Solution 3.1:8c - Revision history

Tek at 14:29, 30 September 2008

Tek — Tue, 30 Sep 2008 14:29:31 GMT

←Older revision		Revision as of 14:29, 30 September 2008
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-	Since	+	Since <math>2\textrm{.}5^{2} = 2\textrm{.}5\cdot 2\textrm{.}5 = 6\textrm{.}25\,</math>, then <math>2\textrm{.}5 = \sqrt{6\textrm{.}25}\,</math>, and then we see that <math>\sqrt{7}</math> is greater than 2.5 since <math>7^{1/2} > 6\textrm{.}25^{1/2}\,</math>.
-	<math>2.5^{2}=2.5\~~centerdot~~ 2.5=6.25</math>, then	+
-	<math>2.5=\sqrt{6.25}</math>, and then we see that	+
-	<math>\sqrt{7}</math>	+
-	is greater than 2.5 since	+
-	<math>7^{{1}/{2~~}\;~~}>6.25^{{1}/{2}\;}</math>	+

Ian — Tue, 23 Sep 2008 11:17:13 GMT

←Older revision		Revision as of 11:17, 23 September 2008
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-	{~~{NAVCONTENT_START}~~}	+	Since
-	<~~center~~> ~~[[Image:3_1_8c~~.~~gif]]~~ </~~center~~>	+	<math>2.5^{2}=2.5\centerdot 2.5=6.25</math>, then
-	{{~~NAVCONTENT_STOP~~}}	+	<math>2.5=\sqrt{6.25}</math>, and then we see that
		+	<math>\sqrt{7}</math>
		+	is greater than 2.5 since
		+	<math>7^{{1}/{2}\;}>6.25^{{1}/{2}\;}</math>

Tekbot — Tue, 09 Sep 2008 09:16:43 GMT

←Older revision

Revision as of 09:16, 9 September 2008

Tekbot — Thu, 21 Aug 2008 06:55:29 GMT

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Oskar — Tue, 01 Apr 2008 14:15:41 GMT

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