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DochEvi [55]
3 years ago
12

Simplify. √20⋅√3·√18 5√30 5√10 6√10 6√30

Mathematics
2 answers:
Lilit [14]3 years ago
8 0

Answer:Simplify:   " √20 ⋅ √3 · √18 " :

1)  Simplify the "first term" :

√20   = √4 √5 = 2√5

Step 2):  Consider the "second term:  

√3  = √3  

Step 3)  Simplifly the "third term" :

√18    =  √9 * √2 =  " 3√2 " ;

We have:  " 2√5 * √3 * 3√2 " ;

                 =   (2* 3) *  (√5 * √3 * √2) ;

                 =  6 * (√5 √3 √2)  ;

Note:  " (√5 * √3 * √2 )"  =   " √(5 * 3 * 2) "  =  " √ 30 " .

So:    " 6 * (√5 * √3 * √2)  = 6 √30  

ddd [48]3 years ago
5 0

The correct answer is:   [D]:  " 6√30 " .


_____________________________________


Explanation:


_____________________________________


Simplify:   " √20 ⋅ √3 · √18 " :


___________________________


Step 1)  Simplify the "first term" :

"√20  " = √4 √5 = 2√5  ;


___________________________


Step 2):  Consider the "second term:  

                  "√3 " = √3  (already simplified); 


___________________________


Step 3)  Simplifly the "third term" :

" √18  "  =  √9 * √2 =  " 3√2 " ;


____________________________

We have:  " 2√5 * √3 * 3√2 " ;

                 =   (2* 3) *  (√5 * √3 * √2) ;

                 =  6 * (√5 √3 √2)  ;

Note:  " (√5 * √3 * √2 )"  =   " √(5 * 3 * 2) "  =  " √ 30 " .

So:    " 6 * (√5 * √3 * √2)  =

______________________________________

          →   " 6 √30 " .

______________________________________

The answer is:  " 6√30 " ;

       →  which is:  "Answer choice:  [D]:  " 6√30 " .

_____________________________________

Hope this answer helps!

Best wishes!

_____________________________________

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===========================

we want to find all possible values of α+β+γ when <u>tanα+tanβ+tanγ = tanαtanβtanγ</u><u> </u>to do so we can use algebra and trigonometric skills first

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\rm\displaystyle   \tan( \gamma ) =  \frac{ \tan( \alpha )  +  \tan( \beta ) }{ \tan( \alpha )  \tan( \beta )    - 1}

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\rm\displaystyle   \tan( \gamma ) =   -  \tan( \alpha  +  \beta )

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\rm\displaystyle   \tan( \gamma ) =   -  \tan( t)

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\rm\displaystyle   \tan( \gamma ) =    -\tan(   \alpha   +\beta\pm k\pi )

therefore <u>when</u><u> </u><u>k </u><u>is </u><u>1</u> we obtain:

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isolate -α-β to left hand side and change its sign:

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<u>when</u><u> </u><u>i</u><u>s</u><u> </u><u>0</u>:

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recall that if we have common trigonometric function in both sides then the angle must equal therefore:

\rm\displaystyle  \gamma  =      -   \alpha   -  \beta \pm 0

isolate -α-β to left hand side and change its sign:

\rm\displaystyle \alpha  +  \beta  +   \gamma  =  \boxed{ 0  }

and we're done!

8 0
3 years ago
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