Square root of -30 over square root of -10

 The answer is:  " √3 " .

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           →   The simplified expression is:  " √3 " .

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Step-by-step explanation:

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We are given:

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  →      ;

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Note the "imaginary number" —  " √(-1) " —

               which is represented by the symbol:  " *i* "  .

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So, let us factor out Both the numerator And the denominator:

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Start with the "numerator" ; as follows:

The numerator is:  √(-30) ;  which factors into:

                              →   √-1 * √30 ;

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Note that:  " √-1 " ; is an imaginary number; which is represented

                              by the symbol:  " *i* "  .

As such, we can rewrite the numerator as:

                             →   *i* √30  .

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The denominator is:  √(-10) ; which factors into:

                              →  √-1  * √10  ;

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As noted above:

           →    " √-1 "  ;   is an imaginary number; which is represented

                                    by the symbol:  " *i* "  .

As such, we can rewrite the denominator as:

                             →   *i* √10   .

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Now, let us rewrite the entire expression:

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          ( *i* √30)

       ____________

          ( *i* √10)    

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Note:  The  *i*  symbols "cancel out" to:  " 1 " ;  

   →   { since:  any "non-zero value" ;  divided by "that same value" ;

                    is equal to:  "1 " .} ;

   →   { i.e.,  in our case:  [ *i*  ÷  *i* ] ;  that is;  " [*i* / *i* ], equals : " 1 " .}.

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And we are left with:

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   →  " " .

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Note that both the numerator and the denominator are "square roots" —    

           and furthermore — are square roots of a "positive number".

Specifically, the "numerator" is:  " √30 " ;

        And  the "denominator is:  " √10 " .

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  →    Note that the numerator:  " √30 " ;

                →    can be factored into:  "√10 * √3 " ;

  →    And that the "denominator" ;  which is:  " √10 " ;  is one of the aforementioned 'factors' of the "numerator" .

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Let us rewrite the expression — and further simplify:

          →       ;

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   =    ;

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Note:  The "(√10)" values,  "cancel out"  to:  "1" ;

→  {since:  any "non-zero value" ;  divided by "that same value" ;

                    is equal to:  "1 " .} ;

         →   { i.e.,  in our case:  [" √10 ÷ √10 " ] ;  

         →   that is;  [ " " ]" ;

                                       →   equals : " 1 " .}.

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→   and we are left with:  " (√3) / 1 " ;

           →  which equals:   " √3  "  ;  

           →  which is our answer.

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Hope this is helpful to you.

     Best wishes in your academic endeavors

             — and within the "" community!

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answer: ur v dumb

step-by-step explanation: