Question

How many halves are in 3 5/10

Answer 1

The answer:  3 ⁵/⏨  = 7/2 ;  There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Explanation:
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Method 1):  3 ⁵/⏨   x/2; solve for "x".
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→ 3 ⁵/⏨ = 3 ½ =  [ [(2*3) + 1] / 2 = (6+1)/2 = 7/2 = x/2 ; → x = 7;
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 → {Note: We know that "3 ⁵/⏨ =  3 ½ " because "⁵/⏨" = "½" ;
The fraction, "⁵/⏨"; can be simplified (reduced) to "½", since:
 
  → ⁵/⏨ = 5/10 = (5÷5) / (10÷5) = 1/2 = ½ ; → So, 3 ⁵/⏨ =  3 ½ .}.
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→  x = 7 ; → 3 ⁵/⏨ = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨.
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Method 2):  → 3 ⁵/⏨ = x/2; solve for "x".
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→3 ⁵/⏨ = 3.5 = x/2 ; {Note: "3 ⁵/⏨" can be written in decimal form as: "3.5".}.
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→ (3.5)*2 = x  ↔ x = (3.5)*2 = 7 ↔ x = 7 ;
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→ 3 ⁵/⏨  = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Method 3): → 3 ⁵/⏨ = x/2; solve for "x".
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→ 3 ⁵/⏨ = 35/10 = x/2 ;
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→{We change the "3 ⁵/⏨" to an "improper fraction", which is "35/10";
→We do this as follows: "3 ⁵/⏨ = [(10*3) + 5] / 10 = (30 + 5) / 10 = 35/10"}.

→We have: 35/10 = x/2
          → Change the "10" to a "5"; and change the "2" to a "1" ;
                 → Since "5*2 = 10"; and since, "10÷2 = 5"; and "2÷2 =1" ;

→ Rewrite as: 35/5 = x ; & solve:  → 7 = x ↔ x = 7 ;
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 → 3 ⁵/⏨ = 7/2 ;  → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Method 4): (variant of "Method 3"):  → 3 ⁵/⏨ = x/2; solve for "x".
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→ 3 ⁵/⏨ = 35/10 = x / 2 ; 
→ {Note: We write: "3 ⁵/⏨" as "35/10", the "improper fraction" form.
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Here is the variant:  We shall continue WITHOUT simplifying the "10" and the "2" to a "5" and a "1"; rather, we will simply "cross-multiply":
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→ We have: → 35/10 = x/2 ; → Cross-multiply, to get:
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→ 10x = 35*2 ; From this point we have a few options:
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    Option 1): We have: → 10x = 35*2 ;
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→ Multiply "(35*2)" ; → to get "70" ; and rewrite the equation:   
→ 10x = (35*2) ; → 10x = 70 ;
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Divide EACH side of the equation by "10"; to isolate "x" on one side of the equation; &to solve for "x":
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→ 10x = 70 ;  → 10x / 10 = 70 / 10 ;  → x = 7 ;
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 → 3 ⁵/⏨ = 7/2 ;  → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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    Option 2):  We have: → 10x = 35 * 2 ;
 
→Divide each side of the equation by "10" to isolate "x" on one side of the equation (and ultimately, to solve for "x");
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→ Rewrite, as follows:  10x / 10 = $\frac{35*2}{10}$  :

→ Now, on the "right-hand side of the equation, "cancel out" the "2" (that is, change the "2" to a "1");& change the "10" to a "5" ; (since "2/10", or "2 ÷ 10", = "1/5", or, "⅕").  The "1" can be eliminated, since anything multiplied by "1" is equal to that same value.

 → We now have "x" = $\frac{35}{5}$   =  7;  →  x = 7 ;
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 → 3 ⁵/⏨ = 7/2 ;  → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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    Option 2):  → We have: 10x = 35*2 ;

→  Divide EACH side of the equation by "(35*2)" ;
 
→  10x / (35*2) = (35*2) / (35*2) ;

→ Rewrite as: 10x / (35*2) = 1 ;

{Note: The "right-hand side" of the equation; "(35 * 2) / (35 *2)" ; or,  
"(35 * 2) ÷ (35 *2)" = "1" ;
      → since any [non-zero value], divided by itself, is equal to: "1".}.
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→ So, we have:
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→ 10x / (35*2) = 1 ; ↔ 10x ÷ (35*2) = 1 ;

→ 1*(35*2) = 10x; → 35*2 = 10x ; →  70 = 10x  ↔  10x = 70 ;
       
→ We have: 10x = 70; → Now, divide EACH side of the equation by "10", to isolate "x" on one side of the equation; & to solve for "x" ; 

→ 10x/10 = 70/10 ; → x = 7 ;
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 → 3 ⁵/⏨  = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ . 
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    Option 3): very similar to {option 2}.
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We have: → 10x = 35*2 ;
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Here is the variant from "Option 2" :
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We multiply the: (35*2) on the right-hand side of our equation:
 
→ 35*2 = 70 ; & rewrite our equation as:
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→ 10x = 70 ; → Now, we divide each side of the equation by "70"; 

→ 10x /70 = 70/70 ; → to get:  10x/70 = 1;
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WIth 10x /70 = 1; we can EITHER: 
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         1) Reduce the "10x/70" on the left-hand side of the equation, to: 

 → "1x/7", or "x/7 " ; [Since (10x÷10) / (70÷10) = 1x/7 = x/7] ;
  
 → AND THEN; Rewrite the equation as:
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  → x/7 = 1; → Then solve for "x" ;

→ 1*7 = x → 7 = x ↔  x = 7 ;
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 → 3 ⁵/⏨  = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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OR:    2)  Keep the equation as: 10x/70 = 1 ; and solve for "x"
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→ 10x /70 = 1 ; → 10x = 1*70 ;
 
→ to get: → 10x = 70 ; Solve for "x" by dividing each side of the equation by "10"; to isolate "x" on one side of the equation; & to solve for "x" ;

→ 10x = 70 ; → 10x/10 = 70/10 ;  → x = 7 ;
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 → 3 ⁵/⏨  = 7/2 ; → There are 7 (seven) “halves” in 3 ⁵/⏨ .
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Answer 2

3 5/10 divided by 1/2

Convert 3 5/10 to an improper fraction: 

7/2 

We now have this problem: 

7/2 divided by 1/2

Make the division problem into a multiplication problem using the rule Same-Change-Change or Keep-Change-Flip: 

7/2 multiplied by 2/1 

Multiply the numerators together and the denominators together: 

7*2= 14 2*1= 2

We are left with this fraction as our answer: 

14/2 

This fraction must be simplified: 

14/2=7 

The final answer is 7. 

There are 7 halves in 3 5/10. 

Hope this helps! :)