Solution 4.1:7c - Revision history

Tek at 11:31, 8 October 2008

Tek — Wed, 08 Oct 2008 11:31:19 GMT

←Older revision		Revision as of 11:31, 8 October 2008
Line 1:		Line 1:
-	By completing the square, we can rewrite the	+	By completing the square, we can rewrite the ''x''- and ''y''-terms as quadratic expressions,
-	~~<math>~~x~~</math>~~	+
-	- and	+
-	~~<math>~~y~~</math>~~	+
-	-terms as quadratic expressions,	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>x^{2}-2x=~~\left~~( x-1 ~~\right~~)^{2}-1^{2~~}</math>~~	+	x^2 - 2x &= (x-1)^2 - 1^2\,,\\[5pt]
-		+	y^2 + 6y &= (y+3)^2 - 3^2\,,
-		+	\end{align}</math>}}
-	~~<math>~~y^{2}+6y=~~\left~~( y+3 ~~\right~~)^{2}-3^{2}</math>	+

	and the whole equation then has standard form,		and the whole equation then has standard form,

		+	{{Displayed math\|\|<math>\begin{align}
		+	(x-1)^2 - 1 + (y+3)^2 - 9 &= -3\,,\\[5pt]
		+	(x-1)^2 + (y+3)^2 &= 7\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	From this, we see that the circle has its centre at (1,-3) and radius <math>\sqrt{7}\,</math>.
-	~~& \left~~( x-1 ~~\right)^{2}~~-~~1+\left( y+~~3 ~~\right~~)~~^{2}-9=-3~~ \\	+
-	~~& \Leftrightarrow \quad \left( x-1 \right)^~~{2}+\~~left( y+3 \right)^{2}=7 \\~~	+
-	~~\end{align}~~</math>	+

-
-	From this, we see that the circle has its centre at
-	<math>\left( 1 \right.,\left. -3 \right)</math>
-	and radius
-	<math>\sqrt{7}</math>.
-
-
-	{{NAVCONTENT_START}}

	<center> [[Image:4_1_7_c.gif]] </center>		<center> [[Image:4_1_7_c.gif]] </center>
-
-	{{NAVCONTENT_STOP}}

Ian — Sat, 27 Sep 2008 12:06:52 GMT

←Older revision		Revision as of 12:06, 27 September 2008
Line 1:		Line 1:
		+	By completing the square, we can rewrite the
		+	<math>x</math>
		+	- and
		+	<math>y</math>
		+	-terms as quadratic expressions,
		+
		+
		+	<math>x^{2}-2x=\left( x-1 \right)^{2}-1^{2}</math>
		+
		+
		+	<math>y^{2}+6y=\left( y+3 \right)^{2}-3^{2}</math>
		+
		+	and the whole equation then has standard form,
		+
		+
		+	<math>\begin{align}
		+	& \left( x-1 \right)^{2}-1+\left( y+3 \right)^{2}-9=-3 \\
		+	& \Leftrightarrow \quad \left( x-1 \right)^{2}+\left( y+3 \right)^{2}=7 \\
		+	\end{align}</math>
		+
		+
		+	From this, we see that the circle has its centre at
		+	<math>\left( 1 \right.,\left. -3 \right)</math>
		+	and radius
		+	<math>\sqrt{7}</math>.
		+
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}

	<center> [[Image:4_1_7_c.gif]] </center>		<center> [[Image:4_1_7_c.gif]] </center>

-	<center> [[Image:4_1_7c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Tekbot — Tue, 09 Sep 2008 09:38:23 GMT

←Older revision

Revision as of 09:38, 9 September 2008

Tekbot — Thu, 21 Aug 2008 07:06:19 GMT

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

Martin — Fri, 15 Aug 2008 14:25:06 GMT

Oskar — Wed, 02 Apr 2008 13:31:36 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_1_7c.gif </center> {{NAVCONTENT_STOP}}

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