Solution 1.2:5a - Revision history

Tek at 11:24, 19 September 2008

Tek — Fri, 19 Sep 2008 11:24:40 GMT

←Older revision		Revision as of 11:24, 19 September 2008
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-	We begin by calculating the numerator in the main fraction:	+	We begin by calculating the numerator in the main fraction,

		+	{{Displayed math\|\|<math>\frac{2}{\,\dfrac{1}{7}-\dfrac{1}{15}\vphantom{\Biggl(}\,} = \frac{2}{\,\dfrac{1\cdot 15}{7\cdot 15}-\dfrac{1\cdot 7}{15\cdot 7}\vphantom{\Biggl(}\,} = \frac{2}{\,\dfrac{15-7}{7\cdot 15}\vphantom{\Biggl(}\,} = \frac{2}{\,\dfrac{8}{7\cdot 15}\vphantom{\Biggl(}\,}\,</math>.}}

-	<math>~~\frac{2}{\frac{1}{~~7}-\~~frac{1}{~~15~~}}=\frac{2}{\frac{1\centerdot 15}{7\centerdot 15}~~-~~\frac{1\centerdot~~ 7~~}{15\centerdot 7}}=\frac{2}{~~\~~frac{~~15~~-7}{7\centerdot 15}}=\frac{2}{\frac{~~8~~}{7\centerdot 15}}~~</math>	+	Note that we keep <math>7\cdot 15</math> as it is, and do not multiply it to give 105, because this will make the task of cancellation later simpler. We calculate the fraction on the right-hand side by multiplying top and bottom by <math>7\cdot 15/8</math>,

		+	{{Displayed math\|\|<math>\frac{2}{\,\dfrac{8}{7\cdot 15}\vphantom{\Biggl(}\,} = \frac{\,2\cdot \dfrac{7\cdot 15}{8}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{/}8}{\rlap{/}7\cdot{}\rlap{\,/}15}\cdot \dfrac{\rlap{/}7\cdot{}\rlap{\,/}15}{\rlap{/}8}\vphantom{\Biggl(}\,} = \frac{2\cdot 7\cdot 15}{8}\,</math>.}}

-	~~Note that~~ we ~~keep~~	+	If we now divide up 8 and 15 into their smallest possible integer factors <math>8=2\cdot 2\cdot 2</math> and <math>15=3\cdot 5</math>, we see that the answer in simplified form will be
-	<math>7\~~centerdot 15~~</math>	+
-	~~as it is,~~ and ~~do not multiply it to give~~	+
-	<math>~~105~~</math>	+
-	, ~~because this~~ will ~~make the task~~	+
-	~~of cancellation later simpler. We calculate the fraction on the right-hand side by multiplying top and bottom by~~	+
-	~~<math>7\centerdot {15}/{8}\;</math>~~	+
-	:	+

-		+	{{Displayed math\|\|<math>\frac{2\cdot 7\cdot 15}{8}=\frac{\rlap{/}2\cdot 7\cdot 3\cdot 5}{\rlap{/}2\cdot 2\cdot 2}=\frac{7\cdot 3\cdot 5}{2\cdot 2}=\frac{105}{4}\,</math>.}}
-	<math>\frac{2}{\~~frac{8}{~~7\~~centerdot~~ 15~~}}=\frac{2\centerdot \frac{7\centerdot 5~~}{8}}{\frac{~~8}{7~~\~~centerdot 15}\centerdot \frac~~{~~7\centerdot 5~~}~~{8}}=\frac{~~2\~~centerdot~~ 7\~~centerdot 15}{8}</math>~~	+
-		+
-		+
-	~~If we now divide up~~	+
-	~~<math>8</math>~~	+
-	~~and~~	+
-	~~<math>15</math>~~	+
-	~~into their smallest possible integer factors,~~	+
-	~~<math>8=2\centerdot 2\centerdot 2</math>~~	+
-	~~and~~	+
-	~~<math>15=~~3\~~centerdot~~ 5~~</math>~~	+
-	~~, we see that the answer in simplified form will be~~	+
-		+
-		+
-	~~<math>\frac{2\centerdot 7\centerdot 15~~}{~~8}=~~\~~frac~~{~~2\centerdot 7\centerdot 3\centerdot 5~~}{2\~~centerdot~~ 2\~~centerdot~~ 2}=\frac{7\~~centerdot~~ 3\~~centerdot~~ 5}{2\~~centerdot~~ 2}=\frac{105}{4}</math>	+

Ian at 14:51, 11 September 2008

Ian — Thu, 11 Sep 2008 14:51:27 GMT

←Older revision		Revision as of 14:51, 11 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We begin by calculating the numerator in the main fraction:
-	<~~center~~> ~~[[Image:1_2_5a~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\frac{2}{\frac{1}{7}-\frac{1}{15}}=\frac{2}{\frac{1\centerdot 15}{7\centerdot 15}-\frac{1\centerdot 7}{15\centerdot 7}}=\frac{2}{\frac{15-7}{7\centerdot 15}}=\frac{2}{\frac{8}{7\centerdot 15}}</math>
		+
		+
		+	Note that we keep
		+	<math>7\centerdot 15</math>
		+	as it is, and do not multiply it to give
		+	<math>105</math>
		+	, because this will make the task
		+	of cancellation later simpler. We calculate the fraction on the right-hand side by multiplying top and bottom by
		+	<math>7\centerdot {15}/{8}\;</math>
		+	:
		+
		+
		+	<math>\frac{2}{\frac{8}{7\centerdot 15}}=\frac{2\centerdot \frac{7\centerdot 5}{8}}{\frac{8}{7\centerdot 15}\centerdot \frac{7\centerdot 5}{8}}=\frac{2\centerdot 7\centerdot 15}{8}</math>
		+
		+
		+	If we now divide up
		+	<math>8</math>
		+	and
		+	<math>15</math>
		+	into their smallest possible integer factors,
		+	<math>8=2\centerdot 2\centerdot 2</math>
		+	and
		+	<math>15=3\centerdot 5</math>
		+	, we see that the answer in simplified form will be
		+
		+
		+	<math>\frac{2\centerdot 7\centerdot 15}{8}=\frac{2\centerdot 7\centerdot 3\centerdot 5}{2\centerdot 2\centerdot 2}=\frac{7\centerdot 3\centerdot 5}{2\centerdot 2}=\frac{105}{4}</math>

Tekbot: Lösning 1.2:5a moved to Solution 1.2:5a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:19:23 GMT

Lösning 1.2:5a moved to Solution 1.2:5a: Robot: moved page

←Older revision

Revision as of 08:19, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:26:49 GMT

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

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Lina — Wed, 26 Mar 2008 10:07:50 GMT

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