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AleksAgata [21]
3 years ago
15

Eddie drilled into a piece of wood and I joist to attach them if he drilled 3 inches in the wood was one and 1 1/3 inches thick

how far did you drill into the joist
Mathematics
1 answer:
NARA [144]3 years ago
7 0
Let JD = Distance drilled in joist 
<span>JD = 3 - 1 1/3 </span>
<span>JD = 9/3 - 4/3 </span>
<span>========================== </span>
<span>JD = 5/3 or 1 2/3 in ◄ Ans 

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

</span>
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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 <span>− 32) " ; in which "x = f" ; and "y = c(f) " ;
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</span>→  <span>Write the original function as:  " y = </span>(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 ;  <u><em>AND</em></u>:

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 ; 

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→  5y − 160 = 9x ; 

Now, add "160" to each side of the equation; as follows:
_______________________________________________________

→  5y − 160 + 160 = 9x + 160 ; 

to get:

→  5y  =  9x + 160 ; 

Now,  divided Each side of the equation by "5" ; 
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→  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" ;
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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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