Respuesta :
[tex]1.\\\\\bold{v_1}=\langle2,4\rangle\qquad\bold{v_2}=\langle-1,5\rangle\\\\
\bold{v_1}\circ\bold{v_2}=2\cdot(-1)+4\cdot5=-2+20=18\qquad [\text{C}]\\\\\\
2.\\\\
\bold{v_1}= \langle2, 5\rangle\qquad \bold{v_2}= \langle4, -3\rangle\\\\\\
\bold{v_1}\circ\bold{v_2}=2\cdot4+5\cdot(-3)=8-15=-7\\\\||\bold{v_1}||=\sqrt{2^2+5^2}=\sqrt{4+25}=\sqrt{29}\\\\
||\bold{v_2}||=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=\sqrt{25}=5\\\\\\
[/tex]
[tex]\cos\Theta=\dfrac{\bold{v_1}\circ\bold{v_2}}{||\bold{v_1}||\cdot||\bold{v_2}||}=\dfrac{-7}{5\cdot\sqrt{29}}\\\\\\ \cos\Theta=-\dfrac{7}{5\sqrt{29}}\quad\implies\quad\Theta=\arccos\left(-\dfrac{7}{5\sqrt{29}}\right)\\\\\\\boxed{\Theta\approx105,1^\circ}[/tex]
[tex]\cos\Theta=\dfrac{\bold{v_1}\circ\bold{v_2}}{||\bold{v_1}||\cdot||\bold{v_2}||}=\dfrac{-7}{5\cdot\sqrt{29}}\\\\\\ \cos\Theta=-\dfrac{7}{5\sqrt{29}}\quad\implies\quad\Theta=\arccos\left(-\dfrac{7}{5\sqrt{29}}\right)\\\\\\\boxed{\Theta\approx105,1^\circ}[/tex]