Solution 3.1:4b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.1:4b moved to Solution 3.1:4b: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
By writing
-
<center> [[Image:3_1_4b.gif]] </center>
+
<math>0.0\text{27 }</math>
-
{{NAVCONTENT_STOP}}
+
as
 +
<math>\text{27}\cdot \text{1}0^{-\text{3}}</math>, where
 +
<math>\text{27}=\text{3}\cdot \text{3}\cdot \text{3}=\text{3}^{\text{3}}</math>
 +
and
 +
<math>10^{-3}=\left( 10^{-1} \right)^{3}=0.1^{3}</math>
 +
we see that
 +
 
 +
 
 +
<math>\begin{align}
 +
& \sqrt[3]{0.027}=\sqrt[3]{27\centerdot 10^{-3}}=\sqrt[3]{27}\centerdot \sqrt[3]{10^{-3}}=\sqrt[3]{3^{3}}\centerdot \sqrt[3]{0.1^{3}} \\
 +
& =3\centerdot 0.1=0.3 \\
 +
\end{align}</math>
 +
 
 +
 
 +
where we have used
 +
<math>\sqrt[3]{a^{3}}=\left( a^{3} \right)^{\frac{1}{3}}=a^{3\centerdot \frac{1}{3}}=a^{1}=a</math>

Revision as of 13:58, 22 September 2008

By writing \displaystyle 0.0\text{27 } as \displaystyle \text{27}\cdot \text{1}0^{-\text{3}}, where \displaystyle \text{27}=\text{3}\cdot \text{3}\cdot \text{3}=\text{3}^{\text{3}} and \displaystyle 10^{-3}=\left( 10^{-1} \right)^{3}=0.1^{3} we see that


\displaystyle \begin{align} & \sqrt[3]{0.027}=\sqrt[3]{27\centerdot 10^{-3}}=\sqrt[3]{27}\centerdot \sqrt[3]{10^{-3}}=\sqrt[3]{3^{3}}\centerdot \sqrt[3]{0.1^{3}} \\ & =3\centerdot 0.1=0.3 \\ \end{align}


where we have used \displaystyle \sqrt[3]{a^{3}}=\left( a^{3} \right)^{\frac{1}{3}}=a^{3\centerdot \frac{1}{3}}=a^{1}=a