Solution 3.1:4a
From Förberedande kurs i matematik 1
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- | {{ | + | The decimal number <math>0\textrm{.}16</math> can also be written as <math>16\cdot 10^{-2}</math> and then it is easier to see that, since <math>16 = 4\cdot 4 = 4^2</math> and <math>10^{-2} = (10^{-1})^2 = 0\textrm{.}1^2</math>, |
- | < | + | |
- | {{ | + | {{Displayed math||<math>\begin{align} |
+ | \sqrt{0\textrm{.}16} &= \sqrt{16\cdot 10^{-2}} = \sqrt{16}\cdot \sqrt{10^{-2}} = \sqrt{4^2}\cdot \sqrt{0\textrm{.}1^2}\\[5pt] | ||
+ | &= 4\cdot 0\textrm{.}1 = 0\textrm{.}4\,\textrm{.} | ||
+ | \end{align}</math>}} | ||
+ | |||
+ | Another alternative is, of course, to see directly that <math>0\textrm{.}16 = 0\textrm{.}4\cdot 0\textrm{.}4 = 0\textrm{.}4^2</math>, and then that <math>\sqrt{0\textrm{.}16} = \sqrt{0\textrm{.}4^2} = 0\textrm{.}4\,\textrm{.}</math> |
Current revision
The decimal number \displaystyle 0\textrm{.}16 can also be written as \displaystyle 16\cdot 10^{-2} and then it is easier to see that, since \displaystyle 16 = 4\cdot 4 = 4^2 and \displaystyle 10^{-2} = (10^{-1})^2 = 0\textrm{.}1^2,
\displaystyle \begin{align}
\sqrt{0\textrm{.}16} &= \sqrt{16\cdot 10^{-2}} = \sqrt{16}\cdot \sqrt{10^{-2}} = \sqrt{4^2}\cdot \sqrt{0\textrm{.}1^2}\\[5pt] &= 4\cdot 0\textrm{.}1 = 0\textrm{.}4\,\textrm{.} \end{align} |
Another alternative is, of course, to see directly that \displaystyle 0\textrm{.}16 = 0\textrm{.}4\cdot 0\textrm{.}4 = 0\textrm{.}4^2, and then that \displaystyle \sqrt{0\textrm{.}16} = \sqrt{0\textrm{.}4^2} = 0\textrm{.}4\,\textrm{.}