2.
[tex]a_1 = -4\\a_2 = -1\\a_3 = 2\\a_4 = 5\\S_{n} = 95\\n = ?\\\\S_{n} = \frac{2a_1+(n-1)r}{2} \cdot n\\\\95 = \frac{2\cdot(-4)+(n-1)\cdot3}{2}\cdot n\\\\95 = \frac{-4+3n-3}{2}\cdot n\\\\95 = \frac{3n-11}{2}\cdot n \ \ /\cdot2\\\\(3n-11)n = 190\\\\3n^{2}-11n - 190 = 0\\\\\Delta} = (-11)^{2}-4\cdot3\cdot(-190) = 121+2280 = 2401\\\\\sqrt{\Delta} = \sqrt{2401} = 49\\\\n \in N+\\\\n_1 = \frac{11-49}{6} = -\frac{38}{6} < 0 \ \ \notin D\\\\n_2 = \frac{11+49}{6} = \frac{60}{6} = 10[/tex]
[tex]n = 10[/tex] - ilość początkowych wyrazów
3.
[tex]a_1 = 3\\q = \frac{1}{2}\\a_3, a_5, a_6 = ?\\\\a_{n} = a_1\cdot q^{n-1}\\\\\\a_3 = 3\cdot(\frac{1}{2})^{3-1}=3\cdot(\frac{1}{2})^{2} = 3\cdot\frac{1}{4} =\frac{3}{4}\\\\a_5 = 3\cdot(\frac{1}{2})^{5-1} = 3\cdot(\frac{1}{2})^{4} = 3\cdot\frac{1}{16} = \frac{3}{16}\\\\a_6 = 3\cdot(\frac{1}{2})^{6-1} = 3\cdot(\frac{1}{2})^{5} = 3\cdot\frac{1}{32}=\frac{3}{32}[/tex]
4.
[tex]K=(-5,2), \ \ \ \ \ L=(3,5)\\K = (x_{k}, y_{K}), \ \ \ L = (x_{L}, y_{L})\\\\|KL| = \sqrt{(x_{L}-x_{K})^{2}+(y_{L}-y_{K})^{2}}\\\\|KL| = \sqrt{(3-(-5)^{2} + (5-2)^{2}} =\sqrt{8^{2}+3^{2}} = \sqrt{64+9} = \sqrt{73}[/tex]
5.
Równanie okręgu w postaci kanonicznej:
(x - a)² + (y - b)² = r²
gdzie:
S = (a,b) - środek okręgu
r - promień okręgu
a)
[tex](x - 3)^{2} + y^{2} = 100\\\\(x -3)^{2} + (y+0)^{2} = 10^{2}\\\\S = (3, 0)\\r = 10[/tex]
b)
[tex](x+2)^{2} + (y-1)^{2} = 4\\\\(x-2)^{2}+(y-1)^{2} = 2^{2}\\\\S = (-2,1)\\r = 2[/tex]
