Question

Doar 1, 3, 5 si 6.
Măcar unul din astea dacă știe cineva.

Answer 1

1.

$2^n \leq 3 + log_2 n\\\\\textrm{Daca n = 1:}\\\\ 2^1 \leq 3 + log_2 1\\\\ 2 \leq 3+0\\\\ 2 \leq 3\textrm{ adevarat, 1 numar}\\\\\textrm{Daca n = 2:}\\\\ 2^2 \leq 3 + log_2 2 \\\\ 4 \leq 3 + 1\\\\ 4\leq 4\textrm{ adevarat, 2 numere}\\\\\textrm{Daca n = 4:}\\\\ 2^4 \leq 3 + log_2 4\\\\ 16 \leq 3 + 2\\\\ 16 \leq 5\textrm{ fals}\\\\\textrm{Deoarece }2^n \textrm{ creste mult mai rapid decat }log_2 n\textrm{, exista doar 2 numere care verifica inegalitatea}\\\\p = \frac{2}{5} = 40\%$

3.

$S = 1 + 11 + 21 + 31 + \cdots + 91\Bigg | - 10\\\\S-10 = 0 + 10 + 20 + 30 + \cdots + 90\\\\S-10 = 10(1+2+3+4+\cdots + 9)\\\\ S-10 = 10\cdot \frac{9\cdot 10}{2}\\\\ S-10 = 5\cdot 90\\\\ S = 450 + 10\\\\ S = 460$

5.

$\cos{1^{\circ}} + \cos{2^{\circ}} + \cos{3^{\circ}} + \cdots + \cos{179^{\circ}}\\\\\cos{(180-x)} = \cos{180^{\circ}} \cos{x} + \sin{180^{\circ}}\sin{x} = -1\cos{x} + 0 = -\cos{x}\\\\\textrm{Putem scrie suma astfel: }\\\\\cos{1^{\circ}} + \cos{2^{\circ}} + \cos{3^{\circ}} + \cdots + \cos{(180 - 3)} + \cos{(180 - 2)} + \cos{(180 - 1)}\\\\\cos{1^{\circ}} + \cos{2^{\circ}} + \cos{3^{\circ}} + \cdots -\cos{3^{\circ}} - \cos{2^{\circ}} - \cos{1^{\circ}} = \cos{90^{\circ}} = 0$

6.

${\Big(\frac{5}{22}\Big)}^{2x-3} = (4,4)^{3x-2}\\\\\textrm{Se observa ca } 4,4 = \frac{22}{5}\\\\{\Big(\frac{5}{22}\Big)}^{2x-3} = {\Bigg({\Big(\frac{5}{22}\Big)}^{-1}\Bigg)}^{3x-2}\\\\ {\Big(\frac{5}{22}\Big)}^{2x-3} = {\Big(\frac{5}{22}\Big)}^{-3x+2}\\\\\implies 2x-3 = -3x+2 \\\\ 5x = 5\\\\ x = 1$