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S_A_V [24]
3 years ago
12

What is the sum of StartRoot negative 2 EndRoot and StartRoot negative 18 EndRoot?

Mathematics
1 answer:
Darya [45]3 years ago
3 0

Answer:

4 \sqrt{2} i

Step-by-step explanation:

We want to find the sum of

\sqrt{ - 2}  +  \sqrt{ - 18}

We can rewrite this as

\sqrt{ 2 \times  - 1}  +  \sqrt{ 18 \times  - 1}

This becomes;

\sqrt{2}  \times  \sqrt{ - 1}  +  \sqrt{18}  \times  \sqrt{ - 1}

Recall that;

i =  \sqrt{ - 1}

This implies that;

\sqrt{2} i + 3 \sqrt{2} i

Combine like terms:

4 \sqrt{2} i

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Identify the conic section represented by the equation:<br> x2 + 2x + 2y2 – 12y + 11 = 0<br> 2
wlad13 [49]

Answer:

Equation of the Ellipse

 \frac{(x+1)^{2} }{8} + \frac{(y-3)^{2} }{4} = 1

Step-by-step explanation:

<u><em>Step(i):-</em></u>

Given that the equation

                 x² + 2 x + 2y² - 12 y +11 = 0

      ⇒       x² + 2 x + 1 - 1 + 2(y² - 6 y )+ 11 = 0

                 x² + 2 x + 1 - 1 + 2(y² - 2(3) y+9-9 )+ 11 = 0

     ⇒        x² + 2 x + 1 - 1 + 2(y² - 2(3 y ) +  3²- 3² ) + 11 = 0

By using (a +b)² = a² + 2 a b + b²

               (a -b)² = a² - 2 a b + b²

<u><em>Step(ii):-</em></u>

          x² + 2 x + 1 - 1 + 2(y² - 2(3 y ) +  3²- 3² ) + 11 = 0

    ⇒  ( x+1)² +2( y-3 )² - 1 - 2(9) +11 =0

   ⇒   ( x+1)² +2( y-3 )² - 8 =0

           ( x+1)² +2( y-3 )² = 8

Dividing '8' on both sides , we get

         \frac{(x+1)^{2} }{8} + \frac{2(y-3)^{2} }{8} = 1

      \frac{(x+1)^{2} }{8} + \frac{(y-3)^{2} }{4} = 1

This equation represents the Ellipse

 

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Heyy i need ur help plz
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Step-by-step explanation:

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