Question

cine ma poate ajuta va rog frumos e o tema pentru trecerea anului scolar​

Answer 1

$\it I)\ 1)\ 2\sqrt5x=2\sqrt{45} \Rightarrow x=\dfrac{2\sqrt{45}^{(2}}{2\sqrt5}=\dfrac{\sqrt{45}}{\sqrt5}=\sqrt{\dfrac{45}{5}}=\sqrt9=3$

$\it 2)\ \mathcal{V}=\mathcal{A}_b\cdot h =3^2\cdot5=9\cdot5=45\ cm^3$

$\it 3)\ CC'\perp(A'B'C'D')\ \ \ \ (1)\\ \\ B'D'\subset (A'B'C'D')\ \ \ \ \ \ (2)\\ \\ (1),\ (2) \Rightarrow\ CC'\perp B'D' \Rightarrow m(\widehat{CC',\ B'D'})=90^o$

$\it 4)\ A(0,\ 2) \Rightarrow\ ordonata\ este\ 2.$

Answer 2

Răspuns:

Vezi atasamente