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anonymous

  • 5 years ago

Please HELP! :)mr. Duncan traveled to a city 180 miles from him home to attend a meeting. Due to car trouble, his average speed returning was 13 miles less than his speed going. If the total time for the road trip was 7 hours, at what rate of speed did he travel to the city? (round to nearest tenth)

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  1. anonymous
    • 5 years ago
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    Formula is d=rt. Make two equations and set them equal to each other to find the rate, and then find the average, which is the first plus the second divided by 2

  2. anonymous
    • 5 years ago
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    so 180=r7 so 173/2..so 86.5?

  3. anonymous
    • 5 years ago
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    distance equal rate x time there are two equations we can use here. Going and coming. let t= time going let 7-t= time coming let x = rate going let x-13 = rate coming distance going and coming is 90 (180/2) so remember d=rt coming 90 =(x-13)(7-t) going 90 = xt solve this system of equations.

  4. anonymous
    • 5 years ago
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    wait, did you just solve it? ha, i'm confused. That all kind of makes sense. I just am not sure what the last step is?

  5. anonymous
    • 5 years ago
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    @junwagh I think u r making a mistake The question says "Duncan traveled to a city 180 miles from his home" This means one way distance is 180 miles and not 90 miles

  6. anonymous
    • 5 years ago
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    I set up the problem. We set up two equations using d =rt. One for going and one for coming. Then you solve that system of equations. I won't do that for you, the hard part is setting up the equation. 90 = xt 90 = (x-13)(7-t)

  7. anonymous
    • 5 years ago
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    yep the total distance is 180

  8. anonymous
    • 5 years ago
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    yes the total distance is 180, which means the distance going there is 90 and the distance coming back is also 90.

  9. anonymous
    • 5 years ago
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    oh wait mi bad lol replace 90with 180 then.

  10. anonymous
    • 5 years ago
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    The total distance has to be 360 miles as the "city is 180 miles from his house" so he has to go 180 miles and come back 180 miles

  11. anonymous
    • 5 years ago
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    it says from his house to attend the meeting he drove 180 miles.. hmm yeah. you might be right. Ahh. does anyone have an answer. Now all the numbers are confusing me. want to make sure I know how to solve with the last steps

  12. anonymous
    • 5 years ago
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    someone suggested 59 mph. anyone else get that?

  13. anonymous
    • 5 years ago
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    Okay here is how u do it Let his speed while going = x miles /hour Then his speed while coming back = (x-13) miles/hr Distance each side is 180 miles we know distance = speed x time or time = distance/speed So time taken while going = 180/x and time taken while coming back = 180/(x-13)

  14. anonymous
    • 5 years ago
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    Now the total time taken is 7 hours. This means (180/x) + 180/(x-13) = 7 or 180(x-13) + 180 (x) ------------------ = 7 x(x-13) Simplifying the above we get 180x - 2340 + 180x = 7 (x^2 - 13x)

  15. anonymous
    • 5 years ago
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    so how do we find the rate of speed he traveled to the city Thank you for writing that all out for me. I just want to make sure I put in the correct answer.

  16. anonymous
    • 5 years ago
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    DOes anyone have a final answer for this problem? I have been trying to work it out for over an hour so I would really appreciate any help with the solution.

  17. anonymous
    • 5 years ago
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    what answer did you get?

  18. anonymous
    • 5 years ago
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    Further this gives us 360x - 2340 = 7x^2 - 91x Further simplifying we get 0 = 7x^2 - 91x - 360x + 2340 or 7x^2 - 451x + 2340 = 0 This is a quadratic eqn. U can resolve it and get value of x

  19. anonymous
    • 5 years ago
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    59 mph

  20. anonymous
    • 5 years ago
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    that got me 2 solutions. 5.69 and 58.74...Is there a way to make this into 1 answer?

  21. anonymous
    • 5 years ago
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    maybe i'll just try 58.74.. Think this will be right?

  22. anonymous
    • 5 years ago
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    i got it right. THanks everyone

  23. anonymous
    • 5 years ago
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    yes it is 59.1 mph

  24. anonymous
    • 5 years ago
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    58.74 is correct as your speed cannot be 13 miles less than 5.69!!

  25. anonymous
    • 5 years ago
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    i actually did 58.7 and this was correct.

  26. anonymous
    • 5 years ago
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    It is a long winded question for sure

  27. anonymous
    • 5 years ago
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    http://www.mathsisfun.com/quadratic-equation-solver.html you can use that to solve your quadratic equation exactly. Just plug in your values.

  28. anonymous
    • 5 years ago
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    x=speed out, t=time out, x-13=speed back, 7-t=time back {x t == 180, (x-13)*(7-t)==180} {{x->5.69119,t->31.6278},{x->58.7374,t->3.06449}} Using second pair of x, y values. Speed, Time Out = {58.7374,3.06449} Speed, Time Back = {45.7374,3.93551}

  29. anonymous
    • 5 years ago
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    @robtobey how did u get th values {{x->5.69119,t->31.6278},{x->58.7374,t->3.06449}}

  30. anonymous
    • 5 years ago
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    Running the Mathematica Solve function: Solve[ { x t == 180, (x-13)*(7-t)==180}, {x,t} ]

  31. anonymous
    • 5 years ago
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    The syntax is somewhat weird. Those are, two consecutive = symbols.

  32. anonymous
    • 5 years ago
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    I am not familiar with this... Anyway thanks for th reply!!

  33. anonymous
    • 5 years ago
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    \[\left\{x\to \frac{1}{14} \left(451+\sqrt{137881}\right),t\to \frac{1}{26} \left(451-\sqrt{137881}\right)\right\} \] The above is the actual output returned from Mathematica for the x, t, values used. Placing "//N" at the end converts to decimal numbers.

  34. anonymous
    • 5 years ago
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    Is this Mathematica some program to calculate answer or is it a mathematical theory like quadratic equations etc The answer seems like solution to the quadratic eqn

  35. anonymous
    • 5 years ago
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    I calculated using the formula -b +- root b^2 - 4ac x = --------------------- 2a

  36. anonymous
    • 5 years ago
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    Harkirat, mathematica is a mathematics software tool which helps you do calculations.

  37. anonymous
    • 5 years ago
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    OK!!??!! Thanks dhatraditya !! Is it free and available on net??

  38. anonymous
    • 5 years ago
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    Mathematica is a very large application system. About 3Gb of data to install it. It has been years in development and is mathematically very comprehensive in scope. It is about $2500 to a the common mom and popper who belly up to the cash counter to buy. For students and teachers they almost give it away. On the order of $125 one time cost as long as the student remains a student. The teachers probably would not allow it to be used in a class test but it will always show the correct answers for home work as long as the input is correct. If you give me a minute I'll find their web site.

  39. anonymous
    • 5 years ago
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    sure , thanks

  40. anonymous
    • 5 years ago
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    The Mathematica site is: http://www.wolfram.com/ I just purchasee Mathematica version 8 Home edition. It cost me about $280 for a box edition delivered to California. By the way the installation comes on a DVD not a CD. Because of low price, the use of the Home is very restricted. For instance I can not use in to make any money.

  41. anonymous
    • 5 years ago
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    Thanks robtobey, I am in India and this is beyond my reach

  42. anonymous
    • 5 years ago
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    They may ship to India or have an agent there. I have to break off now. Been good talking to you. Adios.

  43. anonymous
    • 5 years ago
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    thank friend n bye till we chat again

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