Lösung 1.2:3a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K (Lösning 1.2:3a moved to Solution 1.2:3a: Robot: moved page)
Zeile 1: Zeile 1:
-
{{NAVCONTENT_START}}
+
The denominator in the expression has
-
<center> [[Image:1_2_3a.gif]] </center>
+
<math>10</math>
-
{{NAVCONTENT_STOP}}
+
as a common factor,
 +
 
 +
 
 +
<math>\frac{3}{2\centerdot 10}+\frac{7}{5\centerdot 10}-\frac{1}{10}</math>
 +
 
 +
 
 +
and it is therefore sufficient to multiply the top and bottom of each fraction by the other factors in the denominators in order to obtain a common denominator,
 +
 
 +
 
 +
<math>\frac{3\centerdot 5}{20\centerdot 5}+\frac{7\centerdot 2}{50\centerdot 2}-\frac{1\centerdot 5\centerdot 2}{10\centerdot 5\centerdot 2}=\frac{15}{100}+\frac{14}{100}-\frac{10}{100}</math>
 +
 
 +
 
 +
The lowest common denominator (LCD) is therefore
 +
<math>100</math>
 +
, and the expression is equal to
 +
 
 +
 
 +
<math>\frac{15}{100}+\frac{14}{100}-\frac{10}{100}=\frac{15+14-10}{100}=\frac{19}{100}</math>

Version vom 12:58, 11. Sep. 2008

The denominator in the expression has \displaystyle 10 as a common factor,


\displaystyle \frac{3}{2\centerdot 10}+\frac{7}{5\centerdot 10}-\frac{1}{10}


and it is therefore sufficient to multiply the top and bottom of each fraction by the other factors in the denominators in order to obtain a common denominator,


\displaystyle \frac{3\centerdot 5}{20\centerdot 5}+\frac{7\centerdot 2}{50\centerdot 2}-\frac{1\centerdot 5\centerdot 2}{10\centerdot 5\centerdot 2}=\frac{15}{100}+\frac{14}{100}-\frac{10}{100}


The lowest common denominator (LCD) is therefore \displaystyle 100 , and the expression is equal to


\displaystyle \frac{15}{100}+\frac{14}{100}-\frac{10}{100}=\frac{15+14-10}{100}=\frac{19}{100}

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_1.2:3a“