Processing Math: Done
To print higher-resolution math symbols, click the
Hi-Res Fonts for Printing button on the jsMath control panel.
jsMath

Solution 1.1:1d

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 1.1:1d moved to Solution 1.1:1d: Robot: moved page)
Current revision (06:41, 19 September 2008) (edit) (undo)
 
(One intermediate revision not shown.)
Line 1: Line 1:
{{NAVCONTENT_START}}
{{NAVCONTENT_START}}
-
Precis som tidigare börjar vi med att räkna ut det innersta parentesuttrycket först
+
Just as in exercise c, we calculate the innermost bracket
:<math>3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5 = 3-(7-\bbox[#FFEEAA;,1.5pt]{\,10\,})-5</math>
:<math>3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5 = 3-(7-\bbox[#FFEEAA;,1.5pt]{\,10\,})-5</math>
{{NAVCONTENT_STEP}}
{{NAVCONTENT_STEP}}
-
och arbetar sedan successivt utåt,
+
and work our way successively outwards,
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 3-\firstcbox{#FFEEAA;}{\,(7-10)\,}{(-3)}-5</math>
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 3-\firstcbox{#FFEEAA;}{\,(7-10)\,}{(-3)}-5</math>
{{NAVCONTENT_STEP}}
{{NAVCONTENT_STEP}}
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 3-\secondcbox{#FFEEAA;}{\,(7-10)\,}{(-3)}-5\,</math>.
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 3-\secondcbox{#FFEEAA;}{\,(7-10)\,}{(-3)}-5\,</math>.
-
{{NAVCONTENT_STEP}}
+
All that remains is to combine the terms from left to right
-
Nu återstår bara att lägga ihop termerna från vänster till höger
+
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = \firstcbox{#FFEEAA;}{\,3-(-3)\,}{6}-5</math>
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = \firstcbox{#FFEEAA;}{\,3-(-3)\,}{6}-5</math>
{{NAVCONTENT_STEP}}
{{NAVCONTENT_STEP}}
Line 16: Line 15:
{{NAVCONTENT_STEP}}
{{NAVCONTENT_STEP}}
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 1</math>.
:<math>\phantom{3-(7-\bbox[#FFEEAA;,1.5pt]{\,(4+6)\,})-5}{} = 1</math>.
-
{{NAVCONTENT_START}}
+
{{NAVCONTENT_STOP}}

Current revision

Just as in exercise c, we calculate the innermost bracket

3−(7−(4+6))−5=3−(7−10)−5

and work our way successively outwards,

=3−(7−10)−5
=3−(−3)−5.

All that remains is to combine the terms from left to right

=3−(−3)−5
=3+3−5
=6−5
=1.
Retrieved from "http://wiki.math.se/wikis/2009/bridgecourse1-ImperialCollege/index.php/Solution_1.1:1d"