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anonymous
 one year ago
Find the angle between the given vectors to the nearest tenth of a degree.
u = <2, 4>, v = <3, 8>
anonymous
 one year ago
Find the angle between the given vectors to the nearest tenth of a degree. u = <2, 4>, v = <3, 8>

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I got this much so far, is it right? u=sqrt. 2^2+4^2 v=sqrt.3^2+8^2 u=4.5 v=8.5 dot=38.25

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1look at your dot again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0isnt dot multiplying u and v?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0okay and now what do i do?

Michele_Laino
 one year ago
Best ResponseYou've already chosen the best response.1better is if we write the lengths of our vectors, like this: \[\Large \begin{gathered} \left\ u \right\ = \sqrt {4 + 16} = \sqrt {20} \hfill \\ \left\ v \right\ = \sqrt {9 + 64} = \sqrt {73} \hfill \\ \end{gathered} \]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1\(\large cos \theta = \frac{\vec u \bullet \vec v}{\vec u\vec v} \)

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1find arccos use calculator!

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im so confused lol what do I put for cos theta= like what goes in the u and v?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1calculate the RHS first then, ie using your numbers: \(\large \frac{38}{4.5 \times 8.5}\) what do you get we NEED TO optimise later but do this for now

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0.9934640523 arccos=6.55

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yeah! that's a correct answer for "your" numbers but as @Michele_Laino pointed out, your rounding is going to let you down use these instead \(\huge \frac{38}{\sqrt{20}\times \sqrt{73}}\) your calculator will now compute these square roots for you to a very high dgeree of accuracy, and you will get a much more accurate answer

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1yep that should give a very accurate answer if you do not round yourself into oblivion! what do you get for \(\theta\) now?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What how do I find theta again?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1you say \(cos \theta = .9945054529\) so ask your calculator for the inverse cosine of \( .9945054529\) like you did last time....

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1the issue here might be calculator this is what mine looks like and it is a fairly basic one

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1see how all information is in the calculator, so i am not rounding anything. that guarantees accuracy

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1i've also inadvertently given you the answer (!!) .... but anyways compare that to the 6.55 you originally calculated. the question wants you to be accurate to within 1/10 th of a degree, so 6.55 would have failed...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yeah whoa accuracy haha yeah my calculator doesn't read what yours reads I wonder why

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thank you so much can you help me in a last question though?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1sure but make a new thread for each quetsion
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