Question

va roggg,dau coroana,plzzzzzz​

Answer 1

Mijlocul lui BC este (0,0) , atunci AO= Ax=4 u.m.

Answer 2

$5.~~A(4,0),~B(0,2),~C(0,-2)\\\\ Notam~cu~M(x_M,y_M)~mijlocul~segmentului~BC.\\\\ x_M=\cfrac{x_B+x_C}{2} =\cfrac{0+0}{2} =\cfrac{0}{2} =0\\~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~M(0,0)\\Y_M=\cfrac{y_B+y_C}{2}=\cfrac{2+(-2)}{2} =\cfrac{0}{2} =0\\\\ AM=\sqrt{(x_M-x_A)^2+(y_M-y_A)^2} \Rightarrow AM=\sqrt{(0-4)^2+(0-0)^2}\Rightarrow\\\\\Rightarrow AM=\sqrt{4^2-0^2} \Rightarrow AM=\sqrt{16} \Rightarrow AM=4$

$6.~~x \in \Bigg(0,\cfrac{\pi}{2}\Bigg),~cosx=\cfrac{1}{4} \\\\ sin^2x+cos^2x=1 \Rightarrow sinx=\pm \sqrt{1-cos^2x} \\\\ x\in\Bigg(0,\cfrac{\pi}{2} \Bigg) \Rightarrow sinx=\sqrt{1-cos^2x} \\\\ sinx=\sqrt{1-cos^2x}\Rightarrow sinx= \sqrt{1-\Bigg(\cfrac{1}{4} \Bigg)^2} \Rightarrow sinx=\sqrt{1-\cfrac{1}{16} } \\\\\Rightarrow sinx=\sqrt{\cfrac{16-1}{16} } \Rightarrow sinx=\sqrt{\cfrac{15}{16} } \Rightarrow sinx=\cfrac{\sqrt{15} }{\sqrt{16} }\Rightarrow sinx=\cfrac{\sqrt{15} }{4}$