Question

Bună! Vă rog să mă ajutați şi pe mine la exercițiile 5 şi 6 . Mulțumesc!

Answer 1

[tex]\\A(2,4) \:\ B(6,8) \:\ C(8,2) \\ \\ \\Notam \:\ cu \:\ M \:\ mijlocul \:\ lui \:\ AB \\ \\ \\M= \frac{x1+x2}{2} ; \frac{y1+y2}{2} = \frac{2+6}{2} ; \frac{4+8}{2} = \frac{8}{2} ; \frac{12}{2} =\ \textgreater \ \boxed{ M(4,6)} \\ \\ \\CM = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} = \sqrt{(4-8)^2 + (6-2)^2} = \sqrt{16 + 16} \\ =\ \textgreater \ \boxed{4 \sqrt{2}}[/tex]

[tex]\\5.) (cos120 + cos 60)(sin135-sin45) \\=\left(2\cos \left(\frac{120^{\circ \:}+60^{\circ \:}}{2}\right)\cos \left(\frac{120^{\circ \:}-60^{\circ \:}}{2}\right)\right)\left(\sin \left(135^{\circ \:}\right)-\sin \left(45^{\circ \:}\right)\right) \\= 2\cos \left(90^{\circ \:}\right)\cos \left(30^{\circ \:}\right)\left(\sin \left(135^{\circ \:}\right)-\sin \left(45^{\circ \:}\right)\right) \\ =2\cdot \:0\cdot \frac{\sqrt{3}}{2}\left(\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}\right) = \boxed{0}[/tex]