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love_to_love_you

  • one year ago

Write an equation for the translation of y = 2/x that has the given asymptotes. x = 4 and y = -8 Show all work

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  1. love_to_love_you
    • one year ago
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    @zepdrix

  2. jim_thompson5910
    • one year ago
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    what is the vertical asymptote for y = 2/x ?

  3. love_to_love_you
    • one year ago
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    oh nvm I don't even have to do this problem. Could you help me with something else?

  4. love_to_love_you
    • one year ago
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    @jim_thompson5910

  5. jim_thompson5910
    • one year ago
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    sure

  6. love_to_love_you
    • one year ago
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    Okay so what I have to do is explain why for each problem. I have the answers already but I do not know the explanation.

  7. jim_thompson5910
    • one year ago
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    ok

  8. love_to_love_you
    • one year ago
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    a. If sin theta = sqrt 2/2, which could not be the value of theta? 225 degrees

  9. jim_thompson5910
    • one year ago
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    where is 225 degrees? which quadrant?

  10. love_to_love_you
    • one year ago
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    Um is it quadrant 3?

  11. jim_thompson5910
    • one year ago
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    yes, and sine is negative in Q3 and Q4

  12. jim_thompson5910
    • one year ago
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    so it's impossible for theta to be 225 degrees

  13. love_to_love_you
    • one year ago
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    ok

  14. love_to_love_you
    • one year ago
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    b. For which value of theta is tan theta equal to sin theta? 2pi

  15. jim_thompson5910
    • one year ago
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    I'm guessing you had a list of choices?

  16. love_to_love_you
    • one year ago
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    It's a problem. I have 2 problems. They have a and b parts to them

  17. love_to_love_you
    • one year ago
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    So what we have to do is explain why this is the correct answer/ show work.

  18. jim_thompson5910
    • one year ago
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    well tan(theta) is the same as sin(theta)/cos(theta) it's one of the many identities

  19. jim_thompson5910
    • one year ago
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    \[\Large \tan(\theta) = \sin(\theta)\] \[\Large \frac{\sin(\theta)}{\cos(\theta)} = \sin(\theta)\] \[\Large \frac{\sin(\theta)}{\cos(\theta)}*{\color{red}{\frac{1}{\sin(\theta)}}} = \sin(\theta)*{\color{red}{\frac{1}{\sin(\theta)}}}\] \[\Large \frac{\cancel{\sin(\theta)}}{\cos(\theta)}*{\color{black}{\frac{1}{\cancel{\sin(\theta)}}}} = \cancel{\sin(\theta)}*{\color{black}{\frac{1}{\cancel{\sin(\theta)}}}}\] \[\Large \frac{1}{\cos(\theta)} = 1\] in step 3, I'm multiplying both sides by 1/sin(theta) since both sides have a sine that cancels making sense so far?

  20. love_to_love_you
    • one year ago
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    Yeah

  21. love_to_love_you
    • one year ago
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    so what happens next?

  22. jim_thompson5910
    • one year ago
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    multiply both sides by cos(theta) to get cos(theta) = 1 then you'll use arccosine to isolate theta

  23. love_to_love_you
    • one year ago
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    Idk how to use arccosine very well

  24. jim_thompson5910
    • one year ago
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    what kind of calculator do you have?

  25. love_to_love_you
    • one year ago
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    just an online graphing calc

  26. jim_thompson5910
    • one year ago
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    what's the link to it? so I can have a look

  27. love_to_love_you
    • one year ago
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    it's a download from my school

  28. jim_thompson5910
    • one year ago
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    I gotcha

  29. jim_thompson5910
    • one year ago
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    ok I'm going to use this calculator here http://web2.0calc.com/

  30. love_to_love_you
    • one year ago
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    ok

  31. jim_thompson5910
    • one year ago
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    click the "rad" button (next to "deg") to convert over to radian mode

  32. jim_thompson5910
    • one year ago
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    then type in "arccos(1)" without quotes http://web2.0calc.com/#arccos(1) what do you get?

  33. love_to_love_you
    • one year ago
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    cos^-1(1) = 0

  34. love_to_love_you
    • one year ago
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    and then there's a tiny 2 pi under it

  35. love_to_love_you
    • one year ago
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    @jim_thompson5910

  36. jim_thompson5910
    • one year ago
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    correct, so 0 radians is one answer notice how 2pi radians and 0 radians are coterminal angles

  37. jim_thompson5910
    • one year ago
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    so since theta = 0 is one answer, theta = 2pi is also an answer there are infinitely many other answers. They only want theta = 2pi for some reason

  38. love_to_love_you
    • one year ago
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    So that makes 2 pi the answer as well?

  39. jim_thompson5910
    • one year ago
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    yeah there are infinitely many answers, but the computer or teacher is only accepting 2pi

  40. love_to_love_you
    • one year ago
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    ok

  41. love_to_love_you
    • one year ago
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    I'm gonna try to figure out a for the next problem on my own but can you help me with b?

  42. love_to_love_you
    • one year ago
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    A man stands on his balcony, 140 feet above the ground. He looks at the ground, with his sight line forming an angle of 75 degrees with the building, and sees a bus stop. The function d = 140 sec theta models the distance from the man to any object given his angle of sight theta. How far is the bus stop from the man? Round your answer. 541 ft.

  43. jim_thompson5910
    • one year ago
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    140*sec(theta) = 140*(1/cos(theta)) plug in theta = 75 make sure you are in degree mode use this calculator if needed http://web2.0calc.com/

  44. love_to_love_you
    • one year ago
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    541

  45. jim_thompson5910
    • one year ago
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    yeah I'm getting 540.9184627208766586 which rounds to 541 (assuming you round to the nearest foot)

  46. love_to_love_you
    • one year ago
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    yeah i rounded

  47. love_to_love_you
    • one year ago
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    Thanks so much

  48. jim_thompson5910
    • one year ago
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    you're welcome

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