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sarver1995

in a 45 45 90 triangle i need to find what x is

  • one year ago
  • one year ago

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  1. sarver1995
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    |dw:1342917395925:dw|

    • one year ago
  2. Romero
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    |dw:1342917641735:dw|

    • one year ago
  3. sarver1995
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    would it be 6 square root of 2

    • one year ago
  4. Romero
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    Use Pythagoras theorem \[x^2+x^2=12^2\]

    • one year ago
  5. Romero
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    yes

    • one year ago
  6. Australopithecus
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    you cant solve this with pythagoras theorem you have to use trig I'm pretty sure

    • one year ago
  7. Romero
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    \[2x^2=12^2\]\[2x^2=144\]\[x^2=\frac{144}{2}\]\[x^2=72\]\[x= \sqrt{72}\]\[x= \sqrt{36*2}\]\[x=6 \sqrt{2}\]

    • one year ago
  8. Romero
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    Yes you can you have two sides that are equal

    • one year ago
  9. Calcmathlete
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    Hey @sarver1995 Been a while :) 45-45-90 triangles: |dw:1342917784578:dw|

    • one year ago
  10. Australopithecus
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    oh yeah just noticed :)

    • one year ago
  11. Calcmathlete
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    \[x\sqrt{2} = 12\]\[x = \frac{12}{\sqrt{2}}\]\[x = \frac{12}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\]\[x = \frac{12\sqrt{2}}{2}\]\[x = 6\sqrt{2}\]There you go :)

    • one year ago
  12. Australopithecus
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    SOH CAH TOA Sin = opposite/hypotenuse cos = adjacent/hypotenuse tan = opposite/adjacent cos(45) = x/12 cos(45)(12) = x

    • one year ago
  13. Romero
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    from this I noticed that you can simply get one of the sides by multiplying the hypotenuse by \[\frac{1}{\sqrt{2}}\]

    • one year ago
  14. Romero
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    Only if the triangle is a 45 45 90

    • one year ago
  15. sarver1995
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    |dw:1342917788335:dw|

    • one year ago
  16. Calcmathlete
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    Same thing...use the ratio above... |dw:1342918025299:dw|

    • one year ago
  17. Romero
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    same thing like I said \[\frac{1}{\sqrt{2}}*x=10\]

    • one year ago
  18. Romero
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    solve for x

    • one year ago
  19. sarver1995
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    to calcmathlete yeah i am now reviewing and oh course still confused i only have a few more lessons and i am finished with 11th grade. its been really rough

    • one year ago
  20. sarver1995
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    |dw:1342918127585:dw| would x be 10 square root of 2

    • one year ago
  21. Calcmathlete
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    Yes :) \[x = 10\sqrt{2}\]

    • one year ago
  22. Romero
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    Good job mate!

    • one year ago
  23. sarver1995
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    |dw:1342918257888:dw| okay not sure how to do this one

    • one year ago
  24. sarver1995
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    it cant be 1/2 square root of 2

    • one year ago
  25. Calcmathlete
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    Does it tell you if it's a right triangle?

    • one year ago
  26. sarver1995
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    45 45 90

    • one year ago
  27. Calcmathlete
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    It tells you that?

    • one year ago
  28. sarver1995
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    yes

    • one year ago
  29. Calcmathlete
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    Ok. The ratio again. x, x, x√2 x = 1.2 x√2 = ?

    • one year ago
  30. sarver1995
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    i will try to draw it again maybe i ddin't do it right

    • one year ago
  31. Calcmathlete
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    By the way, if the x's get confusing because they are using x as a variable, I'll just use y.

    • one year ago
  32. sarver1995
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    |dw:1342918455394:dw|

    • one year ago
  33. Calcmathlete
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    Ok. Same situation. The legs are always congruent in a 45-45-90 triangle. The hypotenuse is always in this ratio: \[Leg \times \sqrt{2}\]Can you figure it out since you have the leg?

    • one year ago
  34. sarver1995
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    1.2 x square root of 2

    • one year ago
  35. sarver1995
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    not sure on this one

    • one year ago
  36. Calcmathlete
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    Yup. which can be rewritten as \(1.2\sqrt{2}\)

    • one year ago
  37. sarver1995
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    okay i have one more like these can i try it with us

    • one year ago
  38. Calcmathlete
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    Alright shoot away :)

    • one year ago
  39. sarver1995
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    |dw:1342918731847:dw|

    • one year ago
  40. Calcmathlete
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    \[x = \frac{\sqrt{24}}{\sqrt{2}} = \sqrt{\frac{24}{2}} = \sqrt{12} = ?\]

    • one year ago
  41. sarver1995
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    2 square root of 3

    • one year ago
  42. Calcmathlete
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    yup :)

    • one year ago
  43. sarver1995
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    okay i have a different one i just need u to help me understand the problem it will take a little while to draw

    • one year ago
  44. Calcmathlete
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    ok.

    • one year ago
  45. sarver1995
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    |dw:1342918993022:dw| Refer to the figure to complete the following item. Given: If m arc vu =80° and marc st =40°, then 1 =

    • one year ago
  46. Calcmathlete
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    Oh gosh...a proof?

    • one year ago
  47. sarver1995
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    yeah lol

    • one year ago
  48. Calcmathlete
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    Do you have to like fill in a blank or is it like write an entire proof?

    • one year ago
  49. sarver1995
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    its just asking what angle 1 would equal

    • one year ago
  50. sarver1995
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    just an answer

    • one year ago
  51. Calcmathlete
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    I think that it's 50...

    • one year ago
  52. sarver1995
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    can you tell me how you got that because i have three more problems like this one

    • one year ago
  53. Calcmathlete
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    \[m\angle1 = \frac 12(arcVT - arcST)\]

    • one year ago
  54. sarver1995
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    If arc vu m = 70° and m arc st = 30°, then 2 = would it be the same formula for this one

    • one year ago
  55. Calcmathlete
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    I kind of have to go...sorry :/ If it maintains the same shape, it should be the same sort of formula...

    • one year ago
  56. sarver1995
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    thanks for your help its really appreciated and God bless you

    • one year ago
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