Explain what happens to the decimal point in a whole number when you divide it by a power of 10 use this problem with your expla
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The answer is: [C]: " f(c) = 
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 = " (
) * (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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" () * (y − 32) " ;
= [ () * y ] − [ () * (32) ] ;
= [ () y ] − [ () * (
" ;
= [ () y ] − [ (
] ;
= [ () y ] − [ (
] ;
= [ (
) ] − [ ( ] ;
= [
] ;
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And rewrite as:
→ " x = " ;
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 = " ;
↔ " = x ;
Multiply both sides of the equation by "9" ;
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) = c + 32 " ;
_____________________________________________________
→ which is:
→ Answer choice: [C]: " f(c) = c + 32 " .
_____________________________________________________
________________________________________________________
Explanation:
________________________________________________________
Given the original function:
" c(y) = (5/9) (x <span>− 32) " ; in which "x = f" ; and "y = c(f) " ;
________________________________________________________
</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:
________________________________________________________
x = (5/9) (y − 32) ;
Now, rewrite THIS equation; by solving for "y" ; in terms of "x" ;
_____________________________________________________
→ That is, solve this equation for "y" ; with "c" as an "isolated variable" on the
"left-hand side" of the equation:
We have:
→ x = " (
Let us simplify the "right-hand side" of the equation:
_____________________________________________________
Note the "distributive property" of multiplication:
__________________________________________
a(b + c) = ab + ac ; <u><em>AND</em></u>:
a(b – c) = ab – ac .
__________________________________________
As such:
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" (
= [ (
= [ (
= [ (
= [ (
= [ (
= [
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And rewrite as:
→ " x =
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 =
↔ "
Multiply both sides of the equation by "9" ;
9 *
to get:
→ 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" ;
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" ;
_____________________________________________________
So, the inverse function is: " f(c) = (9/5) c + 32 " .
_____________________________________________________
The answer is: " f(c) =
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→ which is:
→ Answer choice: [C]: " f(c) =
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