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?

- anonymous

- chestercat

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- anonymous

It's the same idea as before. The distance to work doesn't change.

- anonymous

Remember the formula \(d = vt\).

- anonymous

Since \(d\) is the same no matter how he gets there, \((vt)_{jog} = (vt)_{bike}\).

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## More answers

- anonymous

but I don't know why in sometimes it becomes d1+d2

- anonymous

All I can think for that would be \(d_{1} + d_{2} = d_{total}\)

- anonymous

But that formula wouldn't apply for this question.

- anonymous

\[vt 1=vt2\]
\[\frac{ 1 }{ 7 }t=\frac{ 1 }{ 6 } (12-t) \]

- 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

So V for both will be 12 ?

- 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

No, V cannot be 12 for both.

- anonymous

Okay so v bike = 12 and v jop= 12-t

- anonymous

No.

- anonymous

He doesn't bike at a velocity of 12.

- anonymous

he rides 12 miles per hour faster than he jogs

- anonymous

Yes! There you go.

- anonymous

He will ride a bike 12 miles per hour
and he will jog 12-t

- 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

" rides 12 miles per hour FASTER THAN" he jogs

- anonymous

Have you plugged in your values for these?
\[\begin{align*}
v_{jog} &=
\\ v_{bike} &=
\\t_{jog} &=
\\t_{bike} &=
\end{align*}\]

- anonymous

nope

- anonymous

If he bikes 12 mph faster than he jogs, \(v_{bike} = v_{jog} + 12\), right?

- anonymous

That should get you started. There are two more values you can plug in.

- anonymous

@zello then you'd better fill it in, keep in mind the unit match with variable !

- anonymous

Yes! Keeping track of units is crucial, as is remembering what your variable means when you're done.

- 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

V bike= V jog+12
V jog = V bike-12
T jog =7/12
T bike = 1/6

- anonymous

Your times are right, but not your velocities.

- anonymous

V-jog is wrong.

- anonymous

It's just V-jog. Don't equate it to anything. It will be the variable you solve for.

- anonymous

Okay

- anonymous

\[vt Bike = vt Jog\]
\[\frac{ 1 }{ 6 }(vJog+12) =\frac{ 7 }{ 12 } v\]

- anonymous

Yes! Good.

- anonymous

now what ?

- anonymous

Now solve for \(v_{jog}.\)

- anonymous

And rememberâ€”the v on the right side is also \(v_{jog}\).

- 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

Ok

- 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

I left my answers in fractions, so I'm not sure. What do you get for your final answer?

- anonymous

I just solve V jog

- anonymous

@zello something isn't right with your calculation !

- anonymous

No, that's not what the question wants. Go back and read the last sentence.

- anonymous

@Chlorophyll It's just rounding error. I got \(v_{jog} = \frac{24}{5}\), so I think he's okay.

- anonymous

Although I only looked at his answer, not his work.

- anonymous

I'm a girl -_-

- anonymous

Oh, I'm sorry. I judge too quickly by avatars.

- anonymous

Although that's obviously not you... :S

- anonymous

Yes, it's very neat V = 24/5 ( 4.8 mi/hr)
@zello Better keep it in fraction form!

- 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

I have a headache because from 9 A.M until now, I still working on my math >.<

- 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

how can i found the distance then

- anonymous

find *

- anonymous

You know your distance formula. We've used it a lot.

- anonymous

d= vt
d jog= d bike

- anonymous

Top one.

- anonymous

vt jog = vt bike

- anonymous

No, just \(d = vt\). You know \(v\) now, and you already knew \(t\), so solve for \(d\).

- anonymous

d bike = v bike * t bike
d= 4.76 * 0.58
d= 2.7608

- 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

Excellent job working through this.

- anonymous

@zello I wonder if you know how to say thanks by distributing the medal (?)

- anonymous

Thank you so much and I'm sorry that I took a lot of your time. I really appreciate that

- Hero

I got 2.8 exactly with my methods.

- 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

Yup, that's why I say zello's calculation is incorrect!

- anonymous

It' not incorrect, it's just because I didn't round my answer

- anonymous

It's exact, so there's no need to round!

- anonymous

it's not exact in my calculator

- anonymous

So you're unable to calculate the fraction?

- anonymous

I know how to calculate the fraction but the question says that they want the answer rounded to the nearest tenth

- 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

There's nothing to round off :)

- anonymous

when I calculate it, I didn't use the fraction

- anonymous

However, thank you

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