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anonymous

  • one year ago

3tanx3-10tanx=0 , in the domain 180<x<360, the measure of x is to the nearest degree?

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  1. anonymous
    • one year ago
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    excuse me is it tan3x or tanx3

  2. anonymous
    • one year ago
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    tanx^3

  3. anonymous
    • one year ago
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    sorry i mean tanx^2

  4. anonymous
    • one year ago
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    so 2tanx^2 -10tanx

  5. anonymous
    • one year ago
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    omggg 3tanx^2 -10tanx

  6. anonymous
    • one year ago
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    @nincompoop

  7. anonymous
    • one year ago
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    @abb0t

  8. jim_thompson5910
    • one year ago
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    Is it \[\Large 3\tan(x^2) - 10\tan(x)=0\] or Is it \[\Large 3\left(\tan(x)\right)^2 - 10\tan(x)=0\] ??

  9. anonymous
    • one year ago
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    the second one

  10. jim_thompson5910
    • one year ago
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    you can let z = tan(x) that gives \[\Large 3z^2 - 10z = 0\] \[\Large z(3z - 10) = 0\] agreed?

  11. anonymous
    • one year ago
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    actually the x is the angle sign

  12. anonymous
    • one year ago
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    yep i get those steps dk what to di next

  13. jim_thompson5910
    • one year ago
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    so if z(3z-10) = 0, then either z = 0 or 3z-10 = 0 by the zero product property

  14. jim_thompson5910
    • one year ago
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    re-substitute in z = tan(x) z = 0 turns into tan(x) = 0 3z-10 = 0 turns into 3*tan(x) - 10 = 0

  15. anonymous
    • one year ago
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    jim we need to consider the xrange of x i also got here but the range got me confused

  16. jim_thompson5910
    • one year ago
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    true, that's a good point

  17. anonymous
    • one year ago
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    ok i get this whats next

  18. jim_thompson5910
    • one year ago
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    solve tan(x) = 0 and 3*tan(x) - 10 = 0 and make sure 180<x<360

  19. jim_thompson5910
    • one year ago
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    use the arctan function on a calculator to undo the tan function

  20. anonymous
    • one year ago
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    or we can put the range

  21. anonymous
    • one year ago
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    tan 0 i get 0 , for other i get 73.300

  22. jim_thompson5910
    • one year ago
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    is x = 0 in the interval 180<x<360 ?

  23. jim_thompson5910
    • one year ago
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    is x = 73.3 in the interval 180<x<360 ?

  24. anonymous
    • one year ago
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    umm

  25. anonymous
    • one year ago
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    yea that the part i am stuck on

  26. jim_thompson5910
    • one year ago
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    if it is, then you can keep the solution if not, then add/subtract 180 until you get in the interval

  27. jim_thompson5910
    • one year ago
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    being in the interval means that the number is between 180 and 360

  28. anonymous
    • one year ago
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    ok 73 is

  29. jim_thompson5910
    • one year ago
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    nope, 73 is not between 180 and 360

  30. anonymous
    • one year ago
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    no i mean when i add 180

  31. jim_thompson5910
    • one year ago
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    oh ok

  32. anonymous
    • one year ago
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    excuse since its periodic the range can be taken in 0<x<180 or 0<x<2pie

  33. anonymous
    • one year ago
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    k got the answer thank you

  34. jim_thompson5910
    • one year ago
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    no problem

  35. anonymous
    • one year ago
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    thanks jim

  36. anonymous
    • one year ago
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    u guy can stick around ? i got more

  37. anonymous
    • one year ago
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    oky woul love to in any way

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