Mohon bantuannya kk, 30 mnit lgi di kumpulkan

Mohon bantuannya kk, 30 mnit lgi di kumpulkan

Jawab:

c. $frac{2}{3}$

Penjelasan dengan langkah-langkah:

$tan alpha =-frac{2}{3}\ frac{y}{x}=-frac{2}{3}\ y=-2, x=3\ r=sqrt{x^2+y^2}=sqrt{3^2+(-2)^3} =sqrt{9+4}=sqrt{13}$

$sin alpha =frac{y}{r}= frac{-2}{sqrt{13} }=frac{-2}{13}sqrt{13} \cos alpha =frac{x}{r}= frac{3}{sqrt{13} }=frac{3}{13}sqrt{13}$

$frac{5 sin alpha +6cos alpha }{2 cos alpha -3sin alpha} =frac{5.frac{-2}{13}sqrt{13}+6.frac{3}{13}sqrt{13}}{2.frac{3}{13}sqrt{13}-3.frac{-2}{13}sqrt{13}} \=frac{5(-2)+6(3)}{2(3)-3(-2)} \=frac{-10+18}{6+6}\ =frac{8}{12}\ =frac{2}{3}$ (c.)