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Zina [86]
3 years ago
9

N/5+0.6=2 solve for n

Mathematics
2 answers:
Pavlova-9 [17]3 years ago
5 0
Well, to solve this problem, you have to isolate x. The first step of doing this is subtracting 0.6 from each side to leave the equation as N/5=1.4. Then, you need to multiply both sides by 5 to find N. N/5*5=N, and 1.4*5=7. That leaves the equation as N=7, so the value of N is 7.
Hope this helps!
-Raiden
matrenka [14]3 years ago
3 0
The answer is:  " 7 "  . 
_________________________________________________
       →    " n = 7 " .
_________________________________________________

Explanation:
_________________________________________________

Given:  
_________________________________________________
" (n/5) + 0.6 = 2 " ;   Solve for "n" ;

Subtract "0.6" from EACH SIDE of the equation:

(n/5) + 0.6 − 0.6 = 2 <span>−0.6 ; 

to get:

(n/5) = 1.4 ; 

 Multiply EACH SIDE by "5" ; to isolate "n" on one side of the equation; 
                                                & to solve for "n" ;


5 * (n/5) = 1.4 * 5  ;

to get:
__________________________________________
   " n = 7 ".
__________________________________________ 

Method 2)
__________________________________________
</span>Given:  " (n/5) + 0.6 = 2 " ;   Solve for "n" ;<span>
   
Note that we have "fifths" (the "fraction, (n/5)" ;

           </span>→  <span>and:  "10ths" (the decimal, "0.6" = "6/10" ) ; 

Multiply the entire equation (both sides) by "10" ; to get rid of both the "fraction" and the "decimal" ; 

Note:  We choose "10"; since "10" is the highest "fraction" (that is represented by a "decimal value"; that is:  "6/10" ;  {i.e.,  "0.6 = 6/10" ; as aforementioned} ;
                                    </span>→ and since:  "5" is a factor of "10" ; 

So, multiplying the entire equation by "10"; will get rid of both the "fraction" AND the "decimal" values:

   →  10 * { (n/5) + 0.6) } = 2 * (10) ; 
___________________________________________________
Note the "distributive property" of multiplication:___________________________________________________a(b + c) = ab + ac ;
a(b – c) = ab – ac .___________________________________________________As such:  

  →  Consider the  "left-hand side" of the equation:
__________________________________________

   →  10 * { (n/5) + 0.6) } ;
 
                       =    [10 * (n/5) ] + [10 * 0.6] ; 

                       =    (10n / 5) + 6 ; 

                       =    (10/5)n + 6 ;

                       =     2n + 6  ;

So we rewrite the equation:

  →  2n + 6 =  2 * (10) ; 

Let us consider the "right-hand side of the equation: 

 →  2 * (10) = 20 .
________________________________________________
Rewrite the equation as:

  →  2n + 6 =  20 ;

Now, solve for "n" ;
________________________________________________
Variant # 1 ):
________________________________________________
   →  2n + 6 =  20 ;

Subtract "6" from EACH SIDE of the equation; as follows:

   →  2n + 6 − 6 = 20 <span>− 6 ; 

to get:  

   </span>→  2n = 14 ; 

Now, divide each side of the equation by "2" ; 
     to isolate "n" on one side of the equation; & to solve for "n" ;

   →  2n / 2 = 14 / 2 ; 

   →  " n = 7 ".
____________________________________________________
Variant # 1 ):
____________________________________________________
   →  2n + 6 =  20 ;
____________________________________________________
Divide the entire question by "2" ; to simplify; as follows:
____________________________________________________
   →  {2n + 6} / 2 =  20 / 2  ;

  →  (2n/2)  +  (6/2) =  (20/2 ) ; 

  →   n  +  3 = 10  ;

Subtract "3" from EACH SIDE of the equation;
       to isolate "n" on one side of the equation; & to solve for "n" ;

 
    →   n  +  3 − 3 = 10 <span>− 3 </span>; 

to get:

             →    " n = 7 " .
 _____________________________________________________ 
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Given the sequence 2, 5, 8, 11, 14, ..., which term is 59? (Hint: find n.)
____ [38]
N = 3  ( the next number is 3 plus the previous number
 sequence formula = an = 3(n-1) +2

so for term the term that =59

59 = 3(n-1) +2

distribute:
59 = 3n -3+2

59 = 3n -1

60 = 3n
n = 60/3
n = 20

59 is the 20th term



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