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wampominater

  • one year ago

Find the angle between the given vectors to the nearest tenth of a degree. u = <-5, -4>, v = <-4, -3>

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  1. wampominater
    • one year ago
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    so far, i have followed the formula \[\cos(\theta) = \frac{ u*v }{\left| v \right| \left| u \right| }\] and i have gotten 1, but that isnt an answer choice...

  2. wampominater
    • one year ago
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    and then when I take the arc cos of 1, I get 0, which is not a choice

  3. anonymous
    • one year ago
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    \( \huge |u| = \sqrt{(-5)^2+(-4)^2} = \sqrt{41} \) \( \huge |v| = \sqrt{(-4)^2+(-3)^2} = 5 \) That what you have for |v| and |u| ?

  4. wampominater
    • one year ago
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    yeah, i believe i know what i did wrong now though

  5. anonymous
    • one year ago
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    you sure? I can work it out if you want

  6. wampominater
    • one year ago
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    well, i think that its because i rounded the denom to 32

  7. wampominater
    • one year ago
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    but sure, if you want to help me do it just to make sure, thatd be great! :)

  8. wampominater
    • one year ago
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    so i got approx 1.8 when i didnt round the denominator

  9. anonymous
    • one year ago
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    We should have \( \huge \cos(\theta) = \frac{ u*v }{\left| v \right| \left| u \right| }\) \( \huge \cos(\theta) = \frac{ 32}{\left| 5 \right| \left| \sqrt{41} \right| } = .99951 = 1.78 degrees\) That is what I got.

  10. wampominater
    • one year ago
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    awesome, that is what i got too, thank you for the help!

  11. anonymous
    • one year ago
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    YW!!!

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