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Atkinsoha

  • one year ago

PreCalc & Trig Help, Picture Below!

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  1. Atkinsoha
    • one year ago
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  2. ajprincess
    • one year ago
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    Using pythagoras' theorem u can find the length of the hypotenuse. a^2+b^2=c^2 Here a=2, b=3, c=? |dw:1362407511223:dw||dw:1362407540894:dw| Does that help? @Atkinsoha

  3. Atkinsoha
    • one year ago
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    I did all of that.. I'm stuck on maybe how to solve it, since I can't seem to come up with any of the answers that are given.

  4. ajprincess
    • one year ago
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    What is the value of hypotenuse?

  5. Atkinsoha
    • one year ago
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    (√13)? I hope..

  6. ajprincess
    • one year ago
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    yup right:)

  7. ajprincess
    • one year ago
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    nw what is \(\cos\theta\)

  8. Atkinsoha
    • one year ago
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    3/√13

  9. ajprincess
    • one year ago
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    yup:) what is \(\cos^2\theta\)?

  10. Atkinsoha
    • one year ago
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    That's were I'm lost apparently. 3^2 = 9 and the (√13)^2 = 13?

  11. Atkinsoha
    • one year ago
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    WAIT. Is it A? 5/13? because then it would be 2(9/13)-1 which would be (18/13)-1 and one would be 13/13 so it would be 5/13? A?

  12. ajprincess
    • one year ago
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    yup:)

  13. Atkinsoha
    • one year ago
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    HOLY BEJEBOUS! Thank you so much! I guess I just needed someone to break it down!

  14. Atkinsoha
    • one year ago
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    Can you help me with this one? I'm sure it's really easy, but I've been gone from school.. so haven't learned it :P

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  15. ajprincess
    • one year ago
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    let us say arcsinx=theta nd arccosx=beta so sin theta=x nd cos beta =x sin(arcsinx+arccosx)=sin(theta+beta) =sin theta*cos beta+sin beta * cos theta

  16. ajprincess
    • one year ago
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    cos theta=sqrt(1-sin^2theta) sin beta=sqrt(1-cos^2 beta) plug in the values and u will get the algebraic expression:)

  17. Atkinsoha
    • one year ago
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    Yeahh.. I don't understand at all. haha.

  18. ajprincess
    • one year ago
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    Really sorry. Hav to go nw. Please post this as a new ques. Hope some will sre help u.:) Really sorry

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