Solution 3.1:4b
From Förberedande kurs i matematik 1
(Difference between revisions)
(Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_4b.gif </center> {{NAVCONTENT_STOP}}) |
m |
||
(3 intermediate revisions not shown.) | |||
Line 1: | Line 1: | ||
- | {{ | + | By writing <math>0\textrm{.}027</math> as <math>27\cdot 10^{-3}</math>, where |
- | < | + | <math>27 = 3\cdot 3\cdot 3 = 3^3</math> and <math>10^{-3} = (10^{-1})^{3} = 0\textrm{.}1^3</math> we see that |
- | {{ | + | |
+ | {{Displayed math||<math>\begin{align} | ||
+ | \sqrt[3]{0\textrm{.}027} &= \sqrt[3]{27\cdot 10^{-3}} = \sqrt[3]{27}\cdot\sqrt[3]{10^{-3}} = \sqrt[3]{3^{3}}\cdot\sqrt[3]{0\textrm{.}1^3}\\[5pt] | ||
+ | &= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | where we have used <math>\sqrt[3]{a^{3}} = \bigl(a^{3}\bigr)^{1/3} = a^{3\cdot \frac{1}{3}} = a^{1} = a\,\textrm{.}</math> |
Current revision
By writing \displaystyle 0\textrm{.}027 as \displaystyle 27\cdot 10^{-3}, where \displaystyle 27 = 3\cdot 3\cdot 3 = 3^3 and \displaystyle 10^{-3} = (10^{-1})^{3} = 0\textrm{.}1^3 we see that
\displaystyle \begin{align}
\sqrt[3]{0\textrm{.}027} &= \sqrt[3]{27\cdot 10^{-3}} = \sqrt[3]{27}\cdot\sqrt[3]{10^{-3}} = \sqrt[3]{3^{3}}\cdot\sqrt[3]{0\textrm{.}1^3}\\[5pt] &= 3\cdot 0\textrm{.}1 = 0\textrm{.}3\,\textrm{,} \end{align} |
where we have used \displaystyle \sqrt[3]{a^{3}} = \bigl(a^{3}\bigr)^{1/3} = a^{3\cdot \frac{1}{3}} = a^{1} = a\,\textrm{.}