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sarahi_smiles

Solve 2 - 3 cos x = 5 + 3 cos x for 0 degrees lessthan x lessthan 180 degrees (1 point) can anyone help me...

  • one year ago
  • one year ago

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  1. sarahi_smiles
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    The possible answers are 150 degrees, 30 degrees, 60 degrees, and 120 degrees...?

    • one year ago
  2. BrittanyFrench
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    Hey I'm tryin to figure this one out!!

    • one year ago
  3. sarahi_smiles
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    Oh ok...if you could help me I would greatly appreciate it!!

    • one year ago
  4. BrittanyFrench
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    :) Sure thing. This is geomtry right? Or Algebra 2?

    • one year ago
  5. sarahi_smiles
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    I think algebra 2...I'm in precal but shes going over several subject at one time...

    • one year ago
  6. sarahi_smiles
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    Like an over view of algebra...

    • one year ago
  7. yrelhan4
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    solve the equation. it will give you cosx=-1/2 --> x=120 degrees

    • one year ago
  8. sarahi_smiles
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    thanks so much...could you help me answer another question @yrelhan4

    • one year ago
  9. yrelhan4
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    cheers! only if i know it.

    • one year ago
  10. sarahi_smiles
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    I have to find the angle the normal makes with the positive x-axis....& I have to rewrite it each equation in the normal for and solve the length of the normal... 1. 3x-2y-1=0 2. 5x+y-12=0 3. 4x+3y-4=0

    • one year ago
  11. sarahi_smiles
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    If not its ok...but thanks anyway!!!

    • one year ago
  12. yrelhan4
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    normal to these lines?

    • one year ago
  13. sarahi_smiles
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    Yea..idk what it means either. :/ It said: find the angle the normal makes with the positive x–axis. Hint: You should first rewrite each equation in normal form and solve for the length of the normal.

    • one year ago
  14. yrelhan4
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    see that normal from of a line is xsino + ysino =P whee o is angle made with positive x axis and p is the length of normal from origin to the line.

    • one year ago
  15. yrelhan4
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    can you find a line perpendicular to the given lines and passing through origin?

    • one year ago
  16. sarahi_smiles
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    Ok...no?

    • one year ago
  17. yrelhan4
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    slope of a line is -(coefficient of x)/(coefficient of y). and m18m2 =-1 where m1 is the slope of the given line and m2 of the normal. write the normal line in the form of y=mx+c. substitute x,y=0,0 as it passes through the origin. this will give you value of c. plug this value back into the equation. suppose that the line is ax+by+c=0. thelength of perpendicular from x1,y1 is ax1 +by1+c/sqrt(a^2 + b^2). susbtitute 0,0 in place of x1,y1. this will give you value of perpendicular. i am still working on the angle. i seem to know it but cant recall it.

    • one year ago
  18. yrelhan4
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    in the first line it is m2*m1=-1

    • one year ago
  19. sarahi_smiles
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    If its confusing you & you can't figure it out its ok....I'm just not that good in math lol

    • one year ago
  20. yrelhan4
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    just tell me if you get the length of the normal. or i'll you refer you to a site.

    • one year ago
  21. sarahi_smiles
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    Can you just refer me....because I'm really confused

    • one year ago
  22. sarahi_smiles
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    Sorry...

    • one year ago
  23. yrelhan4
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    its alright sweety. i was as poorer at math. chill!! http://answers.yahoo.com/question/index?qid=20090817231959AASOXEf look at this to find the line perpendicular to the given line. and then look at my explanation. and i am not following the hint you gave me. i dont follow it. this method of mine makes more sense to me.

    • one year ago
  24. yrelhan4
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    feel free to back to me if any problem.

    • one year ago
  25. yrelhan4
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    slope of a line=tano. so arctan[-(coefficient of x)/coefficient of y)] will give you the angle.

    • one year ago
  26. sarahi_smiles
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    ok thanks!!!

    • one year ago
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