Solution 3.1:4c - Revision history

Tek at 11:05, 30 September 2008

2008-09-30T11:05:52Z

←Older revision		Revision as of 11:05, 30 September 2008
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	Each term in the expression can be simplified by breaking down the number under the root sign into its factors,		Each term in the expression can be simplified by breaking down the number under the root sign into its factors,

		+	{{Displayed math\|\|<math>\begin{align}
		+	50 &= 5\cdot 10 = 5\cdot 5\cdot 2 = 2\cdot 5^{2}\,,\\[5pt]
		+	20 &= 2\cdot 10 = 2\cdot 2\cdot 5 = 2^{2}\cdot 5\,,\\[5pt]
		+	18 &= 2\cdot 9 = 2\cdot 3\cdot 3 = 2\cdot 3^{2}\,,\\[5pt]
		+	80 &= 8\cdot 10 = (2\cdot 4)\cdot (2\cdot 5) = (2\cdot 2\cdot 2)\cdot (2\cdot 5) = 2^{4}\cdot 5\,,
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	and then taking the squares out from under the root sign,
-	~~& 50=5\centerdot 10=5\centerdot 5\centerdot 2=2\centerdot 5^{2} \\~~	+
-	~~& 20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5 \\~~	+
-	~~& 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2} \\~~	+
-	~~& 80=8\centerdot 10=\left( 2\centerdot 4 \right)\centerdot \left( 2\centerdot 5 \right)=\left( 2\centerdot 2\centerdot 2 \right)\centerdot \left( 2\centerdot 5 \right)=2^{4}\centerdot 5 \\~~	+
-	~~\end{align}</math>~~	+
-		+
-	and then taking the squares out from under the root sign.	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \sqrt{50}=\sqrt{2\centerdot 5^{2}}=5\sqrt{2} \\~~	+
-	~~& \sqrt{20}=\sqrt{2^{2}\centerdot 5}=2\sqrt{5} \\~~	+
-	~~& \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\sqrt{2} \\~~	+
-	~~& \sqrt{80}=\sqrt{2^{4}\centerdot 5}=2^{2}\sqrt{5}=4\sqrt{5} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\sqrt{50} &= \sqrt{2\cdot 5^2} = 5\sqrt{2}\,,\\
		+	\sqrt{20} &= \sqrt{2^2\cdot 5} = 2\sqrt{5}\,,\\
		+	\sqrt{18} &= \sqrt{2\cdot 3^2} = 3\sqrt{2}\,,\\
		+	\sqrt{80} &= \sqrt{2^4\cdot 5} = 2^{2}\sqrt{5} = 4\sqrt{5}\,\textrm{.}
		+	\end{align}</math>}}

	All together, we get		All together, we get

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\sqrt{50} + 4\sqrt{20} - 3\sqrt{18} - 2\sqrt{80}
-	& \sqrt{50}+4\sqrt{20}-3\sqrt{18}-2\sqrt{80} \\	+	&= 5\sqrt{2} + 4\cdot 2\sqrt{5} - 3\cdot 3\sqrt{2} - 2\cdot 4\sqrt{5}\\[5pt]
-	& =5\sqrt{2}+4\~~centerdot~~ 2\sqrt{5}-3\~~centerdot~~ 3\sqrt{2}-2\~~centerdot~~ 4\sqrt{5} \\	+	&= 5\sqrt{2} + 8\sqrt{5} - 9\sqrt{2} - 8\sqrt{5}\\[5pt]
-	& =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\	+	&= (5-9)\sqrt{2} + (8-8)\sqrt{5} = -4\sqrt{2}\,\textrm{.}
-	& =~~\left~~( 5-9 ~~\right~~)\sqrt{2}+~~\left~~( 8-8 ~~\right~~)\sqrt{5}=-4\sqrt{2} \\	+	\end{align}</math>}}
-	\end{align}</math>	+

Ian at 14:11, 22 September 2008

2008-09-22T14:11:17Z

←Older revision		Revision as of 14:11, 22 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Each term in the expression can be simplified by breaking down the number under the root sign into its factors,
-	<~~center~~> ~~[[Image:3_1_4c~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& 50=5\centerdot 10=5\centerdot 5\centerdot 2=2\centerdot 5^{2} \\
		+	& 20=2\centerdot 10=2\centerdot 2\centerdot 5=2^{2}\centerdot 5 \\
		+	& 18=2\centerdot 9=2\centerdot 3\centerdot 3=2\centerdot 3^{2} \\
		+	& 80=8\centerdot 10=\left( 2\centerdot 4 \right)\centerdot \left( 2\centerdot 5 \right)=\left( 2\centerdot 2\centerdot 2 \right)\centerdot \left( 2\centerdot 5 \right)=2^{4}\centerdot 5 \\
		+	\end{align}</math>
		+
		+	and then taking the squares out from under the root sign.
		+
		+
		+	<math>\begin{align}
		+	& \sqrt{50}=\sqrt{2\centerdot 5^{2}}=5\sqrt{2} \\
		+	& \sqrt{20}=\sqrt{2^{2}\centerdot 5}=2\sqrt{5} \\
		+	& \sqrt{18}=\sqrt{2\centerdot 3^{2}}=3\sqrt{2} \\
		+	& \sqrt{80}=\sqrt{2^{4}\centerdot 5}=2^{2}\sqrt{5}=4\sqrt{5} \\
		+	\end{align}</math>
		+
		+
		+	All together, we get
		+
		+
		+	<math>\begin{align}
		+	& \sqrt{50}+4\sqrt{20}-3\sqrt{18}-2\sqrt{80} \\
		+	& =5\sqrt{2}+4\centerdot 2\sqrt{5}-3\centerdot 3\sqrt{2}-2\centerdot 4\sqrt{5} \\
		+	& =5\sqrt{2}+8\sqrt{5}-9\sqrt{2}-8\sqrt{5} \\
		+	& =\left( 5-9 \right)\sqrt{2}+\left( 8-8 \right)\sqrt{5}=-4\sqrt{2} \\
		+	\end{align}</math>

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

2008-09-09T09:11:43Z

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

←Older revision

Revision as of 09:11, 9 September 2008

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

2008-08-21T06:52:59Z

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

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2008-04-01T14:04:22Z

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