What are the excluded values? m+5/mn+3m
1 answer:
values that are <u>excluded from the domain</u> of a rational expression are values that make the denominator 0, since if that's so, the rational will be undefined. That happens when the denominator is zero out, let's do so
so, if ever m = 0, the denominator will become 0 and the rational becomes undefined, and whenever n = -3, the same will happen to the rational, thus those values are excluded.
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The correct answer is: [C]: " 
" .
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<span>Explanation:
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</span>
Let us begin by converting the:
"0.42 (with the "repeating bar on the digits, "42") " ;
into a fraction, as follows:
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Let x = 0.42424242424242424242424242424242....
100x = 42.42424242424242424242424242424242....
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100x = 42.42424242424242424242424242424242....
– x = 0.42424242424242424242424242424242...
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99x = 42.000000000000000000000000000000.... ;
→ 99x = 42 ;
Divide each side of the equation by " 99 " ;
to isolate "x" on one side of the equation; & to solve for "x" ;
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→ 99x / 99 = 42/99 = (42 ÷ 3) / (99 ÷ 3) = 14/ 33;
→ x = "
" .
So; "0.42 (with a repeating bar over the digits, "42" ;
is equal to: ""
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So, rewrite the question/problem being asked:
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"Find the quotient:
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÷ 
;
= * 
;
Note: The "14" cancels to a "7" <span>; and</span> the "2" cancels to a "1" ;
→ {since: " 14 ÷ 2 = 7 " ; and since: " 2 ÷ 2 = 1 "} ;
Note : The "33" cancels to an "11" ; and the "9" cancels to a "3" ;
→ {since: " 33 ÷ 3 = 11 " ; and since: " 9 ÷ 3 = 3 "} ;
And we can rewrite the problem as:
____________________________________________________
"
* 
" ;
=
= ;
→ which is: Answer choice: [C]: " " .
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_____________________________________________________
<span>Explanation:
_____________________________________________________
</span>
Let us begin by converting the:
"0.42 (with the "repeating bar on the digits, "42") " ;
into a fraction, as follows:
______________________________________________________
Let x = 0.42424242424242424242424242424242....
100x = 42.42424242424242424242424242424242....
______________________________________________________
100x = 42.42424242424242424242424242424242....
– x = 0.42424242424242424242424242424242...
______________________________________________________
99x = 42.000000000000000000000000000000.... ;
→ 99x = 42 ;
Divide each side of the equation by " 99 " ;
to isolate "x" on one side of the equation; & to solve for "x" ;
______________________________________________________
→ 99x / 99 = 42/99 = (42 ÷ 3) / (99 ÷ 3) = 14/ 33;
→ x = "
So; "0.42 (with a repeating bar over the digits, "42" ;
is equal to: "
________________________________________________
So, rewrite the question/problem being asked:
________________________________________________
"Find the quotient:
________________________________________________
=
Note: The "14" cancels to a "7" <span>; and</span> the "2" cancels to a "1" ;
→ {since: " 14 ÷ 2 = 7 " ; and since: " 2 ÷ 2 = 1 "} ;
Note : The "33" cancels to an "11" ; and the "9" cancels to a "3" ;
→ {since: " 33 ÷ 3 = 11 " ; and since: " 9 ÷ 3 = 3 "} ;
And we can rewrite the problem as:
____________________________________________________
"
=
→ which is: Answer choice: [C]: "
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