Solution 1.3:4a - Förberedande kurs i matematik 1

                       Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Numerical calculations
                       
                        1.1 Different types of numbers 
                        1.2 Fractions 
                        1.3 Powers
                      
                    
                      2. Algebra
                       
                        2.1 Algebraic expressions 
                        2.2 Linear expressions 
                        2.3 Quadratic expressions
                      
                    
                      3. Roots and logarithms
                       
                        3.1 Roots 
                        3.2 Equations with roots 
                        3.3 Logarithms 
                        3.4 Logarithmic equations
                      
                    
                      4. Trigonometry
                       
                        4.1 Angles and circles 
                        4.2 Functions 
                        4.3 Relations 
                        4.4 Equations
                      
                    
                      5. Written mathematics
                                            
                    
                    
		       Search
		       
		         

                           
                            
			 
		       
		    
		    
		  
		  
		  
		    
		      
		        
			
			Solution 1.3:4a
			
			  From Förberedande kurs i matematik 1
			  (Difference between revisions)
			  			                            Jump to: navigation, search
                          
		          
			
			
			
			
			
			
				Revision as of 11:45, 15 September 2008 (edit)
Ian  (Talk | contribs)
← Previous diff
				Current revision (13:51, 22 September 2008) (edit) (undo)
Tek  (Talk | contribs) 
m 
 
			
		Line 1:
Line 1:
 Because the base is the same in both factors, the exponents can be combined according to the power rules  Because the base is the same in both factors, the exponents can be combined according to the power rules
  
-   + {{Displayed math||<math>2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,</math>.}}
- <math>2^{9}\centerdot 2^{-7}=2^{9-7}=2^{2}=4</math> + 
-   + 
  
 Alternatively, the expressions for the powers can be expanded completely and then cancelled out,  Alternatively, the expressions for the powers can be expanded completely and then cancelled out,
  
-   + {{Displayed math||<math>\begin{align}
- <math>\begin{align} + 2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] 
- & 2^{9-7}=2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot \frac{1}{2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2} \\ + &= 2\cdot 2 = 4\,\textrm{.}\end{align}</math>}}
- &  \\ + 
- & =2\centerdot 2=4 \\ + 
- \end{align}</math> + 
Current revision
Because the base is the same in both factors, the exponents can be combined according to the power rules





\displaystyle 2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,.


Alternatively, the expressions for the powers can be expanded completely and then cancelled out,





\displaystyle \begin{align}
2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] 
&= 2\cdot 2 = 4\,\textrm{.}\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_1.3:4a"
                       Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Numerical calculations
                       
                        1.1 Different types of numbers 
                        1.2 Fractions 
                        1.3 Powers
                      
                    
                      2. Algebra
                       
                        2.1 Algebraic expressions 
                        2.2 Linear expressions 
                        2.3 Quadratic expressions
                      
                    
                      3. Roots and logarithms
                       
                        3.1 Roots 
                        3.2 Equations with roots 
                        3.3 Logarithms 
                        3.4 Logarithmic equations
                      
                    
                      4. Trigonometry
                       
                        4.1 Angles and circles 
                        4.2 Functions 
                        4.3 Relations 
                        4.4 Equations
                      
                    
                      5. Written mathematics
                                            
                    
                    
		       Search
		       
		         

                           
                            
			 
		       
		    
		    
		  
		  
		  
		    
		      
		        
			
			Solution 1.3:4a
			
			  From Förberedande kurs i matematik 1
			  (Difference between revisions)
			  			                            Jump to: navigation, search
                          
		          
			
			
			
			
			
			
				Revision as of 11:45, 15 September 2008 (edit)
Ian  (Talk | contribs)
← Previous diff
				Current revision (13:51, 22 September 2008) (edit) (undo)
Tek  (Talk | contribs) 
m 
 
			
		Line 1:
Line 1:
 Because the base is the same in both factors, the exponents can be combined according to the power rules  Because the base is the same in both factors, the exponents can be combined according to the power rules
  
-   + {{Displayed math||<math>2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,</math>.}}
- <math>2^{9}\centerdot 2^{-7}=2^{9-7}=2^{2}=4</math> + 
-   + 
  
 Alternatively, the expressions for the powers can be expanded completely and then cancelled out,  Alternatively, the expressions for the powers can be expanded completely and then cancelled out,
  
-   + {{Displayed math||<math>\begin{align}
- <math>\begin{align} + 2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] 
- & 2^{9-7}=2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot \frac{1}{2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2} \\ + &= 2\cdot 2 = 4\,\textrm{.}\end{align}</math>}}
- &  \\ + 
- & =2\centerdot 2=4 \\ + 
- \end{align}</math> + 
Current revision
Because the base is the same in both factors, the exponents can be combined according to the power rules





\displaystyle 2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,.


Alternatively, the expressions for the powers can be expanded completely and then cancelled out,





\displaystyle \begin{align}
2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] 
&= 2\cdot 2 = 4\,\textrm{.}\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_1.3:4a"
-                      Welcome to the course
                       
                        About the course 
                        Examinations
                      
                    
                      1. Numerical calculations
                       
                        1.1 Different types of numbers 
                        1.2 Fractions 
                        1.3 Powers
                      
                    
                      2. Algebra
                       
                        2.1 Algebraic expressions 
                        2.2 Linear expressions 
                        2.3 Quadratic expressions
                      
                    
                      3. Roots and logarithms
                       
                        3.1 Roots 
                        3.2 Equations with roots 
                        3.3 Logarithms 
                        3.4 Logarithmic equations
                      
                    
                      4. Trigonometry
                       
                        4.1 Angles and circles 
                        4.2 Functions 
                        4.3 Relations 
                        4.4 Equations
                      
                    
                      5. Written mathematics
                                            
                    
                    
		       Search
+			Solution 1.3:4a
			
			  From Förberedande kurs i matematik 1
			  (Difference between revisions)
			  			                            Jump to: navigation, search
                          
		          
			
			
			
			
			
			
				Revision as of 11:45, 15 September 2008 (edit)
Ian  (Talk | contribs)
← Previous diff
				Current revision (13:51, 22 September 2008) (edit) (undo)
Tek  (Talk | contribs) 
m 
 
			
		Line 1:
Line 1:
 Because the base is the same in both factors, the exponents can be combined according to the power rules  Because the base is the same in both factors, the exponents can be combined according to the power rules
  
-   + {{Displayed math||<math>2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,</math>.}}
- <math>2^{9}\centerdot 2^{-7}=2^{9-7}=2^{2}=4</math> + 
-   + 
  
 Alternatively, the expressions for the powers can be expanded completely and then cancelled out,  Alternatively, the expressions for the powers can be expanded completely and then cancelled out,
  
-   + {{Displayed math||<math>\begin{align}
- <math>\begin{align} + 2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] 
- & 2^{9-7}=2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot \frac{1}{2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2} \\ + &= 2\cdot 2 = 4\,\textrm{.}\end{align}</math>}}
- &  \\ + 
- & =2\centerdot 2=4 \\ + 
- \end{align}</math> + 
Current revision
Because the base is the same in both factors, the exponents can be combined according to the power rules





\displaystyle 2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,.


Alternatively, the expressions for the powers can be expanded completely and then cancelled out,





\displaystyle \begin{align}
2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] 
&= 2\cdot 2 = 4\,\textrm{.}\end{align}




Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_1.3:4a"
@@ Line 1: / Line 1: @@
 Because the base is the same in both factors, the exponents can be combined according to the power rules
+{{Displayed math||<math>2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,</math>.}}
-<math>2^{9}\centerdot 2^{-7}=2^{9-7}=2^{2}=4</math>
 Alternatively, the expressions for the powers can be expanded completely and then cancelled out,
+{{Displayed math||<math>\begin{align}
-<math>\begin{align}
+^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt]
-& 2^{9-7}=2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot \frac{1}{2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2\centerdot 2} \\
+&= 2\cdot 2 = 4\,\textrm{.}\end{align}</math>}}
-&  \\
-& =2\centerdot 2=4 \\
-\end{align}</math>
 \displaystyle 2^{9}\cdot 2^{-7} = 2^{9-7} = 2^{2} = 4\,.
 \displaystyle \begin{align}
2^{9-7} &= 2\cdot 2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot \frac{1}{{}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2\cdot {}\rlap{/}2}\\[5pt] 
&= 2\cdot 2 = 4\,\textrm{.}\end{align}