[tex]\\1)\\x = 2,(3) : [1,25 - 0,2(7)] = \frac{7}{3}:[\frac{5}{4} - \frac{27-2}{90}] = \frac{7}{3} : [\frac{5}{4} - \frac{5}{18}] = \frac{7}{3} : [\frac{45}{36} - \frac{10}{36}] = \frac{7}{3} : \frac{35}{36} = \frac{7}{3} \cdot \frac{36}{35} = \frac{12}{5} = 2\frac{2}{5}[/tex]
[tex]2)\\x = 4^{29} \ \ \ y = 8^{19} = 4^{19}\cdot2^{19}.\\$Ca sa demonstram ca x ${data-answer}lt; $ y sau vicersa trrebuie sa aratam ca x/y sau y/x ${data-answer}lt; c $ unde $c \in \mathbb{Z}$.\\$\frac{x}{y} = \frac{4^{29}}{4^{19}\cdot2^{19}} = \frac{4^{10}}{2^{19}} = \frac{2^{20}}{2^{19}} = 2 => y < x.\\\\$ \overline{X} - $ media aritmetica $.\\\overline{X} = \frac{x+y}{2} = \frac{4^{29} + 4^{19}\cdot2^{19}}{2} = 4^{19}\cdot\frac{4^{10} + 2^{19}}{2} = 2^{37} [4^{10} + 2^{19}] \approx 2.162 \cdot 10^{17}[/tex]
[tex]L = 6 m \ \ l = \frac{2}{3} L \ \ \ h = 25\% \ din \ l =>\\L = 6m \ \ \ l = 4 m \ \ \ h = 1m\\V_p = L\cdot l\cdot h = 6 \cdot 4 \cdot 1 m^3 = 24m^3\\$Stiind ca un litru = 1 dm^3 = 10^{-3} m^3 =>\\$ V_p = 24 \cdot 1000 l = 24000 l.[/tex]
