at a velocity of 12.
he rides 12 miles per hour faster than he jogs
He will ride a bike 12 miles per hour and he will jog 12-t
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.
" rides 12 miles per hour FASTER THAN" he jogs
Have you plugged in your values for these? \[\begin{align*} v_{jog} &= \\ v_{bike} &= \\t_{jog} &= \\t_{bike} &= \end{align*}\]
If he bikes 12 mph faster than he jogs, \(v_{bike} = v_{jog} + 12\), right?
That should get you started. There are two more values you can plug in.
@zello then you'd better fill it in, keep in mind the unit match with variable !
Yes! Keeping track of units is crucial, as is remembering what your variable means when you're done.
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
V bike= V jog+12 V jog = V bike-12 T jog =7/12 T bike = 1/6
Your times are right, but not your velocities.
It's just V-jog. Don't equate it to anything. It will be the variable you solve for.
\[vt Bike = vt Jog\] \[\frac{ 1 }{ 6 }(vJog+12) =\frac{ 7 }{ 12 } v\]
Now solve for \(v_{jog}.\)
And remember—the v on the right side is also \(v_{jog}\).
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.
\[\frac{ 1 }{ 6 }(v+12)=\frac{ 7 }{ 12 }v\] \[\frac{ 1 }{ 6 }v+2=\frac{ 7 }{ 12 }\] \[\frac{ 1 }{ 6}v-\frac{ 7 }{ 12 }v=-2\] \[0.16v-0.58=-2\] \[-0.42 v=-2\] \[v=4.76\]
I left my answers in fractions, so I'm not sure. What do you get for your final answer?
@zello something isn't right with your calculation !
No, that's not what the question wants. Go back and read the last sentence.
@Chlorophyll It's just rounding error. I got \(v_{jog} = \frac{24}{5}\), so I think he's okay.
Although I only looked at his answer, not his work.
Oh, I'm sorry. I judge too quickly by avatars.
Although that's obviously not you... :S
Yes, it's very neat V = 24/5 ( 4.8 mi/hr) @zello Better keep it in fraction form!
@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.
I have a headache because from 9 A.M until now, I still working on my math >.<
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.
how can i found the distance then
You know your distance formula. We've used it a lot.
No, just \(d = vt\). You know \(v\) now, and you already knew \(t\), so solve for \(d\).
d bike = v bike * t bike d= 4.76 * 0.58 d= 2.7608
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.
Excellent job working through this.
@zello I wonder if you know how to say thanks by distributing the medal (?)
Thank you so much and I'm sorry that I took a lot of your time. I really appreciate that
I got 2.8 exactly with my methods.
Not a problem at all. Glad I could help. Thank you for making the effort to solve it on your own and learn.
Yup, that's why I say zello's calculation is incorrect!
It' not incorrect, it's just because I didn't round my answer
It's exact, so there's no need to round!
it's not exact in my calculator
So you're unable to calculate the fraction?
I know how to calculate the fraction but the question says that they want the answer rounded to the nearest tenth
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
There's nothing to round off :)
when I calculate it, I didn't use the fraction