Solution 3.1:4d - Revision history

Tek at 11:13, 30 September 2008

2008-09-30T11:13:32Z

←Older revision		Revision as of 11:13, 30 September 2008
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	We start by factorizing the numbers under the root sign,		We start by factorizing the numbers under the root sign,

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	48 &= 2\cdot 24 = 2\cdot 2\cdot 12 = 2\cdot 2\cdot 2\cdot 6 = 2\cdot 2\cdot 2\cdot 2\cdot 3 = 2^{4}\cdot 3\,,\\
-	& 48=2\~~centerdot~~ 24=2\~~centerdot~~ 2\~~centerdot~~ 12=2\~~centerdot~~ 2\~~centerdot~~ 2\~~centerdot~~ 6=2\~~centerdot~~ 2\~~centerdot~~ 2\~~centerdot~~ 2\~~centerdot~~ 3=2^{4}\~~centerdot~~ 3, \\	+	12 &= 2\cdot 6 = 2\cdot 2\cdot 3 = 2^{2}\cdot 3\,,\\
-	& 12=2\~~centerdot~~ 6=2\~~centerdot~~ 2\~~centerdot~~ 3=2^{2}\~~centerdot~~ 3, \\	+	3 &= 3\,,\\
-	& 3=3, \\	+	75 &= 3\cdot 25 = 3\cdot 5\cdot 5 = 3\cdot 5^{2}\,\textrm{.}
-	& 75=3\~~centerdot~~ 25=3\~~centerdot~~ 5\~~centerdot~~ 5=3\~~centerdot~~ 5^{2} \\	+	\end{align}</math>}}
-	\end{align}</math>	+

	Now, we can take the squares out from under the root signs,		Now, we can take the squares out from under the root signs,

		+	{{Displayed math\|\|<math>\begin{align}
		+	\sqrt{48} &= \sqrt{2^4\cdot 3} = 2^2\sqrt{3} = 4\sqrt{3}\,,\\[5pt]
		+	\sqrt{12} &= \sqrt{2^2\cdot 3} = 2\sqrt{3},\\[5pt]
		+	\sqrt{3} &= \sqrt{3}\,,\\[5pt]
		+	\sqrt{75} &= \sqrt{3\cdot 5^{2}} = 5\sqrt{3}\,,
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	and then simplify the whole expression
-	~~& \sqrt{48}=\sqrt{2^{4}\centerdot 3}=2^{2}\sqrt{3}=4\sqrt{3} \\~~	+
-	~~& \sqrt{12}=\sqrt{2^{2}\centerdot 3}=2\sqrt{3} \\~~	+
-	~~& \sqrt{3}=\sqrt{3} \\~~	+
-	~~& \sqrt{75}=\sqrt{3\centerdot 5^{2}}=5\sqrt{3} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	and then simplify the whole expression:	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \sqrt{48}+\sqrt{12}+\sqrt{3}-\sqrt{75}=4\sqrt{3}+2\sqrt{3}+\sqrt{3}-5\sqrt{3} \\	+	\sqrt{48} + \sqrt{12} + \sqrt{3} - \sqrt{75}
-	& =~~\left~~( 4+2+1-5 ~~\right~~)\sqrt{3}=2\sqrt{3} \\	+	&= 4\sqrt{3} + 2\sqrt{3} + \sqrt{3} - 5\sqrt{3}\\[5pt]
-	\end{align}</math>	+	&= (4+2+1-5)\sqrt{3}\\[5pt]
		+	&= 2\sqrt{3}\,\textrm{.}
		+	\end{align}</math>}}

Ian at 14:20, 22 September 2008

2008-09-22T14:20:44Z

←Older revision		Revision as of 14:20, 22 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We start by factorizing the numbers under the root sign,
-	<~~center~~> ~~[[Image:3_1_4d.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& 48=2\centerdot 24=2\centerdot 2\centerdot 12=2\centerdot 2\centerdot 2\centerdot 6=2\centerdot 2\centerdot 2\centerdot 2\centerdot 3=2^{4}\centerdot 3, \\
		+	& 12=2\centerdot 6=2\centerdot 2\centerdot 3=2^{2}\centerdot 3, \\
		+	& 3=3, \\
		+	& 75=3\centerdot 25=3\centerdot 5\centerdot 5=3\centerdot 5^{2} \\
		+	\end{align}</math>
		+
		+	Now, we can take the squares out from under the root signs,
		+
		+
		+	<math>\begin{align}
		+	& \sqrt{48}=\sqrt{2^{4}\centerdot 3}=2^{2}\sqrt{3}=4\sqrt{3} \\
		+	& \sqrt{12}=\sqrt{2^{2}\centerdot 3}=2\sqrt{3} \\
		+	& \sqrt{3}=\sqrt{3} \\
		+	& \sqrt{75}=\sqrt{3\centerdot 5^{2}}=5\sqrt{3} \\
		+	\end{align}</math>
		+
		+
		+	and then simplify the whole expression:
		+
		+
		+	<math>\begin{align}
		+	& \sqrt{48}+\sqrt{12}+\sqrt{3}-\sqrt{75}=4\sqrt{3}+2\sqrt{3}+\sqrt{3}-5\sqrt{3} \\
		+	& =\left( 4+2+1-5 \right)\sqrt{3}=2\sqrt{3} \\
		+	\end{align}</math>

Tekbot: Lösning 3.1:4d moved to Solution 3.1:4d: Robot: moved page

2008-09-09T09:12:03Z

Lösning 3.1:4d moved to Solution 3.1:4d: Robot: moved page

←Older revision

Revision as of 09:12, 9 September 2008

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

2008-08-21T06:53:09Z

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

←Older revision		Revision as of 06:53, 21 August 2008
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_4d.gif {{NAVCONTENT_STOP}}

2008-04-01T14:04:55Z

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