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anonymous

  • 5 years ago

tan∠A - csc∠C = 3x

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  1. anonymous
    • 5 years ago
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    How do you solve this?

  2. anonymous
    • 5 years ago
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    At some point I got 3x sqrt74/5= 7/sqrt 74

  3. anonymous
    • 5 years ago
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    i am confused because you have 3 variables, A, C and x. what are you trying to solve?

  4. anonymous
    • 5 years ago
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    I am solving for this :tan∠A - csc∠C = 3x

  5. anonymous
    • 5 years ago
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    here are the variables

  6. anonymous
    • 5 years ago
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    tangent A= 7/sqrt 74 and csc C= sqrt74/5

  7. anonymous
    • 5 years ago
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    Hello?

  8. anonymous
    • 5 years ago
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    idk what you are saying

  9. anonymous
    • 5 years ago
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    Its tangent a subtracted from cosecant C and those are the values of them

  10. anonymous
    • 5 years ago
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    What the bannana are you talking about?

  11. anonymous
    • 5 years ago
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    haha

  12. shadowfiend
    • 5 years ago
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    Wait, I'm confused. So you have: \[\tan A - \csc C = 3x\] And: \begin{align} \tan A &= \frac{7}{\sqrt{74}}\\ \csc C &= \frac{\sqrt{74}}{5} \end{align} Right?

  13. shadowfiend
    • 5 years ago
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    jonh—please be respectful.

  14. anonymous
    • 5 years ago
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    Yes that is right

  15. shadowfiend
    • 5 years ago
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    Ok, so you really have: \[\frac{7}{\sqrt{74}} - \frac{\sqrt{74}}{5} = 3x\]

  16. anonymous
    • 5 years ago
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    Yes exactly

  17. shadowfiend
    • 5 years ago
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    First things first, I would do the subtraction on the left. Do you know how to do that?

  18. anonymous
    • 5 years ago
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    do you subtract the square?

  19. anonymous
    • 5 years ago
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    So, 3x = -39/(5*SQRT(74)) => x = -13/(5*SQRT(74))

  20. anonymous
    • 5 years ago
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    did you use the values GT?

  21. anonymous
    • 5 years ago
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    I am kind of shaky on how to subtract with squares shadowfield

  22. anonymous
    • 5 years ago
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    |dw:1326987711543:dw|

  23. anonymous
    • 5 years ago
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    |dw:1326987783481:dw|

  24. shadowfiend
    • 5 years ago
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    You can make the denominators the same by multiplying each fraction by the other fraction's denominator: \[\] \begin{align} \frac{7}{\sqrt{74}} &- \frac{\sqrt{74}}{5}\\ \frac{5}{5}\frac{7}{\sqrt{74}} &- \frac{\sqrt{74}}{\sqrt{74}}\frac{\sqrt{74}}{5}\\ \frac{35}{5\sqrt{74}} &- \frac{74}{5\sqrt{74}}\\ \end{align} So you get: \[\frac{35 - 74}{5\sqrt{74}}\] GT has the rest of the work :)

  25. anonymous
    • 5 years ago
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    is that the simplified answer? Because I don't see it on my answer choices?

  26. anonymous
    • 5 years ago
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    |dw:1326988065164:dw|

  27. anonymous
    • 5 years ago
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    There are other ways to write it. What are the choices?

  28. anonymous
    • 5 years ago
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    74/15 , 7-5sqrt 74/5, 7-sqrt74/15 or 74/21

  29. anonymous
    • 5 years ago
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    Then, the ratios you gave for tan and csc must be wrong. Where did you get those from?

  30. anonymous
    • 5 years ago
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    |dw:1326988372290:dw|

  31. anonymous
    • 5 years ago
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    I first had to find the missing side then use the ratios

  32. anonymous
    • 5 years ago
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    What is A and B and C?

  33. anonymous
    • 5 years ago
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    Well, before I say that I found C by using the pythagorean therom and got sqrt 74?

  34. anonymous
    • 5 years ago
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    I understand and that is right. BTW - from the picture, I can already tell your Tan(a) is wrong. There is no hypotenuse involved in the tangent ratio.

  35. anonymous
    • 5 years ago
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    Ok wait... let me post the letters

  36. anonymous
    • 5 years ago
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    |dw:1326988584560:dw|

  37. anonymous
    • 5 years ago
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    Sorry for the messiness lol

  38. anonymous
    • 5 years ago
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    Good. Now, we can do it right. Tan A = BC/AB (opposite / adjacent) = 7/5 Csc C = AC/AB (hypotenuse / opposite) = SQRT(74)/5

  39. anonymous
    • 5 years ago
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    Now subtract and you will get the right answer.

  40. anonymous
    • 5 years ago
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    |dw:1326988606286:dw|

  41. anonymous
    • 5 years ago
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    That is equal to 3x. Now, x will be equal to 1/3 of that value.

  42. anonymous
    • 5 years ago
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    |dw:1326988707325:dw|

  43. anonymous
    • 5 years ago
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    So you multiplied both sides by 5?

  44. anonymous
    • 5 years ago
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    WOOHOOO!! Thank you so much I got a 100!

  45. anonymous
    • 5 years ago
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    Thanks everyone

  46. anonymous
    • 5 years ago
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    Great. Hope you also now can solve such problems with ease!

  47. anonymous
    • 5 years ago
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    I do to lol!

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