Question

MATH HELP PLEASE!!! PLEASE HELP!!!

Answer 1

The answer is:  [C]:  " f(c) = $\frac{9}{5}$ c  + 32 " .
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Explanation:
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Given the original function:  

" c(y) = (5/9) (x − 32) " ; in which "x = f" ; and "y = c(f) " ;
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→  Write the original function as:  " y = (5/9) (x − 32) " ; 

Now, change the "y" to an "x" ; and the "x" to a "y"; and rewrite; as follows:
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    x = (5/9) (y − 32) ; 

Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ; 
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→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
 "left-hand side" of the equation:

We have:

→  x  =  " (  $\frac{5}{9}$  ) * (y − 32) " ;

Let us simplify the "right-hand side" of the equation:
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Note the "distributive property" of multiplication:
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a(b + c) = ab + ac ;  AND:

a(b – c) = ab – ac .
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As such:
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" ($\frac{5}{9}$) * (y − 32) " ; 

=  [ ($\frac{5}{9}$) * y ]   −  [ ($\frac{5}{9}$) * (32) ] ; 


=  [ ($\frac{5}{9}$) y ]  − [ ($\frac{5}{9}$) * ($\frac{32}{1})$" ;

=  [ ($\frac{5}{9}$) y ]  − [ ($\frac{(5*32)}{(9*1)}$ ] ; 

=  [ ($\frac{5}{9}$) y ]  −  [ ($\frac{(160)}{(9)}$ ] ; 

= [ ($\frac{5y}{9}$) ]  −  [ ($\frac{(160)}{(9)}$ ] ; 

= [ $\frac{(5y-160)}{9}$ ] ;  
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And rewrite as:  

→  " x  =  $\frac{(5y-160)}{9}$ "  ;

We want to rewrite this; solving for "y";  with "y" isolated as a "single variable" on the "left-hand side" of the equation ;

We have:

→  " x  =  $\frac{(5y-160)}{9}$ "  ; 

↔  " $\frac{(5y-160)}{9}$ = x ; 

Multiply both sides of the equation by "9" ; 

 9 * $\frac{(5y-160)}{9}$  =  x * 9 ; 

to get:

→  5y − 160 = 9x ; 

Now, add "160" to each side of the equation; as follows:
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→  5y − 160 + 160 = 9x + 160 ; 

to get:

→  5y  =  9x + 160 ; 

Now,  divided Each side of the equation by "5" ; 
      to isolate "y" on one side of the equation; & to solve for "y" ; 

→  5y / 5  = (9y + 160) / 5 ; 

to get: 
 
→  y = (9/5)x + (160/5) ; 

→  y =  (9/5)x + 32 ; 

 →  Now, remember we had substituted:  "y" for "c(f)" ; 

Now that we have the "equation for the inverse" ;
     →  which is:  " (9/5)x  + 32" ; 

Remember that for the original ("non-inverse" equation);  "y" was used in place of "c(f)" .  We have the "inverse equation";  so we can denote this "inverse function" ; that is, the "inverse" of "c(f)" as:  "f(c)" .

Note that "x = c" ; 
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So, the inverse function is: "  f(c) = (9/5) c  + 32 " .
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 The answer is:  " f(c) = $\frac{9}{5}$ c  + 32 " ;
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 →  which is:  

→  Answer choice:  [C]:  " f(c) = $\frac{9}{5}$ c  + 32 " .
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