Solution 2.1:3d - Revision history

Tek at 08:33, 23 September 2008

Tek — Tue, 23 Sep 2008 08:33:58 GMT

←Older revision		Revision as of 08:33, 23 September 2008
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	By writing the expression as <math> x^2 -2\cdot 5x +5^5 </math>, applying the squaring rule <math> x^2-2ax +a^2=(x-a)^2 </math> then gives		By writing the expression as <math> x^2 -2\cdot 5x +5^5 </math>, applying the squaring rule <math> x^2-2ax +a^2=(x-a)^2 </math> then gives

-	<math> ~~\qquad~~	+	{{Displayed math\|\|<math>x^2 -10x + 25 = x^2 -2\cdot 5x +5^5 = (x-5)^2\,\textrm{.}</math>}}
-	x^2 -10x +~~5^5~~ =x^2 -2\cdot 5x +5^5 =(x-5)^2	+
-	</math>	+
-	~~<!--<center> [[Image:2_1_3d.gif]] </center>-->~~	+
-	~~{{NAVCONTENT_STOP~~}}	+

Tekbot — Tue, 09 Sep 2008 08:35:43 GMT

←Older revision

Revision as of 08:35, 9 September 2008

Tekbot — Thu, 21 Aug 2008 06:34:59 GMT

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

Martin — Wed, 13 Aug 2008 13:12:51 GMT

←Older revision		Revision as of 13:12, 13 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:2_1_3d.gif]] </center>	+	By writing the expression as <math> x^2 -2\cdot 5x +5^5 </math>, applying the squaring rule <math> x^2-2ax +a^2=(x-a)^2 </math> then gives
		+
		+	<math> \qquad
		+	x^2 -10x +5^5 =x^2 -2\cdot 5x +5^5 =(x-5)^2
		+	</math>
		+	<!--<center> [[Bild:2_1_3d.gif]] </center>-->
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina — Mon, 31 Mar 2008 09:53:27 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_3d.gif </center> {{NAVCONTENT_STOP}}

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