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anonymous

  • one year ago

SIMPLE TRIG PROBLEMS NEED HELP. I WILL GIVE MEDAL AND FAN

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Nnesha

  3. anonymous
    • one year ago
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    Pls help

  4. anonymous
    • one year ago
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    @Preetha

  5. anonymous
    • one year ago
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    @pooja195

  6. anonymous
    • one year ago
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    @dan815

  7. anonymous
    • one year ago
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    Not really sure what the first one is asking for, but I can help you with the second one.

  8. anonymous
    • one year ago
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    That sound good?

  9. anonymous
    • one year ago
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    Well I mean, it's really simple. The angle \(\theta\) can be found using the formula \[\large \sf \theta~=~\frac{arc~length}{radius}\]

  10. anonymous
    • one year ago
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    Here's a nice little cheat-sheet for trig formulas: http://learnix.net/wordpress/wp-content/uploads/Trig-Cheat-Sheet-1.4.pdf

  11. anonymous
    • one year ago
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    ok

  12. anonymous
    • one year ago
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    i do not have arc lenght

  13. anonymous
    • one year ago
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    Ugh, this is what I get for trying to do math while sleep deprived. Give me a minute, sorry.

  14. anonymous
    • one year ago
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    lol its ok

  15. mathstudent55
    • one year ago
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    Let's look at the first problem. You are given an angular velocity, and you asked to find a linear velocity.

  16. anonymous
    • one year ago
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    yes

  17. anonymous
    • one year ago
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    are you still there

  18. mathstudent55
    • one year ago
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    In the formula below, we use \(\omega\) (small-case Greek letter "omega" is usually used to denote an angular velocity). Point A moves in a circular path around point O with an angular velocity of \(\omega\). If point A is r units from point O, then the linear velocity of point A is \(v = r \omega \). \(\omega\) must be expressed in radians per unit time in this formula.

  19. mathstudent55
    • one year ago
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    \(\ \large v = r\omega\) In your case, you have the angular velocity given in revolutions per minute. You need to convert revolutions into radians.

  20. mathstudent55
    • one year ago
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    Since the answer must be in miles per hour, you will also need to convert minutes into hours and inches into miles.

  21. anonymous
    • one year ago
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    so angular velocity is 6500

  22. anonymous
    • one year ago
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    i dont get it, what do i do next

  23. anonymous
    • one year ago
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    @mathstudent55

  24. mathstudent55
    • one year ago
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    We start with this: \(\large v = r \omega = 7.25 ~in. \times ~3250 ~\dfrac{rev}{min}\)

  25. mathstudent55
    • one year ago
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    Now we need to apply all the conversion factors.

  26. anonymous
    • one year ago
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    but isnt angular velocity 6500

  27. mathstudent55
    • one year ago
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    Angular velocity needs to be in radians per minute. You have revolutions per minute. 1 revolution is \(2 \pi\) radians, not 2 radians.

  28. anonymous
    • one year ago
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    ohhhhh

  29. mathstudent55
    • one year ago
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    Let's deal with the conversion of revolutions to radians. It's in red below. \(\large v = r \omega = 7.25 ~in. \times ~3250 ~\dfrac{rev}{min} \color{red}{\times \dfrac{2 \pi ~rad }{1~rev}}\)

  30. anonymous
    • one year ago
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    Since you guys are still working on the first part, I'll just leave this here so you can work it out after you finish. You can find arc length from sector area and radius with for formula \[\large \sf Area~=~\frac{1}{2} \times r \times Arc \huge \rightarrow \large \frac{2 \times Area}{r}~=~Arc\] from there, you can find \(\large \sf \theta\) with the original formula. \[\large \sf \theta~=~\frac{Arc~length}{Radius}\]

  31. mathstudent55
    • one year ago
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    Now we need to deal with the conversion of inches into miles. We can convert using 12 in. = 1 ft, and 5280 ft = 1 mile. This conversion is in green below. \(v = r \omega = 7.25 ~in. \color{green}{\times \dfrac{1~ft}{12~in.} \times \dfrac{1~mile}{5280~ft} } \times ~3250 ~\dfrac{rev}{min} \color{red}{\times \dfrac{2 \pi ~rad }{1~rev}}\) Now we have miles per minute. We need to convert minutes to hours.

  32. mathstudent55
    • one year ago
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    The conversion from minutes to hours is in blue below. 60 minutes = 1 hour \(v = r \omega = 7.25 ~in. \color{green}{\times \dfrac{1~ft}{12~in.} \times \dfrac{1~mile}{5280~ft} } \times ~3250 ~\dfrac{rev}{min} \color{red}{\times \dfrac{2 \pi ~rad }{1~rev}} \color{blue} {\times \dfrac{60~min}{1~hour} }\)

  33. mathstudent55
    • one year ago
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    What do you get? \(v = \dfrac{7.25 \times 3250 \times 2 \times \pi \times 60}{12 \times 5280} \dfrac{miles}{hour} \)

  34. anonymous
    • one year ago
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    one sec

  35. anonymous
    • one year ago
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    140.2

  36. mathstudent55
    • one year ago
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    That's what I got.

  37. anonymous
    • one year ago
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    140.2 miles per hour

  38. anonymous
    • one year ago
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    OK @LegendarySadist

  39. anonymous
    • one year ago
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    so whats my first step

  40. anonymous
    • one year ago
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    That would be to find the arc length \[\large \sf Area~=~\frac{1}{2} \times r \times Arc \huge \rightarrow \large \frac{2 \times Area}{r}~=~Arc\]

  41. anonymous
    • one year ago
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    so arc length is 20/3

  42. anonymous
    • one year ago
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    Correct. Now we can plug that into our other formula to find \(\large \sf \theta\) \[\large \sf \theta~=~\frac{Arc~length}{Radius}\]

  43. anonymous
    • one year ago
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    bit i need theta in radians

  44. anonymous
    • one year ago
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    but*

  45. anonymous
    • one year ago
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    This is the solution in radians. That's why I chose this formula instead of the degrees formula.

  46. anonymous
    • one year ago
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    i got 20/27

  47. anonymous
    • one year ago
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    That would be it.

  48. mathstudent55
    • one year ago
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    Second problem: \(\Large A_{sector} = \dfrac{\theta}{2 \pi ~rad}\pi r^2\) \(\Large \theta = \dfrac{2 \pi A}{\pi r^2} = \dfrac{2A}{r^2} = \dfrac{2 \times 30 ft^2}{(9~ft)^2} = \dfrac{60}{81} = \dfrac{20}{27}\)

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