Lösung 3.1:4a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
-
The decimal number
+
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>,
-
<math>0.\text{16 }</math>
+
-
can also be written as
+
-
<math>\text{16}\centerdot \text{1}0^{-\text{2}}\text{ }</math>
+
-
and then it is easier to see that, since
+
-
<math>\text{16}=\text{4}\centerdot \text{4}=\text{4}^{\text{2}}\text{ }</math>
+
-
and
+
-
<math>10^{-2}=\left( 10^{-1} \right)^{2}=0.1^{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>
-
<math>\begin{align}
+
-
& \sqrt{0.16}=\sqrt{16\centerdot 10^{-2}}=\sqrt{16}\centerdot \sqrt{10^{-2}}=\sqrt{4^{2}}\centerdot \sqrt{0.1^{2}} \\
+
-
& =4\centerdot 0.1=0.4 \\
+
-
\end{align}</math>
+
-
 
+
-
 
+
-
Another alternative is, of course, to see directly that
+
-
<math>0.16=0.4\centerdot 0.4=0.4^{2}</math>, and then that
+
-
<math>\sqrt{0.16}=\sqrt{0.4^{2}}=0.4</math>
+

Version vom 10:47, 30. Sep. 2008

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,

Vorlage:Displayed math

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{.}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:4a“