Решебник по алгебре 9 класс Мерзляк Задание 528

 

\[\boxed{\mathbf{528\ (528).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1156 = 1600 \cdot \left( 1 - \frac{p}{100} \right)^{2}\]

\[\left( 1 - \frac{p}{100} \right)^{2} = \frac{1156}{1600}\]

\[1 - \frac{p}{100} = \sqrt{0,7225}\]

\[1 - \frac{p}{100} = 0,85\]

\[\frac{p}{100} = 0,15\]

\[p = 15\]

\[на\ 15\% - происходило\ \]

\[ежемесячное\ снижение\ цены.\]

\[Ответ:на\ 15\%.\]

\[\boxed{\mathbf{528.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ b_{n} = \frac{10}{n}\]

\[b_{2} = \frac{10}{2} = 5,\ \ b_{7} = \frac{10}{7},\]

\[\text{\ \ }b_{100} = \frac{10}{100} = 0,1\]

\[2)\text{\ b}_{n} = 5 - 2n\]

\[b_{2} = 5 - 4 = 1,\ \ \]

\[b_{7} = 5 - 14 = - 9,\ \ \]

\[b_{100} = 5 - 200 = - 195\]

\[\ 3)\ b_{n} = n² + 2n\]

\[b_{2} = 4 + 4 = 8,\ \ \]

\[b_{7} = 49 + 14 = 63,\ \ \]

\[b_{100} = 10\ 000 + 200 = 10\ 200\]

\[4)\ b_{n} = ( - 1)^{n + 1}\]

\[b_{2} = ( - 1)^{3} = - 1,\ \ \]

\[b_{7} = ( - 1)^{8} = 1,\]

\[\text{\ \ }b_{100} = ( - 1)^{101} = - 1\]