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katherinekc
Group Title
Find the angle between the given vectors to the nearest tenth of a degree.
u = <8, 7>, v = <9, 7>
 one year ago
 one year ago
katherinekc Group Title
Find the angle between the given vectors to the nearest tenth of a degree. u = <8, 7>, v = <9, 7>
 one year ago
 one year ago

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ZeHanz Group TitleBest ResponseYou've already chosen the best response.5
You could use the formula:\[\cos \theta=\frac{ \vec u \cdot \vec v}{ \vec u\vec v }\]assuming you are familiar with the dot product of vectors.
 one year ago

Sshmoo Group TitleBest ResponseYou've already chosen the best response.0
here it is on mathematica.
 one year ago

katherinekc Group TitleBest ResponseYou've already chosen the best response.0
that wouldnt let me open it @Sshmoo
 one year ago

katherinekc Group TitleBest ResponseYou've already chosen the best response.0
im just learning it so im still so comfused
 one year ago

Sshmoo Group TitleBest ResponseYou've already chosen the best response.0
i'll try converting it to a different file, all it is is a graph of the vectors and a dot product. It just looks shinier.
 one year ago

Sshmoo Group TitleBest ResponseYou've already chosen the best response.0
here it is in PDF
 one year ago

katherinekc Group TitleBest ResponseYou've already chosen the best response.0
okk hm so
 one year ago

katherinekc Group TitleBest ResponseYou've already chosen the best response.0
so which of these would the answer be ? 8.3° 1.7° 3.3° 13.3°
 one year ago

ZeHanz Group TitleBest ResponseYou've already chosen the best response.5
If you calculate cos theta, you get: \[\cos \theta=\frac{ 8 \cdot 9 + 7 \cdot 7 }{ \sqrt{8^2+7^2}\sqrt{9^2+7^2} }=\frac{ 72+49 }{\sqrt{113}\sqrt{130} }=\frac{ 121 }{ \sqrt{113} \sqrt{130}}\approx 0.99833\]Now take the inverse cosine (cos^1 on your calculator) to see the answer.
 one year ago

katherinekc Group TitleBest ResponseYou've already chosen the best response.0
THANK YOU!
 one year ago
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