Solution 1.2:5c - Revision history

Tek at 12:34, 22 September 2008

Tek — Mon, 22 Sep 2008 12:34:50 GMT

←Older revision		Revision as of 12:34, 22 September 2008
Line 12:		Line 12:
	{{Displayed math\|\|<math>\frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\vphantom{\Biggl(}\,}{\,\dfrac{11}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\cdot \dfrac{16}{11}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{\,/}11}{\rlap{\,/}16}\cdot \dfrac{\rlap{\,/}16}{\rlap{\,/}11}\vphantom{\Biggl(}\,} = \dfrac{16}{10\cdot 11}</math>}}		{{Displayed math\|\|<math>\frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\vphantom{\Biggl(}\,}{\,\dfrac{11}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\cdot \dfrac{16}{11}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{\,/}11}{\rlap{\,/}16}\cdot \dfrac{\rlap{\,/}16}{\rlap{\,/}11}\vphantom{\Biggl(}\,} = \dfrac{16}{10\cdot 11}</math>}}

-	and because <math>16=2\cdot 2\cdot 2\cdot 2</math> and <math>10=2\~~centerdot~~ 5</math>, the simplified answer is	+	and because <math>16=2\cdot 2\cdot 2\cdot 2</math> and <math>10=2\cdot 5</math>, the simplified answer is

	{{Displayed math\|\|<math>\frac{16}{10\cdot 11} = \frac{\rlap{/}2\cdot 2\cdot 2\cdot 2}{\rlap{/}2\cdot 5\cdot 11} = \frac{8}{55}\,</math>.}}		{{Displayed math\|\|<math>\frac{16}{10\cdot 11} = \frac{\rlap{/}2\cdot 2\cdot 2\cdot 2}{\rlap{/}2\cdot 5\cdot 11} = \frac{8}{55}\,</math>.}}

Tek at 12:26, 19 September 2008

Tek — Fri, 19 Sep 2008 12:26:50 GMT

←Older revision		Revision as of 12:26, 19 September 2008
Line 1:		Line 1:
	Method 1		Method 1

-	We calculate the numerator and denominator first.	+	We calculate the numerator and denominator first

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{3}{10}-\frac{1}{5} &= \frac{3}{10}-\frac{1\cdot 2}{5\cdot 2} = \frac{3-2}{10} = \frac{1}{10}\,,\\[10pt]
-	& \frac{3}{10}-\frac{1}{5}=\frac{3}{10}-\frac{1\~~centerdot~~ 2}{5\~~centerdot~~ 2}=\frac{3-2}{10}=\frac{1}{10} \\	+	\frac{7}{8}-\frac{3}{16} &= \frac{7\cdot 2}{8\cdot 2}-\frac{3}{16} = \frac{14-3}{16} = \frac{11}{16}\,\textrm{.}
-	& \frac{7}{8}-\frac{3}{16}=\frac{7\~~centerdot~~ 2}{8\~~centerdot~~ 2}-\frac{3}{16}=\frac{14-3}{16}=\frac{11}{16} \\	+	\end{align}</math>}}
-	\end{align}</math>	+

	Thus, the expression becomes		Thus, the expression becomes

		+	{{Displayed math\|\|<math>\frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\vphantom{\Biggl(}\,}{\,\dfrac{11}{16}\vphantom{\Biggl(}\,} = \frac{\,\dfrac{1}{10}\cdot \dfrac{16}{11}\vphantom{\Biggl(}\,}{\,\dfrac{\rlap{\,/}11}{\rlap{\,/}16}\cdot \dfrac{\rlap{\,/}16}{\rlap{\,/}11}\vphantom{\Biggl(}\,} = \dfrac{16}{10\cdot 11}</math>}}

-	<math>~~\frac{\frac{3}{10}-\frac{1}{5}}{\frac{7}{8}-\frac{3}{~~16}}=\~~frac{~~\~~frac{1}{10}}{~~\~~frac{11}{16}}=\frac{\frac{1}{~~10~~}\centerdot \frac{16}{11}}{\frac{11}{16}\centerdot \frac{16}{11}}~~=~~\frac{16}{10~~\centerdot ~~11}~~</math>	+	and because <math>16=2\cdot 2\cdot 2\cdot 2</math> and <math>10=2\centerdot 5</math>, the simplified answer is

-		+	{{Displayed math\|\|<math>\frac{16}{10\cdot 11} = \frac{\rlap{/}2\cdot 2\cdot 2\cdot 2}{\rlap{/}2\cdot 5\cdot 11} = \frac{8}{55}\,</math>.}}
-	~~and because~~	+
-	<math~~>16=2\centerdot 2\centerdot 2\centerdot 2</math>~~	+
-	~~and 10=2∙5~~	+
-	~~<math>10=2\centerdot 5</math>~~	+
-	~~, the simplified answer is~~	+
-		+
-		+
-	<math>\frac{16}{10\~~centerdot~~ 11}=\frac{2\~~centerdot~~ 2\~~centerdot~~ 2\~~centerdot~~ 2}{2\~~centerdot~~ 5\~~centerdot~~ 11}=\frac{8}{55}</math>	+

	Method 2		Method 2

-	If we look at the individual fractions	+	If we look at the individual fractions 3/10, 1/5, 7/8 and 3/16, we see that the denominators can be factorized as
-	~~<math>{~~3}/{10~~}\;~~,~~\ \ {~~1}/{5,~~\ \ {~~7}/{8~~}\;}\;</math>~~	+
-	and	+
-	~~<math>{~~3}/{16~~}\;</math>~~	+
-	, we see that the denominators can be factorized as	+
-		+
-		+
-	~~<math>10=2\centerdot 5,\ 8=2\centerdot 2\centerdot 2</math>~~	+
-		+
-	~~<math>10=2\centerdot 5,\quad 8=2\centerdot 2\centerdot 2</math>~~	+
-	~~and~~	+
-	~~<math>16=2\centerdot 2\centerdot 2\centerdot 2</math>~~	+
-		+

-	and ~~therefore 2∙2∙2∙2∙5 is the fractions' lowest common denominator.~~	+	{{Displayed math\|\|<math>10=2\cdot 5\,,\quad 8=2\cdot 2\cdot 2\,\quad\text{and}\quad
		+	16=2\cdot 2\cdot 2\cdot 2</math>}}

-	~~If we multiply the top~~ and ~~bottom of~~ the ~~main fraction by~~	+	and therefore 2∙2∙2∙2∙5 = 80 is the fractions' lowest common denominator.
-	~~<math>80</math>~~	+
-	~~, then it will be possible to eliminate all denominators at once,~~	+

		+	If we multiply the top and bottom of the main fraction by 80, then it will be possible to eliminate all denominators at once,

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \frac{\~~frac~~{3}{10}-\~~frac~~{1}{5}}{\~~frac~~{7}{8}-\~~frac~~{3}{16}}=\frac{\left( \~~frac~~{3}{10}-\~~frac~~{1}{5} \right)\~~centerdot~~ 80}{\left( \~~frac~~{7}{8}-\~~frac~~{3}{16} \right)\~~centerdot~~ 80}=\frac{\~~frac~~{3\~~centerdot~~ 80}{10}-\~~frac~~{1\~~centerdot~~ 80}{5}}{\~~frac~~{7\~~centerdot~~ 80}{8}-\~~frac~~{3\~~centerdot~~ 80}{16}} \\	+	\frac{\,\dfrac{3}{10}-\dfrac{1}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7}{8}-\dfrac{3}{16}\vphantom{\Biggl(}\,}
-	& \\	+	&= \frac{\,\left( \dfrac{3}{10}-\dfrac{1}{5} \right)\cdot 80\vphantom{\Biggl(}\,}{\,\left( \dfrac{7}{8}-\dfrac{3}{16} \right)\cdot 80\vphantom{\Biggl(}\,} = \frac{\dfrac{3\cdot 80}{10}-\dfrac{1\cdot 80}{5}\vphantom{\Biggl(}\,}{\,\dfrac{7\cdot 80}{8}-\dfrac{3\cdot 80}{16}\vphantom{\Biggl(}\,}\\[10pt]
-	& =\frac{\~~frac~~{3\~~centerdot~~ 8\~~centerdot~~ 10}{10}-\~~frac~~{8\~~centerdot~~ 2\~~centerdot~~ 5}{5}}{\~~frac~~{7\~~centerdot~~ 8\~~centerdot~~ 10}{8}-\~~frac~~{3\~~centerdot~~ 16\~~centerdot~~ 5}{16}}=\~~frac~~{3\~~centerdot~~ 8-8\~~centerdot~~ 2}{7\~~centerdot~~ 10-3\~~centerdot~~ }=\frac{8}{55} \\	+	&= \frac{\,\dfrac{3\cdot 8\cdot{}\rlap{\,/}10}{\rlap{\,/}10}-\dfrac{8\cdot 2\cdot{}\rlap{/}5}{\rlap{/}5}\vphantom{\Biggl(}\,}{\,\dfrac{7\cdot{}\rlap{/}8\cdot 10}{\rlap{/}8}-\dfrac{3\cdot{}\rlap{\,/}16\cdot 5}{\rlap{\,/}16}\vphantom{\Biggl(}\,} = \dfrac{3\cdot 8-8\cdot 2}{7\cdot 10-3\cdot } = \frac{8}{55}\,\textrm{.}
-	~~& \\~~	+	\end{align}</math>}}
-	\end{align}</math>	+

Ian at 15:00, 11 September 2008

Ian — Thu, 11 Sep 2008 15:00:20 GMT

←Older revision		Revision as of 15:00, 11 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	Method 1
-	<~~center~~> ~~[[Image:1_2_5c~~-1(2~~).gif]]~~ </~~center~~>	+
-	<~~center~~> ~~[[Image:1_2_5c-~~2(2).~~gif]]~~ </~~center~~>	+	We calculate the numerator and denominator first.
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\begin{align}
		+	& \frac{3}{10}-\frac{1}{5}=\frac{3}{10}-\frac{1\centerdot 2}{5\centerdot 2}=\frac{3-2}{10}=\frac{1}{10} \\
		+	& \frac{7}{8}-\frac{3}{16}=\frac{7\centerdot 2}{8\centerdot 2}-\frac{3}{16}=\frac{14-3}{16}=\frac{11}{16} \\
		+	\end{align}</math>
		+
		+	Thus, the expression becomes
		+
		+
		+	<math>\frac{\frac{3}{10}-\frac{1}{5}}{\frac{7}{8}-\frac{3}{16}}=\frac{\frac{1}{10}}{\frac{11}{16}}=\frac{\frac{1}{10}\centerdot \frac{16}{11}}{\frac{11}{16}\centerdot \frac{16}{11}}=\frac{16}{10\centerdot 11}</math>
		+
		+
		+	and because
		+	<math>16=2\centerdot 2\centerdot 2\centerdot 2</math>
		+	and 10=2∙5
		+	<math>10=2\centerdot 5</math>
		+	, the simplified answer is
		+
		+
		+	<math>\frac{16}{10\centerdot 11}=\frac{2\centerdot 2\centerdot 2\centerdot 2}{2\centerdot 5\centerdot 11}=\frac{8}{55}</math>
		+
		+	Method 2
		+
		+	If we look at the individual fractions
		+	<math>{3}/{10}\;,\ \ {1}/{5,\ \ {7}/{8}\;}\;</math>
		+	and
		+	<math>{3}/{16}\;</math>
		+	, we see that the denominators can be factorized as
		+
		+
		+	<math>10=2\centerdot 5,\ 8=2\centerdot 2\centerdot 2</math>
		+
		+	<math>10=2\centerdot 5,\quad 8=2\centerdot 2\centerdot 2</math>
		+	and
		+	<math>16=2\centerdot 2\centerdot 2\centerdot 2</math>
		+
		+
		+	and therefore 2∙2∙2∙2∙5 is the fractions' lowest common denominator.
		+
		+	If we multiply the top and bottom of the main fraction by
		+	<math>80</math>
		+	, then it will be possible to eliminate all denominators at once,
		+
		+
		+	<math>\begin{align}
		+	& \frac{\frac{3}{10}-\frac{1}{5}}{\frac{7}{8}-\frac{3}{16}}=\frac{\left( \frac{3}{10}-\frac{1}{5} \right)\centerdot 80}{\left( \frac{7}{8}-\frac{3}{16} \right)\centerdot 80}=\frac{\frac{3\centerdot 80}{10}-\frac{1\centerdot 80}{5}}{\frac{7\centerdot 80}{8}-\frac{3\centerdot 80}{16}} \\
		+	& \\
		+	& =\frac{\frac{3\centerdot 8\centerdot 10}{10}-\frac{8\centerdot 2\centerdot 5}{5}}{\frac{7\centerdot 8\centerdot 10}{8}-\frac{3\centerdot 16\centerdot 5}{16}}=\frac{3\centerdot 8-8\centerdot 2}{7\centerdot 10-3\centerdot }=\frac{8}{55} \\
		+	& \\
		+	\end{align}</math>

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

Tekbot — Tue, 09 Sep 2008 08:20:03 GMT

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

←Older revision

Revision as of 08:20, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:27:09 GMT

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

←Older revision		Revision as of 06:27, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:1_2_5c-1(2).gif]] </center>	+	<center> [[Image:1_2_5c-1(2).gif]] </center>
-	<center> [[~~Bild~~:1_2_5c-2(2).gif]] </center>	+	<center> [[Image:1_2_5c-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:1_2_5c-1(2).gif Bild:1_2_5c-2(2).gif {{NAVCONTENT_STOP}}

Lina — Wed, 26 Mar 2008 10:09:09 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:1_2_5c-1(2).gif </center> <center> Bild:1_2_5c-2(2).gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:1_2_5c-1(2).gif]] </center>
<center> [[Bild:1_2_5c-2(2).gif]] </center>
{{NAVCONTENT_STOP}}