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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?

  • one year ago
  • one year ago

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

    • one year ago
  2. geoffb
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    Remember the formula \(d = vt\).

    • one year ago
  3. geoffb
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    Since \(d\) is the same no matter how he gets there, \((vt)_{jog} = (vt)_{bike}\).

    • one year ago
  4. zello
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    but I don't know why in sometimes it becomes d1+d2

    • one year ago
  5. geoffb
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    All I can think for that would be \(d_{1} + d_{2} = d_{total}\)

    • one year ago
  6. geoffb
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    But that formula wouldn't apply for this question.

    • one year ago
  7. zello
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    \[vt 1=vt2\] \[\frac{ 1 }{ 7 }t=\frac{ 1 }{ 6 } (12-t) \]

    • one year ago
  8. zello
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    ?

    • one year ago
  9. geoffb
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    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: \[\begin{align*} v_{jog} &= \\ v_{bike} &= \\t_{jog} &= \\t_{bike} &= \end{align*}\]

    • one year ago
  10. zello
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    So V for both will be 12 ?

    • one year ago
  11. geoffb
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    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.

    • one year ago
  12. geoffb
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    No, V cannot be 12 for both.

    • one year ago
  13. zello
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    Okay so v bike = 12 and v jop= 12-t

    • one year ago
  14. geoffb
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    No.

    • one year ago
  15. geoffb
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    He doesn't bike at a velocity of 12.

    • one year ago
  16. zello
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    he rides 12 miles per hour faster than he jogs

    • one year ago
  17. geoffb
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    Yes! There you go.

    • one year ago
  18. zello
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    He will ride a bike 12 miles per hour and he will jog 12-t

    • one year ago
  19. geoffb
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    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.

    • one year ago
  20. Chlorophyll
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    " rides 12 miles per hour FASTER THAN" he jogs

    • one year ago
  21. geoffb
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    Have you plugged in your values for these? \[\begin{align*} v_{jog} &= \\ v_{bike} &= \\t_{jog} &= \\t_{bike} &= \end{align*}\]

    • one year ago
  22. zello
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    nope

    • one year ago
  23. geoffb
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    If he bikes 12 mph faster than he jogs, \(v_{bike} = v_{jog} + 12\), right?

    • one year ago
  24. geoffb
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    That should get you started. There are two more values you can plug in.

    • one year ago
  25. Chlorophyll
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    @zello then you'd better fill it in, keep in mind the unit match with variable !

    • one year ago
  26. geoffb
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    Yes! Keeping track of units is crucial, as is remembering what your variable means when you're done.

    • one year ago
  27. Bronco101
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    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

    • one year ago
  28. zello
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    V bike= V jog+12 V jog = V bike-12 T jog =7/12 T bike = 1/6

    • one year ago
  29. geoffb
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    Your times are right, but not your velocities.

    • one year ago
  30. geoffb
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    V-jog is wrong.

    • one year ago
  31. geoffb
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    It's just V-jog. Don't equate it to anything. It will be the variable you solve for.

    • one year ago
  32. zello
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    Okay

    • one year ago
  33. zello
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    \[vt Bike = vt Jog\] \[\frac{ 1 }{ 6 }(vJog+12) =\frac{ 7 }{ 12 } v\]

    • one year ago
  34. geoffb
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    Yes! Good.

    • one year ago
  35. zello
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    now what ?

    • one year ago
  36. geoffb
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    Now solve for \(v_{jog}.\)

    • one year ago
  37. geoffb
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    And remember—the v on the right side is also \(v_{jog}\).

    • one year ago
  38. geoffb
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    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.

    • one year ago
  39. zello
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    Ok

    • one year ago
  40. zello
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    \[\frac{ 1 }{ 6 }(v+12)=\frac{ 7 }{ 12 }v\] \[\frac{ 1 }{ 6 }v+2=\frac{ 7 }{ 12 }\] \[\frac{ 1 }{ 6}v-\frac{ 7 }{ 12 }v=-2\] \[0.16v-0.58=-2\] \[-0.42 v=-2\] \[v=4.76\]

    • one year ago
  41. geoffb
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    I left my answers in fractions, so I'm not sure. What do you get for your final answer?

    • one year ago
  42. zello
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    I just solve V jog

    • one year ago
  43. Chlorophyll
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    @zello something isn't right with your calculation !

    • one year ago
  44. geoffb
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    No, that's not what the question wants. Go back and read the last sentence.

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

    • one year ago
  46. geoffb
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    Although I only looked at his answer, not his work.

    • one year ago
  47. zello
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    I'm a girl -_-

    • one year ago
  48. geoffb
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    Oh, I'm sorry. I judge too quickly by avatars.

    • one year ago
  49. geoffb
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    Although that's obviously not you... :S

    • one year ago
  50. Chlorophyll
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    Yes, it's very neat V = 24/5 ( 4.8 mi/hr) @zello Better keep it in fraction form!

    • one year ago
  51. geoffb
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    @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.

    • one year ago
  52. zello
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    I have a headache because from 9 A.M until now, I still working on my math >.<

    • one year ago
  53. geoffb
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    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.

    • one year ago
  54. zello
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    how can i found the distance then

    • one year ago
  55. zello
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    find *

    • one year ago
  56. geoffb
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    You know your distance formula. We've used it a lot.

    • one year ago
  57. zello
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    d= vt d jog= d bike

    • one year ago
  58. geoffb
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    Top one.

    • one year ago
  59. zello
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    vt jog = vt bike

    • one year ago
  60. geoffb
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    No, just \(d = vt\). You know \(v\) now, and you already knew \(t\), so solve for \(d\).

    • one year ago
  61. zello
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    d bike = v bike * t bike d= 4.76 * 0.58 d= 2.7608

    • one year ago
  62. geoffb
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    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.

    • one year ago
  63. geoffb
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    Excellent job working through this.

    • one year ago
  64. Chlorophyll
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    @zello I wonder if you know how to say thanks by distributing the medal (?)

    • one year ago
  65. zello
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    Thank you so much and I'm sorry that I took a lot of your time. I really appreciate that

    • one year ago
  66. Hero
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    I got 2.8 exactly with my methods.

    • one year ago
  67. geoffb
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    Not a problem at all. Glad I could help. Thank you for making the effort to solve it on your own and learn.

    • one year ago
  68. Chlorophyll
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    Yup, that's why I say zello's calculation is incorrect!

    • one year ago
  69. zello
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    It' not incorrect, it's just because I didn't round my answer

    • one year ago
  70. Chlorophyll
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    It's exact, so there's no need to round!

    • one year ago
  71. zello
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    it's not exact in my calculator

    • one year ago
  72. Chlorophyll
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    So you're unable to calculate the fraction?

    • one year ago
  73. zello
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    I know how to calculate the fraction but the question says that they want the answer rounded to the nearest tenth

    • one year ago
  74. Chlorophyll
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    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

    • one year ago
  75. Chlorophyll
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    There's nothing to round off :)

    • one year ago
  76. zello
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    when I calculate it, I didn't use the fraction

    • one year ago
  77. zello
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    However, thank you

    • one year ago
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