Question

5 Si 6, am tot incercat dar nu imi iese. mate bac

Answer 1

$\it \ (d):\  2x - 3y + 1 = 0 \Rightarrow y = \dfrac{2}{3} x+\dfrac{1}{3}$

Panta dreptei d este m=2/3, iar panta dreptei perpendiculare pe d va fi 

m' = -(1/m) = -3/2.

Ecuația perpendicularei este:

[tex]\it (d'):\ y=-\dfrac{3}{2}x+n=0 \ \ \ \ \ (1) \\ \\ \\ A(-3,\ 1) \in d' \ \ \ \ (2) \\ \\ \\ (1),\ (2) \Rightarrow 1 = -\dfrac{3}{2}\cdot(-3) +n = 0 \Rightarrow n = 1-\dfrac{9}{2} = - \dfrac{7}{9} \ \ \ \ \ (3) [/tex]


$\it (1),\ (3) \Rightarrow \ (d'):\ y = -\dfrac{3}{2}x-\dfrac{7}{2} \Rightarrow 3x+2y+7=0$


6)

[tex]\it cosx=\dfrac{2\sqrt2}{3} \Rightarrow cos^2x= \left(\dfrac{2\sqrt2}{3}\right)^2 \Rightarrow cos^2x= \dfrac{8}{9} \\ \\ \\ sin^2x= 1-cos^2x \Rightarrow sin^2x = 1-\dfrac{8}{9} \Rightarrow sin^2x = \dfrac{1}{9} \\ \\ \\ cos2x=cos^2x-sin^2x = \dfrac{8}{9}-\dfrac{1}{9} \Rightarrow cos2x=\dfrac{7}{9}[/tex]