Question

PLEASE HELP PLSS

Find the angle between the given vectors to the nearest tenth of a degree.

u = <2, -4>, v = <3, -8>

Answer 1

$\cos\theta=\dfrac{\vec{u}\circ\vec{v}}{|\vec{u}|\cdot|\vec{v}|}\\\\\vec{u}= \ \textless \ 2;-4 \ \textgreater \ ;\ \vec{v}= \ \textless \ 3;-8 \ \textgreater \ \\\\\vec{u}\circ\vec{v}=2\cdot3+(-4)\cdot(-8)=6+32=38\\\\|\vec{u}|=\sqrt{2^2+(-4)^2}=\sqrt{4+16}=\sqrt{20}=\sqrt{4\cdot5}=\sqrt4\cdot\sqrt5=2\sqrt5\\\\|\vec{v}|=\sqrt{3^2+(-8)^2}=\sqrt{9+64}=\sqrt{73}\\\\|\vec{u}|\cdot|\vec{v}|=2\sqrt5\cdot\sqrt{73}=2\sqrt{365}\\\\\cos\theta=\dfrac{38}{2\sqrt{365}}\approx0.9945\to\theta\approx6^o\\\\Answer:The\ angle\ between\ \vec{u}\ and\ \vec{v}:\theta\approx6^o.$

Answer 2

For those who see what the person wrote above, the answer is 6.0*