Solution 1.3:4e - Revision history

Tek at 14:19, 22 September 2008

2008-09-22T14:19:26Z

←Older revision		Revision as of 14:19, 22 September 2008
Line 1:		Line 1:
-	Because	+	Because <math>5^{9} = 5^{8+1} = 5^{8}\cdot 5^{1} = 5^{8}\cdot 5</math>, the two terms inside the brackets have <math>5^{8}</math> as a common factor and can therefore be taken outside the bracket
-	<math>5^{9}=5^{8+1}=5^{8}\~~centerdot~~ 5^{1}=5^{8}\~~centerdot~~ 5</math>,	+
-	the two terms inside the brackets have	+
-	<math>5^{8}</math>	+
-	as a common factor	+
-	and can therefore be taken outside the bracket.	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\bigl(5^{8}+5^{9}\bigr)^{-1} &= \bigl(5^{8}+5^{8}\cdot 5\bigr)^{-1} = \bigl(5^{8}\cdot (1+5)\bigr)^{-1}\\[5pt]
		+	&= \bigl(5^{8}\cdot 6\bigr)^{-1} = 5^{8\cdot (-1)}\cdot 6^{-1} = 5^{-8}\cdot 6^{-1}.
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	Furthermore, <math>625 = 5\cdot 125 = 5\cdot 5\cdot 25 = 5\cdot 5\cdot 5\cdot 5 = 5^{4}</math> and we obtain
-	~~& \left( 5^{8}+5^{9} \right)^{-1}=\left( 5^{8}+5^{8}\centerdot 5 \right)^{-1}=\left( 5^{8}\centerdot \left( 1+5 \right) \right)^{-1} \\~~	+
-	~~& \\~~	+
-	~~& =\left( 5^{8}\centerdot 6 \right)^{-1}=5^{8\centerdot \left( -1 \right)}\centerdot 6^{-1}=5^{-8}\centerdot 6^{-1}. \\~~	+
-	~~\end{align}</math>~~	+
-		+
-	Furthermore,	+
-	<math>625=5\~~centerdot~~ 125=5\~~centerdot~~ 5\~~centerdot~~ 25=5\~~centerdot~~ 5\~~centerdot~~ 5\~~centerdot~~ 5=5^{4}</math>	+
-	and we obtain	+
-		+

		+	{{Displayed math\|\|
	<math>\begin{align}		<math>\begin{align}
-	& 625\~~centerdot~~ \~~left~~( 5^{8}+5^{9} \~~right~~)^{-1}=5^{4}\~~centerdot~~ 5^{-8}\~~centerdot~~ 6^{-1}=5^{4-8}\~~centerdot~~ 6^{-1} \\	+	625\cdot \bigl(5^{8}+5^{9}\bigr)^{-1} &= 5^{4}\cdot 5^{-8}\cdot 6^{-1} = 5^{4-8}\cdot 6^{-1}\\[5pt]
-	~~& \\~~	+	&= 5^{-4}\cdot 6^{-1} = \frac{1}{5^{4}}\cdot \frac{1}{6}\\[5pt]
-	& =5^{-4}\~~centerdot~~ 6^{-1}=\frac{1}{5^{4}}\~~centerdot~~ \frac{1}{6}=\frac{1}{5^{4}\~~centerdot~~ 6}=\frac{1}{5\~~centerdot~~ 5\~~centerdot~~ 5\~~centerdot~~ 5\~~centerdot~~ 6} \\	+	&= \frac{1}{5^{4}\cdot 6} = \frac{1}{5\cdot 5\cdot 5\cdot 5\cdot 6}\\[5pt]
-	~~& \\~~	+	&= \frac{1}{3750}\,\textrm{.}
-	& =\frac{1}{3750} \\	+	\end{align}</math>}}
-	\end{align}</math>	+

Ian at 11:57, 15 September 2008

2008-09-15T11:57:19Z

←Older revision		Revision as of 11:57, 15 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	Because
-	<~~center~~> ~~[[Image:1_3_4e~~.~~gif]]~~ </~~center~~>	+	<math>5^{9}=5^{8+1}=5^{8}\centerdot 5^{1}=5^{8}\centerdot 5</math>,
-	{{~~NAVCONTENT_STOP~~}}	+	the two terms inside the brackets have
		+	<math>5^{8}</math>
		+	as a common factor
		+	and can therefore be taken outside the bracket.
		+
		+
		+	<math>\begin{align}
		+	& \left( 5^{8}+5^{9} \right)^{-1}=\left( 5^{8}+5^{8}\centerdot 5 \right)^{-1}=\left( 5^{8}\centerdot \left( 1+5 \right) \right)^{-1} \\
		+	& \\
		+	& =\left( 5^{8}\centerdot 6 \right)^{-1}=5^{8\centerdot \left( -1 \right)}\centerdot 6^{-1}=5^{-8}\centerdot 6^{-1}. \\
		+	\end{align}</math>
		+
		+	Furthermore,
		+	<math>625=5\centerdot 125=5\centerdot 5\centerdot 25=5\centerdot 5\centerdot 5\centerdot 5=5^{4}</math>
		+	and we obtain
		+
		+
		+	<math>\begin{align}
		+	& 625\centerdot \left( 5^{8}+5^{9} \right)^{-1}=5^{4}\centerdot 5^{-8}\centerdot 6^{-1}=5^{4-8}\centerdot 6^{-1} \\
		+	& \\
		+	& =5^{-4}\centerdot 6^{-1}=\frac{1}{5^{4}}\centerdot \frac{1}{6}=\frac{1}{5^{4}\centerdot 6}=\frac{1}{5\centerdot 5\centerdot 5\centerdot 5\centerdot 6} \\
		+	& \\
		+	& =\frac{1}{3750} \\
		+	\end{align}</math>

Tekbot: Lösning 1.3:4e moved to Solution 1.3:4e: Robot: moved page

2008-09-09T08:26:03Z

Lösning 1.3:4e moved to Solution 1.3:4e: Robot: moved page

←Older revision

Revision as of 08:26, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T06:30:09Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

←Older revision		Revision as of 06:30, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_3_4e.gif]] </center>	+	<center> [[Image:1_3_4e.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:1_3_4e.gif {{NAVCONTENT_STOP}}

2008-03-26T15:06:56Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_3_4e.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:1_3_4e.gif]] </center>
{{NAVCONTENT_STOP}}