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purplemexican

  • one year ago

help please

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  1. Purplemexican
    • one year ago
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  2. tkhunny
    • one year ago
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    Please provide the Right-Triangle definition of the Cosine.

  3. Purplemexican
    • one year ago
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    everywhere

  4. Purplemexican
    • one year ago
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    the most i can tell you is what im looking for

  5. tkhunny
    • one year ago
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    You cannot have been given this problem without the basic Right-Triangle definitions of Sine, Cosine, and Tangent. What are they?

  6. Purplemexican
    • one year ago
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    do i know what?

  7. Purplemexican
    • one year ago
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    i have been given what you see

  8. Purplemexican
    • one year ago
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    and 4 answer choices

  9. Purplemexican
    • one year ago
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    if i knew earlier today i dont remember now

  10. tkhunny
    • one year ago
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    Hogwash, unless this is a placement test. If this is a placement test, you should get this problem wrong so that you are placed in the correct course this semester.

  11. Purplemexican
    • one year ago
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    well i dont want you to do it for me i would like a step by step tho

  12. Purplemexican
    • one year ago
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    im in summerschool

  13. Purplemexican
    • one year ago
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    ive been doing algebra 2 since 9 am to 4:30 for the past few days im half out of my wits with this stuff now

  14. Purplemexican
    • one year ago
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    i have till 4:30 tomorrow to finish about 50 questions i would love to take a brake i really wold

  15. tkhunny
    • one year ago
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    Did you read it?

  16. Purplemexican
    • one year ago
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    yes it make sence

  17. tkhunny
    • one year ago
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    Okay, now solve. What does it want and how shall you find it?

  18. Purplemexican
    • one year ago
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    i have one minuet im gonna guess on the quiz then come back and figure it out

  19. Purplemexican
    • one year ago
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    didnt even get to guess so im looking for angle FHE right and or cos(theta) right?

  20. tkhunny
    • one year ago
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    \(\cos(\angle FHE) = \dfrac{Adjacent}{Hypotenuse}\) You know neither the Adjacent Side (FH) nor the Hypotenuse (EH). How shall we find them?

  21. Purplemexican
    • one year ago
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    i have 2 angles and a side i could find the other sides of the one triangle and then i would have one of the sides

  22. tkhunny
    • one year ago
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    Good. How about \(\triangle{HGF}\)? Can that help us find FH?

  23. Purplemexican
    • one year ago
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    yes it would

  24. tkhunny
    • one year ago
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    It's a 45-45-90 Right Triangle. What is the relationship between the Leg and the Hypotenuse?

  25. Purplemexican
    • one year ago
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    leg FH is 4 right

  26. tkhunny
    • one year ago
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    How did you get that? One for free: \(\sqrt{8}\cdot\sqrt{2} = \sqrt{16} = 4\) I believe you have it! Now, we know the "Adjacent". Now, off to your Pythagorean Theorem to find the Hypotenuse, EH.

  27. Purplemexican
    • one year ago
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    a^2+b^2=c^2\[4^2+3^2=5^2 \] and or 16+9=25 \[c=\sqrt{25}\] or c=5

  28. Purplemexican
    • one year ago
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    @tkhunny do you concur?

  29. tkhunny
    • one year ago
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    We have the hypotenuse. Now, you can solve the problem. What is the requested cosine?

  30. Purplemexican
    • one year ago
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    indeed

  31. tkhunny
    • one year ago
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    Eventually, you should also recognize "Pythagorean Triples" - at least 3-4-5 and 5-12-13. There are many others.

  32. Purplemexican
    • one year ago
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    i get cos(37)=0.765

  33. tkhunny
    • one year ago
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    No one cares about those decimals or degrees. \(\cos\left(\angle{FHE}\right) = \dfrac{4}{5}\) - Done!

  34. Purplemexican
    • one year ago
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    thank you so much i appreciate the help

  35. tkhunny
    • one year ago
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    You got it. Keep up the good work. Start earlier so you CAN take a break. :-)

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