Odpowiedź:
Proszę bardzo! :)
1) [tex]k=3[/tex]
2)[tex]cos\alpha=-\frac{\sqrt{7} }{4}[/tex]
3) [tex]0[/tex]
Szczegółowe wyjaśnienie:
1)
[tex]f(x)=x^2+3x+k\\f(-2)=1\\f(-2)=(-2)^2+3*(-2)+k\\f(-2)=4-6+k\\f(-2)=1\\1=4-6+k\\3=k[/tex]
2)
[tex]\alpha[/tex]∈[tex](90^*;180^*)[/tex]
[tex]4cos^2\alpha -\frac{3}{4}=1\\4cos^2\alpha=1+\frac{3}{4}\\\\ 4cos^2\alpha =\frac{4}{4}+\frac{3}{4}\\\\ 4cos^2\alpha=\frac{7}{4}\ \ \ /:4\\\\cos^2\alpha =\frac{7}{4}*\frac{1}{4}\\\\ cos^2\alpha =\frac{7}{16}\ \ \ / \sqrt{} } \\\\cos\alpha =\frac{\sqrt{7} }{4}\ \ \ \ v \ \ \ \cos\alpha=-\frac{\sqrt{7} }{4}\\[/tex]
[tex]cos\alpha=\frac{\sqrt{7} }{4}[/tex] ∉
3)
[tex](cos(109^*)+(sin19^*))^2\\cos109^*=cos(90^*+19^*)=-sin19^*\\(-sin19^*+sin19^*)^2=0^2=0[/tex]