ГДЗ по алгебре 9 класс Макарычев Задание 583

 

\[\boxed{\text{583\ (583).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\mathrm{\Delta}OA_{1}B_{1} \sim \bigtriangleup OA_{2}b_{2}\ldots \sim \bigtriangleup OA_{n}B_{n};\]

\[так\ как\ угол\ O - общий\ и\ \]

\[A_{1}B_{1} \parallel A_{2}B_{2}\ldots \parallel A_{n}B_{n}.\]

\[OA_{n} = n \cdot OA_{1},\]

\[OB_{n} = n \cdot OB_{1};\]

\[\frac{OA_{n}}{OA_{1}} = \frac{OB_{n}}{OB_{1}} = \frac{A_{n}B_{n}}{A_{1}B_{1}} = n;\]

\[A_{n}B_{n} = n \cdot A_{1}B_{1};\]

\[A_{5}B_{5} = 5 \cdot 1,5 = 7,5\ (см).\]

\[A_{10}B_{10} = 10 \cdot 1,5 = 15\ (см).\]

\[\boxed{\text{583.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x_{1} = 3,\ \ x_{2} = 5,\]

\[\ \ d = x_{2} - x_{1} = 2:\ \]

\[S_{n} = \frac{2x_{1} + d(n - 1)}{2} \cdot n =\]

\[= \frac{6 + 2 \cdot (n - 1)}{2} \cdot n =\]

\[= (3 + n - 1) \cdot n =\]

\[= (n + 2) \cdot n = n^{2} + 2n \leq 120;\]

\[n^{2} + 2n - 120 \leq 0\]

\[D = 1 + 120 = 121\]

\[так\ как\ n > 0 \Longrightarrow\]

\[\Longrightarrow n = - 1 + 11 = 10.\]

\[Ответ:10\ членов.\]