Solution 3.1:3c - Revision history

Tek at 10:24, 30 September 2008

Tek — Tue, 30 Sep 2008 10:24:50 GMT

←Older revision		Revision as of 10:24, 30 September 2008
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-	We start by looking at ~~the~~ one part of the expression,	+	We start by looking at one part of the expression <math>\sqrt{16}</math>. This subexpression can be simplified since <math>16 = 4\cdot 4 = 4^{2}</math> which gives that <math>\sqrt{16} = \sqrt{4^{2}} = 4</math> and the whole expression becomes
-	<math>\sqrt{16}</math>. This ~~root~~ can be simplified since	+
-	<math>16=4\~~centerdot~~ 4=4^{2}</math>	+
-	which gives that	+
-	<math>\sqrt{16}=\sqrt{4^{2}}=4</math>	+
-	and the whole expression becomes	+

		+	{{Displayed math\|\|<math>\sqrt{16+\sqrt{16}} = \sqrt{16+4} = \sqrt{20}\,\textrm{.}</math>}}

-	<math>~~\sqrt{16+\sqrt{16}}=\sqrt{16+4}=~~\sqrt{20}</math>	+	Can <math>\sqrt{20}</math> be simplified? In order to answer this, we split 20 up into integer factors,

		+	{{Displayed math\|\|<math>20 = 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5</math>}}

-	~~Can~~	+	and see that 20 contains the square <math>2^2</math> as a factor and can therefore be taken outside the root sign,
-	~~<math>\sqrt{~~20}</math>	+
-	~~be simplified? In order to answer this, we split~~	+
-	~~<math>\text{~~2}0</math>	+
-	~~up into integer factors~~,	+

-		+	{{Displayed math\|\|<math>\sqrt{20} = \sqrt{2^{2}\centerdot 5} = 2\sqrt{5}\,\textrm{.}</math>}}
-	~~<math>20=2\centerdot 10=2\centerdot 2\centerdot 5=2^~~{~~2}\centerdot 5</math>~~	+
-		+
-		+
-	~~and see that~~	+
-	~~<math>\text~~{~~2}0\text{ }</~~math>	+
-	~~contains the square~~	+
-	~~<math>\text{2}^{\text{2}}</math>~~	+
-	~~as a factor and can therefore be taken outside the root sign,~~	+
-		+
-		+
-	<math>\sqrt{20}=\sqrt{2^{2}\centerdot 5~~}^{2~~}=2\sqrt{5}</math>	+

Ian at 12:59, 22 September 2008

Ian — Mon, 22 Sep 2008 12:59:26 GMT

←Older revision		Revision as of 12:59, 22 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We start by looking at the one part of the expression,
-	<~~center~~> ~~[[Image:3_1_3c.gif]]~~ </~~center~~>	+	<math>\sqrt{16}</math>. This root can be simplified since
-	{{~~NAVCONTENT_STOP~~}}	+	<math>16=4\centerdot 4=4^{2}</math>
		+	which gives that
		+	<math>\sqrt{16}=\sqrt{4^{2}}=4</math>
		+	and the whole expression becomes
		+
		+
		+	<math>\sqrt{16+\sqrt{16}}=\sqrt{16+4}=\sqrt{20}</math>
		+
		+
		+	Can
		+	<math>\sqrt{20}</math>
		+	be simplified? In order to answer this, we split
		+	<math>\text{2}0</math>
		+	up into integer factors,
		+
		+
		+	<math>20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5</math>
		+
		+
		+	and see that
		+	<math>\text{2}0\text{ }</math>
		+	contains the square
		+	<math>\text{2}^{\text{2}}</math>
		+	as a factor and can therefore be taken outside the root sign,
		+
		+
		+	<math>\sqrt{20}=\sqrt{2^{2}\centerdot 5}^{2}=2\sqrt{5}</math>

Tekbot: Lösning 3.1:3c moved to Solution 3.1:3c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:10:23 GMT

Lösning 3.1:3c moved to Solution 3.1:3c: Robot: moved page

←Older revision

Revision as of 09:10, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:52:19 GMT

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

←Older revision		Revision as of 06:52, 21 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_3c.gif]] </center>	+	<center> [[Image:3_1_3c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_3c.gif {{NAVCONTENT_STOP}}

Oskar — Tue, 01 Apr 2008 14:01:20 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_3c.gif </center> {{NAVCONTENT_STOP}}

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{{NAVCONTENT_START}}
<center> [[Bild:3_1_3c.gif]] </center>
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