Loading http://wiki.math.se/jsMath/extensions/bbox.js
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 1.1:2c

From Förberedande kurs i matematik 1

Revision as of 13:13, 13 September 2008 by Ian (Talk | contribs)
(diff) ←Older revision | Current revision (diff) | Newer revision→ (diff)
Jump to: navigation, search

We can divide up the whole expression into two separate parts

\displaystyle \bbox[#FFEEAA;,1.5pt]{\,3\cdot(-7)\,}-\bbox[#FFEEAA;,1.5pt]{\,4\cdot(6-5)\,}

which we first calculate and then combine.

The first part of the expression equals

\displaystyle 3\cdot(-7) = -21

whilst the second expression becomes.

\displaystyle 4\cdot(6-5) = 4\cdot 1 = 4\,.

Combining, we obtain

\displaystyle \bbox[#FFEEAA;,1.5pt]{\,3\cdot(-7)\,}-\bbox[#FFEEAA;,1.5pt]{\,4\cdot(6-5)\,} = \bbox[#FFEEAA;,1.5pt]{\,-21\,}-\bbox[#FFEEAA;,1.5pt]{\,4\,} = -25\,.
Retrieved from "http://wiki.math.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_1.1:2c"