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stealth61 [152]
3 years ago
11

What is 1/2 x 2/3 x 3/4

Mathematics
2 answers:
Tju [1.3M]3 years ago
8 0
The answer is:  "  \frac{1}{4} " ;  or, write as:  "0.25" .
___________
Explanation:
__________
Method 1)
__________
\frac{1}{2}*\frac{2}{3}*\frac{3}{4} ;

= \frac{1 * 2 * 3}{2* 3 *4} ;

= \frac{6}{24} ; 

= \frac{6/6}{24/6} ;

= \frac{1}{4} ;  or, write as:  "0.25" .
_________________
Method 2)
_________________
\frac{1}{2}*\frac{2}{3}*\frac{3}{4} ;

= \frac{1 * 2 * 3}{2* 3 *4} ;

→ The "2's" & "3's" in both the numerator & denominator "cancel out" to "1" ;  {since:  "(2÷2=1)" ; and "(3÷3=1" ) ; 

→ And we have:

   
→  \frac{1 * 1 * 1}{1 * 1 * 4} ;

            =  \frac{1}{4} ;  or; write as:  "0.25" .
________________
Method 3)  (similar to "Method 2" above):
_________________
\frac{1}{2}*\frac{2}{3}*\frac{3}{4} ;

= \frac{1 * 2 * 3}{2* 3 *4} ;

→ The "2" in numerator cancels out to "1" ; and the "4" in the denominator cancels to "2" ;  {since:  "(4÷2=2)" ; and since: "(2÷2=1)" ;

→  The "3's" in both the numerator AND denominator "cancel out" to "1" ;                      {since:  "(3÷3=1" ) ; 

→ And we have:

   →  \frac{1 * 1 * 1}{2 * 1 * 2} ;

            =  \frac{1}{4} ;  or; write as:  "0.25" .
_______________
Method 4)
_______________
\frac{1}{2}*\frac{2}{3}*\frac{3}{4} ;

→ Cancel out the "2" in the denominator of "\frac{1}{2}" ; to a "1" ; AND cancel out the "2" in the numerator of "\frac{2}{3} " ; to a "1" ;
       {since: "{2÷2=1}" ;
 
 → Cancel out the "3" in the denominator of "\frac{2}{3} " ; to a "1" ;
AND cancel out the "3" in the numerator of: "\frac{3}{4}" ; to a "1" ;
      {since:  "(3÷3=1}" ;

→ And we have:

   →  \frac{1}{1} * \frac{1}{1} * \frac{1}{4} ; 

              = \frac{1 * 1 * 1}{1 * 1 *4} ; 
           
              =  \frac{1}{4} ;  or; write as:  "0.25" .
________________
Variation:
________________
At the point when we have:
________________

   → \frac{1}{1} * \frac{1}{1} * \frac{1}{4} ; 
________________
We can eliminate the: 

"  \frac{1}{1}  *  \frac{1}{1}  " ;

{since: "  \frac{1}{1} = 1 " ; 

{and since:  "  \frac{1}{1}  *  \frac{1}{1}  "  =  1 * 1 = 1 ; 

and since: "1", multiplied by any value, equals that exact same value; 
_____________
THAT IS:
_____________
    → \frac{1}{1}  *  \frac{1}{1}   *  \frac{1}{4}  ;

         =  1  *  1  *  \frac{1}{4} ;

         =  (1 * 1)  *  \frac{1}{4} ;

         =    1 *   \frac{1}{4}  ;
  
         =    \frac{1}{4} ;  or, write as:  "0.25" .
_____________
Method 5)   (similar to "Method 4" above):
____________________
\frac{1}{2}*\frac{2}{3}*\frac{3}{4}  ;

  → Cancel out the "2" in the numerator of "\frac{2}{3}" ; to a " 1 " ; 
AND cancel out the "4" in the denominator of "\frac{3}{4}"; to a "2" ;
       {since: "{4÷2=2}" ; and since: "{2÷2=1}" ; 

  → Cancel out the "3" in the denominator of "\frac{2}{3} " ; to a "1" ;
AND cancel out the "3" in the numerator of: "\frac{3}{4}" ; to a "1" ;
      {since:  "(3÷3=1}" ;

→ And we have:
_____________
   →  \frac{1}{2} * \frac{1}{1} * \frac{1}{2} ; 

              = \frac{1 * 1 * 1}{2 * 1  *2} ; 

                      =  \frac{1}{4} ; or, write as:  "0.25" .
__________
Variation:
__________
At the point when we have:
__________
   →  \frac{1}{2} * \frac{1}{1} * \frac{1}{2} ; 
__________
We can eliminate the: 

"  \frac{1}{1}  " ; 

{since: "  \frac{1}{1}  =  1 " } ; 

{and since: "1", multiplied by any value, equals that exact same value} ; 
___________
THAT IS:
___________
   →  \frac{1}{2} * \frac{1}{1} * \frac{1}{2} ; 

      =    \frac{1}{2}   *    1    *    \frac{1}{2}  ;

          =    \frac{1}{2}   *    \frac{1}{2}  ;

          =    \frac{1*1}{2*2}  ;

              =  \frac{1}{4} ; or, write as:  "0.25" .
__________________________________________________________
inna [77]3 years ago
4 0
       
I believe the answer is 0.25



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