(4.9)^3+(2.1)^3/(4.9)^2-(10.29*1)+4.41
You asked:
MathBot Answer:
\[{4.9}^{3} + \frac{{2.1}^{3}}{{4.9}^{2}} - 10.29 \cdot 1 + 4.41 = \frac{785083}{7000} = 112.154\overline{714285}\]
\[\begin{aligned}{4.9}^{3} + \frac{{2.1}^{3}}{{4.9}^{2}} - 10.29 \cdot 1 + 4.41& = \left(4 + \frac{9}{10}\right)^{3} + \left(2 + \frac{1}{10}\right)^{3} \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2}} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}\\& = \left(\frac{49}{10}\right)^{3} + \left(2 + \frac{1}{10}\right)^{3} \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2}} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}\\& = \frac{117649}{1000} + \left(2 + \frac{1}{10}\right)^{3} \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2}} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}\\& = \frac{117649}{1000} + \left(\frac{21}{10}\right)^{3} \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2}} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}\\& = \frac{117649}{1000} + \frac{9261}{1000} \cdot \frac{1}{\left(4 + \frac{9}{10}\right)^{2}} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}\\& = \frac{117649}{1000} + \frac{9261}{1000} \cdot \frac{1}{\left(\frac{49}{10}\right)^{2}} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}\\& = \frac{117649}{1000} + \frac{9261}{1000} \cdot \frac{1}{\frac{2401}{100}} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}\\& = \frac{117649}{1000} + \frac{27}{70} - \left(10 + \frac{29}{100}\right) \cdot 1 + 4 + \frac{41}{100}\\& = \frac{117649}{1000} + \frac{27}{70} - \frac{1029}{100} \cdot 1 + 4 + \frac{41}{100}\\& = \frac{117649}{1000} + \frac{27}{70} - \frac{1029}{100} + 4 + \frac{41}{100}\\& = \frac{826243}{7000} - \frac{1029}{100} + 4 + \frac{41}{100}\\& = \frac{754213}{7000} + 4 + \frac{41}{100}\\& = \frac{782213}{7000} + \frac{41}{100}\\& = \frac{785083}{7000}\end{aligned}\]