Question

Помогите умоляю..........

Answer 1

$4)y=kx-2\frac{7}{13}\\\\(4 \ ; \ 3\frac{6}{13}) \ \Rightarrow x=4 \ ; \ y=3\frac{6}{13}\\\\3\frac{6}{13} =4k-2\frac{7}{13} \\\\4k=3\frac{6}{13}+2\frac{7}{13}\\\\4k=6\\\\\boxed{k=1,5}\\\\\\5)(a-2)(2-a)+a^{2}+7=2a-a^{2}-4+2a+a^{2}+7=4a+3\\\\a=-0,45 \ \Rightarrow \ 4a+3=4*(-0,45)+3=-1,8+3=\boxed{1,2}$