Solution 3.1:4a - Revision history

Tek at 10:47, 30 September 2008

2008-09-30T10:47:58Z

←Older revision		Revision as of 10:47, 30 September 2008
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-	The decimal number	+	The decimal number <math>0\textrm{.}16</math> can also be written as <math>16\cdot 10^{-2}</math> and then it is easier to see that, since <math>16 = 4\cdot 4 = 4^2</math> and <math>10^{-2} = (10^{-1})^2 = 0\textrm{.}1^2</math>,
-	<math>0.\~~text~~{16 }</math>	+
-	can also be written as	+
-	<math>~~\text{~~16}\~~centerdot \text{1}0~~^{-~~\text{~~2~~}}\text{~~ }</math>	+
-	and then it is easier to see that, since	+
-	<math>~~\text{~~16}=~~\text{~~4}\~~centerdot \text{~~4}=~~\text{~~4}^~~{\text{~~2~~}}\text{ }~~</math>	+
-	and	+
-	<math>10^{-2}=~~\left~~( 10^{-1} ~~\right~~)^{2}=0.1^{2}</math>	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\sqrt{0\textrm{.}16} &= \sqrt{16\cdot 10^{-2}} = \sqrt{16}\cdot \sqrt{10^{-2}} = \sqrt{4^2}\cdot \sqrt{0\textrm{.}1^2}\\[5pt]
		+	&= 4\cdot 0\textrm{.}1 = 0\textrm{.}4\,\textrm{.}
		+	\end{align}</math>}}

-		+	Another alternative is, of course, to see directly that <math>0\textrm{.}16 = 0\textrm{.}4\cdot 0\textrm{.}4 = 0\textrm{.}4^2</math>, and then that <math>\sqrt{0\textrm{.}16} = \sqrt{0\textrm{.}4^2} = 0\textrm{.}4\,\textrm{.}</math>
-	~~<math>\begin{align}~~	+
-	~~& \sqrt{0.16}=\sqrt{16\centerdot 10^{-2}}=\sqrt{16}\centerdot \sqrt{10^{-2}}=\sqrt{4^{2}}\centerdot \sqrt{0.1^{2}} \\~~	+
-	~~& =4\centerdot 0.1=0.4 \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	Another alternative is, of course, to see directly that	+
-	<math>0.16=0.4\~~centerdot~~ 0.4=0.4^{2}</math>, and then that	+
-	<math>\sqrt{0.16}=\sqrt{0.4^{2}}=0.4</math>	+

Ian at 13:50, 22 September 2008

2008-09-22T13:50:44Z

←Older revision		Revision as of 13:50, 22 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The decimal number
-	<~~center~~> ~~[[Image:3_1_4a~~.~~gif]]~~ </~~center~~>	+	<math>0.\text{16 }</math>
-	{{~~NAVCONTENT_STOP~~}}	+	can also be written as
		+	<math>\text{16}\centerdot \text{1}0^{-\text{2}}\text{ }</math>
		+	and then it is easier to see that, since
		+	<math>\text{16}=\text{4}\centerdot \text{4}=\text{4}^{\text{2}}\text{ }</math>
		+	and
		+	<math>10^{-2}=\left( 10^{-1} \right)^{2}=0.1^{2}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& \sqrt{0.16}=\sqrt{16\centerdot 10^{-2}}=\sqrt{16}\centerdot \sqrt{10^{-2}}=\sqrt{4^{2}}\centerdot \sqrt{0.1^{2}} \\
		+	& =4\centerdot 0.1=0.4 \\
		+	\end{align}</math>
		+
		+
		+	Another alternative is, of course, to see directly that
		+	<math>0.16=0.4\centerdot 0.4=0.4^{2}</math>, and then that
		+	<math>\sqrt{0.16}=\sqrt{0.4^{2}}=0.4</math>

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

2008-09-09T09:11:03Z

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

←Older revision

Revision as of 09:11, 9 September 2008

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

2008-08-21T06:52:39Z

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

←Older revision		Revision as of 06:52, 21 August 2008
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2008-04-01T14:03:13Z

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