Question

19.2-19.3 Решите пожалуйста

Answer 1

$19.2.\ \ \ sina=-0,6\\\\\\a\in \Big(\pi ;\dfrac{3\pi }{2}\Big)\ \ \ \to \ \ \ \ cosa<0\ ,\ \ tga>0\ ,\ \ ctga>0\\\\\\cos^2a=1-sin^2a=1-(-0,6)^2=1-0,36=0,64\\\\\\cosa=-\sqrt{0,64}=-0,8<0\\\\\\tga=\dfrac{sina}{cosa}=\dfrac{-0,6}{-0,8}=\dfrac{6}{8}=\dfrac{3}{4}=0,75\\\\\\ctga=\dfrac{1}{tga}=\dfrac{1}{0,75}=\dfrac{4}{3}$

$19.3.\ \ \ f(x)=3x^2-4x+5\\\\f(-3)=3\cdot (-3)^2-4\cdot (-3)+5=3\cdot 9+12+5=44$