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azamat
4 years ago
6

Square root of -30 over square root of -10

Mathematics
1 answer:
Nuetrik [128]4 years ago
5 0

Answer:  The answer is:  " √3 " .

_______________________________________

           →   The simplified expression is:  " √3 " .

_______________________________________

Step-by-step explanation:

_______________________________________

We are given:

_______________________________________

  →    \frac{\sqrt{-30}}{\sqrt{-10}}  ;

_______________________________________

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

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

_______________________________________

So, let us factor out Both the <u>numerator</u> And the <u>denominator</u>:

_______________________________________

Start with the "numerator" ; as follows:

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

                       

                              →   √-1 * √30 ;

_____________________________________

Note that:  " √-1 " ; is an <u>imaginary number</u>; which is represented

                              by the symbol:  " *i* "  .

As such, we can rewrite the <u>numerator</u> as:

                             →   *i* √30  .

______________________________________

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

                              →  √-1  * √10  ;

______________________________________

As noted above:

           →    " √-1 "  ;   is an <u>imaginary number</u>; which is represented

                                    by the symbol:  " *i* "  .

As such, we can rewrite the denominator as:

                             →   *i* √10   .

______________________________________

Now, let us rewrite the entire expression:

______________________________________

          ( *i* √30)

       ____________

          ( *i* √10)    

______________________________________

<u>Note</u>:  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 " .}.

______________________________________

And we are left with:

______________________________________

   →  " \frac{\sqrt{30}}{\sqrt{10}} " .

______________________________________

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 " .

______________________________________

  →    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" .

_____________________________________

Let us rewrite the expression — and further simplify:

          →    \frac{\sqrt{30}}{\sqrt{10}}   ;

_____________________________________

   =   \frac{(\sqrt{10} *\sqrt{3})}{\sqrt{10}}  ;

_____________________________________

<u>Note</u>:  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;  [ " \frac{\sqrt{10}}{\sqrt{10}} " ]" ;

                                       →   equals : " 1 " .}.

_____________________________________

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

           →  which equals:   " √3  "  ;  

           →  which is our answer.

_____________________________________

Hope this is helpful to you.

     Best wishes in your academic endeavors

             — and within the "Brainly" community!

_____________________________________

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Given matrix A below, and that A = B, find the value of the elements in B. A = 9 −2 3 2 17 0 3 22 8 b11 = b12 = b13 = b21 = b22
GuDViN [60]

Answer:

b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8

Step-by-step explanation:

Consider the given matrix

A=\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]

Let matrix B is

B=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]

It is given that

A=B

\left[\begin{array}{ccc}9&-2&3\\2&17&0\\3&22&8\end{array}\right]=\left[\begin{array}{ccc}b_{11}&b_{12}&b_{13}\\b_{21}&b_{22}&b_{23}\\b_{31}&b_{32}&b_{33}\end{array}\right]

On comparing corresponding elements of both matrices, we get

b_{11}=9,b_{12}=-2,b_{13}=3

b_{21}=2,b_{22}=17,b_{23}=0

b_{31}=3,b_{32}=22,b_{33}=8

Therefore, the required values are b_{11}=9,b_{12}=-2,b_{13}=3,b_{21}=2,b_{22}=17,b_{23}=0,b_{31}=3,b_{32}=22,b_{33}=8.

3 0
3 years ago
Read 2 more answers
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