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Flura [38]
3 years ago
12

Find the solution of the square root of the quantity of x plus 2 plus 4 equals 6, and determine if it is an extraneous solution.

Mathematics
2 answers:
almond37 [142]3 years ago
6 0

Answer with Explanation:

 \sqrt{x+2+4}=6\\\\ \sqrt{x+6}=6

Squaring Both sides

→x + 6 =36

→x +6 -6=36-6

→x=30

When we substitute , x=30 in the original equation ,we get

L H S

   =\sqrt{30 +6}\\\\=\sqrt{36}=\pm 6

One value is 6 and other is ,-6.

So, x=30, is not an extraneous solution.    

→→Secondly if your equation is

\sqrt{x} +2+4=6\\\\ \sqrt{x}+6=6\\\\ \sqrt{x}=6-6\\\\ \sqrt{x}=0\\\\ x=0

Substituting ,the value of x, in original equation

L H S

= 0 +2 +4

=6

=R HS    

So, x=0, is also not an extraneous solution.  

Or, if the equation is

\rightarrow \sqrt{x+2} +4=6\\\\\rightarrow \sqrt{x+2}=6-4\\\\\rightarrow \sqrt{x+2}=2\\\\ \text{squaring both sides}\\\\ x +2=4\\\\ x=4-2\\\\x=2

Substituting the value of ,x in original equation

LHS

=\sqrt{2+2}+4\\\\= \sqrt{4}+4\\\\=2 +4=6

=RHS

So, x=2 , is not an Extraneous Solution.  

morpeh [17]3 years ago
4 0

minus 4 both sides

square both sides

x+2=4
minus 2
x=2
 
<span>the answer is 
 x = 2; not extraneous</span>
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Step-by-step explanation:

The given data is  

                                          x`             Std. Dev      

R. disagrees                  4.16       0.854        

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From this data we see that the R. disagrees has a mean of 4.16 and standard deviation of 0.854

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C. The R. Disagrees condition has a mean of 4.16 and a standard deviation of 0.85 while the P. Disagrees condition has a mean of 3.82 and standard deviation of 0.97

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Find the values of the sine, cosine, and tangent for ZA.<br><br> (TOP OF TRIANGLE IS (A))
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\bigstar\:{\underline{\sf{In\:right\:angled\:triangle\:ABC\::}}}\\\\

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  • BC = 4 m

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\bf{\dag}\:{\underline{\frak{By\:using\:Pythagoras\: Theorem,}}}\\\\

\star\:{\underline{\boxed{\frak{\purple{(Hypotenus)^2 = (Perpendicular)^2 + (Base)^2}}}}}\\\\\\ :\implies\sf (AB)^2 = (AC)^2 + (BC)^2\\\\\\ :\implies\sf (AB)^2 = (AB)^2 = (7)^2 = (4)^2\\\\\\ :\implies\sf (AB)^2 = 49 + 16\\\\\\ :\implies\sf (AB)^2 = 65\\\\\\ :\implies{\underline{\boxed{\pmb{\frak{AB = \sqrt{65}}}}}}\:\bigstar\\\\

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

☆ Now Let's find value of sin A, cos A and tan A,

⠀⠀⠀

  • sin A = Perpendicular/Hypotenus = \sf \dfrac{4}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{4 \sqrt{65}}{65}}

⠀⠀⠀

  • cos A = Base/Hypotenus = \sf \dfrac{7}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{7 \sqrt{65}}{65}}

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\therefore\:{\underline{\sf{Hence,\: {\pmb{Option\:A)}}\:{\sf{is\:correct}}.}}}

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